Class 7

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 11 Perimeter and Area MCQ Questions with Answers.

PSEB 7th Class Maths Chapter 11 Perimeter and Area MCQ Questions

Multiple Choice Questions :

Question 1.
Out of the following figures which has the greatest perimeter ?
(The length and breadth of each rectangle is equal)
(i) (ii) (iii)

(a) (i)
(b) (ii)
(c) (iii)
(d) (iv).
Answer:
(c) (iii)

Question 2.
Perimeter of a rectangle = 2 (…………)
(a) Length + Breadth
(b) Length × Breadth
(c) Length ÷ Breadth
(d) Length – Breadth.
Answer:
(a) Length + Breadth

Question 3.
Area of rectangle = …………….
(a) Length + Breadth
(b) Length × Breadth
(c) Length – Breadth
(d) Length ÷ Breadth.
Answer:
(b) Length × Breadth

Question 4.
Circumference of a circle = …………….
(a) πr
(b) 2πr
(c) πr²
(d) 2πr².
Answer:
(b) 2πr

Question 5.
Area of a circle =
(a) 2πr
(b) πr
(c) πr²
(d) 2πr²
Answer:
(c) πr²

Question 6.
The area of a square park whose perimeter is 320 m will be :
(a) 3200 m²
(b) 640 m²
(c) 6400 m²
(d) 6400 m
Answer:
(c) 6400 m2

Question 7.
The area of a rectangular plot is 440 m² and its length is 22 m. Its breadth will be :
(a) 20 m
(b) 40 m
(c) 44 m
(d) 21 m.
Answer:
(a) 20 m

Question 8.
The radius ofa circle is 14 cm. Its area will be:
(a) 88 cm²
(b) 616 cm²
(c) 196 cm²
(d) 56 cm²
Answer:
(b) 616 cm²

Fill in the blanks :

Question 1.
Perimeter of a rectangle = 2 (………….)
Answer:
Length + Breadth

Question 2.
Area of rectangle = ………….
Answer:
Length × Breadth

Question 3.
Perimeter of a square = ………….
Answer:
4 × side

Question 4.
Area of a square = ………….
Answer:
(side)2

Question 5.
Area of a triangle = ………….
Answer:
\(\frac {1}{2}\) × Base × height

Write True/False :

Question 1.
Area of a square with side 10 cm is 20 cm² (True/False)
Answer:
False

Question 2.
One are has 100 m² (True/False)
Answer:
True

Question 3.
One hectare has 1000 m² (True/False)
Answer:
False

Question 4.
Area of circle is πr² square unit (True/False)
Answer:
True

Question 5.
Brahmgupta gave the formule for the area of a cyclic quadrilateral (True/False)
Answer:
True

Punjab State Board PSEB 7th Class Hindi Book Solutions Chapter 2 धूल का फूल Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Hindi Chapter 2 धूल का फूल

Hindi Guide for Class 7 PSEB धूल का फूल Textbook Questions and Answers

(क) भाषा-बोध

1. शब्दार्थ

सरपट – तेज़ चाल से
खलिहान = कटी हुई फसल रखने का स्थान
विपन्नता = गरीबी
मजूरी = मज़दूरी
व्यर्थ = बेकार
मुटियार = नवयुवती
उत्थान = उन्नति, बढ़ती
जन्नत = स्वर्ग
निहारते = देखते
प्रांगण = आंगन
पुष्प-गुच्छ = फूलों का गुलदस्ता
गुंजायमान = गूंजता हुआ

2. इन मुहावरों के अर्थ लिखते हुए वाक्य प्रयोग करें:

खुशी के आँसू छलकना ________________ ___________________________
मन बल्लियों उछलना _____________ _________________________
नाम रोशन करना __________________ ______________________
गले लगाना ________________ _______________________
फूले न समाना _______________ __________________________
उत्तर:
खुशी के आँसू छलकना (बहुत अधिक प्रसन्न होना) – परीक्षा में प्रथम आने का समाचार मिलते ही नीलम की आँखों में खुशी के आँसू छलकने लगे।
मन बल्लियों उछलना(बहुत प्रसन्न होना) – बहुत दिनों बाद गाँव जाते हुए प्रशांत का मन बल्लियों उछलने लगा।
नाम रोशन करना (प्रसिद्धि मिलना) – कलक्टर बन कर प्रशांत ने अपने माता-पिता का नाम रोशन कर दिया।
गले लगना (प्रेम से भेंटना)  -माता ने पुत्र को आशीर्वाद देते हुए बाँहों में भरकर गले लगा लिया।
फूले न समना (बहुत प्रसन्न होना) – एशिया-कप हॉकी में विजय प्राप्त कर भारतीय खिलाड़ी फूले न समा रहे थे।

3. विपरीत शब्द लिखें: 

मौन = ……………..
ऊबड़-खाबड़ = ……………
ज़रूरी = ……………..
शहर = …………….
सुनसान = ……………….
मजदूर = ……………..
आमदन = ………………
उत्थान = ……………
उत्तर:
शब्द विपरीत शब्द
मौन = मुखर
ऊबड़-खाबड़ = सीधी-सपाट
ज़रूरी = गैर-ज़रूरी
शहर = गाँव
सुनसान = आबाद
मजदूर = मालिक
आमदन = खर्च
उत्थान = पतन

4. दो-दो पर्यायवाची शब्द लिखें :

तरक्की = ……………
शिक्षा = ……………
अध्यापक = ……………
मेहनत = …………….
शिष्य = …………….
उत्तर:
शब्द पर्यायवाची शब्द
तरक्की = वृद्धि, बढ़ती
शिक्षा = परामर्श, सलाह, तालीम, सबक
अध्यापक = शिक्षक, गुरु
मेहनत = परिश्रम, श्रम
शिष्य = चेला, विद्यार्थी

प्रयोगात्मक व्याकरण

गुरु की शरण = गुरुशरण
माँ और बाप = माँ-बाप

उपर्युक्त पदों में गुरु की शरण को गुरुशरण तथा माँ और बाप को माँ-बाप रूप में संक्षेप में लिख सकते हैं। इस प्रकार शब्दों के मेल से नए शब्द बन जाते हैं।
अतः परस्पर सम्बन्ध रखने वाले दो या दो से अधिक शब्दों के मेल से जब कोई नया सार्थक शब्द बनता है तो उस मेल को समास कहते हैं।

समास करने के बाद जो शब्द बनता है उसे समस्तपद कहते हैं। समस्तपद को इसके शब्द खण्डों में अलग-अलग करने की विधि को विग्रह कहते हैं। जैसे :

गुरुशरण (समस्त पद) = गुरु की शरण (विग्रह)

विशेष:
समस्त पद के दो पद होते हैं- पूर्व पद और उत्तर पद। पहले पद को पूर्व पद तथा बाद को उत्तर पद कहते हैं। जैसे-गुरु (पूर्व पद), शरण (उत्तर पद)

पदों की प्रधानता के आधार पर समास के चार भेद होते हैं :

1. अव्ययीभाव समास
2. तत्पुरुष समास
3. द्वंद्व समास
4. बहुब्रीहि समास

(क)	समस्त पद	विग्रह	जिस अर्थ में अव्यय यहाँ प्रयुक्त हुआ
(1)	बेरोज़गार	रोज़गार के बिना	‘बे’ का प्रयोग के बिना’ अर्थ में हुआ है।
(2)	आजीवन	जीवन तक	‘आ’ का प्रयोग तक के अर्थ में हुआ है।
(3)	यथानियम	नियम के अनुसार	‘यथा’ का प्रयोग ‘अनुसार’ के अर्थ में हुआ है।

यहाँ समस्त पद में ‘बे’, ‘आ’ तथा ‘यथा’ अव्यय हैं तथा इसके मेल से पूर्ण पद ही अव्यय बन गया है।

अतएव जिस समस्त पद में पूर्वपद प्रधान हो और अव्यय हो और समास होने पर पूर्ण पद ही अव्यय बन जाए, वह अव्ययी भाव समास कहलाता है।
अन्य उदाहरण-आमरण-मरने तक, निडर-डर के बिना, भरपेट-पेट भर कर, आजन्म-जन्म भर, प्रति पल-हर पल, बेखबर-बिना खबर के, अनजान-जाने बिना आदि।
तत्पुरुष समास को समझने के लिए कारक का ज्ञान अपेक्षित है। अत: पहले कारकों को समझते हैं।

हमें साहब ने रहने के लिए घर दिया।

यदि इस वाक्य को इस ढंग से लिखें-‘हमें साहब रहने घर दिया’ तो वाक्य में आए शब्दों का एक-दूसरे से सम्बन्ध नहीं प्रकट होता और न ही अर्थ स्पष्ट होता है।
इसलिए वाक्य में आए ने, के लिए चिह्न वाक्य के अन्य शब्दों का परस्पर सम्बन्ध जोड़ते हैं।

अतएव संज्ञा या सर्वनाम के जिस रूप से उनका सम्बन्ध क्रिया तथा वाक्य के दूसरे शब्दों में जाना जाए, उसे कारक कहते हैं।

विशेष :- वाक्य में प्रयुक्त ‘के, ने, से, के लिए’ कारक चिह्नों को परसर्ग भी कहते हैं।

(1) शरण ने क्षमा माँगी।
इस वाक्य में क्षमा माँगने का काम शरण ने किया अर्थात् कर्ता शरण है। अतः शरण ने में कर्ता कारक है।

(2) प्रशांत उच्च पद को प्राप्त हुआ।
इस वाक्य में प्राप्त हुआ क्रिया है, प्रशांत कर्ता है तथा क्रिया का फल पद पर पड़ रहा है। अतः पद को में कर्म कारक है।

अतएव वाक्य में जिस संज्ञा या सर्वनाम पर क्रिया का फल पड़ता है, उसे कर्म कारक कहते हैं।

(3) प्रशांत गाड़ी से गाँव आया।
इस वाक्य में गया क्रिया का साधन गाड़ी है। अतः गाड़ी से में करण कारक है।

अतएव कर्ता जिस साधन की मदद से क्रिया सम्पन्न करता है, उसे करण कारक कहते हैं।

(4) हम पढ़ने के लिए विद्यालय जाते थे।
इस वाक्य में जाना क्रिया का कार्य पढ़ने के लिए है, अतः यहाँ सम्प्रदान कारक है।

अतएव जिस संज्ञा या सर्वनाम के लिए कुछ किया जाए उसे सम्प्रदान कारक कहते हैं।

(5) हमारा पूरा परिवार गाँव से शहर आ गया।
इस वाक्य में गाँव से पद से अलग होने का अर्थ स्पष्ट हो रहा है, इसलिए यहाँ अपादान कारक है।

अतः जिस संज्ञा से पृथक्ता अर्थात् अलग होने का भाव प्रकट हो, उसे अपादान कारक कहते हैं।

इसके अतिरिक्त किसी से सीखने, लगाने, डरने, बचाने, तुलना करने, माँगने, निकलने तथा दूरी का भाव दर्शाने में भी अपादान कारक होता है।

(6) नसीब का लड़का कलक्टर बन गया।
इस वाक्य में नसीब का लड़का से पिता-पुत्र का सम्बन्ध प्रकट हो रहा है अतः यहाँ सम्बन्ध कारक है।

अतएव जहाँ दो संज्ञाओं या सर्वनामों का आपस में सम्बन्ध प्रकट हो, वहाँ सम्बन्ध कारक होता है।

(7) प्रशांत पहले गाँव में रहता था।
इस वाक्य में गाँव में पद में रहना क्रिया के आधार का पता चलता है, यहाँ अधिकरण कारक है।

अतएव जहाँ संज्ञा या सर्वनाम शब्द के आधार का पता चले उसे अधिकरण कारक कहते हैं।

(8) अरे शरण गाड़ी जल्दी चलाओ।
इस वाक्य में अरे शरण ! को सम्बोधन किया गया है, इसलिए यहाँ सम्बोधन कारक है।

अतएव संज्ञा या सर्वनाम के जिस रूप से किसी को पुकारने, बुलाने, सुनाने या सावधान करने का भाव प्रकट हो, वहाँ सम्बोधन कारक होता है। आइए, अब तत्पुरुष समास को समझते हैं।

(ख) समस्त पद विग्रह
पदप्राप्त पद को प्राप्त

उपर्युक्त समास में समस्त पद बनाते समय पूर्वपद (पद) के साथ आए परसर्ग (को) का लोप हो गया है। इसके उत्तरपद (प्राप्त) प्रधान है।
अतएव जिस समास में उत्तर पद प्रधान हो उसे तत्पुरुष समास कहते हैं। पद बनाते समय पूर्वपद के साथ आने वाले परसर्ग का लोप हो जाता है।

अन्य उदाहरण

समस्त पद = विग्रह
यशप्राप्त = यश को प्राप्त
समस्त पद = विग्रह
भावविह्वल = भाव से विह्वल
पाठशाला = पढ़ने के लिए शाला
धनहीन = धन से हीन
विद्याभ्यास = विद्या का अभ्यास
सिरदर्द = सिर में दर्द

(ख) विचार-बोध

1. प्रश्नों के उत्तर एक या दो वाक्यों में लिखें:

प्रश्न 1.
प्रशान्त का मन बल्लियों क्यों उछल रहा था ?
उत्तर:
प्रशान्त का मन बल्लियों इसलिए उछल रहा था क्योंकि वह कई वर्षों के बाद अपने गाँव जा रहा था।

प्रश्न 2.
गाँव की ओर जाते हुए उसे किन-किन लोगों की याद आने लगी ?
उत्तर:
गाँव की ओर जाते हुए पंच जी के खेत, दीनू ग्वाले की गाय-भैंसें, सब्बू कुम्हार के चाक की याद आने लगी।

प्रश्न 3.
उसके अध्यापक का क्या नाम था ?
उत्तर:
उस के अध्यापक का नाम मास्टर आदित्य प्रकाश था।

प्रश्न 4.
हर माँ-बाप का क्या सपना होता है ?
उत्तर:
हर माँ-बाप का यह सपना होता है कि उसकी संतान पढ़-लिखकर कुछ बन जाए।

प्रश्न 5.
प्रशान्त का क्या सपना था? यह सपना उसने कैसे पूरा किया ?
उत्तर:
प्रशान्त का सपना था कि एक दिन ज़रूर वह कुछ बनेगा। यह सपना उस ने खूब पढ़ कर पूरा किया।

प्रश्न 6.
लड़कियों की शिक्षा के सम्बन्ध में उसके क्या विचार थे ?
उत्तर:
लड़कियों की शिक्षा के सम्बन्ध में उसके विचार थे कि लड़कियाँ घर का श्रृंगार होती हैं। उन्हें खूब पढ़ाना चाहिए, क्योंकि वे परिवार का आधार होती हैं।

प्रश्न 7.
प्रशान्त गाँव में क्यों आया था ?
उत्तर:
प्रशान्त गाँव में गाँव की पाठशाला का दर्जा बढ़ाने का आदेश लेकर आया था।

2. इन प्रश्नों के उत्तर चार-पाँच वाक्यों में लिखें:

प्रश्न 1.
गरीबी में रहते हुए भी प्रशान्त ने अपने लक्ष्य को कैसे प्राप्त किया?
उत्तर:
प्रशान्त गरीबी में रहते हुए भी पढ़ाई की तरफ बहुत ध्यान देता था। वह मास्टर आदित्य प्रकाश जी की बातें सुनकर धन्य हो जाता था। पढ़ाई का खर्च चलाने के लिए वह छुट्टियों में छोटा-मोटा आमदनी वाला काम कर लेता था। वह खूब पढ़ कर अफसर बनना चाहता था। उसे विश्वास था कि बड़े बनने की कुंजी विद्या है। इसलिए वह मेहनत से पढ़ता था और कर्म को पूजा मानता था। इस प्रकार परिश्रमपूर्वक पढ़-लिख कर उसने अपना लक्ष्य प्राप्त किया और कलक्टर बन गया।

प्रश्न 2.
आपका क्या लक्ष्य है ? अपने लक्ष्य को प्राप्त करने के लिए आप क्या करेंगे?
उत्तर:
मेरे जीवन का लक्ष्य आदर्श अध्यापक बनना है, जो अपने विद्यार्थियों को विद्या के प्रति सच्ची लगन पैदा करके उन्हें भावी भारत का सच्चा एवं अनुशासित नागरिक बना सके। इसके लिए मैं खूब मेहनत से पढंगा। बी०ए० करने के बाद अध्यापक के प्रशिक्षण के लिए बी०एड्० की परीक्षा उत्तीर्ण करके किसी अच्छे विद्यालय में शिक्षक का पद ग्रहण कर विद्यार्थियों को सर्वगुण सम्पन्न बनाने का प्रयास करूँगा।

3. इस कहानी में कई बिन्दुओं को छुआ गया है जैसे :

…………………… गाँवों से शहर की ओर पलायन
…………………… ग्रामीण लोगों की दशा/गरीबी/यथास्थिति
…………………… लक्ष्य प्राप्त करना
…………………… लड़कियों की शिक्षा के प्रति सोच
…………………… गाँव के प्रति प्यार
…………………… सम्बन्धों की आत्मीयता
…………………… अध्यापकों का सम्मान
…………………… इन बिंदुओं पर विचार-विमर्श करें।
उत्तर:
विद्यार्थी आपस में विचार-विमर्श करें।

PSEB 7th Class Hindi Guide धूल का फूल Important Questions and Answers

निम्नलिखित प्रश्नों के उत्तर उचित विकल्प चुनकर लिखिए

प्रश्न 1.
‘धूल का फूल’ किसकी कहानी है ?
(क) खेत मज़दूर के पुत्र की
(ख) खेत की
(ग) खेत मजदूर की पुत्री की
(घ) नाले की
उत्तर:
(क) खेत मज़दूर के पुत्र की

प्रश्न 2.
खेत मजदूर का लड़का पढ़-लिखकर क्या बनता है ?
(क) चपड़ासी
(ख) कलक्टर
(ग) अध्यापक
(घ) पुलिस कप्तान
उत्तर:
(ख) कलक्टर

प्रश्न 3.
गाड़ी चलाते हुए कौन सोच रहा था ?
(क) करण
(ख) गुरचरण
(ग) गुरशरण
(घ) गौरव
उत्तर:
(ग) गुरशरण

प्रश्न 4.
कुम्हार का क्या नाम था ?
(क) सब्बू
(ख) चौधरी
(ग) कब्बू
(घ) चेतन
उत्तर:
(क) सब्बू

प्रश्न 5.
बड़ा बनने की कुंजी क्या है ?
(क) शरीर की ताकत
(ख) ज़मीन
(ग) धन
(घ) विद्या
उत्तर:
(घ) विद्या

प्रश्न 6.
कलक्टर प्रशांत क्या आदेश लेकर आया था?
(क) गाँव खाली कराने का
(ख) गाँव के विद्यालय का दर्जा बढ़ाने का
(ग) ज़मीन लेने का
(घ) ज़मीन देने का
उत्तर:
(ख) गाँव के विद्यालय का दर्जा बढ़ाने का

निम्नलिखित रिक्त स्थानों की पूर्ति उचित विकल्पों से कीजिए

प्रश्न 1.
कलक्टर के चाचा का नाम ………… था ।
(क) चरण सिंह
(ख) विक्रम सिंह
(ग) करण सिंह
(घ) चेतन सिंह
उत्तर:
(क) चरण सिंह

प्रश्न 2.
कलक्टर का नाम …………… था।
(क) अविरल
(ख) प्रशांत
(ग) चहल
(घ) वेदांत
उत्तर:
(ख) प्रशांत

प्रश्न 3.
प्रशांत के पिता का नाम …………. था ।
(क) अब्दुल्ला
(ख) कासिम
(ग) सुजान सिंह
(घ) नसीब
उत्तर:
(घ) नसीब

प्रश्न 4.
प्रशांत के मास्टर का नाम ………… था ।
(क) आदित्य
(ख) विश्वास
(ग) विवेक शर्मा
(घ) कर्म सिंह
उत्तर:
(क) आदित्य

दिए गए शब्दों का सही अर्थ मिलान कीजिए

प्रश्न 1.
सरपट:
साँप का पेट
सिर का पेट
तेज चाल से
उत्तर:
तेज़ चाल से

प्रश्न 2.
प्रांगण:
आंगन
पराग का कण
परात
उत्तर:
आंगन

प्रश्न 3.
मजूरी:
मंजूरी
मज़दूरी
मंजर
उत्तर:
मजदूरी

प्रश्न 4.
उत्थान:
उनका
उन्नति
उतना
उत्तरः
उन्नति

धूल का फूल Summary

धूल का फूल पाठ का सार

‘धूल का फूल’ एक ऐसे खेत-मज़दूर गरीब पिता के पुत्र की कहानी है, जो अपने परिश्रम तथा इच्छा शक्ति के बल पर पढ़-लिखकर कलक्टर बन जाता है। ऊबड़-खाबड़ सड़क पर गाड़ी चलाते हुए गुरशरण सोच रहा था कि न मालूम क्यों साहब इधर दौरे पर आए हैं ? जब उसने साहब से पूछा कि अभी और कितनी दूर जाना है तो साहब ने उसे चलते रहने को कहा।

साहब का मन बहुत प्रसन्न था। वे बरसों बाद अपने गाँव जा रहे थे। वे बीस-पच्चीस साल पहले के पंच जी के खेत, दीनू ग्वाले की गाय-भैंसें, सब्बू कुम्हार के चाक आदि की बातें सोच कर भाव-विभोर हो रहे थे। उन्हें गन्ने का रस पीना, गर्म गुड़ खाना, ऊधम मचाना, ककड़ियाँ-खरबूजे खाना, पाठशाला में आदित्य मास्टर जी से पढ़ना आदि याद आ रहा था। उसके पिता दूसरों के खेतों में मजदूरी करते थे तथा माँ के साथ वह खेतों पर खाना ले जाता था। उसके पिता को शहर में चपरासी की नौकरी मिली तो सारा परिवार गाँव से शहर आ गया। जहाँ आकर गाड़ी में बैठे अफसर को देखकर उसका मन भी उन जैसा बनने की इच्छा करता और वह सोचता कि जब लाल बहादुर शास्त्री, लिंकन, एडीसन जैसे बड़े बन सकते हैं, तो वह क्यों नहीं?

सरकारी स्कूल दूर थे फिर भी वह पढ़ता गया क्योंकि उसे विश्वास था कि बड़े बनने की कुंजी विद्या ही है और विद्याधन मेहनत के बिना नहीं मिलता। वह कर्म को पूजा मानने लगा। वह शरण को गाड़ी धीरे चलाने के लिए कहता है और शरण से यह जान कर कि उसका लड़का तो पढ़ने जाता है परन्तु वह अपनी लड़की को स्कूल नहीं भेजता तो उसे समझाता है कि लड़की को भी पढ़ाओ क्योंकि लड़कियाँ घर का श्रृंगार होती हैं, वे ही परिवार का आधार हैं। उसने गाड़ी की खिड़की से मुटियारों को सिर पर बोझ लेकर जाते, खेतों में कम्बाइन चलाते तथा स्त्रियों-बच्चों को झोला लिए अनाज की बालियाँ चुनते देखा तो सोचने लगा कि आजादी के इतने सालों बाद भी गरीब की वही दशा है जो उस के ज़माने में थी। तभी उसे एक बुजुर्ग दिखाई दिए। उस ने गाड़ी रुकवाई और उनके पैर स्पर्श किए। वे चाचा चरण सिंह थे।

उन्होंने उसे नहीं पहचाना तो उसने स्वयं ही अपना परिचय दिया कि वह उनका प्रशांत है। चाचा को भी याद आया नसीब का पुत्र प्रशांत। वह आज यहाँ गाँव के विद्यालय का दर्जा बढ़ाने का आदेश लेकर आया था। पाठशाला रंगोली, रंग-बिरंगी झंडियों से सजी हुई थी। पुष्प-गुच्छों से उसका स्वागत हुआ। अपने प्रिय अध्यापक आदित्य प्रकाश के पैरों को स्पर्श करने के लिए जैसे ही प्रशांत झुका कि उन्होंने उसे बाँहों में भरकर गले लगा लिया और वे फूले नहीं समा रहे थे कि उनका गरीब प्रशांत कलक्टर प्रशांत बन गया है। वही सरकार की ओर से पाठशाला का दर्जा बढ़ाने का आदेश लाया है।

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 11 Perimeter and Area Ex 11.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.4

1. An rectangular park is 80 m long and 65 in wide. A path of 5 m width is constructed outside the park. Find the area of path.
Solution:
Let ABCD be a rectangular park.
Length of the park = 80 m
Breadth of the park = 65 m

Area of the rectangular park
ABCD = Length × Breadth
= 80 m × 65 m
= 5200 m²
Length of rectangular garden EFGH (including park)
= 80 + 5 + 5
= 90 m
Breadth = 65 + 5 + 5
= 75 m
Area of rectangular path EFGH = 90 × 75
= 6750 m²
Area of the path = Area of rectangular park EFGH – Area of rectangle ABCD
= 6750 – 5200
= 1550 m²

2. A rectangular garden is 110 m long and 72 m broad. A path of uniform width 8 m has to be constructed around it. Find the cost of gravelling the path at ₹ 11.50 per m².
Solution:
Let ABCD represents the rectangular garden and the shaded region represents the path of width 8 m around the garden.
Length of rectangular garden l = 110 m

Breadth of rectangular garden b = 72 m
Area of rectangular garden ABCD = (110 × 72) m²
= 7920 m²
Length of rectangular garden including path = 110 m + (8m + 8m)
= 126 m
Breadth of rectangular garden including path = 72m + (8m + 8m) = 88 m
Area of garden including path = (126 × 88) m²
= 11088 m²
Area of path = Area of garden including path – Area of garden
Area of path = (11088 – 7920) m²
= 3168 m²
Cost of gravelling 1 m² of path = ₹ 11.50
Cost of gravelling 2928 m2 of path = ₹ 3168 × 11.50
= ₹ 36432

3. A room is 12 m long and 8 m broad. It is surrounded by a verandah, which is 3 m wide all around it. Find the cost of flooring the verandah with marble at ₹ 275 per m².
Solution:
Let ABCD. represents the rectangular floor of room and shaded region represents the verandah 3 m wide all along the outside of a room.

PQ = (3 + 12 + 3) m
= 18 m
PS = (3 + 8 + 3) m
= 14 m
Area of rectangle ABCD = 1 × b
= AB × AD
= 12 m × 8m
= 96 m²
Area of recangle PQRS = 1 × b
= PQ × PS
= 18 m × 14 m
= 252 m²
Area of verandah = [Area of rectangle PQRS] – [Area of rectangle ABCD]
= (252 – 96) m²
= 156 m²
(Rate of flooting the verandha with marble verandah = ₹ 275 per m²)
Cost of flooring verandah with moble.
= ₹ (156 × 275)
= ₹ 42900.

4. A sheet of paper measures 30 cm × 24 cm. A strip of 4 cm width is cut from it, all around. Find the area of remaining sheet and also the area of cut out strip.
Solution:
Let ABCD represent the sheet of 30 cm × 24 cm and shaded region represents the 4 cm width to be cut

PQ = (30 – 4 – 4) cm
= 22 cm
PS = (24 – 4 – 4) cm
= 16 cm

(i) remaining sheet
[Area of rectangle ABCD] – [Area of rectangle PQRS]
= (30 × 24 – 22 × 16)
= (720 – 352 = 368) cm²
Area of the cut our strip i.e. area of rectangle PQRS = 22 × 16 cm²
= 352 cm²

5. A path of 2 m wide is built along the border inside a square garden of side 40 m. Find :

Question (i).
The Area of path.
Solution:
Let ABCD be the square park of side 40 m and the shaded region represents the path 2 m wide
EF = 40 m – (2 + 2) m
= 36 m

Area of square park ABCD = (Side)²
= 40 × 40
= 1600 m²
Area of EFGH = (Side)²
= 36 × 36
= 1296 m²
Area of path = Area of square park ABCD – Area of EFGH
= (1600 – 1296) m²
= 304 m²

Question (ii).
The cost of planting grass in the remaining portion of the garden at the rate of ₹ 50 per m².
Solution:
Cost of planting grass = 50 per m²
Cost of planting grass 1m² = ₹ 50
Cost of 1296 m² = 1296 × 50
= ₹ 64800

6. A nursery school play ground is 150 m long and 75 m wide. A portion of 75 m × 75 m is kept for see-saw slides and other park equipments. In the remaining portion 3 m wide path parallel to its width and parallel to remaining length (as shown in fig). The remaining area is covered by grass. Find the area covered by grass.

Solution:
Area of school ground
= 150 m × 75 m = 11250 m²
Area kept for see-saw slides and other equipments
= 75 × 75
= 5625 m²
Area of path parallel to width of ground = 75 × 3
= 225 m²
Area common to both paths = 3 × 3
= 9 m²
Total area covered by path
= (225 + 225 – 9)
= 441 m²
Area covered by grass = Area of ground – (Area kept for see-saw slides + area covered by paths)
= 11250 – (5625 + 441)
= (11250 – 6066) m²
= 5184 m²

7. Two cross roads each of width 8 m cut at right angle through the centre of a rectangular park of length 480 m and breadth 250 m and parallel to its sides. Find the area of roads. Also, find the area of park excluding cross roads.
Solution:

ABCD represent the rectangular park of length AB = 480 m and breadth BC = 250 m. Area of shaded portion i.e. area of rectangle EFGH and PQRS represent the area of cross roads, but the area of square KLMN is taken twice, So it will be subtracted.
Now EF = 480, FG = 8 m, PQ = 250 m, QR = 8 m, KL = 8 m.
Area covered by roads = Area of rectangle EFGH + area of rectangle PQRS – Area of square KLMN
= (EF × FG) + (PQ × QR) – (KL)²
= (480 × 8) + (250 × 8) – (8 × 8)
= 3840 + 2000 – 64
Area of the road = 5776 m²
Area of park excluding cross roads = 250 × 480 – (250 × 8 + 480 × 8 – 8 × 8)
= 114224 m²

8. In a rectangular field of length 92 m and breadth 70 m, two roads are constructed which are parallel to the sides and cut each other at right angles through the centre of field. If the width of each road is 4 m, find.
(i) The area covered by roads.
(ii) The cost of constructing the roads at the rate of ₹ 150 per m².
Solution:
Let ABCD represents the rectangular field of length ; AB = 92 m and breadth; AD = 70 m. Let the area of shaded portion
i. e. area of the rectangle PQRS and the area of rectangle EFGH represents the area of cross roads.

But in doing this, area of square KLMN is taken twice which is to be subtracted.
Now PQ = 4 m, PS = 70 m
and EH = 4 m, EF = 92 m
and KL = 4 m, KN = 4 m

= PQ × PS + EF × EH – KL × KN
= [(4 × 70) + (92 × 4) – (4 × 4)] m²
= (280 + 368 – 16) m²
= (648 – 16) m²
= 632 m²

(ii) Cost of constructing 1 m² of roads = ₹ 150
Therefore cost of constructing 632 m² of roads = ₹ (150 × 632)
= ₹ 94800.

9. Find the area of shaded region in each of the following figures.

Question (i).

Solution:
Length of rectangle ABDC
= 3m + 15m + 3m
= 21 m
Breadth of rectangle ABDC
= 2m + 12m + 2m
= 16 m
Area of rectangle ABDC
= length × breadth
= 21 × 16 m²
= 336 m²
Length of rectangle PQRS = 15 m
Breadth of rectangle PQRS = 12 m
Area of rectangle PQRS = 15 × 12 m²
= 180 m²
Area of shaded region = Area of rectangle ABCD – Area of rectangle PQRS
= 336 m² – 180 m²
= 156 m²

Question (ii).

Solution:

SR = PQ = 2.5 m
EH = FG = 4 m
KL = 2.5 m
LM = 4 m
Area of shaded region = [Area of rectangle PQRS] + [Area of rectangle EFGH] – Area of rectangle KLMN
= 40 × 2.5 + 80 × 4 – 2.5 × 4
= 100 + 320 – 10
= 420 – 10
= 410 m²

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 11 Perimeter and Area Ex 11.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3

1. Find the circumference of circle whose
(i) Radius (r) = 21 cm
(ii) Radius (r) = 3.5 cm
(iii) Diameter = 84 cm
Solution:
(i) Given radius (r) = 21 cm
circumference of circle = 2πr
= 2 × \(\frac {22}{7}\) ×21
= 132 cm

(ii) Given radius (r) = 3.5 cm
Circumference = 2πl
= 2 × \(\frac {22}{7}\) × 3.5
= 22 cm

(iii) Given Diameter (d) = 84 cm
radius (r) = \(\frac{d}{2}=\frac{84}{2}\)
= 42 cm
Circumference = 2πr
= 2 × \(\frac {22}{7}\) × 42
= 264 cm

2. If the circumference of a circular sheet is 176 m, find its radius.
Solution:
Given circumference of circular sheet = 176 m
Let radius = r
So 2πr = 176
r = \(\frac{176}{2 \pi}\)
\(\frac{176}{2 \times \frac{22}{7}}\)
= 28 m

3. A circular disc of diameter 8.4 cm is divided into two parts what is the perimeter of each semicircular part ?
Solution:
Given diameter of a circular disc = 8.4 cm
radius (r) = \(\frac{8.4}{2}\) = 4.2 cm
Perimeter of semicircular part = πr + 2r
= \(\frac {22}{7}\) × 4.2 + 2 × 4.2
= 22 × 0.6 + 8.4
= 21.6 cm

4. Find the area of the circle having
(i) Radius r = 49 cm
(ii) Radius r = 2.8 cm
(iii) Diameter = 4.2 cm
Solution:
(i) Given radius (r) = 49 cm
Area of circle = πr²
= \(\frac {22}{7}\) × 49 × 49
= 7546 cm²

(ii) Given radius (r) = 2.8 cm
Area of circle = πr²
= \(\frac {22}{7}\) × 2.8 × 2.8
= 24.64 cm²

(iii) Given diameter (d) = 4.2 cm
radius (r) = \(\frac{d}{2}=\frac{4.2}{2}\)
Area of circle = πr²
= \(\frac {22}{7}\) × 2.1 × 2.1
= 13.86 cm²

5. A gardener wants to fence a circular garden of radius 15 m. Find the length of wire, if he makes three rounds offense. Also, find the cost of wire if it costs ₹ 5 per meter (Take π = 3.14).
Solution:
Given radius of circular garden (r) = 15 m
Circumference of the circular garden = 2πr
= 2 × 3.14 × 15
= 94.2 m
So, length of the wire to make three rounds offense
= 3 × 94.2
= 282.6 cm
Cost of wire= ₹ 5 × 282.6
= ₹ 1413

6. Which of the following has larger area and by how much ?
(a) Rectangle with length 15 cm and breadth 5.4 cm
(b) Circle of diameter 5.6 cm.
Solution:
(a) Given length of rectangle = 15 cm
breadth = 5.4 cm
Area of rectangle = length × breadth
= 15 × 5.4
= 81 cm2

(b) Given diameter of circle (d) = 5.6 cm
radius (r) = \(\frac{d}{2}=\frac{5.6}{2}\)
= 2.8 cm
Area of the circle = πr²
= \(\frac {22}{7}\) × (2.8)²
= 24.64 cm²
Hence, Rectangle has more area = 81 – 24.64
= 56.36 cm²

7. From a rectangular sheet of length 15 cm and breadth 12 cm a circle of radius 3.5 cm is removed. Find the area of remaining sheet.
Solution:
Given length of rectangular sheet = 15 cm
Breadth of rectangle sheet = 12 cm
Area of rectangular sheet = length × breadth
= 15 × 12
= 180 cm²
Given radius of circle (r) = 3.5 cm
Area of circle = πr²
= \(\frac {22}{7}\) × (3.5)²
= 38.5 cm²
Since circle is removed from rectangular sheet.
So, area of remaining sheet = Area of rectangular Sheet – Area of circle
= 180 – 38.5
= 141.5 cm2

8. From a circular sheet of radius 7 cm, a circle of radius 2.1 cm is removed, find the area of remaining sheet.
Solution:
Radius of the circular sheet = 7 cm
Area of the circular sheet = 1 cm
= πr2 = \(\frac {22}{7}\) × 7 × 7 cm²
=154 cm²
Radius of the circle = 2.1 cm
Area of the circle
\(\frac {22}{7}\) × 2.1 × 1.1 = \(\frac{22}{7} \times \frac{21}{10} \times \frac{21}{10}\)
= \(\frac {1386}{100}\)
= 13.86 cm²
Area of the remaining sheet = 154 cm² – 13.86 cm²
= 140.14 cm²

9. Smeep took a wire of length 88 cm and bent it into the shape of a circle, find the radius and area of the circle. If the same wire is bent into a square, what will be the side of the square ? Which figure encloses more area ?
Solution:
Given length of wire = 88 cm
The wire is bent into the shape of circle.
Circumference of circle = length of the
2πr = 88
r = \(\frac{88}{2 \pi}=\frac{44}{\pi} \mathrm{cm}\)
= 14 cm
Area of the circle = πr²
= π × (14)²
= \(\frac {22}{7}\) × 14 × 14
= 616 cm²
If the same wire is bent into the square
Let side of the square = a
Perimeter of square = length of the wire
4 × a = 88
a = \(\frac {88}{4}\)
= 22 cm
Area of the square = (side)²
= (22)²
= 484 cm²
Hence circle enclosed more area.

10. A garden is 120 m long and 85 m broad. Inside the garden, there is a circular pit of diameter 14 m. Find the cost of planting the remaining part of the garden at the rate of ₹ 5.50 per square meter.
Solution:
Given length of garden = 120 m
Breadth of garden = 85 m
Area of garden = length × breadth
= 120 × 85
= 10200 m²
Given diameter of circular pit (d) = 14 m
radius (r) = \(\frac{d}{2}=\frac{14}{2}\)
= 7 m
Area of circular pit = πr²
= \(\frac {22}{7}\) × 7 × 7
= 154 m²
Remaining part of garden = Area of garden for planting – Area of circular pit
= 10200 – 154
= 10046 m²
Cost of planting the remaining part of the garden
= ₹ 5.50 × 10046
= ₹ 55243

11. In the figure PQ = QR and PR = 56 cm. The radius of inscribed circle is 7 cm. Q is centre of semicircle. What is the area of shaded region ?

Solution:
Given PQ = QR
PR = 56 cm
Radius of inscribed circle = 7 cm
So PR = PQ + QR
= PQ + PQ = 2PQ
Hence, PQ = \(\frac{\mathrm{PR}}{2}=\frac{56}{2}\)
= 28 cm
So QR = PQ = 28 cm
Area of shaded region = Area of semicircle of diameter PR – Area of semicircle of diameter PQ – Area of semicircle of diameter QR – Area of inscribe circle

12. The minute hand of a circular clock is 18 cm long. How far does the tip of minute hand move in one hour ?
Solution:
Given minute hand of a circular clock = 18 cm
Distance covered by minute hand in 1 hour = 2πr
= 2 × 3.14 × 18
= 2 × \(\frac {314}{100}\) × 18
= \(\frac {11304}{100}\)
= 113.04 cm.

13. Multiple choice questions :

Question (i).
The circumference of a circle of diameter 10 cm is :
(a) 31.4 cm
(b) 3.14 cm
(c) 314 cm
(d) 35.4 cm
Answer:
(a) 31.4 cm

Question (ii).
The circumference of a circle with radius 14 cm is :
(a) 88 cm
(b) 44 cm
(c) 22 cm
(d) 85 cm
Answer:
(a) 88 cm

Question (iii).
What is the area of the circle of radius 7 cm ?
(a) 49 cm
(b) 22 cm²
(c) 154 cm²
(d) 308 cm²
Answer:
(c) 154 cm²

Question (iv).
Find the diameter of a circle whose area is 154 cm² ?
(a) 4 cm
(b) 6 cm
(c) 14 cm
(d) 12 cm
Answer:
(c) 14 cm

Question (v).
A circle has area 100 times the area of another circle. What is the ratio of their circumferences ?
(a) 10 : 1
(b) 1 : 10
(c) 1 : 1
(d) 2 : 1
Answer:
(a) 10 : 1

Question (vi).
Diameter of a circular garden is 9.8 cm. Which of the following is its area ?
(a) 75.46 cm²
(b) 76.46 cm²
(c) 74.4 cm²
(d) 76.4 cm²
Answer:
(a) 75.46 cm²

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 11 Perimeter and Area Ex 11.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2

1. Estimate the area of the following figures by counting unit squares.

Question (i).

Solution:
In the given figure, number of squares covered completely = 135
Area of a square = 1 sq. unit
Area of (135 square) figure = 135 sq. units, (approx.)

Question (ii).

Solution:
In the given figure number of square covered completely = 114
Area of one square = 1 unit
∴ Area of 114 squares = 114 sq units approx
Thus area of given figure = 114 sq units approx.

2. In the following figures find the area of 

Question (i).
ΔABC

Solution:
Given length of rectangle = 15 cm
Breadth of rectangle = 8 cm
The diagonal AC divides the rectangle into two triangles ΔABC and ΔADC
So, area of ΔABC = \(\frac {1}{2}\) × Area of rectangle ABCD
= \(\frac {1}{2}\) × length × breadth
= \(\frac {1}{2}\) × 15 × 8
= 60 cm²

Question (ii).
ΔCOD

Solution:
Given side of square = 6 cm
The diagonals AC and BD divides the square into four equal posses (triangles)
So, area of ΔCOD = \(\frac {1}{4}\) × Area of square
= \(\frac {1}{4}\) × 6 × 6
= 9 cm²

3. Find the area of following parallelograms.

Question (i).

Solution:
Given base of parallelogram = 9 cm
Height of parallelogram = 6 cm
Area of parallelogram = Base × height
= 9 × 6
= 54 cm²

Question (ii).

Solution:
Given base of parallelogram = 6.5 cm
Height of parallelogram= 8.4 cm
Area of parallelogram = Base × height
= 6.5 × 8.4
= 54.6 cm²

4. Find the value of x in the following parallelograms.

Question (i).

Solution:
Given base (AD) of parallelogram = 5.6 cm
Corresponding height of parallelogram = 9 cm
Area of parallelogram = 5.6 × 9 cm² ….(1)
Also in the paralleogram, base (AB) = x
Corresponding height of parallelogram = 7 cm
Area of parallelogram will be = x × 7 ….(2)
From (1) and (2), we get
x × 7 = 5.6 × 9
x = \(\frac{5.6 \times 9}{7}\)
= 7.2

Question (ii).

Solution:
Given base (AB) of parallelogram = 15 cm
Corresponding height = 6 cm
Area of parallelogram =15 × 6 cm² ….(1)
Also Base (AD || BC) of parallelogram = 9 cm
Corresponding height = x
So area of parallelogram = 9 × x ….(2)
From (1) and (2)
9 × x = 15 × 6
x = \(\frac{15 \times 6}{9}\)
= 10 cm.

5. The adjacent sides of a parallelogram are 28 cm and 45 cm and the altitude on longer side is 18 cm. Find the area of parallelogram.
Solution:
Given base of the parallelogram = 45 cm
Corresponding height = 18 cm
Area of parallelogram = Base × Height
= 45 × 18
= 810 cm2

6. ABCD is a parallelogram given in figure. DN and DM are the altitudes on side AB and CB respectively. If area of the parallelogram is 1225 cm², AB = 35 cm and CB = 25 cm, find DN and DM.

Solution:
In the given parallelogram ABCD
Base (AB) = 35 cm
Let height (DN) = x cm
So area of parallelogram = 35 × x cm²
But given area of parallelogram (ABCD) = 1225 cm²
Therefore 35x = 1225
x = \(\frac {1225}{35}\)
= 35 cm
Similarly, for base (BC) and height (DM)
1225 = BC × DM
\(\frac{1225}{\mathrm{BC}}\) = DM
or DM = \(\frac {1225}{25}\)
= 49 cm.

7. Find the area of the following triangles.

Question (i).

Solution:
Given base of triangle = 7 cm
Height of triangle = 4.8 cm
Area of triangle = \(\frac {1}{2}\) × Base × Height
= \(\frac {1}{2}\) × 7 × 4.8
= 16.8 cm².

Question (ii).

Solution:
Given base of triangle =6 cm
Height of triangle = 9 cm
Area of triangle = \(\frac {1}{2}\) × Base × Height
= \(\frac {1}{2}\) × 6 × 9
= 27 cm²

8. Find the value of x in the following triangles.

Question (i).

Solution:
In ΔABC, BC = 8 cm, AC = 15 cm
Area of triangle ABC = \(\frac {1}{2}\) × Base × height
= \(\frac {1}{2}\) × BC × AC
= \(\frac {1}{2}\) × 8 × 15
= 60 cm² …(1)
Also, in ΔABC, AB = 20 cm
height = x
Area of triangle ABC = \(\frac {1}{2}\) × Base × Height
= \(\frac {1}{2}\) × 20 × x ….(2)
From (1) and (2)
\(\frac {1}{2}\) × 20 × x = 60
x = \(\frac{60 \times 2}{20}\)
x = 6 cm.

Question (ii).

Solution:
In ΔABC, base (AC) = 25 cm
height = 14 cm
Area of triangle ABC = \(\frac {1}{2}\) × Base × height
\(\frac {1}{2}\) × 14 × 25 ….(1)
Also, in ΔABC, base AB = x cm
height = 20 cm
So, area of ΔABC = \(\frac {1}{2}\) × Base × Height
= \(\frac {1}{2}\) × x × 20 ….(2)
From (1) and (2) we get
\(\frac {1}{2}\) × x × 20 = \(\frac {1}{2}\) × 14 × 25
x = 17.5 cm

9. ABCD is a square, M is a point on AB such that AM = 9 cm and area of ΔDAM is 171 cm². What is the area of the square ?

Solution:
Given area of ΔDAM = 171 cm²
Base of triangle = 9 cm
As, area of triangle ΔDAM = \(\frac {1}{2}\) × base × height
171 = \(\frac {1}{2}\) × 9 × (DA)
Hence height (DA) = \(\frac{171 \times 2}{9}\)
= 18 cm
Hence side of square (DA) = 18 cm
Therefore area of square = (side)²
= (18)²
= 324 cm²

10. ΔABC is right angled at A as shown in figure. AD is perpendicular to BC, if AB = 9 cm, BC = 15 cm and AC = 12 cm. Find the area of ΔABC, also find file length of AD.

Solution:
Given AB = 9 cm
BC = 15 cm
AC = 12 cm
Let AD = x cm
Area of triangle = \(\frac {1}{2}\) × Base × height
= \(\frac {1}{2}\) × 12 × 9 cm².
= 54 cm² ….(1)
Since, AD is perpendicular to BC
So, area of triangle = \(\frac {1}{2}\) × BC × AD
= \(\frac {1}{2}\) × 15 × AD ….(2)
From (1) and (2) we get
\(\frac {1}{2}\) × 15 × AD = 54
AD = \(\frac{54 \times 2}{15}\)
AD = 7.2 cm

11. ΔABC is isosceles with AB = AC = 9 cm, BC = 12 cm and the height AD from A to BC is 4.5 cm. Find the area of ΔABC. What will be the height from B to AC i.e. BN ?

Solution:
In triangle ABC, Base (BC) = 12 cm
AD = 4.5 cm
AD is perpendicular to BC
So, Area of ΔABC = \(\frac {1}{2}\) × base × height
= \(\frac {1}{2}\) × 12 × 4.5 cm
= 27 cm ….(1)
Also, in ΔABC, Base (AC) = 9 cm
Let corresponding height (BN) = x
So area of ΔABC = \(\frac {1}{2}\) × base × height
= \(\frac {1}{2}\) × 9 × BN ….(2)
From (1) and (2)
\(\frac {1}{2}\) × 9 × BN = 27
BN = \(\frac{27 \times 2}{9}\)
= 6 cm.

12. Multiple choice questions :

Question (i).
Find the height of a parallelogram whose area is 246 cm² and base is 20 cm.
(a) 1.23 cm²
(b) 13.2 cm²
(c) 12.3 cm²
(d) 1.32 cm²
Answer:
(c) 12.3 cm²

Question (ii).
One of the side and the corresponding height of a parallelogram are 7 cm and 3.5 cm respectively. Find the area of the parallelogram.
(a) 21 cm²
(b) 24.5 cm²
(c) 21.5 cm²
(d) 24 cm²
Answer:
(b) 24.5 cm²

Question (iii).
The height of a triangle whose base is 13 cm and area is 65 cm² is :
(a) 12 cm
(b) 15 cm
(c) 10 cm
(d) 20 cm
Answer:
(c) 10 cm

Question (iv).
Find the area of an isosceles right angled triangle, whose equal sides are of length 40 cm each.
(a) 400 cm²
(b) 200 cm²
(c) 600 cm²
(d) 800 cm²
Answer:
(d) 800 cm²

Question (v).
If the sides of a parallelogram are increased to twice of its original length, how much will be the perimeter of the new parallelogram ?
(a) 1.5 times
(b) 2 times
(c) 3 times
(d) 4 times
Answer:
(b) 2 times

Question (vi).
In a right angled triangle one leg is double the other and area is 64 cm² find the smaller leg.
(a) 8 cm
(b) 16 cm
(c) 24 cm
(d) 32 cm.
Answer:
(a) 8 cm

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 11 Perimeter and Area Ex 11.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.1

1. Find the perimeter and the area of a rectangle having :
(i) Length = 28 cm, Breadth = 15 cm
(ii) Length = 9.4 cm Breadth = 2.5 cm
Solution:
(i) Given length of rectangle = 28 cm
Breadth of rectangle = 15 cm
Perimeter of rectangle = 2 [length + Breadth]
= 2 [28 + 15]
= 2 × 43
= 86 cm

Area of rectangle = length × Breadth
= 28 × 15
= 420 cm²

(ii) Perimeter of rectangle = 2 [9.4 + 2.5]
= 2 × 11.9
= 23.8 cm
Area of rectangle = 9.4 × 2.5
= 23.5 cm²

2. Find the perimeter and the area of a square whose side measures
(i) 29 cm
(ii) 8.3 cm
Solution:
(i) Given side of square = 29 cm
Perimeter of square = 4 × side
= 4 × 29
= 116 cm
Area of square = (side)²
= (29)²
= 841 cm²

(ii) Perimeter of square = 4 × 8.3
= 33.2 cm
Area of square = 8.3 × 8.3
= 68.89 cm²

3. The perimeter of a square park is 148 m. Find its area.
Solution:
Given the perimeter of square park = 148 m
Side of the square park = \(\frac{perimeter}{4}\)
= \(\frac {148}{4}\)
Area of the square park = (side)²
= (37)²
= 1369 m²

4. The area of a rectangle is 580 cm². Its length is 29 cm. Find its breadth and also, the perimeter.
Solution:
Given area of rectangle = 580 cm²
Length of the rectangle = 29 cm
Let breadth of the rectangle = b cm
Area of the rectangle = length × breadth
580 = 29 × b
\(\frac {580}{29}\) = b
b = 20 cm
Perimeter of rectangle = 2 [length + breadth]
= 2 [29 + 20]
= 2 × 49
= 98 cm

5. A wire is in the shape of a rectangle. Its length is 48 cm and breadth is 32 cm. If the same wire is rebent into the shape of a square, what will be the measure of each side. Also, find which shape encloses more area and by how much ?
Solution:
Given length of the rectangle = 48 cm
Breadth of the rectangle = 32 cm
Perimeter of the rectangle = 2 [length + breadth]
= 2 [48 + 32]
= 2 × 80
= 160 cm
Let side of square = a cm
Perimeter of the square = 4 × a
Since wire is rebent into the shape of a square
Perimeter of square = Perimeter of rectangle
4 a = 160
Therefore, a = \(\frac {160}{4}\)
= 40 cm
Area of square = (side)²
= 40 × 40
= 1600 cm²
Area of rectangle = length × breadth
= 48 × 32
= 1536 cm²
∴ Square encloses more area by 64 cm²

6. The area of a square park is the same as that of a rectangular park. If the side of the square park is 75 m and the length of the rectangular park is 125 m, find the breadth of the rectangular park. Also, find the perimeter of rectangular park.
Solution:
Given side of square park = 75 m
Area of square park = (75)²
= 75 × 75
= 5625 m²
Length of rectangular park = 125 m
Let breadth of rectangular park = b m
Area of rectangular park = length × breadth
= 125 × b m²
Given that
Area of rectangular park = Area of square park
125 × b = 5625
= 45 m
Perimeter of rectangular park = 2 [length + breadth]
= 2 [125 + 45]
= 2 × 170
= 340 m

7. A door of length 2.5 m and breadth 1.5 m is fitted in a wall. The length of wall, is 9 m and breadth is 6 m. Find the cost of painting the wall, if the rate of painting the wall is ₹ 30 per m².

Solution:
Length of door = 2.5 m
Breadth of door = 1.5 m
Area of door = length × breadth
= 2.5 × 1.5
= 3.75 m²
Area of wall = 9 × 6
= 54 m²
Area of wall painting = Area of wall including door – Area of door
= 54 – 3.75
= 50.25 m²
Cost of painting 1 m² of wall = ₹ 30
Cost of painting 50.25 m² of wall = ₹ 50.25 × 30
= ₹ 1507.50

8. A door of dimensions 3 m × 2 m and a window of dimensions 2.5 m × 1.5 m is fitted in a wall. The length of the wall is 7.8 m and breadth is 3.9 m. Find the cost of painting the wall, if the rate of painting the wall is ₹ 25 per m².
Solution:
Area of door = 3 × 2 = 6 m²
Area of window = 2.5 m × 1.5 m
= 3.75 m²
Area of wall = 7.8 m × 3.9 m
= 30.42 m²
Area of painting the wall = Area of wall – Area of door – Area of window
= 30.42 – 6 – 3.75
= 20.67 m²
Cost of painting the wall = ₹ 25 × 20.67
= ₹ 516.75

9. Find the area and the perimeter of the following figures.

Solution:
(i) Perimeter of the given figure
= AB + BC + CD + DE + EF + FG + GH + HA
= 2 + 3.5 + 3 + 2 + 5 + 3.5 + 10 + 9
= 38 cm²
Area of the figure = Area of rectangle ABCJ + Area of rectangle JDEI + Area of rectangle IFGH
= 2 × 3.5 + 5 × 2 + 10 × 3.5
= 7 + 10 + 35
= 52 cm²

(ii) Perimeter of the given figure
= 8cm + 5 cm + 1.5 cm + 2.5 cm + 2.5 cm + 1.5 cm + 1.5 cm + 1.5 cm + 2.5 cm + 1.5 cm
= 29 cm
Area of the given figure = Area of rectangle I + Area of rectangle II + Area of rectangle III
= 8 cm × 1.5 cm + 3.5 cm × 1.5 cm + 1.5 cm × 1.5 cm
= 12 cm² + 5.25 cm² + 2.25 cm
= 19.5 cm²

10. Multiple Choice Questions :

Question (i).
What is the area of a rectangle of dimensions 12 cm × 10 cm ?
(a) 44 cm²
(b) 120 cm²
(c) 1200 cm²
(d) 1440 cm²
Answer:
(b) 120 cm²

Question (ii).
Find the breadth of a rectangle whose length is 12 cm and perimeter is 36 cm.
(a) 6 cm
(b) 3 cm
(c) 9 cm
(d) 12 cm
Answer:
(a) 6 cm

Question (iii).
If each side of a square is 1 m then its area is ?
(a) 10 cm²
(b) 100 cm²
(c) 1000 cm²
(d) 10000 cm²
Answer:
(d) 10000 cm²

Question (iv).
Find the area of a square whose perimeter is 96 cm.
(a) 576 cm²
(b) 626 cm²
(c) 726 cm²
(d) 748 cm².
Answer:
(a) 576 cm²

Question (v).
The area of a rectangular sheet is 500 cm². If the length of the sheet is 25 cm, what is its breadth ?
(a) 30 cm
(b) 40 cm
(c) 20 cm
(d) 25 cm.
Answer:
(c) 20 cm

Question (vi).
What happens to the area of a square, if its side is doubled ?
(a) The area becomes 4 times, the area of original square.
(b) The area becomes \(\frac {1}{4}\) times, the area of original square.
(c) The area becomes 16 times, the area of original square.
(d) The area becomes \(\frac {1}{6}\) times, 6 the area of original square.
Answer:
(a) The area becomes 4 times, the area of original square.

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 10 Practical Geometry MCQ Questions with Answers.

PSEB 7th Class Maths Chapter 10 Practical Geometry MCQ Questions

Multiple Choice Questions :

Question 1.
Number of parallel lines that can be drawn passing through a point not lying on the given line is :
(a) 0
(b) 1
(c) 2
(d) 3.
Answer:
(b) 1

Question 2.
The sum of three angles of a Δ is :
(a) 90°
(b) 180°
(c) 360°
(d) None
Answer:
(b) 180°

Question 3.
A triangle can be constructed by taking its sides of these :
(a) 3 cm, 5 cm, 7 cm
(b) 4 cm, 5 cm, 9 cm
(c) 4 cm, 3 cm, 8 cm
(d) 3 cm, 2 cm, 5 cm.
Answer:
(a) 3 cm, 5 cm, 7 cm

Question 4.
Two angles of a triangle are 40° and 50°. Third angle is :
(a) 40°
(b) 50°
(c) 90°
(d) 60°
Answer:
(c) 90°

Question 5.
The angles of a triangle are 30° and 50°, third angle is :
(a) 100°
(b) 60°
(c) 80°
(d) 50°.
Answer:
(a) 100°

Fill in the blanks :

Question 1.
Sum of lengths of any two sides of a triangle is …………….
Answer:
greater than third side

Question 2.
In right angled triangle.
(Hypotenuse)² = (…………….)² + (…………….)²
Answer:
Base, Perpendicular

Question 3.
SAS stands for …………….
Answer:
Side, angle, Side

Question 4.
RHS stands for …………….
Answer:
Right angle hypotenuse side

Question 5.
ASA stands for …………….
Answer:
Angle, side, angle.

Write True or False

Question 1.
Exterior angle of a triangle is equal to the sum of opposite interior angles. (True/False)
Answer:
True

Question 2.
The lengths of three sides can be used to construct a triangle. (True/False)
Answer:
True

Question 3.
The sum of the three angles of a triangle is 160°. (True/False)
Answer:
False

Question 4.
Construction of a triangle is possible when some of too angle is 180°. (True/False)
Answer:
True

Question 5.
Each angle of equilateral triangle is 60°. (True/False)
Answer:
True

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 10 Practical Geometry Ex 10.5 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.5

1. Construct a right angled triangle ABC with ∠C = 90°, AB = 5 cm and BC = 3 cm.
Solution:
Given : Two sides of ΔABC as
AB = 5 cm,
BC = 3 cm
and ∠C = 90°.
To construct : A triangle with these two sides and one right angle.
Steps of Construction :
Step 1. We first draw a rough sketch of the triangle ABC and indicate the measure of these two sides and mark the right angle.

Step 2. Draw BC of length 3 cm.

Step 3. At C, draw CX ⊥ BC. (A should be somewhere on this perpendicular).

Step 4. With B as centre, draw an arc of radius 5 cm. (A must be on this arc since it is at a distance of 5 cm from B).

Step 5. A has to be on the perpendicular line CX as well as on the arc drawn with centre C.
∴ A is the meeting point of these two.
ΔABC is now obtained.

2. Construct an isosceles right angled triangle DEF where ∠E = 90° and EF = 6 cm.
Solution:
Given : An isoscele right angled ΔDEF where ∠E = 90° and EF = 6 cm.
To Construct: A right angled triangle with one side.
Steps of Construction:
Steps 1. Draw a rough sketch of given measures.

Step 2. Draw a line segment EF = 6 cm.

Step 3. With the help of compass taking E as centre, draw a ray EX making an angle of 90° with EF.

Step 4. With E as centre and radius 6 cm (= DE) draw an arc intersecting EX at D.

Step 5. Join D and F. Therefore ΔDEF is required isosceles right triangle.

3. Construct a right-angled triangle PQR in which :
∠Q= 90°, PQ = 3.6 cm and PR = 8.5 cm
Solution:
Given : Right triangle be PQR; right-angled at Q
i. e. ∠Q = 90°
and PQ = 3.6 cm,
PR = 8.5 cm
To construct : A triangle with these two sides and one right angle.
Steps of Construction :
Step 1. We first draw a rough sketch of the triangle PQR and indicate the measure of these two sides and mark the right angle.

Step 2. Draw PQ of length 3.6 cm.

Step 3. At Q, draw QX ⊥ PQ.
(R should be somewhere on this perpendicular).

Step 4. With P as centre, draw arc of radius
(R must be on this arc, since it is at a distance of 8.5 cm from P).

Step 5. R has to be on the perpendicular line QX as well as on the arc drawn with centre P.
∴ R is the meeting point of these two.
ΔPQR is now obtained.

4. Question (i).
Which of the following is a pythagorian triplet ?
(a) 1, 2, 3
(b) 2, 3, 4
(c) 4, 5, 6
(d) 12, 13, 5
Answer:
(d) 12, 13, 5

Question (ii).
Construction of unique triangle is not possible when :
(a) Three sides are given.
(b) Two sides and an included angle are given.
(c) Three angles are given.
(d) Two angles & included side are given.
Answer:
(c) Three angles are given.

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 10 Practical Geometry Ex 10.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.4

1. Construct ΔABC, given AB = 6 cm, ∠A = 30° and ∠B = 75°.
Solution:
Given. One side of ΔABC as AB = 6 cm, m∠A = 30° and m∠B = 75°.
To construct: A triangle with one side and these two angles.
Steps of Construction :
Step 1. We first draw a rough sketch of ΔABC and indicate the measures of side and two angles.

Step 2. Draw a ray AB of length 6 cm.

Step 3. At A; draw a ray AX making an angle 30° with AB.

Step 4. With the help of compass. At B; draw a ray BY making an angle of 75° with AB.

Step 5. Both rays AX and BY intersect, at a point. So the point of intersection of the two rays is C Then. ΔABC is now obtained.

2. Construct an isosceles ΔABC such that AB = 5.3 cm and each base angle = 45°.
Solution:
Given : Isosceles ΔABC with AB = 5.3 cm each base angle = 45°.
To Construct: A triangle with one side and two base angles.
Steps of construction :
Step 1. Draw a rough sketch of ΔABC with given measures

Step 2. Draw a line segment AB = 5.3 cm.

Step 3. Taking A as centre with the help of compass. Draw a ray AX making an angle 45° with AB.

Step 4. With the help of compass and taking B as a centre. Draw a ray BY making an angle 45° with the line segment AB.

Step 5. Rays AX and BY intersect, at a point say C, then ABC is the required triangle.

3. Construct ΔXYZ if XY = 4 cm, ∠X = 45° and ∠Z = 60°.
[Hint : ∠Y = 180° – 45° – 60° = 75°]
Solution:
One side of ΔXYZ as
XY = 4 cm,
∠X = 45°
and ∠Z = 60°.
As we know that by angle-sum property of a triangle; sum of all three angles of a triangle is equal to 180°.
∴ ∠X + ∠Y + ∠Z = 180°
⇒ 45° + ∠Y + 60° = 180°
⇒ 105° + ∠Y = 180°
⇒ ∠Y = 75°.
Now it will be easy to construct triangle with side
XY = 4 cm,
∠X = 45°
and ∠Y = 75°.
Steps of Construction :
Step 1. We first draw a rough sketch of ΔXYZ and indicate the measure of side and two angles.

Step 2. Draw a ray XY of length 4 cm.

Step 3. At X draw a ray XA making an angle of 45° with XY.

Step 4. At Y; draw a ray YB making an angle of 75° with XY.

Step 5. Z has to lie on both rays XA and YB. So, the point of intersection of two rays is Z.
ΔXYZ is now obtained.

4. Examine whether you can construct ΔPQR such that ∠P = 100°, ∠Q = 90° and PQ = 4.3 cm If not possible given reason.
Solution:
No, we cannot construct given ΔPQR.
Reason :
As we know that by angle sum property of a triangle; sum of all three angles a triangle is equal to 180°. But in given question sum of two angles;
m∠P + m∠Q
= 100° + 90°
= 190°
The sum of these two angles should be less than 180°. So triangle with given measures cannot be constructed as it violates the angle sum property of a triangle.

5. Question (i).
In which of the following cases a unique triangle can be drawn ?
(a) BC = 5 cm, ∠B = 90° and ∠C = 100°
(b) AB = 4 cm, BC = 7 cm and CA = 2 cm
(c) XY = 5 cm, ∠X = 45°, ∠Y = 60°
(d) An isosceles triangle with length of each equal side equal to 5 cm.
Answer:
(c) XY = 5 cm, ∠X = 45°, ∠Y = 60°

Question (ii).
A triangle can be constructed by taking two of its angles as.
(a) 110°, 40°
(b) 70°, 115°
(c) 135°, 45°
(d) 90°, 90°
Answer:
(a) 110°, 40°

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 10 Practical Geometry Ex 10.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.3

1. Construct ΔABC such that AB = 4 cm, ∠B = 30°, BC = 4 cm. Also name the type of triangle on the basis of sides.
Solution:
Given : Two sides of ΔABC as AB = 4 cm, BC = 4 cm and ∠B = 30°.
To construct: A triangle with these two sides and included angle.
Step of Construction :
Step 1. We first draw a rough sketch of the ΔABC and indicate the measure of these two sides and included angle.

Step 2. Draw a line segment BC of length 4 cm.

Step 3. At B draw BX making an angle of 30° with BC (The point A must be somewhere on this ray of the angle).

Step 4. (To fix A, the distance AB has been given) With B as centre, draw an arc of radius 3 cm. It cuts BX at the point A.

Step 5. Join AC.
ΔDEF is now obtained.

Since two sides of triangle are equal.
Therefore ΔABC is an isosceles triangle.

2. Construct ΔABC with AB = 7.5 cm, BC = 5 cm and ∠B = 30°.
Solution:
Given. Two sides of ΔABC as AB = 7.5 cm,
BC = 5 cm
and ∠B = 30°
To construct A triangle with these two sides and included angle.
Steps of Construction.
Step 1. We first draw a rough sketch of the ΔABC and indicate the measures of these two sides and included angle.

Step 2. Draw a line segment BC of length 5 cm.

Step 3. At B draw BX making an angle of 30° with BC. (The point A must be somewhere on this ray of the angle)

Step 4. (To fix A; the distance BC has been given) With B as centre draw an arc of radius 7.5 cm. It cuts CX at the point A.

Step 5 : Join AC.
ΔABC is now obtained.

3. Construct a triangle XYZ, such that XY = 6 cm, YZ = 6 cm and ∠Y = 60°. Also name the type of this triangle.
Solution:
Step 1. Draw a rough sketch of XYZ with given measures.

Step 2. Draw a line segment XY of length 6 cm.

Step 3. With the help of compass, at Y, draw a ray YA making an angle 60°

Step 4. With Y as centre and radius 6 cm. draw an arc intersecting the ray YX at point Z.

Step 5. Join XZ.ΔXYZ is required triangle, Measure the third side. We see that ZX = 6 cm
∴ In Δ XYZ
XY = YZ = ZX = 6 cm
Therefore XYZ is an equilateral triangle.

4. Which of the following triangle can be constructed using SAS criterion.
(a) AB = 5 cm, BC = 5 cm, CA = 6 cm
(b) AB = 5 cm, BC = -5 cm, ∠B = 40°
(c) ∠A = 60°, ∠B = 60°, ∠C = 60°
(d) BC = 5 cm, ∠B = ∠C = 45°
Answer:
(b) AB = 5 cm, BC = -5 cm, ∠B = 40°