Class 10

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

PSEB Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.6

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

PSEB Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5

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

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

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

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

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

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

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

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

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

PM² = QM × RM Hence proved.

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

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

[If two triangles are similar corresponding sides are proportional]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

⇒ CE² = OC² – OE² ……………(3)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

PSEB Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3

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

(i)

(ii)

(iii)

(iv)

(v)

(vi)

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

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

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

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

∴ ΔABC ~ ΔQRP [By SSS similarity criterion]

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

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

\(\)
∴ Two Triangles are not similar.

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

∠M = ∠Q (each 70°)

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

∴ ΔMNL ~ ΔPQR [By SAS similarity cirterion]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

PSEB Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.4

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

△ABC ~ △DEF

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

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

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

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

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

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

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

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

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

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

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

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

[∵ ∆ = \(\frac{1}{2}\) × b × p]

Hence proved.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

∆ABC ~ ∆DEF (given)

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

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

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

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

PSEB Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.2

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

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

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

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

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

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

∴ EF is not parallel to QR.

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

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

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

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

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

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

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

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

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

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

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

Solution:
In ∆ABC, DE || AC(given)

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

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

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

∴ \(\frac{P D}{D O}=\frac{P E}{E Q}\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 6 Triangles Ex 6.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.1

Question 1.
Fill in the blanks using the correct word given in brackets:
(i) All circles are ………………. (congruent, similar).
Solution:
All circles are similar.

(ii) All squares are ………………. (similar, congruent).
Solution:
MI squares are similar.

(iii) All ………………. triangles are similar. (isosceles, equilateral).
Solution:
All equilateral triangles are similar.

(iv) Two polygons of the same number of sides are similar, if
(a) their corresponding angles are __________ and
Solution:
equal

(b) their corresponding sides are ………………. (equal, proportional).
Solution:
proportional.

Question 2.
Give two different examples of pair
(i) similar figures
(ii) non-similar figures.
Solution:
(i) 1. Pair of equilateral triangle are similar figures.
2. Pair of squares are similar figures.

(ii) 1. A triangle and quadrilateral form a pair of non-similar figures.
2. A square and rhombus form pair of non – similar figures.

Question 3.
State whether the following quadrilaterals are similar or not :-

Solution:
The two quadrilaterals in the figure are not similar because their corresponding angles are not equal.