Class 10

Punjab State Board PSEB 10th Class English Book Solutions English Literature Book Chapter 1 Bed Number-29 Textbook Exercise Questions and Answers.

Class 10th English Literature Book Chapter 1 Bed Number-29 Question Answers

Bed Number-29 Class 10 Questions and Answers

Question 1.
How did the author lose his eyesight?
Answer:
One day the author was going on a busy street. Suddenly he was struck by a fast-moving vehicle. As a result of this accident, the author lost his eyesight.

एक दिन लेखक एक व्यस्त सड़क पर चला जा रहा था। अचानक एक तेज गति से आ रहा वाहन उससे आ टकराया। इस दुर्घटना के परिणामस्वरूप लेखक अपनी दृष्टि खो बैठा।

Question 2.
What did the author do before he lost his eyesight ?
Answer:
The author was a painter. He used to paint pictures before he lost his eyesight.

लेखक एक चित्रकार था। अपनी दृष्टि खोने से पहले वह चित्र बनाया करता था।

Question 3.
Who did the author meet in the hospital ward ? Why was he there?
Answer:
The author met a person named Naeem in the hospital ward. Like author,Naeem too was blind. Both of them had been admitted there for the operation of their eyes.

लेखक अस्पताल के वार्ड में नईम नामक एक व्यक्ति से मिला। लेखक की भान्ति नईम भी अन्धा था। उन दोनों को वहां आंखों के ऑपरेशन के लिए भर्ती किया गया था।

Question 4.
When did the author regain his confidence and how ?
Answer:
Naeem urged the author to start painting again. He would describe a scene and the author would paint it on a canvas. Naeem would praise the paintings highly. Thus the author regained his confidence.

नईम ने लेखक से दुबारा पेन्टिंग शुरू करने का आग्रह किया। वह किसी दृश्य का वर्णन किया करता और लेखक इसे चित्रपट पर चित्रित किया करता। नईम चित्रों की बहुत प्रशंसा किया करता। इस प्रकार
लेखक को अपना विश्वास दोबारा प्राप्त हो गया।

Question 5.
What happened when the author’s second operation failed ? Who consoled him then ?
Answer:
The author was deeply depressed at the failure of his second operation. He had spent his last rupee on it. It was Naeem who consoled him and offered to help him.

लेखक अपने दूसरे आप्रेशन की असफलता पर गहरा निराश हुआ। उसने इस पर अपना अन्तिम रुपया खर्च कर दिया था। यह नईम था जिसने उसे ढांढस बन्धाया और उसकी सहायता करने की
पेशकश की।

Question 6.
How did the author get the money to get operated the third time ? Who helped him ?
Answer:
It was Naeem who helped the author. He would keep with himself each painting made by the author, and would give him the money. He would say that a rich man liked the paintings and bought them.

यह नईम था जिस ने लेखक की मदद की। वह लेखक के द्वारा बनाई गई प्रत्येक पेन्टिंग को अपने पास रख लेता और उसे पैसे दे दिया करता। वह ऐसा कह दिया करता कि एक धनी आदमी को पेन्टिंगैं
पसन्द आ गई थीं और उसने उन्हें खरीद लिया था।

Question 7.
Where was Naeem when the author regained his eyesight ?
Answer:
Naeem had left the hospital by then. He had spent all his money in paying the author for his paintings. Now he had no money to pay for his third operation. So he had to leave the hospital.

नईम तब तक अस्पताल छोड़ कर जा चुका था। उसने अपने सब पैसे लेखक को उसके चित्रों का भुगतान करने में खर्च कर दिए थे। अब उसके पास अपने तीसरे आपरेशन के लिए कोई पैसे नहीं बचे थे। इसलिए उसे अस्पताल छोड़ कर जाना पड़ा।

Question 8.
Did the author know that Naeem was also blind like him ? Give reasons to support your answer.
Answer:
No, the author did not know that Naeem was also blind like him. Naeem would describe for the author the scenes outside the window. He used to admire the author’s paintings. So the author could never imagine that Naeem was blind like him.

नहीं, लेखक नहीं जानता था कि नईम भी उसकी भान्ति अन्धा था। नईम लेखक के लिए खिड़की से बाहर के दृश्यों का वर्णन किया करता। वह लेखक के चित्रों की प्रशंसा किया करता। इस प्रकार लेखक
कभी यह अनुमान भी न लगा पाया कि नईम उसकी भान्ति अन्धा था।

Question 9.
Why could Naeem not get his treatment done ?
Answer:
Naeem had spent all his money in paying for the author’s paintings. He was left with no money for his own treatment. He had to leave the hospital.

नईम ने अपने सारे पैसे लेखक के चित्रों का भुगतान करने में खर्च कर दिए थे। उसके पास अपने खुद के उपचार के लिए कोई पैसे नहीं बचे थे। उसे अस्पताल छोड़ कर जाना पड़ा।

Question 10.
How did the author feel when he learnt that Naeem had left the hospital because he had no money for the treatment ?
Answer:
The author was deeply shocked. He was unable to move for some time. His eyes were filled with tears.

लेखक को गहरा सदमा महसूस हुआ। वह कुछ समय के लिए हिल भी न सका। उसकी आंखें आंसुओं

Question 11.
How could Naeem describe different seasons in detail ?
Answer:
Naeem himself was an artist. He had painted four paintings of different seasons. He had described the colours of these very paintings to the author.

नईम स्वयम् एक कलाकार था। उसने चार विभिन्न मौसमों की पेन्टिंगें बनाईं थीं। उसने इन्हीं पेन्टिंगों के रंगों का वर्णन लेखक को किया था।

Question 12.
What does the message “The goddess of hope smiled on me ……….. and then it vanished.’ mean ?
Answer:
The author had regained his eyesight after the operation. Now he hoped that he would be able to see Naeem who had helped him so much. But when he read Naeem’s letter, all his hope was dashed.

लेखक को आपरेशन के बाद आंखों की ज्योति फिर से प्राप्त हो गई थी। अब उसे आशा थी कि वह नईम को देख पायेगा जिसने उसकी इतनी मदद की थी। परन्तु जब उसने नईम का पत्र पढ़ा तो उसकी सब आशा खण्डित हो गई।

Objective Type Questions

Question 1.
Naeem was in the hospital when the author regained his eyesight. (True/False)
Answer:
False

Question 2.
Who did the author meet in the hospital ward ?
(i) Naeem
(ii) Bobby Gillian
(iii) Subbiah
(iv) Della.
Answer:
(i) Naeem

Question 3.
The author regained his eyesight after the ………..
(i) first operation
(ii) second operation
(iii) third operation
(iv) fourth operation.
Answer:
(iii) third operation

Question 4.
The author was a ……….. before he lost his eyesight.
Answer:
painter

Question 5.
The author knew that Naeem was also blind ……….. (True/False)
Answer:
False

Question 6.
Naeem urged the author to restart …………..
(i) dancing
(ii) singing
(iii) studying
(iv) painting.
Answer:
(iv) painting.

Answer each of the following in one word / phrase / sentence :

Question 1.
What did the author use to do before he lost his eyesight ?
Answer:
He used to paint pictures.

Question 2.
What did Naeem ask the author to do?
Answer:
He asked him to start painting again.

Question 3.
Who consoled the author when his second operation failed ?
Answer:
Naeem.

Question 4.
Who bought the paintings made by the author when he was in the hospital ?
Answer:
Naeem.

Question 5.
What would Naeem describe for the author ?
Answer:
He would describe the scenes outside the window for the author.

Question 6.
Why did Naeem have no money for his third operation ?
Answer:
He had spent all his money in buying the author’s paintings.

Question 7.
When did the author get deeply shocked ?
Answer:
When he learnt that Naeem had left the hospital because he had no money left for the treatment.

Question 8.
What does the story, ‘Bed Number-29 describe ?
Answer:
The supreme sacrifice made by Naeem for the sake of the author.

Question 9.
Did the author know that Naeem was also blind like him ?
Answer:
No, the author did not know that Naeem was also blind like him.

Complete the following :

1. The author was an amateur ……………….
2. The author met a person named ………………. in the hospital ward.
3. Naeem himself was an ……………….
4. The author could never imagine that ……………….
5. Naeem had no money to pay for
6. It was Naeem who ……………….
Answer:
1. painter
2. Naeem
3. artist
4. Naeem was blind like him
5. his third operation
6. helped the author.

Write True or False against each statement :

1. Naeem was a short story writer.
Answer:
False

2. The author first met Naeem in the operation theatre in the hospital.
Answer:
False

3. Naeem urged the author to start singing again.
Answer:
False

4. Naeem used to admire the author’s paintings.
Answer:
True

5. Naeem would sell each painting made by the author to a rich man.
Answer:
False

6. The author was deeply depressed at the failure of his third operation.
Answer:
True

Choose the correct option for each of the following:

Question 1.
Who said, “The goddess of hope smiled on me ………. and then it vanished.” ?
(a) Naeem.
(b) The author.
(c) Naeem’s doctor.
(d) The author’s friend.
Answer:
(b) The author.

Question 2.
Naeem too was ………………… like the author.
(a) deaf
(b) blind
(c) dumb
(d) lame.
Answer:
(b) blind

Question 3.
The unknown customer of the author’s paintings was :
(a) Naeem himself
(b) the author himself
(c) a rich man of the city
(d) Naeem’s brother.
Answer:
(a) Naeem himself

Question 4.
Naeem had provided the author everything needed for
(a) photography
(b) painting
(c) colouring
(d) none of these three.
Answer:
(b) painting

Bed Number-29 Summary & Translation in English

Bed Number-29 Introduction:
This story describes a supreme sacrifice made by Naeem for the sake of the author. The author is an amateur painter. One day he meets with an accident and loses his eyesight. In the hospital, Naeem is his wardmate. He is also known as Number Twenty-nine, which is the number of his bed. He is a very cheeerful person. He describes beautiful scenes to the author and encourages him to start painting again. In the beginning, the author hesitates but agrees at last. Naeem provides him everything for painting.

He praises author’s paintings and tells him that nobody could believe they are painted by a blind man. The doctors operate on the author for the second time. The author has spent his last rupee on this operation. But unfortunately, this operation is also unsuccessful. Naeem offers the author some money, but the author refuses. Naeem finds a rich customer to buy the author’s paintings. With the sale of his paintings, the author gets sufficient money for his third operation. This operation is successful and the author is able to see again.

He wants to meet Naeem, but Naeem has already left the hospital. The author comes to know that the unknown customer of his paintings was Naeem himself. His paintings are just masses of haphazard lines. Naeem was himself blind, but he spent all his money to buy the author’s paintings. He could not have his third operation as he was left with no money. For the same reason, he had to leave the hospital. Under Naeem’s pillow, the author finds four paintings. These paintings were drawn by Naeem, before he was blind. These showed he was a great artist.

Bed Number-29 Summary & Translation in Hindi

Bed Number-29 Introduction:
यह कहानी नईम द्वारा लेखक के लिए किए गए महान् बलिदान का वर्णन करती है। लेखक एक शौकिया पेन्टर है। एक दिन उसके साथ एक दुर्घटना हो जाती है और वह अपनी दृष्टि खो बैठता है। अस्पताल में नईम उसके वार्ड में ही है। उसे नम्बर उन्तीस भी कहा जाता है जोकि उसके पलंग का नम्बर है। वह एक बहुत खुश-तबीयत व्यक्ति है। वह लेखक को सुन्दर दृश्यों का वर्णन करता है और उसे दोबारा पेन्टिंग शुरू करने के लिए प्रोत्साहित करता है। शुरू में लेखक हिचकिचाता है परन्तु अन्त में मान जाता है। नईम पेन्टिंग के लिए उसे हर चीज़ उपलब्ध करवाता है। वह लेखक की पेन्टिंगों की प्रशंसा करता है और उसे बताता है कि कोई भी इस बात पर विश्वास नहीं करेगा कि इन्हें एक अंधे आदमी ने बनाया है।

डॉक्टर दूसरी बार लेखक का ऑपरेशन करते हैं। लेखक ने इस आपरेशन पर अपना अन्तिम पैसा भी खर्च कर दिया है। परन्तु दुर्भाग्यवश यह ऑपरेशन भी असफल रहता है। नईम लेखक को कुछ पैसे देने की पेशकश करता है, परन्तु लेखक इन्कार कर देता है। नईम लेखक की पेन्टिंगों को खरीदने के लिए एक धनी ग्राहक ढूंढ लेता है। अपनी पेन्टिंगों की बिक्री से लेखक को अपने तीसरे ऑपरेशन के लिए काफ़ी पैसे मिल जाते हैं। यह ऑपरेशन सफल रहता है और लेखक को फिर से दिखने लगता है। वह नईम से मिलना चाहता है, परन्तु नईम पहले ही अस्पताल छोड़ कर जा चुका है। लेखक को पता चलता है कि उसकी पेन्टिंगों का अज्ञात खरीददार स्वयम् नईम था। उसकी पेन्टिंगें आड़ी-तिरछी रेखाओं का मात्र पुंज हैं।

नईम खुद अन्धा था, परन्तु उसने अपने सारे पैसे लेखक की तस्वीरें खरीदने में खर्च कर दिए। वह अपना तीसरा ऑपरेशन नहीं करवा सका, क्योंकि उसके पास कोई पैसा नहीं बचा था। इसी वजह से उसे अस्पताल छोड़ना पड़ा था। नईम के तकिये के नीचे से लेखक को चार पेन्टिंगें मिलती हैं। ये पेन्टिंगें नईम ने अन्धा होने से पहले बनाईं थीं। इनसे पता चलता था कि वह एक महान् कलाकार था।

Bed Number-29 Summary & Translation in Hindi:

(page 5-6)
Brakes shrieked,…………. hope to me.

Word-meanings : 1. shriek—चीखना, (यहां) ब्रेक लगने की तीखी आवाज़; 2. struck-आ कर लगी; 3. leapt-कूदा, होने लगा; 4. recall—-याद आना; 5. predawn glow–प्रभात के पहले वाला प्रकाश; 6. dew-bathed grass-ओस से नहाई हुई घास; 7. masterpiece-सर्वोत्तम कृति; 8. crash–टक्कर; 9. nightmare-बुरा स्वप्न; 10. dreadful-भयानक; 11. depression—-अवसाद; 12. stare– टकटकी लगा कर देखना; 13. limped-लंगड़ा कर चलता था।

अनुवाद- ब्रेक लगने की तीखी आवाज़ आई, कोई चीज़ आकर लगी, कोई चिल्लाया और मेरे चारों ओर अंधेरा घिरता आ रहा था। मेरे पूरे शरीर में पीड़ा होने लगी और एक कोमल आवाज़ ने कहा, “श्रीमान, कृपया हिलिए मत। ऐसा करना खतरनाक हो सकता है।” मैंने समझने की कोशिश की जो भी हुआ था — मैंने प्रभात के पहले होने वाली रोशनी को याद किया; पेड़ और फूल, ओस से नहाई हुई घास, मानो सभी सूर्योदय का इंतज़ार कर रहे थे – मैंने इन सब को अपने कैनवस में कैद कर लिया था, जो मेरी सर्वोत्तम कृति थी, मेरे जीवन की खुशी थी।

इसलिए मैंने उसका नाम ‘जीवन’ रख दिया — उसके बाद मुझे व्यस्त सड़क का दृश्य याद आया, ट्रैफिक का ऊंचा शोर, कार – और वह टक्कर। मेरे हाथ ने आंखों पर बंधी पट्टियों को छुआ। “नहीं, ईश्वर,” मैं कराह उठा, “नहीं, यह नहीं।” । मेरा जीवन आवाजों, भावनाओं, गंधों, स्वादों और भयानक अवसाद का एक दुःस्वप्न बन कर रह गया था। वह अंधेरे का एक पिंजरा था जिसने मुझे कैदी बना कर रखा हुआ था — अंधेरा और मैं, बस हम दोनों थे। समय अब स्थिर हो गया था, अब मेरे लिए सूर्य उदय होना बन्द हो गया था; फूलों का खिलना, जल-धाराएं और निर्मल आकाश अब सिर्फ यादें बन कर रह गए थे। जीवन मेरे साथ ही मरता जा रहा था, एक घंटे से दूसरे घंटे तक मैं पलंग पर लेटा रहता मानो छत की ओर टकटकी लगा कर देख रहा होऊं।

“कहो, क्या हाल है ?” मेरे वार्ड का साथी, नईम, पूछ रहा था जिसे ‘उन्तीस नंबर’ के नाम से जाना जाता था जो कि उसके बैड का नंबर था। वह एक मधुरभाषी और प्रसन्नचित्त व्यक्ति था जो कहानियां सुना कर मुझे सांत्वना देता रहता था और इस प्रकार मेरे मन को जीवन की भयानक वास्तविकताओं से परे ले जाता था। सिवाए इसके कि वह लंगड़ा कर चलता था और उन्तीस नंबर के बैड पर था, मुझे उसके बारे में मुश्किल से ही कुछ पता था। वह फूलों की क्यारियों में फुदकती हुई चिड़ियों का और प्रातःकाल की चमक का वर्णन इतने आश्चर्यजनक ढंग से करता था कि मैं कल्पना करने लगता था मानो उस दृश्य को स्वयं ही देख रहा होऊं। “बोलते रहो,” मैं उससे आग्रह करता जब उसकी आवाज़ आनी बंद हो जाती। इस प्रकार वह सारा दिन अपनी खिड़कियों के बाहर के दृश्य का वर्णन बहुत विस्तार से करता रहता था। उसका ऐसा करना मुझे आशा देता था।

(Page 6)
“Listen,” he said ……….. happened with myself.”

Word-meanings :
1. hobby-शौक के लिए किया गया काम; 2. lash out-फटकारना; 3. hysterical —मानो दौरा पड़ गया हो; 4. moved—बहुत प्रभावित; 5. brightened—जोश आ गया; 6. haystack-घास का ढेर; 7. ablaze चमकदार रंगों से भरी; 8. bask-धूप सेंकना; 9. gasp-लंबी सांसें लेना; 10. uneasy -बेचैनी-भरी; 11. embarrassed—लज्जित कर दिया; 12. after all-कुछ भी हो; 13. plead-विनती करना; 14. attendants—कर्मचारी; 15. miracle-चमत्कार।

अनुवाद- “सुनो,” उसने एक दिन सवेरे कहा। “तुम चित्र बनाना शुरू कर दो, जो जैसा तुमने कहा था, पहले तुम्हारा शौकिया काम हुआ करता था।” मैंने उसे बहुत फटकारा। मैं चिल्लाया, मानो दौरा पड़ गया हो, कि उसे मेरी कला के बारे में मज़ाक करने का कोई अधिकार नहीं था। वह लंगड़ाता हुआ अपने बैड पर चला गया।

कई दिन बीत गए। फिर एक दिन मैंने उससे पूछा कि क्या वह किसी चीज़ से गहरे रूप से प्रभावित हुआ था। “हां,” उसने धीरे-धीरे कहना शुरू किया, “हां, कई चीज़ों से।” अचानक उसकी आवाज़ में जोश आ गया। “हां, बिल्कुल।” “एक बार अक्तूबर की एक सुनहरी शाम के वक्त मैं एक फार्महाउस के पास से गुजर रहा था और मैंने घास का एक ढेर देखा। वह भूसा नहीं था, वह शुद्ध सोना था। चारों तरफ दुनिया रंगों से भरी हुई थीलाल पत्तियां, सफ़ेद बत्तखें जो सुर्ख लाल पश्चिम दिशा में डूबते हुए सूर्य की आखिरी किरणों में धूप सेंक रही थीं। वहां से हिलने में असमर्थ हुआ, मैं वहां खड़ा हो गया और गहरी-गहरी सांसें लेने लगा। “क्या,” मैं चिल्लाया, “तुमने चित्र नहीं पेंट किया ?” एक बेचैनी-भरी खामोशी छा गई जिसने मुझे ऐसा प्रश्न पूछने की लज्जा से भर दिया। कुछ भी हो, मैंने सोचा, हर आदमी कलाकार नहीं होता। “मेरा मतलब है, मैंने तो उसका चित्र पेंट कर लिया होता,” जल्दी-जल्दी मैंने कहा। “तुम क्यों नहीं पेंट कर लेते ? वह दृश्य मेरे मन में है और मैं जानता हूं कि तुम पेंट कर सकते हो। कृपा करके ‘हां’ कह दो, मान जाओ,” उसने विनती की, और इससे पहले कि मैं जान पाऊं कि मैं क्या कर बैठा था, मैंने कह दिया, “हां।”

मेरे जीवन ने एक नया मोड़ ले लिया। उसने मुझे हर वह चीज़ दे दी थी जिसकी ज़रूरत मुझे पेंटिंग में पड़ सकती थी और जब अस्पताल के कर्मचारी आश्चर्य में भर कर चिल्ला उठे जब वे कमरे में आए, तो नईम ने उन्हें चुप करा दिया। फिर चमत्कार शुरू हुआ और उत्सुक, पर लगभग कांपती हुई, उंगलियों से मैं एक उस दृश्य का चित्र बनाने लगा जो मुझे किसी समय बहुत अच्छा लगा था। मैं अपनी याद्दाश्त के कैनवस पर से, लगातार मेहनत करते हुए कागज़ पर उस दृश्य का रेखाचित्र बनाता रहा; मैं अपने काम में इतना डूबा हुआ था कि अपने अंधेपन के विषय में कुछ भी नहीं सोच पा रहा था। मैंने दृश्य का चित्र समाप्त किया और कांपती हुई आवाज़ में नईम को बुलाया। लंबे-लंबे कदमों से वह मेरे बैड की तरफ आया और कुछ समय के लिए मुझे कुछ भी सुनाई नहीं दिया। मेरा दिल डूब गया। “अवश्य ही मैंने कोई भारी गड़बड़ कर दी होगी,” मैं सोचने लगा। फिर उसकी आवाज़ ने खामोशी को भंग कर दिया। “यह बहुत अद्भुत है। यह अविश्वसनीय है, तुम प्रतिभाशाली हो, तुममें विलक्षण प्रतिभा है, कौन कह सकता है कि तुम अंधे हो।” मेरे मन को आराम मिला और मैं कह उठा, “सचमुच! मैं कभी विश्वास नहीं कर सकता था यदि यह मेरे स्वयं के साथ न हुआ होता।”

(Page 7-8)
Every morning, …………. living for something.

Word-meanings : 1. dreamy-स्वप्निल, मानो सपने में हो; 2. magic dreamland sceneryजादुई स्वप्नदेश का दृश्य; 3. create-रचना करना, (यहां) चित्र बनाना; 4. anxiety-चिंता; 5. lump in one’s throat गला रुंध जाना, मानो गले में कुछ अटक गया हो; 6. unwound-(पट्टियां) खोली जा रही थीं; 7. buried—गड़ा लिया; 8. console-सांत्वना देना; 9. shock—सदमा लगना; 10. persuadeमनाना; 11. awful-भद्दा, बेकार; 12. strike a bargain-सौदा कर सकना।

अनुवाद- हर सुबह, नाश्ते के बाद, नईम मेरे बैड पर आता, एक स्वप्निल आवाज़ में एक दृश्य का वर्णन करता जिस पर मैं सवेरे से शाम तक काम करता रहता, मानो कि दिन कभी खत्म नहीं होगा। एक कैनवस खत्म होता और दूसरा शुरू होता। यह सब बहुत अद्भुत था। नईम मुझसे जादुई स्वप्नदेश का सारा दृश्य बनवाता। स्वयं को उसके रंगों की दुनिया में गुम करके, और अपने अंधेपन को भूल कर, मैं कागज़ पर वह सब चित्रित करता जो वह कहता। वह हमेशा मेरी प्रशंसा करता और मुझे अपनी प्रतिभा पर ज्यादा, और ज्यादा विश्वास होता गया। वह स्वयं रंगों का मिश्रण करता, और किसी जगह हल्की शेड और किसी और जगह गहरी रेखा का इस्तेमाल करने का सुझाव देता। यह वही समय था जब डाक्टरों ने एक बार फिर मेरा ऑपरेशन किया।

नईम ने पढ़ कर मुझे सुनाने का या मेरी खिड़की के बाहर के दृश्य का वर्णन करने का काम अपने ऊपर ले लिया क्योंकि मैं बैड से हिलने में असमर्थ था। जैसे-जैसे दिन बीते, मेरी चिंता बढ़ती गई – आंशिक रूप से इसलिए क्योंकि मैं अपनी स्वयं की आंखों से रंगों की दुनिया को देखना चाहता था, पर ज्यादा इसलिए क्योंकि मैंने इसी ऑपरेशन में अपना अंतिम रुपया भी खर्च कर दिया था, और ऑपरेशन के असफल हो जाने की स्थिति में मैंने अंधेरों और कष्टों के विपत्तिपूर्ण जीवन से डरते रहना था। मैं नर्स के साथ डाक्टरों के कमरे की तरफ जा रहा था जब नईम आया और बोला, “आज का दिन बहुत सुंदर है, मुझे पूरी उम्मीद है कि तुम इसे जल्दी देख सकोगे।” मैंने जवाब देना चाहा पर मानो गले में कुछ अटक गया। मुझे ऑपरेशन थियेटर से आने वाली गंध आई और मैंने एक दस्ताने वाले हाथ को कोमलता से स्वयं को छूते हुए अनुभव किया। मेरी पट्टियां खोली जा रही थीं। दीवार-घड़ी की टिक-टिक सुनाई दे रही थी और एक आवाज़ ने कहा, “अपनी आँखें खोलो,” और मैंने अपनी आँखें खोलीं। वही समाप्त न होने वाला अंधेरा अभी भी मौजूद था।

नर्स की मदद से मैं वापस अपने कमरे में आ गया। तो मेरे सामने ऐसा जीवन था – अंधेरे से भरा हुआ। मैंने अपना सिर तकिए में गड़ा लिया। नईम मेरी बगल में था और मुझे सांत्वना दे रहा था। “मैं जल्दी ही यहां से चला जाऊंगा, नईम,” मैंने एक दिन उदास हो कर कहा। “मेरे पास अब पैसे नहीं बचे, इस ऑपरेशन में मेरा सब कुछ खर्च हो गया।” उसे सदमा लगा। “ओह, नहीं! मेरे पास कुछ पैसे हैं, तुम उन्हें ले सकते हो,” उसने कोमलतापूर्वक कहा। मैंने दृढ़ता से उत्तर दिया, “धन्यवाद, नईम। मैंने कभी भी किसी से कुछ नहीं मांगा, न ही भविष्य में मांगूंगा, फिर भी (पेशकश के लिए) मैं तुम्हारा धन्यवाद करता हूं।” उसने मुझे मनाने की कोशिश की, पर मैंने सुनने से इन्कार कर दिया। एक दिन दोपहर में नईम दौड़ता हुआ मेरे बैड पर आया और कहने लगा, “सुनो यार, मेरा एक दोस्त है जो कला-प्रेमी है। वह तुम्हारे पेंट किए हुए चित्र खरीदना चाहता है।” “ऐसा कैसे हो सकता है, वह तो बहुत भद्दे होंगे!” मैंने सोचा। “वह अमीर है, हम उससे सौदा कर सकते हैं।” मैंने हामी भर दी और नईम खुशी से कमरे में इधर-उधर नाचने लगा। अगले दिन उसने मुझे कुछ कड़कड़ाते हुए नोट पकड़ाए। मेरी प्रसन्नता की कोई सीमा न रही। आशा फिर से ऊंची उठ गई। एक बार फिर मैं किसी उद्देश्य के लिए जीने लगा था।

(Page 8)
I took to ……… paintings in my hand.

Word-meanings :
1. tale—कहानी; 2. off and on—कभी-कभी; 3. purple-जामुनी; 4. splash —छींटे मारना, छितरा जाना; 5. mind’s eye—मन की आँखें ; 6. regained consciousness होश में आया; 7. flash-चमक, कौंध; 8. wheeled back-पहिएदार कुसी में बिठाकर वापिस ले गए; 9. pale -पीला (बीमारी या घबराहट से); 10. vanished—गायब हो गई।

अनुवाद- मैंने फिर से चित्र पेंट करना शुरू कर दिया। हर सुबह नईम मेरी बगल में बैठ जाता और अपनी स्वप्निल कहानी शुरू कर देता। मैं चित्र बनाता और बनाता ही गया। वह अपरिचित खरीददार कभी-कभी आता और मेरे चित्र खरीद लेता। नईम चारों ऋतुओं के दृश्यों का वर्णन इतने भावनात्मक ढंग से करता, विशेषकर सूर्यास्त के दृश्यों का, कि गुलाबी, जामुनी, सफ़ेद, बैंगनी और सुनहरे रंग के सारे शेड मेरे ‘मन की आंखों’ के सामने छितरा जाते।

मेरे सारे चित्र खरीद लिए गए थे और मैंने स्वयं को तीसरे ऑपरेशन के लिए तैयार पाया। ऑपरेशन के बाद जब मुझे होश आया, मुझसे कहा गया कि मैं न तो हिलूं और न ही किसी से बात करूं। बैड नंबर उन्तीस, नर्स ने बताया, खाली हो चुका था। जब मेरी पट्टी खोली जाने वाली थी तो मैंने नईम को बुला लाने को कहा किन्तु नर्स ने कहा कि वह बीमार था और आने में असमर्थ था। डाक्टर ने पट्टी खोली और जब मैंने अपनी आंखें खोली तो रोशनी की एक चमक मेरी आंखों में घुसती चली गई – मैं देख पा रहा था। वे पहिएदार कुर्सी में बिठाकर मुझे वापिस कमरे में ले गए।

मैं ऊंची आवाज़ में बोला, “नईम, नईम।” “सिस्टर, नईम कहां है ?” मैंने पूछा। नर्स का चेहरा पीला पड़ गया जब उसने मुझे नईम का पत्र पकड़ाया – आशा की देवी मुझ पर मुस्कराई और फिर उसी शीघ्रता से, जितनी शीघ्रता से वह आई थी, वह गायब हो गई। दुःख से पागल होकर मैं अलमारी की तरफ भागा, और वहां मेरे सारे चित्र पड़े हुए थे। वे केवल अव्यविस्थत रूप से खींची गई लकीरों के पुंज थे जिनमें कोई रंग नहीं था। नर्स बोली, “वह बहुत महान् व्यक्ति था। उसने अपने सारे पैसों से यह चित्र खरीदे और अस्पताल छोड़ कर चला गया जब उसके पास चित्र खरीदने के लिए पैसे नहीं रहे। वह अपना तीसरा ऑपरेशन भी नहीं करवा सका।”

“क्या ?” मैं चिल्ला पड़ा, “ऑपरेशन ? कौन सा ऑपरेशन ? “क्यों ? बेशक उसकी आंखों का ऑपरेशन, वह अंधा था,” वह बोली। कुछ देर के लिए मैं हिलने में असमर्थ हो गया। आंसुओं ने मेरी आंखों को धुंधला कर दिया था। उसके तकिए के नीचे वे चार चित्र पड़े हुए थे जिनमें उसने अंधा होने से पहले चार ऋतुओं की पेटिंग की थी। उसने उन्हीं चित्रों का वर्णन मेरे सामने किया था – और इस प्रकार मेरे ‘कैनवस पर वे चित्र पेंट करने की कोशिश की थी। आंसुओं ने मेरी आंखों को धुंधला कर दिया जब मैंने उसके चित्रों को अपने हाथों में लिया।

Class 10th English Literature Book PSEB Supplementary Reader

• Bed Number-29 Question Answer
• Half A Rupee Worth Question Answer
• One Thousand Dollars Question Answer
• The Dying Detective Question Answer
• How Much Land Does A Man Need? Question Answer
• Return to Air Question Answer

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 7 Coordinate Geometry Ex 7.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.3

Question 1.
Find the area of the triangle whose vertices are:
(i) (2, 3); (- 1, 0); (2, – 4)
(ii) (- 5, – 1); (3, – 5); (5, 2)
Solution:
(i) Let vertices of the ∆ABC are A (2, 3); B(- 1, 0) and C (2, – 4)
Here x₁ = 2, x₂ = – 1 x₃ = 2
y₁ = 3, y₂ = 0, y₃ = – 4 .
∴ Area of ∆ABC = \(\frac{1}{2}\) [x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)
= \(\frac{1}{2}\) [2 × (0 + 4) – 1 × (- 4 – 3) + 2 × (3 – 0)]
= \(\frac{1}{2}\) [8 + 7 + 6] = \(\frac{21}{2}\)
= 10.5 sq units.

(ii) Let vertices of the ∆ABC are A (- 5, – 1); B (3, – 5) and C (5, 2)
Here x₁ = – 5, x₂ = 3, x₃ = 5
y₁ = – 1, y₂ = – 5, y₃ = 2
∴ Area of ∆ABC = \(\frac{1}{2}\) [x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)]
= \(\frac{1}{2}\) [- 5 (- 5 – 2) + 3 (2 + 1) + 5 (- 1 + 5)]
= \(\frac{1}{2}\) [35 + 9 + 20]
= \(\frac{1}{2}\) × 64 = 32 sq units.

Question 2.
In each of the following find the value of ‘k’ for which the points are coimear.
(i) (7, – 2); (5, 1); (3, k)
(ii) (8, 1); (k, – 4); (2, – 5)
Solution:
(i) Let given points be A (7, – 2); B (5, 1) and C (5, k)
Here x₁ = 7, x₂ = 5, x₃ = 3
y₁ = – 2, y₂ = 1 y₃ = k
Three points are collinear iff
\(\frac{1}{2}\) [x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)] = 0
or \(\frac{1}{2}\) [7 (1 – k) + 5(k + 2) + 3(- 2 – 1)] = 0
or 7 – 7k + 5k +10 – 9 = 0
or – 2k + 8 = 0
or – 2k = – 8
or – k = \(\frac{-8}{-2}\) = 4 .
Hence k = 4.

(ii) Let given points be A (8, 1); B (k, – 4) and C(2, – 5)
Here x₁ = 8 x₂ = k, x₃ = 2
y₁ = 1, y = – 4, y = – 5
Three points are collinear iff
\(\frac{1}{2}\) [x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)] = 0
or \(\frac{1}{2}\) [8 (- 4 + 5) + k (- 5 – 1) + 2 (1 + 4) = 0]
or 8 – 6k + 10 = 0
or – 6k = – 18 .
or k = \(\frac{-18}{-6}\) = 3.
Hence k = 3.

Question 3.
Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, – 1), (2, 1) and (0, 3). FInd the ratio of the area of the triangle formed to the area of the given triangle.
Solution:
Let vertices of given triangle ABC are A(0, – 1); B (2, 1) and C (0, 3).
Also, D, E, F be the mid points of AB, BC, CA respectively.
Using mid point formula,
Coordinates of D = \(\left(\frac{0+2}{2}, \frac{-1+1}{2}\right)\) = (1, 0)

Coordinates of E = \(\left(\frac{2+0}{2}, \frac{1+3}{2}\right)\) = (1, 2)

Coordinates of F = \(\left(\frac{0+0}{2}, \frac{3-1}{2}\right)\) = (0, 1)

∴ Co-ordinates of the vertices of DEF are D (1, 0); E (1, 2); F (0,1).
Here x₁ = 1, x₂ = 1, x₃ = 0
y₁ = 0, y₂ = 2, y₃ = 1.
Area of ∆DEF = \(\frac{1}{2}\) [x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)]
= \(\frac{1}{2}\) [1 (2 – 1) + 1 (1 – 0) + 0 (0 – 2)]
= \(\frac{1}{2}\) [1 + 1 + 0] = \(\frac{2}{2}\) = 1.

In ∆ABC,
x₁ = 0, x₂ = 2, x₃ = 0
y₁ = – 1, y₂ = 1, y₃ = 3.
Area of ∆ABC = \(\frac{1}{2}\) [x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)]
= \(\frac{1}{2}\) [0 (1 – 3) + 2 (3 + 1) + 0 (- 1 – 1)]
= \(\frac{1}{2}\) [0 + 8 + 0] = \(\frac{8}{2}\) = 4
Required ratio = \(\frac{\text { Area of } \triangle \mathrm{DEF}}{\text { Area of } \triangle \mathrm{ABC}}\)
= \(\frac{1}{4}\)

Question 4.
Find the area of the quadrilateral whose vertices taken in order, are (- 4, – 2); (- 3, – 5); (3, – 2); (2, 3).
Solution:
Let co-ordinates of the given quadrilateral ABCD are A(- 4, – 2); B(-3, – 5); C(3, – 2) and D (2, 3).
Join AC then Quad. ABCD divides in two triangles
i.e. ∆ABC and ∆CDA

In ∆ABC
Here x₁ = – 4, x₂ = – 3, x₃ = 3
y₁ = – 2, y₂ = – 5, y₃ = – 2
Area of ∆ABC = \(\frac{1}{2}\) [x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)]
= \(\frac{1}{2}\) [- 4 (5 + 2) + (- 3) (- 2 + 2) + 3 (- 2 + 5)]
= \(\frac{1}{2}\) [12 + 0 + 9] = \(\frac{21}{2}\) sq. units.

In ∆CDA
x₁ = 3, x₂ = 2, x₃ = – 4
y₁ = – 2, y₂ = 3, y₃= – 2
Area of ∆CDA = \(\frac{1}{2}\) [x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)]
= \(\frac{1}{2}\) [3 (3 + 2) + 2 (- 2 + 2) + (-4) (- 2 – 3)]
= \(\frac{1}{2}\) [20 + 15 + 0] = \(\frac{35}{2}\) sq. units.

Now, Area of quadritateral ABCD = (Area of ∆ABC) + (Area of ∆ACD)
= \(\frac{21}{2}+\frac{35}{2}=\frac{21+35}{2}\)
= \(\frac{56}{2}\) = 28 sq. units.

Question 5.
You have studied in Class IX, (Chapter 9, Q. 3) that a median of a triangle divides it into two triangles of equal areas. Verify this result for ∆ABC whose vertices are A(4, – 6), B(3, – 2) and C(5, 2).
Solution:
Given that coordinates of the vertices of ∆ABC are A(4, – 6); B (3, – 2) and C (5, 2)
Let CD is the median i.e. D is the mid point of AB which divides AABC into two pails i.e.

∆ADC and ∆CDB
Coordinates of D = \(\left(\frac{4+3}{2}, \frac{-6-3}{2}\right)\)
= \(\left(\frac{7}{2}, \frac{-8}{2}\right)\) = (3.5,- 4).

In ∆ADC
x₁ = 4, x₂ = 3.5, x₃ = 5
y₁ = – 6, y₂ = -4, y₃ = 2
Area of ∆ADC = \(\frac{1}{2}\) [x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)]
= \(\frac{1}{2}\) [4(—4—2)+3.5(2+6)÷5(—6+4)]
= \(\frac{1}{2}\) [- 24 + 28 – 101]
= \(\frac{1}{2}\) × -6
= 3 sq. units (∵ area cannot be negative).

In ∆CDB
x = 5, x = 35, x = 3
y = 2, y = – 4, y = – 2
Area of ∆CDB = \(\frac{1}{2}\) [x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)]
= \(\frac{1}{2}\) [5 (- 4 + 2) + 3.5 (- 2 – 2) + 3 (2 + 4)]
= \(\frac{1}{2}\) [- 10 – 14 + 18]
= \(\frac{1}{2}\) × – 6 = – 3
= 3 sq. units(∵ area cannot be negalive)
From above discussion it is clear that area of ∆ADC = area of ∆CDB = 3 sq. units
Hence, a median of a triangle divides it into two triangles of equal areas.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 7 Coordinate Geometry Ex 7.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.2

Q.uestion 1.
Find the co-ordinates of the point which divides the join (- 1, 7) and (4, – 3) in the ratio 2 : 3.
Solution:
Let required point be P (x, y) which divides the join of given points A (- 1, 7)
and B (4, – 3) in the ratio of 2 : 3.
(-1, 7) (x, y) (4, – 3)

∴ x = \(\frac{2 \times 4+3 \times-1}{2+3}=\frac{8-3}{5}=\frac{5}{5}=1\)

and y = \(\frac{2 \times-3+3 \times 7}{2+3}=\frac{-6+21}{5}=\frac{15}{5}=3\)
Hence, required point be (1, 3).

Question 2.
Find the co-ordinates of the points of trisection of the line segment joining (4, – 1) and (2, – 3).
Solution:
Let P (x₁, y₁) and Q (x₂, y₂) be the required points which trisect the line segment joining A (4, – 1)and B (- 2, – 3) i.e., P(x₁, y₁) divides AB in ratio 1: 2 and Q divides AB in ratio 2 : 1.

∴ x₁ = \(\frac{1 \times-2+2 \times 4}{1+2}=\frac{-2+8}{3}=\frac{6}{3}=2\)

and y₁ = \(\frac{1 \times-3+2 \times-1}{1+2}=\frac{-3-2}{3}=-\frac{5}{3}\)
∴ P(x₁, y₁) be (2, \(-\frac{5}{3}\))

Now, x₂ = \(\frac{2 \times-2+1 \times 4}{2+1}\)
= \(\frac{-4+4}{3}\) = 0

y₂ = \(\frac{2 \times-3+1 \times-1}{2+1}=\frac{-6-1}{3}=-\frac{7}{3}\)

∴ Q(x₂, y₂) be (0, \(-\frac{7}{3}\))
Hence, required points be (2, \(-\frac{5}{3}\)) and (0, \(-\frac{7}{3}\)).

Question 3.
To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1m from each other along AD, as shown in fig. Niharika runs \(\frac{1}{4}\) th the distance AD on the 2nd line and posts a green flag.

Preet runs \(\frac{1}{5}\) th the distance AD on the eighth line and posts a red flag. What is the distance betweenboth the flags? If Rashmi has to post a blue flag exactly half way between the line (segment) joining the two flags, where should she post her flag?

Solution:
In the given figure, we take A as origin. Taking x-axis along AB and y-axis along AD.
Position of green flag = distance covered by Niharika
= Niharika runs \(\frac{1}{4}\)th distance AD on the 2nd line
= \(\frac{1}{4}\) × 100 = 25 m
∴ Co-ordinates of the green flag are (2, 25)
Now, position of red flag = distance covered by Preet = Preet runs \(\frac{1}{5}\)th the distance AD on the 8th line
= \(\frac{1}{5}\) × 100 = 20 m.
Co-ordinates of red flag are (8, 20)
∴ distance between Green and Red flags = \(\sqrt{(8-2)^{2}+(20-25)^{2}}\)
= \(\sqrt{36+25}=\sqrt{61}\) m.
Position of blue flag = mid point of green flag and red flag
= \(\left(\frac{2+8}{2}, \frac{25+20}{2}\right)\)
= (5, 22.5).
Hence, blue flag is in the 5th line and at a distance of 22.5 m along AD.

Question 4.
Find the ratio in which (he segment joining the points (- 3, 10) and (6, – 8) is divided by (- 1, 6).
Solution:
Let point P (- 1, 6) divides the line segment joining the points A (- 3, 10) and B (6, – 8) the ratio K : 1.

∴ -1 = \(\frac{6 \times \mathrm{K}-3 \times 1}{\mathrm{~K}+1}\)
or – K – 1 = 6K – 3
or – K – 6K = – 3 + 1
or – 7K = – 2
K : 1 = \(\frac{2}{7}\) : 1 = 2 : 7
Hence, required ratio is 2 : 7.

Question 5.
Find the ratio in which the line segment joining A (1, – 5) and B (- 4, 5) is divided by the x-axis. Also find the co
ordinates of the point of division.
Solution:
Let required point on x-axis is P (x, 0) which divides the line segment joining the points A (1, – 5) and B (- 4, 5) in the
ratio K : 1.

Consider, y co-ordinates of P ¡s:
0 = \(\frac{5 \times \mathrm{K}+(-5) \times 1}{\mathrm{~K}+1}\)

or 0 = \(\frac{5 \mathrm{~K}-5}{\mathrm{~K}+1}\)
or 5K – 5 = 0
or 5K = 5
or K = \(\frac{5}{5}\) = 1
∴ Required ratio is K : 1 = 1 : 1.
Now, x co-ordinate of P is:
x = \(\frac{-4 \times K+1 \times 1}{K+1}\)
Putting the value of K = 1, we get:
x = \(\frac{-4 \times 1+1 \times 1}{1+1}=\frac{-4+1}{2}\)
x = \(-\frac{3}{2}\)
Hence, required point be (\(-\frac{3}{2}\), 0).

Question 6.
If (1, 2); (4, y); (x, 6) and (3, 5)are the vertices of a parallelogram taken in order, find x and y.
Solution:
Let points of parallelogram ABCD are A (1, 2) (4, y) ; C (x, 6) and D (3, 5)
But diagonals of a || gm bisect each other.
Case I. When E is the mid point of A (1, 2) and C (x, 6)
∴ Co-ordinates of E are:
E = \(\left(\frac{x+1}{2}, \frac{6+2}{2}\right)\)
E = (\(\frac{x+1}{2}\), 4) …………..(1)

Case II. When E is the mid point B (4, y) and D (3, 5)
∴ Co-ordinates of E are:

E = \(\left(\frac{3+4}{2}, \frac{5+y}{2}\right)\)

E = \(\left(\frac{7}{2}, \frac{5+y}{2}\right)\) …………….(2)
But values of E in (1) and (2) are same, so comparing the coordinates, we get
\(\frac{x+1}{2}=\frac{7}{2}\)
or x + 1 = 7
or x = 6.

and 4 = \(\frac{5+y}{2}\)
or 8 = 5 + y
or y = 3
Hence, values of x and y are 6 and 3.

Question 7.
Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, – 3) and B is (1, 4).
Solution:
Let, coordinates of A be (x, y). But, centre is the’ niij ioint of the vertices of the diameter.

∴ O is the mid point of A(x, y) and B(1, 4)
∴ \(\left(\frac{x+1}{2}, \frac{y+4}{2}\right)\) = (2, -3)
On comparing, we get
\(\frac{x+1}{2}\)
or x + 1 = 4
or x = 3

and \(\frac{y+4}{2}\) = – 3
or y + 4 = – 6
or y = – 10
Hence, required point A be (3, – 10).

Question 8.
If A and B are (- 2, – 2) and (2, – 4) respectively, find the coordinates of P such that AP = AB and P lies ¡n the line segment AB.
Solution:
Let required point P be (x, y)
Also AP = \(\frac{3}{7}\) AB …(Given)
But, PB = AB – AP
= AB – \(\frac{3}{7}\) AB = \(\frac{7-3}{7}\) AB
PB = \(\frac{4}{7}\) AB
∴ \(\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{\frac{3}{7} \mathrm{AB}}{\frac{4}{7} \mathrm{AB}}=\frac{3}{4}\).

∴ P divides given points A and B in ratio 3 : 4.
Now,
x = \(\frac{3 \times 2+4 \times-2}{3+4}\)

or x = \(\frac{6-8}{7}=-\frac{2}{7}\)

and y = \(\frac{3 \times-4+4 \times-2}{3+4}\)
= \(\frac{-12-8}{7}=-\frac{20}{7}\)

Hence, coordinates of P be (\(-\frac{2}{7}\), \(-\frac{20}{7}\)).

Question 9.
Find the coordinates of the points which divides the line segment joining A (- 2, 2) and B (2, 8) into four equal parts.
Solution:
Let required points are C, D and E which divide the line segment joming the points A (- 2, 2) and B (2, 8) into four equal parts. Then D is mid point of A and B ; C is the mid point of A and D ; E is the mid point of D and B such that
AC = CD = DE = EB

Now, mid point of A and B (i.e., Coordinates of D)
= \(\left(\frac{-2+2}{2}, \frac{2+8}{2}\right)\) = (0, 5)

Mid point of A and D (i.e., Coordinates of C)
= \(\left(\frac{-2+0}{2}, \frac{2+5}{2}\right)=\left(-1, \frac{7}{2}\right)\)

Mid point of D and B (i.e., Coordinates of E)
= \(\left(\frac{2+0}{2}, \frac{8+5}{2}\right)=\left(1, \frac{13}{2}\right)\)

Hence, requned points be (0, 5), (-1, \(\frac{7}{2}\)), (1, \(\frac{13}{2}\)).

Question 10.
Find the area of a rhombus if the vertices are (3, 0); (4, 5); (- 1, 4) and(- 2, – 1) taken in order.
[Hint: Areas of a rhombus = \(\frac{1}{2}\) (Product of its diagonals)]
Solution:
Let coordinates of rhombus ABCD are A (3, 0); B(4, 5); C(-1, 4) and D(- 2, – 1).
Diagonal, AC = \(\sqrt{(-1-3)^{2}+(4-0)^{2}}\)
= \(\sqrt{16+16}=\sqrt{32}=\sqrt{16 \times 2}\) = 4√2

and diagonal BD
BD = \(\sqrt{(-2-4)^{2}+(-1-5)^{2}}\)
= \(\sqrt{36+36}=\sqrt{72}=\sqrt{36 \times 2}\) = 6√2.

∴ Area of rhombus ABCD = \(\frac{1}{2}\) × AC × BD
ABCD = [\(\frac{1}{2}\) × 4√2 × 6√2] sq. units
(\(\frac{1}{2}\) × 24 × 2) sq. units
= 24 sq. units
Hence, area of rhombus is 24 sq. units.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 7 Coordinate Geometry Ex 7.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.1

Question 1.
Find the distance between the following pairs of points:
(i) (2, 3); (4, 1)
(ii)(-5, 7); (-1, 3)
(iii) (a, b); (-a, -b).
Solution:
(i) Given points are: (2, 3); (4, 1)
Required distance = \(\sqrt{(4-2)^{2}+(1-3)^{2}}\)
\(\sqrt{4+4}=\sqrt{8}=\sqrt{4 \times 2}\)
= 2√2.

(ii) Given points are: (-5, 7); (-1, 3)
Required distance = \(\sqrt{(-1+5)^{2}+(3-7)^{2}}\)
\(\sqrt{16+16}=\sqrt{32}\)
= \(\sqrt{16 \times 2}\)
= 4√2.

(iii) Given points are: (a, b); (-a, -b)
Required distance = \(\sqrt{(-a-a)^{2}+(-b-b)^{2}}\)
= \(\sqrt{(-2 a)^{2}+(-2 b)^{2}}\)
= \(\sqrt{4 a^{2}+4 b^{2}}\)
= √4 \(\sqrt{a^{2}+b^{2}}\)
= \(2 \sqrt{a^{2}+b^{2}}\)

Question 2.
Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B
discussed in section 7.2.
Solution:
Given points are: A (0, 0) and B (36, 15)
Distance, AB = \(\sqrt{(0-36)^{2}+(0-15)^{2}}\)
\(\sqrt{1296+225}=\sqrt{1521}\) = 39.
According to Section 7.2
Draw the distinct points A (0, 0) and B (36, 15) as shown in figure.

Draw BC ⊥ on X-axis.
Now, In rt. ∠d ∆ACB,
AB = \(\sqrt{\mathrm{AC}^{2}+\mathrm{BC}^{2}}\)
= \(\sqrt{(36)^{2}+(15)^{2}}\)
= \(\sqrt{1296+225}=\sqrt{1521}\)
= 39.
Hence, required distance between points is 39.

Question 3.
Determine if the points (1, 5), (2, 3) and (- 2, – 11) are collinear.
Solution:
Given point are : A (1. 5); B (2.3) and C (- 2, – 11).
AB = \(\sqrt{(2-1)^{2}+(3-5)^{2}}\)
= \(\sqrt{1+4}=\sqrt{5}\)

BC = \(\sqrt{(-2-2)^{2}+(-11-3)^{2}}\)
= \(\sqrt{16+196}=\sqrt{212}\)

CA = \(\sqrt{(1+2)^{2}+(5+11)^{2}}\)
= \(\sqrt{9+256}=\sqrt{265}\)
From above distances, it is clear that sum of any two is not equal to third one.
Hence, given points are not collinear

Question 4.
Check whether (5, – 2); (6, 4) and (7, – 2) are the Vertices of an isosceles triangle.
Solution:
Given points be A (5, – 2); B (6, 4) and C (7, – 2).
AB = \(\sqrt{(5-6)^{2}+(-2-4)^{2}}\)
= \(\sqrt{1+36}=\sqrt{37}\)

BC = \(\sqrt{(6-7)^{2}+(4+2)^{2}}\)
= \(\sqrt{1+36}=\sqrt{37}\)

CA = \(\sqrt{(7-5)^{2}+(-2+2)^{2}}\)
= \(\sqrt{4+0}=2\)
From above discussion, it is clear that AB = BC = √37.
Given points are vertices of an isosceles triangle.

Question 5.
In a classroom, 4 friends are seated at the points A, B, C and D as shown in fig. Champa and Charnel walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square”? Chameli disagrees. Using distance formula, find which of them is correct, and why?

Solution:
In the given diagram, the vertices of given points are : A (3, 4); B (6, 7); C (9, 4) and D (6, 1).
Now,
AB = \(\sqrt{(6-3)^{2}+(7-4)^{2}}\)
= \(\sqrt{9+9}=\sqrt{18}\)

BC = \(\sqrt{(9-6)^{2}+(4-7)^{2}}\)
= \(\sqrt{9+9}=\sqrt{18}\)

CD = \(\sqrt{(6-9)^{2}+(1-4)^{2}}\)
= \(\sqrt{9+9}=\sqrt{18}\)

DA=\(\sqrt{(3-6)^{2}+(4-1)^{2}}\)
= \(\sqrt{9+9}=\sqrt{18}\)

AC = \(\sqrt{(9-3)^{2}+(4-4)^{2}}\)
= \(\sqrt{36+0}=6\)

BD = \(\sqrt{(6-6)^{2}+(1-7)^{2}}\)
= \(\sqrt{0+36}\) = 6
From above discussion, it is clear that
AB = BC = CD = DA = √18 and AC = BD = 6.
ABCD formed a square and Champa is correct about her thinking.

Question 6.
Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(i) ( 1,- 2), (1, 0),(- 1, 2), (- 3, 0)
(ii) ( 3, 5), (3, 1), (0, 3), (- 1, – 4)
(iii) (4, 5), (7, 6), (4, 3), (1, 2).
Solution:
(i) Given points be A (- 1, – 2); B(1, 0); C(- 1, 2) and D(- 3, 0).
AB = \(\sqrt{(1+1)^{2}+(0+2)^{2}}\)
= \(\sqrt{4+4}=\sqrt{8}\)

BC = \(\sqrt{(-1-1)^{2}+(2-0)^{2}}\)
= \(\sqrt{4+4}=\sqrt{8}\)

CD = \(\sqrt{(-3+1)^{2}+(0-2)^{2}}\)
= \(\sqrt{4+4}=\sqrt{8}\)

DA = \(\sqrt{(-1+3)^{2}+(-2+0)^{2}}\)
= \(\sqrt{4+4}=\sqrt{8}\)

AC = \(\sqrt{(-1+1)^{2}+(2+2)^{2}}\)
= \(\sqrt{0+16}=4\)

BD = \(\sqrt{(-3-1)^{2}+(0-0)^{2}}\)
= \(\sqrt{16+0}=4\)

From above discussion, it is clear that
AB = BC = CD = DA = √8 and AC = BD = 4.
Hence, given quadrilateral ABCD is a square.

(ii) Given points be A (- 3, 5); B (3, 1); C (0, 3) and D (- 1,- 4)
AB = \(\sqrt{(-3-3)^{2}+(5-1)^{2}}\)
= \(\sqrt{36+16}=\sqrt{52}=\sqrt{4 \times 13}\)
= 2√13

BC = \(\sqrt{(3-0)^{2}+(1-3)^{2}}\)
= \(\sqrt{9+4}=\sqrt{13}\)

CA = \(\sqrt{(0+3)^{2}+(3-5)^{2}}\)
= \(\sqrt{9+4}=\sqrt{13}\)
Now, BC + CA = \(\sqrt{13}+\sqrt{13}\) = 2√13 = AB
∴A, B and C are collinear then A, B, C and D do not form any quadrilateral.

(iii) Given points are A (4, 5); B (7, 6); C (4, 3) and D (1, 2)
AB = \(\sqrt{(7-4)^{2}+(6-5)^{2}}\)
= \(\sqrt{9+1}=\sqrt{10}\)

BC = \(\sqrt{(4-7)^{2}+(3-6)^{2}}\)
= \(\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}\)

CD = \(\sqrt{(1-4)^{2}+(2-3)^{2}}\)
= \(\sqrt{9+1}=\sqrt{10}\)

DA = \(\sqrt{(4-1)^{2}+(5-2)^{2}}\)
= \(\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}\)

AC = \(\sqrt{(4-4)^{2}+(3-5)^{2}}\)
= \(\sqrt{0+4}\) = 2

BD = \(\sqrt{(1-7)^{2}+(2-6)^{2}}\)
= \(\)

From above discussion, it is clear that AB = CD and BC = DA. and AC ≠ BD.
i.e., opposite sides are equal but their diagonals are not equal.
Hence, given quadrilateral ABCD is a parallelogram.

Question 7.
Find the points on the x-axis which is equidistant from (2, – 5) and (- 2, 9).
Solution:
Let required point be P (x, 0) and given points be A (2, – 5) and B (- 2, 9).
According to question,
PA = PB
(PA)² = (PB)²
or (2 – x)² + (- 5- 0)² = (- 2 – x)² + (9 – 0)²
or 4 + x² – 4x + 25 = 4 + x²+ 4x + 81
-8x = 56
x = \(\frac{4}{4}\) = – 7
Hence, required point be (- 7, 0).

Question 8.
Find the values of y for which the distance between the points P (2, – 3) and Q (10, y) is 10 units.
Solution:
Given points are P (2, – 3) and Q (10, y)
PQ = \(\sqrt{(10-2)^{2}+(y+3)^{2}}\)
= \(\sqrt{64+y^{2}+9+6 y}\)
= \(\sqrt{y^{2}+6 y+73}\)
According to question,
PQ = 10
or \(\sqrt{y^{2}+6 y+73}\) = 10
Squaring
or y² + 6y + 73 = 100
or y² + 6y – 27 = 0
or y² + 9y – 3y – 27 = 0
S = 6 P = – 27
or y (y + 9) – 3 (y + 9) = 0
or (y + 9) (y – 3) = 0
Either y + 9 = 0 or y – 3 = 0
y = – 9 or y = 3
Hence, y = – 9 and 3.

Question 9.
If Q (0, 1) is equidistant from P (5, – 3) and R (x, 6), find the values of x. Also find the distances QR and PR.
Solution:
Given points Q (0, 1); P (5, – 3) and R (x, 6)
QP = \(\sqrt{(5-0)^{2}+(-3-1)^{2}}\)
= \(\sqrt{25+16}=\sqrt{41}\)

and QR = \(\sqrt{(x-0)^{2}+(6-1)^{2}}\)
= \(\sqrt{x^{2}+25}\)

According to question,
QP = QR
or \(\sqrt{41}=\sqrt{x^{2}+25}\)
Squaring
or 41 = x² + 25
or x² = 16
or x = ± √16 = ± √4.

When x = 4 then R (4, 6).
QR = \(\sqrt{(4-0)^{2}+(6-1)^{2}}\)
= \(\sqrt{16+25}=\sqrt{41}\)

PR = \(\sqrt{(4-5)^{2}+(6+3)^{2}}\)
= \(\sqrt{1+81}=\sqrt{82}\)

When x = – 4 then R (- 4, 6).
QR = \(\sqrt{(-4-0)^{2}+(6-1)^{2}}\)
= \(\sqrt{16+25}=\sqrt{41}\)

PR = \(\sqrt{(-4-5)^{2}+(6+3)^{2}}\)
= \(\sqrt{81+81}=\sqrt{162}\).

Question 10.
Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (- 3, 4).
Solution:
Let required points be P (x, y) and given points are A (3, 6) and B (- 3, 4)
PA = \(\sqrt{(3-x)^{2}+(6-y)^{2}}\)
= \(\sqrt{9+x^{2}-6 x+36+y^{2}-12 y}\)
= \(\sqrt{x^{2}+y^{2}-6 x-12 y+45}\)

and PB = \(\sqrt{(-3-x)^{2}+(4-y)^{2}}\)
= \(\sqrt{9+x^{2}+6 x+16+y^{2}-8 y}\)
= \(\sqrt{x^{2}+y^{2}+6 x-8 y+25}\)

According to question,
PA = PB
\(\sqrt{x^{2}+y^{2}-6 x-12 y+45}\) = \(\sqrt{x^{2}+y^{2}+6 x-8 y+25}\)
sq,. both sides, we have,
or x² + y² – 6x – 12y + 45 = x² + y² + 6x – 8y – 25
or -12x – 4y + 20 = 0
or 3x + y – 5 = 0 is the required relation.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 6 Triangles Ex 6.6 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.6

Question 1.
In figure, PS is bisector of ∠QPR of ∆PQR. Prove that = \(\frac{\mathrm{QS}}{\mathrm{SR}}=\frac{\mathrm{PQ}}{\mathrm{PR}}\).

Solution:
Given: ∆PQR. PS is bisector of ∠QPR
i.e., ∠1 = ∠2
To prove: \(\frac{\mathrm{QS}}{\mathrm{SR}}=\frac{\mathrm{PQ}}{\mathrm{PR}}\)
Construction : Through R draw a line parallel to PS to meet QP produced at T.
Proof: In ∆QRT, PS || TR

∠1 = ∠4 (Corresponding angle)
but ∠1 = ∠2 (given)
∴ ∠3 = ∠4
In ∆PRT,
∠3 = ∠4 (Proved)
PT = PR
[Equal side have equal angle opposite to it]
In ∆QRT,
PS || TR
∴ \(\frac{\mathrm{QP}}{\mathrm{PT}}=\frac{\mathrm{QS}}{\mathrm{SR}}\)
[By Basic Proportionality Theorem]
\(\frac{\mathrm{QP}}{\mathrm{PR}}=\frac{\mathrm{QS}}{\mathrm{SR}}\) (PT = PR)
\(\frac{\mathrm{PQ}}{\mathrm{PR}}=\frac{\mathrm{QS}}{\mathrm{SR}}\)
Which is the required result.

Question 2.
In the given fig., D is a point on hypotenuse AC of ∆ABC, DM ⊥ BC, DN ⊥ AB, prove that:
(i) DM² = DN.MC
(ii) DN² = OMAN.

Solution:
Given: ∆ABC, DM ⊥ BC, DN ⊥ AB
To prove: DM² = DN . AC
DN² = DM . AN.
Proof: BD ⊥ AC (Given)
⇒ ∠BDC = 90°
⇒ ∠BDM + ∠MDC = 90°
In ∠DMC, ∠DMC = 90°
[∵ DM ⊥ BC (Given)]
⇒ ∠C + ∠MDC = 90°
From (1) and (2),
∠BDM + ∠MDC = ∠C + ∠MDC
∠BDM =∠C
[Cancelling ∠MDC from both sides]
Now in ∆BMD and ∆MDC,
∠BDM = ∠C [Proved)
∠BMD = ∠DMC [Each 90°]
∆BMD ~ ∆MDC [By AA criterion of similarity]
⇒ \(\frac{\mathrm{DM}}{\mathrm{BM}}=\frac{\mathrm{MC}}{\mathrm{DM}}\)
[∵ Corresponding sides of similar triangles are proportional]
⇒ DM² = BM × MC
⇒ DM² = DN × MC [∵ BM = DN]
Similarly, ∆NDA ~ ∆NBD
⇒ \(\frac{\mathrm{DN}}{\mathrm{BN}}=\frac{\mathrm{AN}}{\mathrm{DN}}\)
[∵ Corresponding sides of similar triangles are próportional]
⇒ DN² = BN × AN
⇒ DN² = DM × AN .
Hence proved.

Question 3.
In fig., ABC is triangle in which ∠ABC > 90° and AD ⊥ BC produced, prove that AC² = AB² + BC² + 2BC.BD.

Solution:
Given: ∠ABC, AD ⊥ BC when produced, ∠ABC > 90°.
To prove : AC² = AB² + BC² + 2BC. BD.
Proof: Let BC = a,
CA = b,
AB = c,
AD = h
and BD = x.
In right-angled ∆ADB,
Using Pythagoras Theorem.
AB² = BD² + AD²
i.e., c² = x² + h²
Again, in right-angled AADC,
AC² = CD² + AD²
i.e.. b² = (a + x)² + h²
= a² + 2ax + x² + h²
= a² + 2ax + c²; [Using (1)]
b² = a² + b² + 2w.
Hence, AB² = BC² + AC² + 2BC × CD.

Question 4.
In fig., ABC is a triangle in which ∠ABC < 90°, AD ⊥ BC, prove that AC² = AB² + BC² – 2BC.BD.

Solution:
Given: ∆ ABC, ∠ABC < 90°, AD ⊥ BC.
To prove : AC² = AB² + BC² – 2BC BD.
Proof: ADC is right-angled z at D
AC² = CD² + DA² (Pythagora’s Theorem) ……………..(1)
Also, ADB is right angled ∆ at D
AB² = AD² + DB² ……………….(2)
From (1), we get:
AC² = AD² + (CB – BD)²
= AD² + CB² + BD² – 2CB × BD
or AC² = (BD² + AD²) + CB² – 2CB × BD
AC² = AB² + BC² – 2BC × BD. [Using (2)]

Question 5.
In fig., AD is a median of a triangle ABC and AM ⊥ BC. Prove that:
(i) AC² = AD² + BC. DM + \(\left(\frac{B C}{2}\right)^{2}\)
(ii) AB² = AD² ± BC.DM + \(\left(\frac{B C}{2}\right)^{2}\)
(iii) AC² + AB² = 2 AD² + \(\frac{1}{2}\) BC²

Solution:
Given: ∆ABC, AM ⊥ BC,
AD is median of ¿ABC.
To prove:
(i) AC² = AD² + BC. DM + \(\left(\frac{B C}{2}\right)^{2}\)
(ii) AB² = AD² ± BC.DM + \(\left(\frac{B C}{2}\right)^{2}\)
(iii) AC² + AB² = 2 AD² + \(\frac{1}{2}\) BC²
Proof: In ∆AMC.
AC² = AM² + MC²
= AM² + (MD + DC)²
AC² = AM² + MD² + DC² + 2MD × DC
AC² = (AM² + MD²) + \(\left(\frac{\mathrm{BC}}{2}\right)^{2}\) + 2 . MD \(\left(\frac{\mathrm{BC}}{2}\right)\)
AC² = AD² + BC . MD + \(\frac{\mathrm{BC}^{2}}{4}\) …………(1)

(ii) In right angled triangle AME,
AB² = AM² + BM²
= AM² + (BD – MD)²
=AM² + BD² + MD² – 2BD × MD
= (AM² + MD²) + BD² – 2(\(\frac{1}{2}\) BC) MD
= AD² + (\(\frac{1}{2}\) BC)² – BC . MD
[∵ In ∆ AMD; AD² = MA² + MD²]
AB² + AD² (\(\left(\frac{\mathrm{BC}}{2}\right)^{2}\)) – BC . MD ………….(2)

(iii) Adding (1) and (2),
AB² + AC² = AD² + BC.MD + (\(\left(\frac{\mathrm{BC}}{2}\right)^{2}\)) + AD² + (\(\left(\frac{\mathrm{BC}}{2}\right)^{2}\)) – BC . MD
= 2 AD² + \(\frac{\mathrm{BC}^{2}}{4}+\frac{\mathrm{BC}^{2}}{4}\)
= 2AD² + 2 \(\frac{\mathrm{BC}^{2}}{4}\)
AB² + AC² = 2AD² + \(\frac{\mathrm{BC}^{2}}{2}\)
Which is the required result.

Question 6.
Prove that sum of squares of the diagonals of a parallelogram is equal to sum of squares of its sides.
Solution:

Given:
Let ABCD be a parallelogram in which diagonalš AC and BD intersect at point M.
To prove: AB² + BC² + CD² + DA² = AC² + BD²
Solution:
Proof: Diagonals of a parallelogram bisect each other.
∴ In || gm ABCD,
Diagonal BD and AC bisect each other.
Or MB and MD are medians of ∆ABC and ∆ADC respectively.
We know that, if AD is a medians of ¿ABC,
then AB² + AC² = 2AD² + BC²
Using this result, we get:
AB² + BC² = 2 BM² + \(\frac{1}{2}\) AC² ………..(1)
and AD² + CD² = 2 DM² + \(\frac{1}{2}\) AC² ………….(2)
Adding (1) and (2), we get:
AB² + BC² + AD² + CD² = 2 (BM² + DM²) + (AC² + AC²)
AB² + BC² + AD² + CD² = 2 (\(\frac{1}{2}\) BD² + \(\frac{1}{4}\) BD²) + AC²
AB² + BC² + AD² + CD² = BD² + AC²
Hence, sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

Question 7.
In fig., two chords AB and CD intersect each other at the point P prove that:
(i) ∆APC ~ ∆DPB
(ii) AP.PB = CP.DP.

Solution:
Given: Circle, AB and CD are two chords intersects each other at P.

To prove:
(i) ∆APC ~ ∆DPB
(ii) AP.PB = CP.DP.
Proof:
(i) In ∆APC and ∆DPB,
∠1 = ∠2 (Vertically opposite angle)
∠3 = ∠4 (angle on same segment)
∴ ∆APC ~ ∆DPB [AA similarity criterion]

(ii) ∆APC ~ ∆DPB (Proved above)
\(\frac{\mathrm{AP}}{\mathrm{DP}}=\frac{\mathrm{PC}}{\mathrm{PB}}\)
(If two triangles are sitnilar corresponding sides are proportional)
AP.PB = PC.DP
Which is the required result.

Question 8.
In fig., two chords AB and CD of a circle intersect each other at point P (when produced) outside the circle prove:
(i) ∆PAC ~ ∆PDB
(ii) PA.PB = PC.PD.

Solution:
Given: AB and CD are two chord of circle intersects each other at P (when produced)
To prove:
(i) ∆PAC ~ ∆PDB
(ii) PA.PB = PC.PD.
Proof:
(i) In ∆PAC and ∆PDB,
∠P = ∠P (Common)
∠PAC = ∠PDB.
(Exterior angle of cyclic quadrilqteral is equal to interior opposite angle)
∴ ∆PAC ~ ∆PDB [AA similarity criterion]

(ii) ∆PAB ~ ∆WDB
∴ \(\frac{\mathrm{PA}}{\mathrm{PD}}=\frac{\mathrm{PC}}{\mathrm{PB}}\)
[If two triangles are similar corresponding sides are proportional]
PA × PB = PC × PD.

Question 9.
In fig., D is a point on side BC of BD AB ∆ABC such that \(\frac{\mathbf{B D}}{\mathbf{D C}}=\frac{\mathbf{A B}}{\mathbf{A C}}\). Prove that: AD is bisector of ∠BAC.

Solution:
Given: A ∆ABC, D is a point on BC such that \(\frac{\mathbf{B D}}{\mathbf{D C}}=\frac{\mathbf{A B}}{\mathbf{A C}}\)
To prove: AD bisects ∠BAC
i.e., ∠1 = ∠2
Construction: Through C draw CE || DA meeting BA produced at E.

Proof:
In ∆BCE, we have:
AD || CE ………(const.)
So, by Basic Proportionality Theorem,
But \(\frac{\mathrm{BD}}{\mathrm{DC}}=\frac{\mathrm{AB}}{\mathrm{AE}}\)
\(\frac{\mathrm{BD}}{\mathrm{DC}}=\frac{\mathrm{AB}}{\mathrm{AC}}\)

⇒ \(\frac{\mathrm{AB}}{\mathrm{AE}}=\frac{\mathrm{AB}}{\mathrm{AC}}\)
⇒ AE = AC

In ∆ACE, we have:
AE = AC
⇒ ∠3 = ∠4 ………. (∠s opp. to equal sides)
Since CE || DA and AC cuts them, then:
∠2 = ∠4 ……….(alt ∠s)
Also CE || DA and BAE cuts them, then:
∠1 = ∠3 …………(Corr. ∠s)
Thus we have:
∠3 = ∠4
⇒ ∠3 = ∠1
But ∠4 = ∠2
⇒ ∠1 = ∠2.
HenCe AD bisects ∠BAC.

Question 10.
Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out? If she pulls in the string at the rate of 5 cm per second, what will the horizontal distance of the fly from her after 12 seconds?

Solution:
A right angled triangle, ABC, in which,
AB = 1.8 cm,
BC = 2.4 cm.
∠B = 90°
By Pythagoras Theorem,
AC² = AB² + BC²
AC² = (1.8)² + (2.4)²
AC² = 3.24 + 5.76 =9
AC² = (3)²
AC = 3 cm
Now, when Nazima pulls in the string at the rate of 5 cm/sec ; then the length of the string decrease = 5 × 12 = 60 cm
= 0.6 m in 12 seconds.
Let after 12 seconds, position of the fly will be at D.
∴ AD = AC – distance covered in 12 seconds
AD = (3 – 0.6) m
AD = 2.4 m
Also, in right angled ∆ABD,
Using Pythagoras Theorem,
AD² = AB² + BD²
(2.4)² = (1.8)² + BD²
BD² = 5.76 – 3.24
BD² = 2.52 m
BD = 1.587 m.
∴ Horizontal distance of the fly from Nazima = BD + 1.2 m
= (1.587 + 1.2) m
= 2.787 m
= 2.79 m
Hence, original length of string and horizontal distance of the fly from Nazima is 3 m and 2.79 m.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 6 Triangles Ex 6.5 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5

Question 1.
Sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse.
(i) 7 cm, 24 cm, 25 cm
(ii) 3 cm, 8 cm, 6 cm
(iii) 50 cm, 80 cm, 100 cm
(iv) 13 cm, 12 cm, 5 cm.
Solution:
(i) Let ∆ABC, with AB = 7 cm BC = 24 cm, AC = 25 cm
AB² + BC² = (7)² + (24)²
= 49 + 576 = 625
AC² = (25)² = 625
Now AB² + BC² = AC²
∴ ∆ABC is right angled triangle. Hyp. AC = 25cm.

(ii) Let ∆PQR with PQ = 3 cm, QR = 8 cm PR = 6 cm
PQ² + PR² = (3)² + (6)²
= 9 + 36 = 45
QR² = (8)² = 64.
Here PQ² + PR² ≠ QR²
∴ ∆PQR is not right angled triangle.

(iii) Let ∆MNP, with MN =50 cm, NP = 80 cm, MP = 100 cm
MN ²+ NP² = (50)² + (80)²
= 2500 + 6400 = 8900
MP² = (100)² = 10000
Here MP² ≠ MN² + NP².
∴ ∆MNP is not right angled triangle.

(iv) Let ∆ABC, AB = 13 cm, BC = 12 cm, AC = 5 cm
BC² + AC² = (12)² + (5)²
= 144 + 25 = 169
AB² = (13)² = 169
∴ AB² = BC² + AC²
∆ABC is right angled triangle.
Hyp. AB = 13 cm.

Question 2.
PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM² = QM . MR.
Solution:
Given: ∆PQR is right angled at P and M is a point on QR such that PM ⊥ QR.
To prove : PM² = QM × MR

Proof: ∠P = 90° (Given)
∴ ∠1 + ∠2 = 90°
∠M = 900 (Given)
In ∆PMQ,
∠1 + ∠3 + ∠5 = 180°
=> ∠1 + ∠3 = 90° [Angle Sum Property] ………….(2) [∠5 = 90°]
From (1) and (2),
∠1 + ∠2 = ∠1 + ∠3
∠2 = ∠3
In ∆QPM and ∆RPM,
∠3 = ∠2 (Proved)
∠5 = ∠6 (Each 90°)
∴ ∆QMP ~ ∆PMR [AA similarity]
\(\frac{{ar} .(\Delta \mathrm{QMP})}{{ar} .(\Delta \mathrm{PMR})}=\frac{\mathrm{PM}^{2}}{\mathrm{MR}^{2}}\)

[If two triangles are similar, ratio o their areas is equal to square of corresponding sides]
\(\frac{\frac{1}{2} \mathrm{QM} \times \mathrm{PM}}{\frac{1}{2} \mathrm{RM} \times \mathrm{PM}}=\frac{\mathrm{PM}^{2}}{\mathrm{MR}^{2}}\)

[area of ∆ = \(\frac{1}{2}\) Base × Altitude]

\(\frac{\mathrm{QM}}{\mathrm{RM}}=\frac{\mathrm{PM}^{2}}{\mathrm{RM}^{2}}\)

PM² = QM × RM Hence proved.

Question 3.
In fig., ABD is a triangle right angled at A and AC ⊥ BD. Show that

(i) AB² = BC.BD
(ii) AC² = BC.DC
(üi) AD² = BD.CD.
Solution:
Given. A right angled ∆ABD in which right angled at A and AC ⊥ BD.
To Prove:
(i) AB² = BC.BD
(ii) AC² = BC.DC .
(iii) AD² = BD.ÇD .
Proof. In ∆DAB and ∆DCA,
∠D = ∠D (common)
∠A = ∠C (each 90°)
∴ ∆DAB ~ ∆DCA [AA similarity]
In ∆DAB and ∆ACB,
∠B = ∠B (common)
∠A = ∠C . (each 90°)
∴ ∆DAB ~ ∆ACB, .
From (1) and (2),
∆DAB ~ ∆ACB ~ ∆DCA.
(i) ∆ACB ~ ∆DAB (proved)
∴ \(\frac{{ar} .(\Delta \mathrm{ACB})}{{ar} .(\Delta \mathrm{DAB})}=\frac{\mathrm{AB}^{2}}{\mathrm{DB}^{2}}\)

[If two triangles are similar corresponding sides are proportional]

\(\frac{\frac{1}{2} \mathrm{BC} \times \mathrm{AC}}{\frac{1}{2} \mathrm{DB} \times \mathrm{AC}}=\frac{\mathrm{AB}^{2}}{\mathrm{DB}^{2}}\)
[Area of triangle = \(\frac{1}{2}\) Base × Altitude]
BC = \(\frac{\mathrm{AB}^{2}}{\mathrm{BD}}\)
AB² = BC × BD.

(iii) ∆ACB ~ ∆DCA (proved)
\(\frac{{ar} .(\Delta \mathrm{DAB})}{{ar} .(\Delta \mathrm{DCA})}=\frac{\mathrm{DA}^{2}}{\mathrm{DB}^{2}}\)
[If two triangles are similar corresponding sidec are proportional]

\(\frac{\frac{1}{2} \mathrm{CD} \times \mathrm{AC}}{\frac{1}{2} \mathrm{BD} \times \mathrm{AC}}=\frac{\mathrm{AD}^{2}}{\mathrm{BD}^{2}}\)

CD = \(\frac{\mathrm{AD}^{2}}{\mathrm{BD}}\)
⇒ AD² = BD × CD.

Question 4.
ABC is an isosceles triangle right angled at C. Prove that AB² = 2AC².
Solution:
Given: ABC is an isosceles triangle right angled at C.
To prove : AB² = 2AC².
Proof: In ∆ACB, ∠C = 90° & AC = BC (given)

AB² = AC² + BC²
[By using Pythagoras Theorem]
=AC² + AC² [BC = AC]
AB² = 2AC²
Hence proved.

Question 5.
ABC is an isosceles triangle with AC = BC. If AB² = 2AC², prove that ABC is right triangle.
Solution:
Given: ∆ABC is an isosceles triangle AC = BC
To prove: ∆ABC is a right triangle.

Proof: AB² = 2AC² (given)
AB² = AC² + AC²
AB² = AC² + BC² [AC = BC]
∴ By Converse of Pythagoras Theorem,
∆ABC is right angled triangle.

Question 6.
ABC is an equilateral triangle of side 2a. Find each of its altitudes.
Solution:
∆ABC is equilateral triangle with each side 2a
AD ⊥ BC
AB = AC = BC = 2a
∆ADB ≅ ∆ADC [By RHS Cong.]
∴ BD = DC = a [c.p.c.t]

In right angled ∆ADB
AB² = AD² + BD²
(2a)² = AD² + (a)²
4a² – a² = AD².
AD² = 3a²
AD = √3a.

Question 7.
Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals. [Pb. 2019]
Solution:
Given: Rhombus, ABCD diagonal AC and BD intersect each other at O.
To prove:
AB² + BC² + CD² + AD² = AC² + BD²
Proof:The diagonals of a rhombus bisect each other at right angles.

∴ AO = CO, BO = DO
∴ ∠s at O are rt. ∠s
In ∆AOB, ∠AOB = 90°
∴ AB² = AO² + BO² [By Pythagoras Theorem] …………..(1)
Similarly, BC² = CO² + BO² ……………..(2)
CD² = CO² + DO² ……………(3)
and DA² = DO² + AO² ……………….(4)
Adding. (1), (2), (3) and (4), we get
AB² + BC² + CD² + DA² = 2AO² + 2CO² + 2BO² + 2DO²
= 4AO² + 4BO²
[∵ AO = CO and BO = DO]
= (2AO)² + (2BO)² = AC² + BD².

Question 8.
In fig., O is a point In the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that
(i) OA² + OB² + OC² – OD² – OE² – OF² = AF² + BD² + CE²

(ii)AF² + BD² + CE² = AE² + CD² + BF².

Solution:
Given: A ∆ABC in which OD ⊥ BC, 0E ⊥ AC and OF ⊥ AB.

To prove:
(i) AF² + BD² + CE² = OA² + OB² + OC² – OD² – OE² – OF²
(ii) AF² + BD² + CE² = AE² + CD² + BF².
Construction: Join OB, OC and OA.
Proof: (i) In rt. ∠d ∆AFO, we have
OA² = OF² + AF² [By Pythagoras Theorem]
or AF² = OA² – OF² …………..(1)

In rt. ∠d ∆BDO, we have:
OB² = BD²+ OD² [By Pythagoras Theorem]
⇒ BD² = OB² – OD² …………..(2)

In rt. ∠d ∆CEO, we have:
OC² = CE² + OE² [By Pythagoras Theorem]

⇒ CE² = OC² – OE² ……………(3)

∴ AF² + BD² + CE² = OA² – OF² + OB² – OD² + OC² – 0E²
[On adding (1), (2) and (3)]
= OA² + OB² + OC² – OD² – OE² – OF²
which proves part (1).
Again, AF² + BD² + CE² = (OA² – OE²) + (OC² – OD²) + (OB² – OF²)
= AE² + CD² + BF²
[∵AE² = AO² – OE²
CD² = OC² – OD²
BF² = OB² – OF²].

Question 9.
A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.
Solution:
Height of window from ground (AB) = 8m.

Length of ladder (AC) = 10 m
Distance between foot of ladder and foot of wall (BC) = ?
In ∆ABC,
AB² + BC² = AC² [By Pythagoras Theorem]
(8)² + (BC)² = (10)²
64 + BC² = 100
BC² = 100 – 64
BC = √36
BC = 6 cm.
∴ Distance between fóot of ladder and foot of wall = 6 cm.

Question 10.
A guy wire attached to a vertical pole of height 18 m Is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?
Solution:
Let AB is height of pole (AB) = 18 m
AC is length of wire = 24 m

C is position of stake AB at ground level.
In right angle triangle ABC,
AB² + BC² = AC² [By Pythagoras Theorem]
(18)² + (BC)² = (24)²
324 + (BC)² = 576
BC² = 576 – 324
BC = \(\sqrt{252}=\sqrt{36 \times 7}\)
BC = 6√7 m.

Question 11.
An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. How far apart will be the two pLanes after 1\(\frac{1}{2}\) hours?
Solution:
Speed of first aeroplane = 1000km/hr.

Distance covered by first aeroplane due north in 1\(\frac{1}{2}\) hours =1000 × \(\frac{3}{2}\)
OA = 1500 km
Speed of second aeroplane = 1200 km/hr.
Distance covered by second aeroplane in 1\(\frac{1}{2}\) hours = 1200 × \(\frac{3}{2}\)
OB = 1800 km.
In right angle ∆AOB
AB² = AO² + OB² [By Phyrhagoras Theorem]
AB² = (1500)² + (1800)²
AB = \(\sqrt{2250000+3240000}\)
= \(\sqrt{5490000}\)
= \(\sqrt{61 \times 90000}\)
AB = 300√61 km.
Hence, Distance between two aeroplanes = 300√61 km.

Question 12.
Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the
distance between their tops.
Solution:
Height of pole AB = 11 m
Height of pole (CD) = 6 m

Distance between foot of pole = 12 m
from C draw CE ⊥ AB. such that
BE = DC = 6 m
AE = AB – BE = (11 – 6) m = 5 m.
and CE = DB = 12 m.
In rt. ∠d ∆AEC,
AC² = AE² + FC²
[By Phythagoras Theorem)
AC = \(\sqrt{(5)^{2}+(12)^{2}}\)
= \(\sqrt{25+144}\)
= \(\sqrt{169}\) = 13.
Hence, Distance between their top = 13m.

Question 13.
D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C.
Prove that AE² + BD² = AB² + DE².
Solution:
Given: In right angled ∆ABC, ∠C = 90° ;
D and E are points on sides CA & CB respectively.
To prove: AE² + BD² = AB² + DE²
Proof: In rt. ∠d ∆BCA,
AB² = BC² + CA² …………..(1) [By Pythagoras Theorem]

In rt. ∠d ∆ECD,
DE² = EC² + DC² ……………….(2) [By Pythagoras Theorem]
In right angled triangle ∆ACE,
AE² = AC² + CE² ……………….(3)
In right angled triangle ∆BCD
BD² = BC² + CD² ……………….(4)
Adding (3) and (4),
AE² + BD² = AC² + CE² + BC² + CD²
= [AC² + CB²] + [CE² + DC²]
= AB² + DE²
[From (1) and (2)]
Hence 2 + BD² = AB² + DE².
Which is the required result.

Question 14.
The perpendicular from A on side BC of a ∆ABC intersects BC at D such that DB = 3 CD. Prove that 2AB² = 2AC² + BC².

Solution:
Given: ∆ABC, AD ⊥ BC
BD = 3CD.
To prove: 2AB² = 2AC² + BC².

Proof: In rt. ∠d triangles ADB and ADC, we have
AB² = AD² + BD²;
AC² = AD² + DC² [By Pythagoras Theorem]
∴ AB² – AC² = BD² – DC²
= 9 CD² – CD²; [∵ BD = 3CD]
= 8CD² = 8 (\(\frac{\mathrm{BC}}{4}\))²
[∵ BC = DB + CD = 3 CD + CD = 4 CD]
∴ CD = \(\frac{1}{4}\) BC
∴ AB² – AC² = \(\frac{\mathrm{BC}^{2}}{2}\)
⇒ 2(AB² – AC²) = BC²
⇒ 2AB² – 2AC² = BC²
∴ 2AB² = 2AC² + BC².
Which is the required result.

Question 15.
In an equilateral triangle ABC, D is a point on side BC such that BD = \(\frac{1}{3}\) BC. Prove that 9 AD² = 7 AB².
Solution:
Given: Equilateral triangle ABC, D is a point on side BC such that BD = \(\frac{1}{3}\) BC.
To prove: 9AD² = 7 AB².
Construction: AB ⊥ BC.
Proof: ∆AMB ≅ ∆AMC [By R.HS. Rule since AM = AM and AB = AC]

∴ BM = MC = \(\frac{1}{2}\) BC [c.p.c.t.]
Again BD = \(\frac{1}{3}\) BC and DC = \(\frac{1}{3}\) BC (∵ BC is trisected at D)
Now in ∆ADC, ∠C is acute
∴ AD² = 2AC² + DC² – 2 DC × MC
= AC² + \(\left[\frac{2}{3} \mathrm{BC}\right]^{2}\) – 2 \(\left[\frac{2}{3} \mathrm{BC}\right] \frac{1}{2} \mathrm{BC}\)

[∵ DC = \(\frac{2}{3}\) BC and MC = \(\frac{1}{2}\) BC]
= AB² + \(\frac{4}{9}\) AB² – \(\frac{2}{3}\) AB²
[∵ AC = BC = AB]
= (1 + \(\frac{4}{9}\) – \(\frac{2}{3}\)) AB²

= \(\left(\frac{9+4-6}{9}\right) \mathrm{AB}^{2}=\frac{7}{9} \mathrm{AB}^{2}\)

∴ AD² = \(\frac{7}{9}\) AB²
⇒ 9 AD² = 7 AB².

Question 16.
In an equilateral triangle, prove that three times the square of one side Ls equal to four times the square of one of its
altitudes.
Solution:
Given:
ABC is equilateral ∆ in which AB = BC = AC

To prove: 3 AB² = 4 AD²
Proof: In right angled ∆ABD,
AB² = AD² + BD² (Py. theorem)
AB² = A BD² (Py. theorem)

AD² = \(\frac{3}{4}\) AB²
⇒ 4 AD² = 3 AB²
Hence, the result.

Question 17.
Tick the correct answer and justify: In ∆ABC, AB = 6 cm, AC = 12 cm and BC = 6√3 cm. [The angles of B are respectively
(A) 120°
(B) 64°
(C) 90°
(D) 45°
Solution.
AC = 12 cm
AB = 6√3 cm
BC = 6 cm
AC² = (12)² = 144 cm
AB² + BC² = (6√3)² + (6)²
= 108 + 36
AB√3 + BC√3 = 144
∴ AB√3 + BC√3 = AC√3
Hence by converse of pythagoras theorem ∆ABC is right angred triangle right angle at B
∴ ∠B = 90°
∴ correct option is (C).

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 4 Quadratic Equations Ex 4.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.1

Question 1.
Check whether the following are quadratic equations:
(i) (x + 1)² = 2(x – 3)
(ii) x² – 2x = (-2) (3 – x)
(iii) (x – 2) (x + 1) = (x – 1) (x + 3)
(iv) (x – 3)(2x + 1) = x (x + 5)
(v) (2x – 1) (x – 3) = (x + 5) (x – 1)
(vi) x² + 3x + 1 = (x – 2)
(vii) (x + 2)³ = 2x(x² – 1)
(viii) x³ – 4x² – x + 1 = (x – 2)³

Solution:
(i) Given that
(x + 1)² = 2(x – 3)
Or x² + 1 + 2x = 2x – 6
Or x² + 1 + 2x – 2x + 6 = 0
Or x² + 7 = 0
Or x² + 0x + 7 = 0
which is in the formof ax² + bx + c = 0;
∴ It is a quadratic equation.

(ii) Given that
x² – 2x = (-2) (3 – x)
Or x² – 2x = -6 + 2x
Or x² – 2x + 6 – 2x = 0
Or x² – 4x + 6 = 0
which is the form of ax² + bx + c = 0; a ≠ 0
∴ It is the quadratic equation.

(iii) Given that ,
(x – 2) (x + 1) = (x – 1) (x + 3)
Or x² + x – 2x – 2 = x² + 3x – x – 3
Or x² – x – 2 = x² + 2x – 3
Or x² – x – 2 – x² -2x + 3 = 0
Or -3x + 1 = 0 which have no term of x².
So it is not a quadratic equation.

(iv) Given that
(x – 3)(2x + 1) = x(x + 5)
Or 2x² + x – 6x – 3 = x² + 5x
Or 2x² – 5x – 3 – x² – 5x = 0
Or x² – 10x – 3 = 0
which is a form of ax² + bx + c = 0; a ≠ 0
∴ It is a quadratic equation.

(v) Given that ,
(2x – 1) (x – 3) = (x + 5) (x – 1)
0r2x² – 6x – x + 3 = x² – x + 5x – 5
Or 2x² – 7x + 3 = x² + 4x – 5
Or 2x² – 7x + 3 – x² – 4x + 5 = 0
Or x² – 11x + 8 = 0
which is a form of ax² + bx + c = 0; a ≠ 0
∴ It is a quadratic equation.

(vi) Given that
x²+3x+1 = (x – 2)²
Or x² + 3x + 1 = x² + 4 – 4x
Or x² + 3x + 1 – x² – 4 + 4x = 0
Or 7x – 3 = 0
which have no term of x².
So it is not a quadratic equation.

(vii) Given that
(x + 2)³ = 2x(x² – 1)
Or x³ + (2)³ + 3 (x)² 2 + 3(x)(2)² = 2x³ – 2x
Or x³ + 8 + 6x² + 12x = 2x³ – 2x
Or x³ + 8 + 6x² + 12x – 2x³ + 2x = 0
Or -x³ + 6x² + 14x + 8 = 0
Here the highest degree of x is 3. which is a cubic equation.
∴ It is not a quadratic equation.

(viii) Given that
x³ – 4x² – x+ 1= (x – 2)³
Or x³ – 4x² – x + 1 = x³ – (2)³ + 3(x)² (-2) + 3 (x) (-2)²
Or x³ – 4x² – x + 1 = x³ – 8 – 6x² + 12x
Or x³ – 4x² – x + 1 – x³ + 8 + 6x² – 12x = 0
Or 2x² – 13x + 9 = 0
which is in the form of ax² + bx +c = 0; a ≠ 0
∴ It is a quadratic equation.

Question 2.
Represent the following situations in the form of quadratic equations:
(i) The area of a rectangular plot is 528 m². The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.

(ii) The product of two consecutive positive integers is 306. We need to find the integers.

(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.

(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.

Solution:
(i) Let Breadth of rectangular plot = x m
Length of rectangular plot= (2x + 1) m
∴ Area of rectangular plot = [x (2x + 1)] m² = (2x² + x) m²
According to question,
2x² + x = 528
S = 1
P = -528 × 2 = -1056
0r 2x² + x – 528 = 0
Or 2x² – 32x + 33x – 528 = 0
Or 2x(x – 16) + 33(x – 16) = 0
Or (x – 16) (2x + 33) = 0
Either x – 16 = 0 Or 2x + 33 = 0
x = 16 Or x = 2
∵ breadth of any rectangle cannot be negative, so we reject x = \(\frac{-33}{2}\), x = 16
Hence, breadth of rectangular plot = 16 m
Length of rectangular plot = (2 ×16 + 1)m = 33m
and given problem in the form of Quadratic Equation are 2x² + x – 528 = 0.

(ii) Let two consecutive positive integers are x and x + 1.
Product of Integers = x (x + 1) = x² + x
According to question,
Or x² + x – 306 = 0
S = 1, P = – 306
Or x² + 18x – 17x – 306 = 0
Or x(x + 18) -17 (x + 18) = 0
Or (x + 18) (x – 17) = 0
Either x + 18 = 0 Or x – 17 = 0
x = -18 Or x = 17
∵ We are to study about the positive integers, so we reject x = – 18.
x = 17
Hence, two consecutive positive integers are 17, 17 + 1 = 18
and given problem in the form of Quadratic Equation is x² + x – 306 = 0.

(iii) Let present age of Rohan = x years
Rohan’s mother’s age = (x + 26) years
After 3 years, Rohan’s age = (x + 3) years
Rohan’s mother’s age = (x + 26 + 3) years = (x + 29) years
∴ Their product = (x + 3) (x + 29)
= x² + 29x + 3x + 87
= x² + 32x + 87
According to question,
x² + 32x + 87 = 360
Or x² + 32x + 87 – 360 = 0
Or x² + 32x – 273 = 0
Or x² + 39x – 7x – 273 = 0
S = 32, P = – 273
Or x(x + 39) – 7(x + 39) = 0
Or (x + 39) (x – 7) =
Either x + 39 = Or x – 7 = 0
x = -39 Or x = 7
∵ age of any person cannot be negative so, we reject x = -39
∴ x = 7
Hence, Rohans present age = 7 years
and given problem in the form of Quadratic Equation is x² + 32x – 273 = 0.

(iv) Let u km/hour be the speed of train.
Distance covered by train = 480 km
Time taken by train = \(\frac{480}{u}\) hour
[ Using, Speed = \(\frac{\text { Distance }}{\text { Time }}\)
or Time = \(=\frac{\text { Distance }}{\text { Speed }}\) ]

If speed of train be decreased 8km/hr.
∴ New speed of train = (u – 8) km/hr.
and time taken by train = \(\frac{480}{u-8}\) hour
According to question.

or 3840 = 3 (u² – 8u)
or u² – 8u = 1280
or u² – 8u – 1280=0
or u² – 40u + 32u – 1280 = 0
S = -8, P = – 1280
or u(u – 40) + 32 (u – 40) = 0
or (u – 40)(u + 32) = 0
Either u – 40 = 0
or u + 32 = 0
u = 40 or u = -32
But, speed cannot be negative so we reject
u = – 32
∴ u = 40.
Hence speed of train is 40 km/hr Ans.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 6 Triangles Ex 6.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3

Question 1.
State which pairs of triangles in Fig. are similar. Write the similarity criterion used by you for answering the queStion and also write the pairs of similar triangles in the symbolic form:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

Solution:
(i) In ∆ABC and ∆PQR,
∠A = ∠P (each 60°)
∠B = ∠Q (each 80°)
∠C = ∠R (each 40°)
∴ ∆ABC ~ PQR [AAA Similarity criterion]

(ii) In ∆ABC and ∆PQR,
\(\frac{\mathrm{AB}}{\mathrm{RQ}}=\frac{2}{4}=\frac{1}{2}\) …………….(1)

\(\frac{\mathrm{AC}}{\mathrm{PQ}}=\frac{3}{6}=\frac{1}{2}\) ……………..(2)

\(\frac{\mathrm{BC}}{\mathrm{PR}}=\frac{2.5}{5}=\frac{1}{2}\) ……………(3)
From (1), (2) and (3),
\(\frac{\mathrm{AB}}{\mathrm{RQ}}=\frac{\mathrm{AC}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{PR}}=\frac{1}{2}\)

∴ ΔABC ~ ΔQRP [By SSS similarity criterion]

(iii) In ΔLMP and ΔDEF,
\(\frac{\mathrm{MP}}{\mathrm{DE}}=\frac{2}{4}=\frac{1}{2}\)

\(\frac{\mathrm{PL}}{\mathrm{DF}}=\frac{3}{6}=\frac{1}{2}\) \(\frac{\mathrm{LM}}{\mathrm{EF}}=\frac{2.7}{5}=\frac{27}{50}\)

\(\)
∴ Two Triangles are not similar.

(iv) In ΔMNL and ΔPQR,
\(\frac{\mathrm{ML}}{\mathrm{QR}}=\frac{5}{10}=\frac{1}{2}\)

∠M = ∠Q (each 70°)

\(\frac{\mathrm{MN}}{\mathrm{PQ}}=\frac{2.5}{5}=\frac{1}{2}\)

∴ ΔMNL ~ ΔPQR [By SAS similarity cirterion]

(v) In ΔABC and ΔDEF,
\(\frac{\mathrm{AB}}{\mathrm{DF}}=\frac{2.5}{5}=\frac{1}{2}\)

\(\frac{\mathrm{BC}}{\mathrm{EF}}=\frac{3}{6}=\frac{1}{2}\)

But ∠B ≠ ∠F
∴ ΔABC and ΔDEF are not similar.

(vi) In ΔDEF, ∠D = 70°, ∠E = 80°
∠D + ∠E + ∠F = 180°
70° + 80° + ∠F = 180° [Angle Sum Propertyl
∠F= 180° – 70° – 80°
∠F = 30°
In ΔPQR,
∠Q = 80°, ∠R = 30°
∠P + ∠Q + ∠R = 180°
(Sum of angles of triangle)
∠P + 80° + 30° = 180°
∠P = 180° – 80° – 30°
∠P = 70°
In ΔDEF and ΔPQR,
∠D = ∠P (70° each)
∠E = ∠Q (80° each)
∠F = ∠R (30° each)
∴ ΔDEF ~ ΔPQR (AAA similarity criterion).

Question 2.
In Fig., ΔODC ~ ΔOBA, ∠BOC = 125° and ∠CDO = 70°. FInd ∠DOC, ∠DCO and ∠OAB.

Solution:
Given that: ∠BOC = 125°
∠CDO = 70°
DOB is a straight line
∴ ∠DOC + ∠COB = 180°
[Linear pair Axiom]
∠DOC + 125° = 180°
∠DOC = 180°- 125°
∠DOC = 55°
∠DOC = ∠AOB = 55°
[Vertically opposite angle]
But ΔODC ~ ΔOBA
∠D = ∠B = 70°
In ΔDOC, ∠D + ∠O + ∠C = 180°
70° + 55° + ∠C = 180°
∠C= 180° – 70° – 55°
∠C = 55°
∠C = ∠A = 55°
Hence ∠DOC = 55°
∠DCO = 55°
∴ ∠OAB = 55°.

Question 3.
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that \(\frac{\mathbf{O A}}{\mathbf{O C}}=\frac{\mathbf{O B}}{\mathbf{O D}}\).
Solution:

Given: In Trapezium ABCD, AB || CD, and diagonal AC and BD intersects each other at O.
To Prove = \(\frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{OB}}{\mathrm{OD}}\) (Given)
Proof: AB || DC (Given)
In ΔDOC and ΔBOA,
∠1 = ∠2 (alternate angle)
∠5 = ∠6 (vertical opposite angle)
∠3 = ∠4 (alternate angle)
∴ ΔDOC ~ ΔBOA [AAA similarity criterion]
∴ \(\frac{\mathrm{DO}}{\mathrm{BO}}=\frac{\mathrm{OC}}{\mathrm{OA}}\)
[If two triangle are similar corresponding sides are Proportional }
⇒ \(\frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{BO}}{\mathrm{DO}}\)
Hence Proved.

Question 4.
In Fig., \(\) and ∠1 = ∠2. Show that ∆PQS ~ ∆TQR.

Solution:
Given that,
\(\frac{\mathrm{QR}}{\mathrm{QS}}=\frac{\mathrm{QT}}{\mathrm{PR}}\) and
∠1 = ∠2

To Prove. PQS – ITQR
Proof: In ΔPQR,
∠1 = ∠2 (given)
∴ PR = PQ
[Equal angle have equal side opposite to it]
and = \(\frac{\mathrm{QR}}{\mathrm{QS}}=\frac{\mathrm{QT}}{\mathrm{PR}}\) (given)
or \(\frac{\mathrm{QR}}{\mathrm{QS}}=\frac{\mathrm{QT}}{\mathrm{PQ}}\) [PR = PQ]
⇒ \(\frac{\mathrm{QS}}{\mathrm{QR}}=\frac{\mathrm{PQ}}{\mathrm{QT}}\)
In ΔPQS and ΔTQR,
\(\frac{\mathrm{QS}}{\mathrm{QR}}=\frac{\mathrm{PQ}}{\mathrm{QT}}\)
∠1 = ∠1 (common)
∴ ∆PQS ~ ∆TQR [SAS similarity criterion]
Hence proved.

Question 5.
S and T are points on skies PR and QR of ∆PQR such that ∠P = ∠RTS. Show that ∆RPQ ~ ∆RTS.
Solution:
S and T are the points on side PR and QR such that ∠P = ∠RTS.

To Prove. ∆RPQ ~ ∆RTS
Proof: In ∆RPQ and ∆RTS
∠RPQ = ∠RTS (given)
∠R = ∠R [common angle]
∴ RPQ ~ ARTS
[By AA similarity critierion which is the required result.]

Question 6.
In figure ∆ABE ≅ ∆ACD show that ∆ADE ~ ∆ABC.

Solution:
Given. ∆ABC in which ∆ABE ≅ ∆ACD
To Prove. ∆ADE ~ ∆ABC
Proof. ∆ABE ≅ ∆ACD (given)
AB = AC (cpct) and AE = AD (cpct)
\(\frac{A B}{A C}=1\) ……………..(1)
\(\frac{A E}{A D}=1\) …………….(2)
From (1) and (2).
\(\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{AD}}{\mathrm{AE}}\)
In ∆ADE and ∆ABC,
\(\frac{\mathrm{AD}}{\mathrm{AE}}=\frac{\mathrm{AB}}{\mathrm{AC}}\)
∠A = ∠A (common)
∴ ∆ADE ~ ∆ABC [By SAS similarity criterion].

Question 7.
In Fig., altitudes AD and CE of ∆ABC intersect each other at the point P.

Show that:
(i) ∆AEP ~ ∆CDP
(ii) ∆ABD ~ ∆CBE
(iii) ∆AEP ~ ∆ADB
(iv) ∆PDC ~ ∆BEC
Solution:
Given. ∆ABC, AD ⊥ BC CE⊥AB,

To Prove. (i) ∆AEP ~ ∆CDP
(ii) ∆ABD ~ ∆CBE
(iii) ∆AEP ~ ∆ADB
(iv) ∆PDC ~ ∆BEC
Proof:
(i) In ∆AEP and ∆CDP,
∠E = ∠D (each 90°)
∠APE = ∠CPD (vertically opposite angle)
∴ ∆AEP ~ ∆CDP [By AA similarity criterion].

(ii) In ∆ABD and ∆CBE,
∠D = ∠E (each 90°)
∠B = ∠B (common)
∴ ∆ABD ~ ∆CBE [AA Similarity criterion]

(iii) In ∆AEP and ∆ADB.
∠E = ∠D (each 90°)
∠A = ∠A (common)
∴ ∆AEP ~ ∆ADB [AA similarity criterion].

(vi) In ∆PDC and ∆BEC,
∠C = ∠C
∠D = ∠E
∴ ∆SPDC ~ ∆BEC [AA similarity criterion].

Question 8.
E is a point on the side AD produced of a parallelogram ABCD and BE Intersects CD at F. Show that AABE – &CFB.
Solution:
Given. Parallelogram ABCD. Side AD is produced to E, BE intersects DC at F.

To Prove. ∆ABE ~ ∆CFB
Proof. In ∆ABE and ∆CFB.
∠A = ∠C (opposite angle of || gm)
∠ABE = ∠CFB (alternate angle)
∴ ∆ABE ~ ∆CFB (AA similarity criterion)

Question 9.
In Fig., ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:
(i) ∆ABC ~ ∆AMP
(ii) \(\frac{\mathbf{C A}}{\mathbf{P A}}=\frac{\mathbf{B C}}{\mathbf{M P}}\)

Solution:
Given. ∆ABC and ∆AMP are two right triangles right angled at B and M.
To Prove. (i) ∆ABC ~ ∆AMP
(ii) \(\frac{\mathbf{C A}}{\mathbf{P A}}=\frac{\mathbf{B C}}{\mathbf{M P}}\)
Proof. In ∆ABC and ∆AMP,
∠A = ∠A (common)
∠B = ∠M (each 90°)
(i) ∴ ∆ABC ~ ∆AMP (AA similarity criterion)

(ii) ∴ \(\frac{\mathrm{AC}}{\mathrm{AP}}=\frac{\mathrm{BC}}{\mathrm{MP}}\)
[If two triangles are similar corresponding sides]
\(\frac{\mathrm{CA}}{\mathrm{PA}}=\frac{\mathrm{BC}}{\mathrm{MP}}\)
Hence Proved.

Q. 10.
CD and GH are respectively the vectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ∆ABC and
∆EFG respectively. If ∆ABC ~ ∆FEG, show
(i) \(\frac{\mathbf{C D}}{\mathbf{G H}}=\frac{\mathbf{A C}}{\mathbf{F G}}\)
(ii) ∆DCB ~ ∆HGE
(iii) ∆DCA ~ ∆HGF

Given. In ∆ABC and ∆EFG, CD and OH are bisector of ∠ACB and ∠EGF
i.e. ∠1 = ∠2
and ∠3 = ∠4
and ∆ABC ~ ∆FEG
To Prove. (i) = \(\frac{\mathbf{C D}}{\mathbf{G H}}=\frac{\mathbf{A C}}{\mathbf{F G}}\)
(ii) ∆DCB ~ ∆HGE
(iii) ∆DCA ~ ∆HGF
Proof.
(i) Given that, ∆ABC ~ ∆FEG
∴ ∠A = ∠F; ∠B = ∠E
and ∠C = ∠C
[∵ The corresponding angles of similar triangles are equal]
Consider, ∠C = ∠C [Proved above]
\(\frac{1}{2}\) ∠C = \(\frac{1}{2}\) ∠G
∠2 = ∠4 or ∠1 = ∠3
Now, in ∆ACD and ∆FGH
∠A = ∠F [Proved above]
∠2 = ∠4 [Proved above]
∴ ∠ACD ~ ∠FGH [∵ AA similarity creterion]
Also, \(\frac{\mathrm{CD}}{\mathrm{GH}}=\frac{\mathrm{AC}}{\mathrm{FG}}\)
[∵ Corresponding sides are in proportion].

(ii) In ∆DCB and ∆HGE,
∠B = ∠E [Proved above]
∠1 = ∠3 [Proved above]
∴ ∆DCB ~ ∆HGE [∵ AA similarity criterion]

(iii) In ∆DCA and ∆HGF
∠A = ∠F [Proved above]
∠2 = ∠4 [Proved above]
∴ ∆DCA ~ ∆HGF [∵ AA similarity criterion].

Question 11.
In Fig., E is a point on side CB produced of an Isosceles triangle ABC with AB = AC. IfAD ⊥BC and EF ⊥ AC, prove that ∆ABD ~ ∆ECF.

Solution:
Given. ∆ABC, isosceles triangle with AB = AC AD ⊥ BC, side BC is produced to E. EF ⊥ AC
To Prove. ∆ABD ~ ∆ECF
Proof. ∆ABC is isosceles (given)
AB = AC
∴ ∠B = ∠C [Equal sides have equal angles opposite to it)
In ∆ABD and ∆ECF,
∠ABD = ∠ECF (Proved above)
∠ADB = ∠EFC (each 90°)
∴ ∠ABD – ∠ECF [AA similarity).

Question 12.
Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ∆PQR (see Fig.). Show that ∆ABC ~ ∆PQR.

Solution:
Given. ∆ABC and ∆PQR, AB, BC, and median AD of ∆ABC are proportional to side PQ; QR and median PM of ∆PQR,
i.e., \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{\mathrm{AD}}{\mathrm{PM}}\)

To prove: ∆ABC ~ ∆PQR
Construction: Produce AD to E such that AD = DE and Produce PM to N such that PM = MN join BE, CE, QN and RN
Proof: \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{\mathrm{AD}}{\mathrm{PM}}\) (given) …………..(1)
BD = DC (given)
AD = DE (construction)
Diagonal bisects each other ¡n quadrilateral ABEC
∴ Quadrilateral ABEC is parallelogram
Similarly PQNR is a parallelogram
∴ BE = AC (opposite sides of parallelogram) and QN = PR
\(\frac{\mathrm{BE}}{\mathrm{AC}}=1\) ……………(i)
\(\frac{\mathrm{QN}}{\mathrm{PR}}=1\) …………..(ii)
From (i) and (ii),
\(\frac{\mathrm{BE}}{\mathrm{AC}}=\frac{\mathrm{QN}}{\mathrm{PR}}\)
⇒ \(\frac{\mathrm{BE}}{\mathrm{QN}}=\frac{\mathrm{AC}}{\mathrm{PR}}\)
But \(\frac{A B}{P Q}=\frac{A C}{P R}\) (Given)
∴ \(\frac{B E}{Q N}=\frac{A B}{P Q}\) …………..(2)
\(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AD}}{\mathrm{PM}}\) From (1)
= \(\frac{2 \mathrm{AD}}{2 \mathrm{PM}}\)
\(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AE}}{\mathrm{PN}}\) …………..(3)
From (2) and (3),
\(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BE}}{\mathrm{QN}}=\frac{\mathrm{AE}}{\mathrm{PN}}\)
∴ ∆ABE ~ ∆PQN [Sides are Proportional]
∴ ∠1 = ∠2 …………….(4) [Corresponding angle of similar triangle]
|| ly ∆ACE ~ ∆PRN ……….(5) [Corresponding angle of similar triangle]
Adding (4) and (5).
∠1 + ∠3 = ∠2 + ∠4
∠A = ∠P
Now in ∆ABC and ∆PQR,
∠A = ∠P (Proved)
\(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}}\) (given)
∴ ∆ABC ~ ∆PQR [By using SA similarity criterion]
Hence Proved.

Question 13.
D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA² = CB. CD.
Solution:
Given. ∆ABC, D is a point on side BC such that ∠ADC = ∠BAC
To Prove. CA² = BC × CD
Proof. In ABC and ADC,

∠C = ∠C (common)
∠BAC = ∠ADC (given)
∴ ∆ABC ~ ∆DAC [by AA similarity criterion]
∴ \(\frac{\mathrm{AC}}{\mathrm{DC}}=\frac{\mathrm{BC}}{\mathrm{AC}}\)
[If two triangles are similar corresponding sides are proportional]
AC² = BC. DC Hence Proved.

Question 14.
Sides AB and AC and median AD of a triangle ABC are proportional to sides PQ and PR and median PM of another
triangle PQR. Prove that ∆ABC ~ ∆PQR.
Solution:

Given: Two ∆s ABC and PQR. D is the mid-point of BC and M is the mid-point of QR. and \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{\mathrm{AD}}{\mathrm{PM}}\) ………..(1)
To Prove: ∆ABC ~ ∆PQR
Construction:
Produce AD to E such that AD = DE
Join BE and CE.
Proof. In quad. ABEC, diagonals AE and
BC bisect each other at D.
∴ Quad. ABEC is a parallelogram.
Similarly it can be shown that quad PQNR is a parallelogram.
Since ABEC is a parallelogram
∴. BE = AC ………….(2)
Similarly since PQNR is a || gm
∴ QN = PR ………….(3)
Dividing (2) by (3), we get:
\(\frac{B E}{Q N}=\frac{A C}{P R}\) …………….(4)
Now \(\frac{\mathrm{AD}}{\mathrm{PM}}=\frac{2 \mathrm{AD}}{2 \mathrm{PM}}=\frac{\mathrm{AE}}{\mathrm{PN}}\)
∴ ∠BAE = ∠QPN ………….(5)
From (1), (4) and (5), we get:
\(\frac{A D}{P Q}=\frac{B E}{Q N}=\frac{A E}{P N}\)
Thus in ∆s ABE and PQN, we get:
\(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BE}}{\mathrm{QN}}=\frac{\mathrm{AE}}{\mathrm{PN}}\)
∴ ∆ABC ~ ∆PQN
∴ ∠BAE = ∠QPN ………..(6)
Similarly it can be proved that
∆AEC ~ ∆PNR
∴ ∠EAC = ∠NPR …………..(7)
Adding (6) and (7), we get:
∠BAE + ∠EAC = ∠QPN + ∠NPR
i.e., ∠BAC = ∠QPR
Now in ∆ABC and ∆PQR.
\(\frac{A B}{P Q}=\frac{A C}{P R}\)
and included ∠A = ∠P
∴ ∆ABC ~ ∆QPR (By SAS criterion of similarity).

Question 15.
A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.
Solution:

Length of vertical stick = 6 m
Shadow of stick = 4 m
Let height of tower be H m
Length of shadow of tower = 28 m
In ∆ABC and ∆PMN,
∠C = ∠N (angle of altitude of sun)
∠B = ∠M (each 90°)
∴ ∆ABC ~ ∆PMN [AA similarity criterion]
∴ \(\frac{\mathrm{AB}}{\mathrm{PM}}=\frac{\mathrm{BC}}{\mathrm{MN}}\)
[If two triangles are similar corresponding sides are proportional]
∴ \(\frac{6}{\mathrm{H}}=\frac{4}{28}\)
H = \(\frac{6 \times 28}{4}\)
H = 6 × 7
H = 42 m.
Hence, Height of Tower = 42 m.

Question 16.
If AD and PM are medians of triangles ABC and PQR, respectively where ∆ABC ~ ∆PQR, prove that \(\frac{\mathbf{A B}}{\mathbf{P Q}}=\frac{\mathbf{A D}}{\mathbf{P M}}\).

Solution:
Given: ∆ABC and ∆PQR, AD and PM are median and ∆ABC ~ ∆PQR
To Prove: \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AD}}{\mathrm{PM}}\)
Proof. ∆ABC ~ ∆PQR (given)
∴ \(\frac{A B}{P Q}=\frac{B C}{Q R}=\frac{A C}{P R}\)
(If two triangles are similar corrosponding sides are Proportional)
∠A = ∠P
(If two triangles are similar corrosponding angles are equal)
∠B = ∠Q
∠C = ∠R
D is mid Point of BC
∴ BD = DC = \(\frac{1}{2}\) BC ……………..(2)
M is mid point of OR
∴ QM = MR = \(\frac{1}{2}\) QR …………….(3)
\(\frac{A B}{P Q}=\frac{B C}{Q R}\)
\(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{2 \mathrm{BD}}{2 \mathrm{QM}}\) (from(2)and(3))
\(\frac{A B}{P Q}=\frac{B D}{Q M}\)
∠ABD = ∠PQM (given)
∆ABC ~ ∆PQM (By SAS similarity criterion)
\(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AD}}{\mathrm{PM}}\)
[If two triangles are similar corresponding sides are proportional].

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 6 Triangles Ex 6.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.4

Question 1.
Let △ABC ~ △DEF and their areas be respectively 64 cm² and 121 cm². If EF = 15.4 cm, find BC.
Solution:
△ABC ~ △DEF ;
area of △ABC = 64 cm²;
area of △DEF = 121 cm²;
EF= 15.4 cm

△ABC ~ △DEF

∴ \(\frac{\text { ar. }(\Delta \mathrm{ABC})}{\text { ar. }(\Delta \mathrm{DEF})}=\frac{\mathrm{AB}^{2}}{\mathrm{DE}^{2}}=\frac{\mathrm{AC}^{2}}{\mathrm{DF}^{2}}=\frac{\mathrm{BC}^{2}}{\mathrm{EF}^{2}}\)
(If two traingles are similar, ratio of their area is square of corresponding sides }

\(\frac{64}{121}=\frac{\mathrm{BC}^{2}}{\mathrm{EF}^{2}}\) \(\left(\frac{8}{11}\right)^{2}=\left(\frac{\mathrm{BC}}{15.4}\right)^{2}\)

⇒ \(\frac{8}{11}=\frac{\mathrm{BC}}{15.4}\)

BC = \(\frac{8 \times 15.4}{11}\)
BC = 8 × 1.4
BC = 11.2 cm.

Question 2.
Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2 CD. Find the ratio of the areas of traingles AOB and COD.
Solution:
ABCD is trapezium AB || DC. Diagonals AC and BD intersects each other at the point O. AB = 2 CD

In △AOB and △COD,
∠1 = ∠2 (alternate angles)
∠3 = ∠4 (alternate angles)
∠5 = ∠6 (vertically opposite angle)
∴ △AOB ~ △COD [AA, A similarity criterion]

(If two triangles are similar ratio of their areas is square of corresponding sides)

= \(\frac{(2 \mathrm{CD})^{2}}{\mathrm{CD}^{2}}\) [∵ AB = 2 CD] (Given)

∴ Required ratio of ar △AOB and △COD = 4 : 1.

Question 3.
In the fig., △ABC and △DBC are two triangles on the same base BC. If AD intersects BC at O show that

Solution:
Given. ∆ABC and ∆DBC are the triangles on same base BC. AD intersects BC at O

To Prove:
Construction: Draw AL ⊥ BC, DM ⊥ BC
Proof: In ∆ALO and ∆DMO.
∠1 = ∠2 (vertically opposite angle)
∠L = ∠M (each 90°)
∴ ∆ALO ~ ∆DMO [AA similarity criterion]
∴ \(\frac{\mathrm{AL}}{\mathrm{DM}}=\frac{\mathrm{AO}}{\mathrm{DO}}\) ……………(1)
[If two triangles are similar, corrosponding sides are proportional

[∵ ∆ = \(\frac{1}{2}\) × b × p]

Hence proved.

Question 4.
If the areas of two similar triangles are equal, prove that they are congruent.
Solution:
Given: Two ∆s ABC and DEF are similar and equal in area.
To Prove : ∆ABC ≅ ∆DEF

Proof: Since ∆ABC ~ ∆DEF,
∴ \(\frac{\text { area }(\Delta \mathrm{ABC})}{\text { area }(\Delta \mathrm{DEF})}=\frac{\mathrm{BC}^{2}}{\mathrm{EF}^{2}}\)

⇒ \(\frac{\mathrm{BC}^{2}}{\mathrm{EF}^{2}}=1\) [∵ area (∆ABC) = area (∆DEF)]
⇒ BC² = EF²
⇒ BC = EF.
Also, since ∆ABC ~ ∆DEF,
therefore they are equiangular and hence
∠B = ∠E
and ∠C = ∠F.
Now in ∆s ABC and DEF,
∠B = ∠E, ∠C = ∠F
and BC = EF
∴ ∆ABC ≅ ∆DEF (ASA congruence).

Question 5.
D, E and F are respectively the mid points of the sides BC, CA and AB of ∆ABC. Determine the ratio of the areas of triangles DEF and ABC.
Solution:

Given. D, E, F are the mid-point of the sides BC, CA and AB respectively of a tABC.
To find : ar (∆DEE) : ar (∆ABC)
Proof: In ∆ABC,
F is the mid-point of AB …(given)
E is the mid-point of AC …(given)
So, by the Mid-Foint Theorem
FE || BC and FE = \(\frac{1}{2}\) BC
⇒ FE || BD and FE = BD [∵ BD = \(\frac{1}{2}\) BC]
∴ BDEF is a || gm.
(∵ Opp. sides are || and equal)
In △s FBD and DEF,
FB = DE (opp. sides of || gm BDEF)
FD = FD .. .(common)
BD = FE
. ..(opp. sides of || gm BDEF)
∴ △FBD ≅ △DEF … (SSS Congruency Theorem)

Similary we can prove that:
△AFE ≅ △DEF
and △EDC ≅ △DEF
if △s are , then they are equal in area.
∴ ar (∆FBD) = ar. (∆DEF) ……………(1)
ar (∆AFE) = ar (∆DEF) ……………(2)
ar (∆EDC) = ar (∆DEF) ……………(3)
Now ar ∆ (ABC)
= ar (∆FBD) + ar (∆DEF) + ar (∆AFE) + ar (∆EDC)
= ar.(∆DEF) + ar (∆DEF) + ar (∆DEF) + ar. (∆DEF) [Using (1), (2) and (3)]
= 4 ar (∆DEF)
⇒ (∆DEF) = \(\frac{1}{4}\) ar(∆ABC)
⇒ \(\frac{{ar} .(\Delta \mathrm{DEF})}{{ar} .(\Delta \mathrm{ABC})}=\frac{1}{4}\)
∴ ar (∆DEF) : ar (∆ABC) = 1 : 4.

Question 6.
Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.
Solution:
Given: ∆ABC ~ ∆DEF.
AX and DY are the medians to the side BC and EF respectively.

To prove: \(\frac{{ar} .(\Delta \mathrm{ABC})}{{ar} .(\Delta \mathrm{DEF})}=\frac{\mathrm{AX}^{2}}{\mathrm{DY}^{2}}\)
Proof: ∆ABC ~ ∆DEF (Given)
∴ \(\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EF}}=\frac{2 \mathrm{BX}}{2 \mathrm{EY}}\)
[∵ AX and DY are medians
∴ BC = 2BX and EF = 2EY]

⇒ \(\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BX}}{\mathrm{EY}}\) ………………(1)
In ∆ABX and ∆DEY, [∵ ∆ABC ~ ∆DEF]
\(\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BX}}{\mathrm{EY}}\) [Prove in (1)]
∴ ∆ABC ~ ∆DEY [By SAS criterion of similarity]
∴ \(\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{AX}}{\mathrm{DY}}\) …………(2)
As the areas of two similar triangles are proportional to the squares of the corresponding sides, so
∴ \(\frac{{ar} .(\Delta \mathrm{ABC})}{{ar} .(\Delta \mathrm{DEF})}=\frac{\mathrm{AB}^{2}}{\mathrm{DE}^{2}}=\frac{\mathrm{AX}^{2}}{\mathrm{D} \mathrm{Y}^{2}}\)
Hence proved.

Question 7.
Prove that the areas of the equilateral triangle described on the side of a square Is half the area of the equilateral triangle described on its diagonal.
Solution:
Given: ABCD is a square. Equilateral ∆ABE is described on the side AB of the square and equilateral ∆ACF is desribed on the diagonal AC.

To prove: \(\frac{{ar} .(\Delta \mathrm{ABE})}{{ar} .(\Delta \mathrm{ACF})}=\frac{1}{2}\)
Proof: In rt. ∆ABC,
⇒ AB² + BC² = AC² [By Pathagoras theorem]
= AB² + AB² = AC² [∵ AB = BC, being the sides of the same square]
⇒ 2AB² = AC² ………….(1)
Now each of ∆ABE and ∆ACF are equilateral and therefore equiangular and hence similar.
i.e., ∆ABE ~ ∆ACF.
Here any side of one ∆ is proportional to any side of other.
∴ \(\frac{\text { ar. }(\Delta \mathrm{ABE})}{\text { ar. }(\Delta \mathrm{ACF})}=\frac{\mathrm{AB}^{2}}{\mathrm{AC}^{2}}\)

[∵ The ratio of the areas of two similar∆s is equal to their corresponding sides]
= \(\frac{\mathrm{AB}^{2}}{2 \mathrm{AB}^{2}}=\frac{1}{2}\) [Using (1)]

Question 8.
Tick the correct answer and justify: ABC and BDE are two equilateral triangles such that D is the mid point of BC. Ratio of the areas of triangles ABC and BDE is
(A) 2 : 1
(B) 1 : 2
(C) 4 : 1
(D) 1 : 4.
Solution:
∆ABC and ∆BDE are two equilateral thangles. D is mid point of BC.
∴ BD = DC = \(\frac{1}{2}\) BC,
Let each side of triangles are 2a
∴ ∆ABC ~ ∆BDE
∴ \(\frac{\text { ar. }(\Delta \mathrm{ABC})}{\text { ar. }(\triangle \mathrm{BDE})}=\frac{\mathrm{AB}^{2}}{\mathrm{BD}^{2}}\)

= \(\frac{(2 a)^{2}}{(a)^{2}}\)
= \(\frac{4 a^{2}}{a^{2}}\)
= \(\frac{4}{1}\) = 4 : 1
∴ Correct option is (C).

Question 9.
Tick the correct answer and justify: Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are ¡n the ratio
(A) 2 : 3
(B) 4 : 9
(C) 81 : 16
(D) 16 : 81.
Solution:

∆ABC ~ ∆DEF (given)

\(\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{AC}}{\mathrm{DF}}=\frac{\mathrm{BC}}{\mathrm{EF}}=\frac{4}{9}\)

∴ \(\frac{\text { ar. }(\Delta \mathrm{ABC})}{\text { ar. }(\Delta \mathrm{DEF})}=\frac{\mathrm{AB}^{2}}{\mathrm{DE}^{2}}\)

[If two triangles are similar ratio of their areas is equal to square of corresponding sides]
\(\frac{{ar} .(\Delta \mathrm{ABC})}{{ar} .(\Delta \mathrm{DEF})}=\left(\frac{4}{9}\right)^{2}\) = \(\frac{16}{81}\) = 16 : 81
∴ Correct option is (D).

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 6 Triangles Ex 6.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.2

Question 1.
In fig. (i) and (ü), DE U BC. Find EC in (i) and AD in (ii).

Solution:
(i) In ∆ABC, DE || BC ……………(given)
∴ \(\frac{\mathrm{AD}}{\mathrm{BD}}=\frac{\mathrm{AE}}{\mathrm{EC}}\)
[By using Basic Proportionality Theorem]
\(\frac{1.5}{3}=\frac{1}{\mathrm{EC}}\)
EC = \(\frac{3}{1.5}\)
EC = \(\frac{3 \times 10}{15}\) = 2
∴ EC = 2 cm.

(ii) In ∆ABC,
DE || BC ……………(given)
∴ \(\frac{\mathrm{AD}}{\mathrm{BD}}=\frac{\mathrm{AE}}{\mathrm{EC}}\)
[By using Basic Proportionality Theorem]
\(\frac{\mathrm{AD}}{7.2}=\frac{1.8}{5.4}\)
AD = \(\frac{1.8 \times 7.2}{5.4}\)
= \(\frac{1.8}{10} \times \frac{72}{10} \times \frac{10}{54}=\frac{24}{10}\)
AD = 2.4
∴ AD = 2.4 cm.

Question 2.
E and F are points on the sides PQ and PR respectively of a APQR. For each of the following cases, state whether EF || QR :
(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
(ii) PE =4 cm, QE = 4.5 cm, PF =8 cm and RF = 9cm.
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm.
Solution:
In ∆PQR, E and F are two points on side PQ and PR respectively.

(i) PE = 3.9 cm, EQ = 3 cm
PF = 3.6 cm, FR = 2.4 cm
\(\frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{3.9}{3}=\frac{39}{30}=\frac{13}{10}=1.3\)

\(\frac{P F}{F R}=\frac{3.6}{2.4}=\frac{36}{24}=\frac{3}{2}=1.5\) \(\frac{\mathrm{PE}}{\mathrm{EQ}} \neq \frac{\mathrm{PF}}{\mathrm{FR}}\)

∴ EF is not parallel to QR.

(ii) PE = 4 cm, QE = 4.5 cm,
PF = 8 cm, RF = 9 cm.
\(\frac{\mathrm{PE}}{\mathrm{QE}}=\frac{4}{4.5}=\frac{40}{45}=\frac{8}{9}\) ………….(1)
\(\frac{P F}{R F}=\frac{8}{9}\) ……………..(2)
From (1) and (2),
\(\frac{\mathrm{PE}}{\mathrm{QE}}=\frac{\mathrm{PF}}{\mathrm{RF}}\)
∴ By converse of Basic Proportionality theorem EF || QR.

(iii) PQ = 1.28 cm, PR = 2.56 cm
PE = 0.18 cm, PF = 0.36 cm.
EQ = PQ – PE = 1.28 – 0.18 = 1.10 cm
ER = PR – PF = 2.56 – 0.36 = 2.20 cm
Here \(\frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{0.18}{1.10}=\frac{18}{110}=\frac{9}{55}\) …………..(1)

and \(\frac{\mathrm{PF}}{\mathrm{FR}}=\frac{0.36}{2.20}=\frac{36}{220}=\frac{9}{55}\) …………….(2)

From (1) and (2), \(\frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{\mathrm{PF}}{\mathrm{FR}}\)
∴ By converse of Basic Proportionality Theorem EF || QR.

Question 3.
In fig., LM || CB; and LN || CD. Prove that \(\frac{\mathbf{A M}}{\mathbf{A B}}=\frac{\mathbf{A N}}{\mathbf{A D}}\).

Solution:
In ∆ABC,
LM || BC (given)
∴ \(\frac{\mathrm{AM}}{\mathrm{MB}}=\frac{\mathrm{AL}}{\mathrm{LC}}\) ………..(1)
(By Basic Proportionality Theorem)
Again, in ∆ACD
LN || CD (given)
∴ \(\frac{A N}{N D}=\frac{A L}{L C}\) …………..(2)
(By Basic Proportionality Theorem)
From (1) and (2),
\(\frac{\mathrm{AM}}{\mathrm{MB}}=\frac{\mathrm{AN}}{\mathrm{ND}}\)

or \(\frac{\mathrm{MB}}{\mathrm{AM}}=\frac{\mathrm{ND}}{\mathrm{AN}}\)

or \(\frac{\mathrm{MB}}{\mathrm{AM}}+1=\frac{\mathrm{ND}}{\mathrm{AN}}+1\)
or \(\frac{\mathrm{MB}+\mathrm{AM}}{\mathrm{AM}}=\frac{\mathrm{ND}+\mathrm{AN}}{\mathrm{AN}}\)

or \(\frac{\mathrm{AB}}{\mathrm{AM}}=\frac{\mathrm{AD}}{\mathrm{AN}}\)

or \(\frac{\mathrm{AM}}{\mathrm{AB}}=\frac{\mathrm{AN}}{\mathrm{AD}}\)
Hence, \(\frac{\mathrm{AM}}{\mathrm{AB}}=\frac{\mathrm{AN}}{\mathrm{AD}}\) is the required result.

Question 4.
In Fig. 6.19, DE || AC and DF || AE. Prove that \(\frac{\mathrm{BF}}{\mathrm{FE}}=\frac{\mathrm{BE}}{\mathrm{EC}}\).

Solution:
In ∆ABC, DE || AC(given)

∴ \(\frac{B D}{D A}=\frac{B E}{E C}\) …………….(1)
[By Basic Proportionality Theorem]
In ∆ABE, DF || AE
\(\frac{\mathrm{BD}}{\mathrm{DA}}=\frac{\mathrm{BF}}{\mathrm{FE}}\) …………….(2)
[By Basic Proportionality Theorem]
From (1) and (2),
\(\frac{\mathrm{BE}}{\mathrm{EC}}=\frac{\mathrm{BF}}{\mathrm{FE}}\)
Hence proved.

Question 5.
In fig. DE || OQ and DF || OR. Show that EF || QR.

Solution:
Given:
In ∆PQR, DE || OQ DF || OR.
To prove: EF || QR.
Proof: In ∆PQO, ED || QO (given)

∴ \(\frac{P D}{D O}=\frac{P E}{E Q}\)

[By Basic Proportionality Theorem]
Again in ∆POR,
DF || OR (given)
∴ \(\frac{P D}{D O}=\frac{P F}{F R}\) ………….(2)
[By Basic Proportionality Theorem]
From (1) and (2),
\(\frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{\mathrm{PF}}{\mathrm{FR}}\)
In ∆PQR, by using converse of Basic proportionaIity Theorem.
EF || QR,
Hence proved.

Question 6.
In flg., A, B and C points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show thatBC || QR.

Solution:
Given : ∆PQR, A, B and C are points on OP, OQ and OR respectively such that AB || PQ, AC || PR.

To prove: BC || QR.
Proof: In ∆OPQ, AB || PQ (given)
∴ \(\frac{\mathrm{OA}}{\mathrm{AP}}=\frac{\mathrm{OB}}{\mathrm{BQ}}\) …………….(1)
[BY using Basic Proportionality Theorem]
Again in ∆OPR.
AC || PR (given)
∴ \(\frac{\mathrm{OA}}{\mathrm{AP}}=\frac{\mathrm{OC}}{\mathrm{CR}}\) ……………….(2)
[BY using Basic Proportionality Theorem]
From (1) and (2),
\(\frac{\mathrm{OB}}{\mathrm{BQ}}=\frac{\mathrm{OC}}{\mathrm{CR}}\)
∴ By converse of Basic Proportionality Theorem.
In ∆OQR, BC || QR. Hence proved.

Question 7.
Using Basic Proportionality theorem, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved ¡t in class IX).
Solution:
Given: In ∆ABC, D is mid point of AB, i.e. AD = DB.
A line parallel to BC intersects AC at E as shown in figure. i.e., DE || BC.

To prove: E is mid point of AC.
Proof: D is mid point of AB.
i.e.. AD = DB (given)
Or \(\frac{\mathrm{AD}}{\mathrm{BD}}\) = 1 ……………..(1)
Again in ∆ABC DE || BC (given)
∴ \(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}\)
[By Basic Proportionality Theorem]
∴ 1 = \(\frac{\mathrm{AE}}{\mathrm{EC}}\) [From (1)]
∴ AE = EC
∴ E is mid point of AC. Hence proved.

Question 8.
Using converse of Basic Proportionality theorem prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done ¡tin Class IX).
Solution:
Given ∆ABC, D and E are mid points of AB and AC respectively such that AD = BD and AE = EC, D and Eare joined

To Prove, DE || BC
Proof. D is mid point of AB (Given)
i.e., AD = BD
Or \(\frac{\mathrm{AD}}{\mathrm{BD}}\) = 1 ………………(1)
E is mid point of AC (Given)
∴ AE = EC
Or \(\frac{\mathrm{AE}}{\mathrm{EC}}\) = 1 ………………(2)
From (1) and (2),
By using converse of basic proportionality Theorem
DE || BC Hence Proved.

Question 9.
ABCD is a trapeiiumin with AB || DC and its diagonals Intersect each other at the point O. Show that \(\frac{A O}{B O}=\frac{C O}{D O}\).
Solution:
Given. ABCD is trapezium AB || DC, diagonals AC and BD intersect each other at O.

To Prove. \(\frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}\)
Construction. Through O draw FO || DC || AB
Proof. In ∆DAB, FO || AB (construction)
∴ \(\frac{\mathrm{DF}}{\mathrm{FA}}=\frac{\mathrm{DO}}{\mathrm{BO}}\) ……………..(1)
[By using Basic Proportionality Theorem]
Again in ∆DCA,
FO || DC (construction)
\(\frac{\mathrm{DF}}{\mathrm{FA}}=\frac{\mathrm{CO}}{\mathrm{AO}}\)
[By using Basic Proportionality Theorem]
From (1) and (2),
\(\frac{\mathrm{DO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{AO}} \quad \frac{\mathrm{AO}}{\mathrm{BO}} \quad \frac{\mathrm{CO}}{\mathrm{DO}}\)
Hence Proved.

Question 10.
The diagonals of a quadrilateral ABCD Intersect each other at the point O such that \(\frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}\).Show that ABCD is a
trapezium.
Solution:
Given: Quadrilateral ABCD, Diagonal AC and BD intersects each other at O
such that = \(\frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}\)

To Prove. Quadrilateral ABCD is trapezium.
Construction. Through ‘O’ draw line EO || AB which meets AD at E.
Proof. In ∆DAB,
EO || AB [Const.]
∴ \(\frac{\mathrm{DE}}{\mathrm{EA}}=\frac{\mathrm{DO}}{\mathrm{OB}}\) ………………(1)
[By using Basic Proportionality Theoremj
But = \(\frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}\) (Given)

or \(\frac{\mathrm{AO}}{\mathrm{CO}}=\frac{\mathrm{BO}}{\mathrm{DO}}\)

or \(\frac{\mathrm{CO}}{\mathrm{AO}}=\frac{\mathrm{DO}}{\mathrm{BO}}\)

⇒ \(\frac{\mathrm{DO}}{\mathrm{OB}}=\frac{\mathrm{CO}}{\mathrm{AO}}\) …………….(2)
From (1) and (2),
\(\frac{\mathrm{DE}}{\mathrm{EA}}=\frac{\mathrm{CO}}{\mathrm{AO}}\)
∴ By using converse of basic
proportionlity Theorem,
EO || DC also EO || AB [Const]
⇒ AB || DC
∴ Quadrilateral ABCD is a trapezium with AB || CD.