Class 10

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 8 Introduction to Trigonometry Ex 8.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.2

Question 1.
Evaluate the following:
(i) sin 60° cos 30° + sin 30° cos 60°

(ii) 2 tan² 45° + cos² 30° – sin² 60°

(iii)

(iv)

(v) \(\frac{5 \cos ^{2} 60^{\circ}+4 \sec ^{2} 30^{\circ}-\tan ^{2} 45^{\circ}}{\sin ^{2} 30^{\circ}+\cos ^{2} 30^{\circ}}\)
Solution:
(i) sin 60° cos 30° + sin 30° cos 60°
= \(\left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right)+\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\)

= \(\left(\frac{\sqrt{3}}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}\)
= \(\frac{3}{4}+\frac{1}{4}\) = 1.

(ii) 2 tan² 45° + cos² 30° – sin² 60° = 2 (tan 45°)² + (cos 30°)² – (sin 60°)²
= 2 (1)² + (\(\frac{\sqrt{3}}{2}\))² – (\(\frac{\sqrt{3}}{2}\))² = 2.

(iii)
= \(\frac{\frac{1}{\sqrt{2}}}{\left(\frac{2}{\sqrt{3}}\right)+(2)}=\frac{\frac{1}{\sqrt{2}}}{\frac{2+2 \sqrt{3}}{\sqrt{3}}}\)

= \(\frac{1}{\sqrt{2}}: \frac{\sqrt{3}}{2+2 \sqrt{3}}=\frac{\sqrt{3}}{2 \sqrt{2}(\sqrt{3}+1)}\)

= \(\frac{\sqrt{3}(\sqrt{3}-1)}{2 \sqrt{2}(\sqrt{3}+1)(\sqrt{3}-1)}\)

= \(\frac{\sqrt{2} \times \sqrt{3} \times(\sqrt{3}-1)}{4(3-1)}=\frac{3 \sqrt{2}-\sqrt{6}}{8}\).

(iv)

= \(\frac{\frac{1}{2}+1-\frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}}+\frac{1}{2}+1}=\frac{\frac{3}{2}-\frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}}+\frac{3}{2}}\)

= \(\frac{3 \sqrt{3}-4}{4+3 \sqrt{3}}\)

= \(\frac{(3 \sqrt{3}-4)(3 \sqrt{3}-4)}{(3 \sqrt{3}+4)(3 \sqrt{3}-4)}\)

= \(\frac{27+16-24 \sqrt{3}}{27-16}\)

= \(\frac{43-24 \sqrt{3}}{11}\)

(v) \(\frac{5 \cos ^{2} 60^{\circ}+4 \sec ^{2} 30^{\circ}-\tan ^{2} 45^{\circ}}{\sin ^{2} 30^{\circ}+\cos ^{2} 30^{\circ}}\)

= \(\begin{array}{r}
5\left(\cos 60^{\circ}\right)^{2}+4\left(\sec 30^{\circ}\right)^{2} \\
\frac{-\left(\tan 45^{\circ}\right)^{2}}{\left(\sin 30^{\circ}\right)^{2}+\left(\cos 30^{\circ}\right)^{2}}
\end{array}\)

= \(\frac{5\left(\frac{1}{2}\right)^{2}+4\left(\frac{2}{\sqrt{3}}\right)^{2}-(1)^{2}}{\left(\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}\)

= \(\frac{\frac{5}{4}+4 \times \frac{4}{3}-1}{\frac{1}{4}+\frac{3}{4}}=\frac{\frac{5}{4}+\frac{1}{3}-1}{\frac{1}{4}+\frac{3}{4}}\)

= \(\frac{5}{4}+\frac{16}{3}-1=\frac{15+64-12}{12}=\frac{67}{12}\).

Question 2.
Choose the correct option and justify your choice.

(i) \(\frac{2 \tan 30^{\circ}}{1+\tan 30^{\circ}}\)
(A) sin 60°
(B) cos 60°
(C) tan 60°
(D) sin 30°

(ii) \(\frac{1-\tan ^{2} 45^{\circ}}{1+\tan 45^{\circ}}\)
(A) tan 90°
(B) 1
(C) sin 45°
(D) 0.

(iii) sin 2A = 2 sin A is true when
(A) 0°
(B) 30°
(C) 45°
(D) 60°

(iv) \(\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}\)
(A) cos 60°
(B) sin 60°
(C) tan 60°
(D) sin 30°.

Solution:
(i) \(\frac{2 \tan 30^{\circ}}{1+\tan 30^{\circ}}=\frac{2\left(\frac{1}{\sqrt{3}}\right)}{1+\left(\frac{1}{\sqrt{3}}\right)^{2}}\)

\(\frac{\frac{2}{\sqrt{3}}}{1+\frac{1}{3}}=\frac{2}{\sqrt{3}} \times \frac{3}{4}=\frac{\sqrt{3}}{2}\) = sin 60°.
So, correct anwer is (A).

(ii) \(\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}=\frac{1-(1)^{2}}{1+(1)^{2}}\) = 0
So, correct anwer is (D).

(iii) Here when A = 0°
L.H.S. = sin 2A = sin 0° = 0
and R.H.S. = 2 sin A = 2 sin 0°
= 2 × 0 = 0
∴ Option (A) is correct.

(iv) \(\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}=\frac{2\left(\frac{1}{\sqrt{3}}\right)}{1-\left(\frac{1}{\sqrt{3}}\right)^{2}}\)

= \(\frac{\frac{2}{\sqrt{3}}}{1-\frac{1}{3}}=\frac{2}{\sqrt{3}} \times \frac{3}{2}=\sqrt{3}\)

= tan 60°
∴ Option (C) is correct.

Question 3.
If tan (A + B) = \(\sqrt{3}\) and tan (A – B) = \(\frac{1}{\sqrt{3}}\); 0° ∠A + B ≤ 90°; A > B. find A and B.
Solution:
tan (A + B) = \(\sqrt{3}\). Given
tan (A + B) = tan 60°
⇒ A + B = 60° ……………..(1)
tan (A – B) = \(\frac{1}{\sqrt{3}}\) (Given)
or tan (A – B) = tan 30°
⇒ A – B = 30° …………….(2)
On adding (1) and (2),

A = 45°

Pu value of A = 45° in (1)
45° + B = 60°
B = 60° – 45°
B = 15°
Hence A = 45° and B = 15°.

Question 4.
State whether the following are true or false. Justify your answer.
(i) sin (A + B) = sin A + sin B.
(ii) The value of sin q increases as q increases.
(iii) The value of cos q Increases as q increases
(iv) sin q = cos q for all value of q.
(v) cot A is not defined for A = 0°.
Solution:
(i) False.
When A = 60°, B = 30°
L.H.S. = sin (A + B) = sin (60° + 30°) = sin 90° = 1
R.H.S. = sin A + sin B
= sin 60° + sin 30°
= \(\frac{\sqrt{3}}{2}+\frac{1}{2}\) ≠ 1
i.e., L.H.S. ≠ R.H.S.

(ii) True, sin 30° = \(\frac{1}{2}\) = 0.5,
Note that sin 0° = 0,
sin 45° = \(\frac{1}{\sqrt{2}}\) = 0.7 (approx.)
sin 60° = \(\frac{\sqrt{3}}{2}\) = 0.87 (approx.)
and sin 90° = 1
i.e., value of sin θ increases as θ increases from 0° to 90°.

(iii) False.
Note that cos 0° = 1,
cos 30° = \(\frac{\sqrt{3}}{2}\) = 0.87(approx.)
cos 45° = \(\frac{1}{\sqrt{2}}\) = 0.7.(approx.)
cos 60° = \(\frac{1}{2}\) = 0.5
and cos 90° = 0.
Hence, value of θ decreases as θ increases from 0° to 90°.

(iv) False
Since sin 30° = \(\frac{1}{2}\)
and cos 30° = \(\frac{\sqrt{3}}{2}\)
or sin 30° ≠ cos 30°
Only we have: sin 45° = cos 45°.
\(\frac{1}{\sqrt{2}}\) = \(\frac{1}{\sqrt{2}}\)

(v) True.
cot 0° = \(\frac{1}{\tan 0^{\circ}}=\frac{1}{0}\), or not defined.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 8 Introduction to Trigonometry Ex 8.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.1

Question 1.
In ∆ABC, right angled at B, AB = 24 cm; BC = 7 cm. Determine
(i) sin A, cos A
(ii) sin C, cos C.
Solution:
(i) We are to find sin A .cos A AB = 24 cm; BC = 7 cm
By using Pythagoras Theorem,

AC² = AB² + BC²
AC² = (24)² + (7)²
AC² = 576 + 49
AC² = 625
AC = \(\sqrt{625}\)
AC = 25 cm.
sin A = \(\frac{\mathrm{BC}}{\mathrm{AC}}\)

sin A = \(\frac{7 \mathrm{~cm}}{25 \mathrm{~cm}}=\frac{7}{25}\)

cos A = \(\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{24 \mathrm{~cm}}{25 \mathrm{~cm}}\)

cos A = \(\frac{24}{25}\)

Hence sin A = \([latex]\frac{7}{25}\)[/latex] and cos A = \([latex]\frac{24}{25}\)[/latex].

(ii) sin C = \(\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{24 \mathrm{~cm}}{25 \mathrm{~cm}}\)

sin C = \(\frac{24}{25}\)

cos C = \(\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{7 \mathrm{~cm}}{25 \mathrm{~cm}}\)

cos C = \(\frac{7}{25}\)

Hence sin C = \(\frac{24}{25}\) and cos C = \(\frac{7}{25}\).

Question 2
In fig., find tan P – cot R.

Solution:
Hyp. PR = 13 cm

By using Pythagoras Theorem,
PR² = PQ² + QR²
or (13)² = (12)² + QR²
or 169 = 144 + (QR)²
or 169 – 144 = (QR)²
or 25 = (QR)²
or QR = ± \(\sqrt{25}\)
or QR = 5, – 5.
But QR = 5 cm.
[QR ≠ – 5, because side cannot be negative]
tan P = \(\frac{R Q}{Q P}=\frac{5}{12}\)

cot R = \(\frac{R Q}{P Q}=\frac{5}{12}\)

∴ tan P – cot R = \(\frac{5}{12}-\frac{5}{12}\) = 0
Hence tan P – cot R = 0.

Question 3.
If sin A = \(\frac{3}{4}\) calculate cos A and tan A.
Solution:
Let ABC be any triangle with right angle at B.

sin A = \(\frac{3}{4}\)
But sin A = \(\frac{\mathrm{BC}}{\mathrm{AC}}\) [From figure]
∴ \(\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{3}{4}\)
But \(\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{3}{4}\) = K
where K, is constant of proportionality.
⇒ BC = 3K, AC = 4K
By using Pythagoras Theorem,
AC² = AB² + BC²
or (4K)² = (AB)² + (3K)²
or 16K² = AB² + 9K²
or 16K² – 9K² = AB²
or 7K² = AB²
or AB = ± \(\sqrt{7 K^{2}}\)
or AB = ± \(\sqrt{7} \mathrm{~K}\)
[AB ≠ \(\sqrt{7 K}\) because side of a triangle cannot be negative]

⇒ AB = \(\sqrt{7} \mathrm{~K}\)
cos A = \(\frac{\mathrm{AB}}{\mathrm{AC}}\)
cos A = \(\frac{\sqrt{7} K}{4 K}=\frac{\sqrt{7}}{4}\)
tan A = \(\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{3 \mathrm{~K}}{\sqrt{7} \mathrm{~K}}=\frac{3}{\sqrt{7}}\)

Hence cos A = \(\frac{\sqrt{7}}{4}\) and tan A = \(\frac{3}{\sqrt{7}}\).

Question 4.
Given 15 cot A = 8, find sin A and sec A.
Solution:
Let ABC be any right angled triangle where A is an acute angle with right angle at B.

15 cot A = 8
cot A = \(\frac{8}{15}\)
But cot A = \(\frac{A B}{B C}\) (fromfig.)
⇒ \(\frac{A B}{B C}=\frac{8}{15}\) = K
where K is constant of proportionality.
AB = 8 K, BC = 15 K
By using Pythagoras Theorem.
AC² = (AB)² + (BC)²
(AC)² = (8 K)² + (15 K)²
(AC)² = 64K² + 225 K²
(AC)² = 289 K²
AC = ± \(\sqrt{289 K^{2}}\)
AC = ± 17 K
⇒ AC = 17K
[AC = – 17 K, Because side cannot be negative]
sin A = \(\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{15 \mathrm{~K}}{17 \mathrm{~K}}=\frac{15}{17}\)

sin A = \(\frac{15}{17}\)

sec A = \(\frac{\mathrm{AC}}{\mathrm{AB}}\)

sec A = \(\frac{17 \mathrm{~K}}{8 \mathrm{~K}}=\frac{17}{8}\)

sec A = \(\frac{17}{8}\)

Hence, sin A = \(\frac{15}{17}\) and sec A = \(\frac{17}{8}\).

Question 5.
Given sec θ = \(\frac{13}{2}\), calculate all other trigonometric ratios.
Solution:
Let ABC be any right angled triangle with right angle at B.
Let ∠BAC = θ

sec θ = \(\frac{13}{12}\)

But sec θ = \(\frac{\mathrm{AC}}{\mathrm{AB}}\) ……….[from fig.]

\(\frac{\mathrm{AC}}{\mathrm{AB}}=\frac{13}{12}\)

But \(\frac{\mathrm{AC}}{\mathrm{AB}}=\frac{13}{12}\) = k where k is constant of proportionality.
AC = 13 k and AB = 12 k
By using Pythagoras Theorem,
AC² = (AB)² + (BC)²
or (13k)² = (12k)² + (BC)²
or 169k² = 144k² + (BC)²
or 169k² – 144k² = (BC)
or (BC)² = 25k²
or BC = ± \(\sqrt{25 k^{2}}\)
or BC = ± 5k
or BC = 5k.
[BC ≠ – 5k because side cannot be negative]

sin θ = \(\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{5 k}{13 k}=\frac{5}{13}\)
cos θ = \(\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{12 k}{13 k}=\frac{12}{13}\)
tan θ = \(\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{5 k}{12 k}=\frac{5}{12}\)
cosec θ = \(\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{13 k}{5 k}=\frac{13}{5}\)
cot θ = \(\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{12 k}{5 k}=\frac{12}{5}\).

Question 6.
If ∠A and ∠B are acute angles such that cos A = cos B, show that LA = LB.
Solution:
Let ABC be any triangle, where ∠A and ∠B are acute angles. To find cos A and cos B.

Draw CM ⊥ AB
∠AMC = ∠BMC = 90°
In right angled ∆AMC,
\(\frac{\mathrm{AM}}{\mathrm{AC}}\) = cos A ……………(1)
In right angled ∆BMC,
\(\frac{\mathrm{BM}}{\mathrm{BC}}\) = cos B ……………(2)
But cos A = cos B [given] ………..(3)
From (1), (2) and (3),
\(\frac{\mathrm{AM}}{\mathrm{AC}}=\frac{\mathrm{BM}}{\mathrm{BC}}\)
\(\frac{\mathrm{AM}}{\mathrm{BM}}=\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{\mathrm{CM}}{\mathrm{CM}}\)
∴ ∆AMC = ∆BMC [By SSS similarity]
⇒ ∠A = ∠B [∵ Corresponding angles of two similar triangles are equal].

Question 7.
If cot θ = \(\frac{7}{8}\) evaluate
(i) \(\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}\)
(ii) cot² θ.
Solution:
(i) ∠ABC = θ.
In right angled triangle ABC with right angle at C.

Given that, cot θ = \(\frac{7}{8}\)
But cot θ = \(\frac{\mathrm{BC}}{\mathrm{AC}}\) [From fig.]
⇒ \(\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{7}{8}\)
Let \(\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{7}{8}\) = k
where k is constant of proportionality.
⇒ BC = 7k, AC = 8k
By using Pythagoras Theorem,
AB² = (BC)² + (AC)²
or (AB)² = (7k)² + (8k)²
or (AB)² = 49k² + 64k²
or (AB)² = 113 k²
or AB = ± \(\)
AB = \(\sqrt{113 k^{2}}\) k
AB = \(\sqrt{113}\) k
[AB ≠ \(\sqrt{113}\) k because side cannot be negative]

sin θ = \(\frac{\mathrm{AC}}{\mathrm{AB}}=\frac{8 k}{\sqrt{113} k}\)
sin θ = \(\frac{8}{\sqrt{113}}\)
cos θ = \(\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{7 k}{\sqrt{113} k}=\frac{7}{\sqrt{113}}\)
cos θ = \(\frac{7}{\sqrt{113}}\)

(1 + sin θ) (1 – sin θ) = (1 + \(\frac{8}{\sqrt{113}}\)) (1 – \(\frac{8}{\sqrt{113}}\))
= (1)² – (\(\frac{8}{\sqrt{113}}\))²
[By using formula (a + b) (a – b) = a² – b²]
= 1 – \(\frac{64}{113}\)
(1 + sin θ) (1 – sin θ) = \(\frac{113-64}{113}=\frac{49}{113}\)
(1 + sin θ)(1 – sin θ) = \(\frac{49}{113}\) ……………..(1)

(1 + cos θ) (1 – cos θ) = (1 + \(\frac{8}{\sqrt{113}}\)) (1 – \(\frac{8}{\sqrt{113}}\))
(1)² – (\(\frac{7}{\sqrt{113}}\))²
[By using formula(a + b) (a – b) = a² – b²]
= 1 – \(\frac{49}{113}\) = \(\frac{113-49}{113}\)
(1 + cos θ) (1 – cos θ) = \(\frac{64}{113}\) ……….(2)

Consider, \(\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}=\frac{\frac{49}{113}}{\frac{64}{113}}\) [From (1) and (2)]

Hence \(\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}=\frac{49}{64}\)

(ii) cot θ = \(\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{7}{8}\)
cot² θ = (cot θ)²
cot² θ= (\(\frac{7}{8}\))²
⇒ cot² θ = \(\frac{49}{64}\).

Question 8.
If 3 cot A = 4 check whether \(\frac{1-\tan ^{2} A}{1+\tan ^{2} A}\) = cos² A – sin² A or not.
Solution:
Let ABC be a right angled triangle with right angled at B.

It is given that 3 cot A = 4
cot A = \(\frac{4}{3}\)
But cot A = \(\frac{\mathrm{AB}}{\mathrm{BC}}\) [From fig.]
⇒ \(\frac{A B}{B C}=\frac{4}{3}\)
But \(\frac{A B}{B C}=\frac{4}{3}\) = k
⇒ AB = 4k, BC = 3k
By using Pythagoras Theorem,
(AC)² = (AB)² + (BC)²
(AC)² = (4k)² + (3k)²
(AC)² = 16 k² + 9 k²
(AC)² = 25 k²
AC=± \(\sqrt{25 k^{2}}\)
AC = ± 5k

But AC = 5k.
[AC ≠ – 5k. because side cannot be negative]
sin A = \(\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{3 k}{5 k}=\frac{3}{5}\)

tan A = \(\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{3 k}{4 k}=\frac{3}{4}\)

cos A = \(\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{4 k}{5 k}=\frac{4}{5}\)

L.H.S. = \(\frac{1-\tan ^{2} \mathrm{~A}}{1+\tan ^{2} \mathrm{~A}}\)

∴ cos² A – sin² A = \(\frac{7}{25}\) ………..(2)

From (1) and (2),
L.H.S = R.H.S
Hence, \(\frac{1-\tan ^{2} \mathrm{~A}}{1+\tan ^{2} \mathrm{~A}}\) = cos² A – sin² A.

Question 9.
In triangle ABC, right angled at B, if tan A = \(\frac{1}{\sqrt{3}}\). Find the value of:
(i) sin A cos C + cos A sin C
(ii) cos A cos C – sin A sin C.
Solution:
(i) Given: ABC with right angled at B

tan A = \(\frac{1}{\sqrt{3}}\) ……………..(1)
But tan A = \(\frac{B C}{A B}\) ……………(2)
From (1) and (2),
\(\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{1}{\sqrt{3}}\)
Let \(\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{1}{\sqrt{3}}\) = k
BC = k, AB = k
where k is constant of proportionality.
In right angled triangle ABC,
By using Pythagoras Theorem,
(AC)² = (AB)² + (BC)²
or (AC)² = (Jk)² + (k)²
or AC² = 3k² + k²
or AC² = 4k²
or AC = ± \(\sqrt{4 k^{2}}\)
AC = ± 2k.
where AC = 2k
[AC ≠ – 2k side cannot be negative]

[sin A = \(\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{k}{2 k}=\frac{1}{2}\)

cos C = \(\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{k}{2 k}=\frac{1}{2}\)

cos A = \(\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\sqrt{3} k}{2 k}=\frac{\sqrt{3}}{2}\)

sin C = \(\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\sqrt{3} k}{2 k}=\frac{\sqrt{3}}{2}\)] …………….(3)

sin A cos C = \(\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)=\frac{1}{4}\)
cos A sin C = \(\left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right)=\frac{3}{4}\)
sin A cos C + cos A sin C = \(\frac{1}{4}+\frac{3}{4}\)
= \(\frac{1+3}{4}\)
= \(\frac{4}{4}\) = 1
∴ sin A cos C + cos A sin C = 1.

(ii) cos A cos C = \(\left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right)=\frac{\sqrt{3}}{4}\) [From (3)]
sin A sin C = \(\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right)=\frac{\sqrt{3}}{4}\) [From (3)]

cos A cos C – sin A sin C = \(\left(\frac{\sqrt{3}}{4}\right)-\left(\frac{\sqrt{3}}{4}\right)\) = 0.

Question 10.
In ∆PQR, right angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.
Solution:
Given: ∆PQR, right angled at Q

PR + QR = 25 cm
PQ = 5 cm
In right angled triangle PQR
By using Pythagoras Theorem,
(PR)² = (PQ)² + (RQ)²
or (PR)² = (5)² + (RQ)²
[∴ PR + QR = 25, QR = 25 – PR]
or (PR)² = 25 + [25 – PR]²
or (PR)² = 25 + (25)² + (PR)² – 2 × 25 × PR
or (PR)² = 25 + 625 + (PR)² – 50
or (PR)² – (PR)² + 50 PR = 650
or 50 PR = 650
or PR = \(\frac{650}{50}\)
or PR = 13 cm
QR = 25 – PR
QR = (25 – 13) cm
or QR = 12 cm.

sin P = \(\frac{Q R}{P R}=\frac{12}{13}\)

cos P = \(\frac{P Q}{P R}=\frac{5}{13}\)

tan P = \(\frac{Q R}{P Q}=\frac{12}{5}\)

Question 11.
State whether the following are true or false. Justify your answer.
(i) The value of tan A is always less than 1
(ii) sec A = \(\frac{12}{5}\) for some value of angle A.
(iii) cos A is abbreviation used for cosecant of angle A.
(iv) cot A is product of cot and A.
(v) sin θ = \(\frac{4}{3}\) for some angle θ.
Solution:
(i) False
∵ tan 60° = √3 = 1.732 > 1.

(ii) True; sec A = \(\frac{12}{5}\) = 240 > 1
∵ Sec A is always greater than 1.

(iii) False.
Because cos A is used for cosine A.

(iv) False.
Because cot A is cotangent of the angle A not the product of cot and A.

(v) False; sin θ = \(\frac{4}{3}\) = 1.666 > 1
Because sin θ is always less than 1.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 7 Coordinate Geometry Ex 7.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

Question 1.
Determine the ratio in which the line it + y – 4 = 0 divides the line segment joining the points A (2, – 2) and B (3, 7).
Solution:
Let line 2x + y – 4 = 0 divides the line segment joining the points A (2,- 2) and B(3, 7) at C (x, y) in the ratio k : 1

∴ Coordinates of C are x = \(\frac{3 k+2 \times 1}{k+1}=\frac{3 k+2}{k+1}\) and y = \(\frac{7 k+(-2) \times 1}{k+1}=\frac{7 k-2}{k+1}\)
∴ C \(\left[\frac{3 k+2}{k+1}, \frac{7 k-2}{k+1}\right]\). must lie on the line 2x + y – 4 = 0

i.e., 2\(\left(\frac{3 k+2}{k+1}\right)+\left(\frac{7 k-2}{k+1}\right)\) – 4 = 0
or \(\frac{6 k+4+7 k-2-4 k-4}{k+1}\) = 0
or 9k – 2 = 0
or 9k = 2
or k = \(\frac{2}{9}\).
∴ ratio k : 1 = \(\frac{2}{9}\) : 1 = 2 : 9.
Hence required ratio is 2 : 9.

Question 2.
Find a relation between x and y if (x, y) ; (1, 2) and (7, 0) are collinear.
Solution:
Let given points are A (x, y); B (1, 2) and C (7, 0).
Here x₁ = x, x₂ = 1, x₃ = 7
y₁ = y, y₂ = 2, y₃ = 0
∵ Three points are collinear
iff \(\frac{1}{2}\) [x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)] = 0
or \(\frac{1}{2}\) x (2 – 0) + 1 (0 – y) + 7 (y – 2)] = 0
or 2x – y + 7y – 14 = 0
or 2x + 6y – 14 = 0
or x + 3y – 7 = 0 is the required relation.

Question 3.
Find the centre of a cirçle passing through the points (6, —6); (3, —7) and (3,3).
Solution:
Let O (x, y) be the required centre of the circle which passes through points P(6, – 6); Q(3, – 7) and R (3, 3).
∴ radii of circle are equal.

∴ OP = OQ = OR
or (OP)² = (OQ)² = (OR)²
Now, (OP)² = (OQ)²
(x – 6)² + (y + 6)² = (x – 3)² + (y + 7)²
or x² + 36 – 12x + y² + 36 + 12y = x² + 9 – 6x + y² + 49 + 14y
or – 12x + 12y + 72 = – 6x + 14y + 58
or – 6x – 2y + 14 = 0
or 3x + y – 7 = 0 ………………(1)
Also, (OQ)² = (OR)²
or (x – 3)² + (y + 7)² = (x – 3)² + (y – 3)²
or (y + 7)² = (y – 3)²
or y² + 49 + 14y = y² + 9 – 6y
or 20y = – 40
y = \(\frac{-40}{20}\) = – 2
Substitute this value of)’ in (1), we get
3x – 2 – 7 = 0
or 3x – 9 = 0
or 3x = 9
or x = \(\frac{9}{3}\) = 3
∴ Required centre is (3, – 2).

Question 4.
The two opposite vertices of a square are (- 1, 2) and (3, 2). Find the coordinates of other two vertices.
Solution:
Let two opposite vertices of a square ACBD are A (- 1, 2) and B (3, 2) and coordinates of C are (x, y)
∵ Length of each sides of square are equal.
∴ AC = BC
or (AC)² = (BC)²
or (x + 1)² + (y – 2)² = (x – 3)² + (y – 2)²

or (x + 1)² = (x – 3)²
or x² + 1 + 2x = x² + 9 – 6x
or 8x = 8
or x = \(\frac{8}{8}\) = 1
Now, in rt ∠d ∆ACB,
Using Pythagoras Theorem,
(AC)² + (BC)² = (AB)²
(x + 1)² + (y – 2)² + (x – 3)² + (y – 2)² = (3 + 1)² + (2 – 2)²
or x² + 1 + 2x + y² + 4 – 4y + x² + 9 – 6x + y² + 4 – 4y = 16
or 2x² + 2y² – 4x – 8y + 2 = 0
or x² + y² – 2x – 4y + 1 = 0
Putting the value of x = 1 in (1), we get
(1)² + y² – 2 (1) – 4y + 1 = 0
or y² – 4y = 0
or y (y – 4) = 0
Either y = 0 or y – 4 = 0
Either y = 0 or y = 4
∴ y = 0, 4
∴ Required points are (1. 0) and (1.4).

Question 5.
The Class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Sapling of Gulmohar are planted on the boundary at a distance of 1m from each other. There ¡s a triangular grassy lawn in the plot as shown in the Fig. The students are to sow seeds of flowering plants on the remaining area of the plot.

(i) Taking A as origin, find the coordinates of the vertices of the triangle.
(ii) What will be the coordinates of the vertices of A PQR if C is the origin? Also calculate the areas of the triangles In these cases. What do you observe?
Solution:
Case I:
When taking A as origin then AD is X-axis and AB is Y-axis.
∴ Coordinates of triangular grassy Lawn
PQR are P (4, 6); Q (3, 2) and R(6, 5).
Here x₁ = 4, x₂ = 3, x₃ = 6
y₁ = 6, y₂ = 2, y₃ = 50
Now, area of ∆PQR = \(\frac{1}{2}\) [x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)]
= \(\frac{1}{2}\) [4 (2 – 5) + 3 (5 – 6) + 6 (6 – 2)]
= \(\frac{1}{2}\) [- 12 – 3 + 24] = \(\frac{9}{2}\)
= 4.5 sq. units.

Case II: When taking C as origin then CB is X – axis and CD is Y – axis.
∴ Coordinates of triangular grassy lawn PQR
are P(12, 2); Q (13,6) and R (10, 3)
Here x₁ = 12, x₂ = 13, x₃ = 10
y₁ = 2, y₂ = 6, y₃ = 3
Now, area of ∆PQR = \(\frac{1}{2}\) [x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)]
= \(\frac{1}{2}\) [12 (6 – 3) + 13 (3 – 2) + 10 (2 – 6)]
= \(\frac{1}{2}\) [36 + 13 – 40]
= \(\frac{9}{2}\) = 4.5 sq. units.
From above two cases, it is clear that area of triangular grassy lawn is same.

Question 6.
The vertices of a ∆ABC are A (4, 6), B (1, 5) and C (7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that \(\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AE}}{\mathrm{AC}}=\frac{1}{4}\) Calculate the area of the ∆ADE and compare it with the area of ∆ABC. (Recall Theorem 6.2 and Theorem 6.6).
Solution:
The vertices of ∆ABC are A (4, 6); B (1, 5) and C (7, 2)

A line is drawn to intersect sides AB and AC at D (x₁, y₁) and E (x₂, y₂) respectively such that \(\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AE}}{\mathrm{AC}}=\frac{1}{4}\).

∴ D and E divides AB and AC in the ratio 1 : 3.
∴ Coordinates of D are
x₁ = \(\frac{1(1)+3(4)}{1+3}=\frac{1+12}{4}=\frac{13}{4}\) and y₁ = \(\frac{1(5)+3(6)}{1+3}=\frac{5+18}{4}=\frac{23}{4}\)

∴ Coordinates of D are (\(\frac{13}{4}\), \(\frac{23}{4}\))
Now, coordinates of E are
x₂ = \(\frac{1(7)+3(4)}{1+3}=\frac{7+12}{4}=\frac{19}{4}\) and y₂ = \(\frac{1(2)+3(6)}{1+3}=\frac{2+18}{4}=\frac{20}{4}=5\)

∴ Coordinates of E are (\(\frac{19}{4}\), 5).

In ∆ADE
x₁ = 4, x₂ = \(\frac{13}{4}\), x₃ = \(\frac{19}{4}\)
y₂ = 6, y₂ = \(\frac{23}{4}\), y₃ = 5
area of ∆ADE = \(\frac{1}{2}\) [x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)]

= \(\frac{1}{2}\left[4\left(\frac{23}{4}-5\right)+\frac{13}{4}(5-6)+\frac{19}{4}\left(6-\frac{23}{4}\right)\right]\)

= \(\frac{1}{2}\left[4\left(\frac{23-20}{4}\right)+\frac{13}{4}(-1)+\frac{19}{4}\left(\frac{24-23}{4}\right)\right]\)

= \(\frac{1}{2}\left[3-\frac{13}{4}+\frac{19}{16}\right]\)

= \(\frac{1}{2}\left[\frac{48-52+19}{16}=\frac{15}{16}\right]\)
= \(\frac{15}{32}\) sq. units.

In ∆ABC
x₁ = 4, x₂ = 1, x₃ = 7
y₂ = 6, 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) + 1 (2 – 6) + 7 (6 – 5)]
= \(\frac{1}{2}\) [12 – 4 + 7] = \(\frac{15}{2}\) sq.units.

Now, \(\frac{\text { area of } \Delta \mathrm{ADE}}{\text { area of } \Delta \mathrm{ABC}}=\frac{\frac{15}{32}}{\frac{15}{2}}=\frac{15}{32} \times \frac{2^{1}}{16_{1}}\)

= \(\frac{1}{16}=\left(\frac{1}{4}\right)^{2}\)

= \(\left(\frac{A D}{A B}\right)^{2} \text { or }\left(\frac{A E}{A C}\right)^{2}\).

Question 7.
Let (4, 2), B (6, 5) and C (1, 4) be the vertices of ∆ABC.
(i) The median from A meets BC at D. Find the coordinates of the point D.
(ii) Find the coordinates of the potnt P on AD such that AP : PD = 2 : 1
(iii) Find the coordinates of points Q and R on medians BE and CF respectively such that BQ : QE = 2 : 1 and CR : RF = 2 : 1.
(iv) What do you observe?
[Note : The point which is common to all the three medians ¡s called centroid and this point divides each median in the ratio 2: 1]
(v) if A (x₁, y₁), B (x₂, y₂) and C (x₃, y₃) are the vertices of ∆ABC, find the coordinates of the centroid of the triangle.
Solution:
Given that vertices of ∆ABC are A (4, 2); B (6, 5) and C (1, 4).
(i) AD is the median from the vertex A.
∴ D is the mid point of BC.

then x = \(\frac{6+1}{2}=\frac{7}{2}\) and y = \(\frac{5+4}{2}=\frac{9}{2}\)
Hence, coordinates of D is (\(\frac{7}{2}\), \(\frac{9}{2}\)).

(ii) Let P(x, y) be point on AD such that AP : PD = 2 : 1

then x = \(\frac{2\left(\frac{7}{2}\right)+1(4)}{2+1}\)
= \(\frac{7+4}{3}=\frac{11}{3}\)

and y = \(\frac{2\left(\frac{9}{2}\right)+1(2)}{2+1}\)
= \(\frac{9+2}{3}=\frac{11}{3}\)

Hence, Coordinates of P is (\(\frac{11}{3}\), \(\frac{11}{3}\)).

(iii) Le BE and CF are the medians of ∆ABC to AC and AB respectively.
∴ E and F are mid points of AC and AB respectively.

Coordinate of E are
x₁ = \(\frac{4+1}{2}=\frac{5}{2}\)
and y₁ = \(\frac{4+2}{2}=\frac{6}{2}\) = 3
Coordinate of E are (\(\frac{5}{2}\), 3)
Coordinate of F are
x₂ = \(\frac{4+6}{2}=\frac{10}{2}\) = 5
and y₂ = \(\frac{5+2}{2}=\frac{7}{2}\)
∴ Coordinate of F are (5, \(\frac{7}{2}\))
Now, Q divides BE such that BQ : QE = 2: 1

∴ Coordinate of Q are \(\left(\frac{2\left(\frac{5}{2}\right)+6(1)}{2+1}, \frac{2(3)+1(5)}{2+1}\right)\)

= \(\left(\frac{5+6}{3}, \frac{6+5}{3}\right)\) = \(\left(\frac{11}{3}, \frac{11}{3}\right)\)

Also, R divides CF such that CR : RF = 2 : 1

∴ Coordinate of R are = \(\left(\frac{2(5)+1(1)}{2+1}, \frac{2\left(\frac{7}{2}\right)+(4)}{2+1}\right)\)

= \(\left(\frac{10+1}{3}, \frac{7+4}{3}\right)\)

= \(\left(\frac{11}{3}, \frac{11}{3}\right)\)

(iv) From above discussion, it is clear that coordinates of P, Q and R are same and coincide at a point, is known as centroid of triangle, which divides each median in the ratio 2: 1.

(v) The vertices of given ∆ABC are
A (x₁, y₁); B (x₂, y₂) and C (x₃, y₃).

Let AD is median of E, ∆ABC.
∴ D is the mid point of BC then coordinates of D are \(\left(\frac{x_{2}+x_{3}}{2}, \frac{y_{2}+y_{3}}{2}\right)\)

Now, G be the centroid of ABC, which divides the median AD in the ratio 2: 1
∴ Coordinates of G are [using (iv)]

= \(\left[\frac{2\left(\frac{x_{2}+x_{3}}{2}\right)+1\left(x_{1}\right)}{2+1}, \frac{2\left(\frac{y_{2}+y_{3}}{2}\right)+1\left(y_{1}\right)}{2+1}\right]\)

= \(\left[\frac{x_{2}+x_{3}+x_{1}}{3}, \frac{y_{2}+y_{3}+y_{1}}{3}\right]\)

= \(\left[\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right]\).

Question 8.
ABCD is a rectangle formed by the points A (- 1, – 1), B (- 1, 4), C (5, 4) and D (5, – 1). P, Q R and S are the mid points
of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square ? a rectangle? or a rhombus ? Justify your answer.
Solution:
Given: The vertices ot’ given rectangle ABCD are
A(- 1, – 1); B(- 1, 4); C(5, 4) and D (5, – 1).

∵ P is the mid point of AB.
∴ Coordinates of P are \(\left(\frac{-1-1}{2}, \frac{-1+4}{2}\right)=\left(-1, \frac{3}{2}\right)\)
∵ Q is the mid point of BC.
∴ Co-ordinates of Q are \(\left(\frac{-5+5}{2}, \frac{4+4}{2}\right)\) = (2, 4)
∵ R is the mid point of CD.
∴ Coordinates of R are \(\left(\frac{5+5}{2}, \frac{4+1}{2}\right)=\left(5, \frac{3}{2}\right)\)

∵ S is the mid point of AD.
∴ Co-ordinates of S are \(\left(\frac{5-1}{2}, \frac{-1-1}{2}\right)\) = (2, -1)

PQ = \(\sqrt{(2+1)^{2}+\left(4-\frac{3}{2}\right)^{2}}\)

= \(\sqrt{9 \times \frac{25}{4}}=\sqrt{\frac{36+25}{4}}\)

PQ = \(\sqrt{\frac{61}{4}}\)

QR = \(\sqrt{(5-2)^{2}+\left(\frac{3}{2}-4\right)^{2}}\)

= \(\sqrt{(3)^{2}+\left(\frac{3-8}{2}\right)^{2}}\)

= \(\sqrt{9+\frac{25}{4}}=\sqrt{\frac{36+25}{4}}\)

QR = \(\sqrt{\frac{61}{4}}\)

RS = \(\sqrt{(2-5)^{2}+\left(-1-\frac{3}{2}\right)^{2}}\)

= \(\sqrt{9+\frac{25}{4}}=\sqrt{\frac{36+25}{4}}\)

RS = \(\sqrt{\frac{61}{4}}\)

and SP = \(\sqrt{(-1-2)^{2}+\left(\frac{3}{2}+1\right)^{2}}\)

SP = \(\sqrt{9+\frac{25}{4}}=\sqrt{\frac{61}{4}}\)

Also PR = \(\sqrt{(5+1)^{2}+\left(\frac{3}{2}-\frac{3}{2}\right)^{2}}\)
PR = \(\sqrt{36+0}=\sqrt{36}\) = 6
QS = \(\sqrt{(2-2)^{2}+(4+1)^{2}}\)
= \(\sqrt{0+25}=\sqrt{25}\) = 5.

Form above discussion it is clear that PQ = QR = RS = SP.
Also, PR ≠ QS.
⇒ All sides of quad. PQRS are equal but their diagonals are not equal.
Quad. PQRS is a rhombus.

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 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 6.3 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].