mathamatics xam idea

20. Prove that the area of a triangle with vertices (t, t – 2) , (t + 2, t + 2) and (t + 3, t) is independent

of t. 1 [CBSE Delhi 2016]
Sol. Area of a triangle 2
= |x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)|

Area of the triangle = 1 |t(t + 2– t) + (t + 2) (t – t + 2) + (t + 3) (t –2– t – 2)|
2

= 1 |2t + 2t + 4– 4t – 12 |
2

= 4 sq. units

which is independent of t.

Hence proved.

Short Answer Questions-II [3 marks]

1. Name the type of quadrilateral formed, if any, by the following points, and give reasons for

your answer: [NCERT]

(i) (– 1, – 2), (1, 0), (– 1, 2), (– 3, 0) (ii) (4, 5), (7, 6), (4, 3), (1, 2)
Sol. (i) Let A(–1, –2), B(1, 0), C(–1, 2) and D(–3, 0) be the four given points.

Then, using distance formula, we have,

AB = (1 + 1)2 + (0 + 2)2 = 4 + 4 = 8 = 2 2

BC = (−1 −1)2 + (2 − 0)2 = 4 + 4 = 8 = 2 2

CD = (−3 + 1)2 + (0 − 2)2 = 4 + 4 = 8 = 2 2
DA = (−1 + 3)2 + (−2 + 0)2 = 4 + 4 = 8 = 2 2

AC = (−1 + 1)2 + (2 + 2)2 = 0 + 16 = 4 and

BD= (−3 − 1)2 + (0 − 0)2 = 16 = 4
Hence, four sides of quadrilateral are equal and diagonals AC and BD are also equal.
\ Quadrilateral ABCD is a square.
(ii) Let A (4, 5), B (7, 6), C (4, 3) and D (1, 2) be the given points. Then,

AB = (7 − 4)2 + (6 − 5)2 = 9 + 1 = 10

BC = (4 − 7)2 + (3 − 6)2 = 9 + 9 = 18 = 3 2

CD = (1 − 4)2 + (2 − 3)2 = 9 + 1 = 10

DA = (4 − 1)2 + (5 − 2)2 = 9 + 9 = 18 = 3 2

AC = (4 − 4)2 + (3 − 5)2 = 0 + 4 = 2 and

BD = (1 − 7)2 + (2 − 6)2 = 36 + 16 = 52 = 2 13

Clearly, AB = CD, BC = DA and AC ≠ BD
\ Quadrilateral ABCD is a parallelogram.

Coordinate Geometry 143

2. Find the value of y for which the distance between the points P(2, –3) and Q(10, y) is 10 units.

[NCERT]

OR

A line segment is of length 10 units. If the coordinates of its one end are (2, –3) and the
abscissa of the other end is 10, find its ordinate.

Sol. We have, PQ = 10

⇒ (10 − 2)2 + ( y + 3)2 = 10

Squaring both sides, we have

⇒ (8)2 + (y + 3)2 = 100 ⇒ (y + 3)2 = 100 – 64

⇒ (y + 3)2 = 36 or y + 3 = ±6

⇒ y + 3 = 6, y + 3 = –6 or y = 3, y = –9

Hence, values of y are – 9 and 3.

3. If Q(0, 1) is equidistant from P(5, –3) and R(x, 6) find the value of x. Also, find the distances of

QR and PR. [NCERT]

Sol. Since, point Q(0, 1) is equidistant from P(5, –3) and R(x, 6).

Therefore, QP = QR

Squaring both sides, we have, QP2 = QR2

⇒ (5 – 0)2 + (–3 –1)2 = (x – 0)2 + (6 – 1)2

⇒ 25 + 16 = x2 + 25

⇒ x2 = 16 \ x = ±4

Thus, R is (4, 6) or (– 4, 6).

Now, QR = (4 – 0)2 + ]6 – 1g2 = 16 + 25 = 41

or, QR = (–4 – 0)2 + ]6 – 1g2 = 16 + 25 = 41

and PR = (4 – 5)2 + ]6 + 3g2 = 1+ 81 = 82

or, PR = ]–4 – 5g2 + ]6 + 3g2 = 81 + 81 = 9 2

4. Find the point on the x-axis which is equidistant from (2, – 5) and (– 2, 9). [NCERT]

OR

Find the coordinates of a point on the x-axis which is equidistant from the points A(2, –5) and

B(–2, 9). [CBSE Delhi 2017 (C)]

Sol. Let P(x, 0) be any point on x-axis.

Now, P(x, 0) is equidistant from point A(2, –5) and B(–2, 9).

\ AP = BP

⇒ (x − 2)2 + (0 + 5)2 = (x + 2)2 + (0 − 9)2

Squaring both sides, we have

(x – 2)2 + 25 = (x + 2)2 + 81
⇒ x2 + 4 – 4x + 25 = x2 + 4 + 4x + 81 ⇒ –8x = 56

\ x = 56 = −7
\ −8
The point on the x-axis equidistant from given point is (– 7, 0).

144 Xam idea Mathematics–X

5. Find the coordinates of the point which divides the line joining of (– 1, 7) and (4, – 3) in the

ratio 2 : 3. [NCERT]

Sol. Let P(x, y) be the required point. Thus, we have

x m1 x2 m2 x1 and y = m1 y2 + m2 y1
m1 m2 m1 + m2
Therefore, Fig. 6.18

x = 2 × 4 + 3 × (−1) = 8−3 = 5 =1
2+3 5 5

y= 2 # (–3)+3 # 7 = –6 + 21 = 15 =3
2+3 5 5

So, the coordinates of P are (1, 3).

6. Find the coordinates of the points of trisection of the line segment joining (4, – 1) and (– 2, – 3).
[NCERT]

Sol. Let the given points be A(4, –1) and B(–2, –3) and points of trisection be P and Q .

Let AP = PQ = QB = k

PB = PQ + QB = k + k = 2k Fig. 6.19
AP : PB = k : 2k = 1 : 2

Therefore, coordinates of P are

d 1# (–2) + 2 # 4 , 1# (–3)+ 2 # (–1) = c2, – 5 m
3 3n 3

Now, AQ = AP + PQ = k + k = 2k

AQ : QB = 2k : k = 2 : 1

and, coordinates of Q are

d 2 # (–2) +1 # 4 , 2 # (–3)+1# (–1) n = c0, – 7 m
3 3 3

Hence, points of trisection are  2, − 5  and  0, − 7  .
 3   3 

7. 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 coordinates of the point of division. [NCERT]

Sol. Let the required ratio be k : 1. Then, the coordinates of the point of division is

P  −4k + 1 , 5k − 5  .
 k +1 k +1 
 

Since, this point lies on x-axis. Therefore its y-coordinate is zero.

i.e., 5k − 5 = 0 ⇒ 5k – 5 = 0
k +1
5
5k = 5 or k= 5 =1

Thus, the required ratio is 1 : 1 and the point of division is P  −4 × 1+ 1 , 5×1− 5
 1+1 
 3   1+1 
 2 
i.e., P − , 0

Coordinate Geometry 145

8. The points A(4, –2), B(7, 2), C(0, 9) and D(–3, 5) form a parallelogram. Find the length of the
altitude of the parallelogram on the base AB.

Sol. Let DE = h be the height of parallelogram ABCD w.r.t. base AB.

Now, area (∆ABD) = 1 {4 (2 – 5) + 7 (5 + 2) + (–3) (–2 – 2)} D (–3, 5) C (0, 9)
2 h

= 1 {– 12 + 49 + 12}
2
1 49
= 2 ×49 = 2 sq. units

Also, AB = (4 – 7)2 + (– 2 – 2)2 = 9 +16 AE B
(4, –2) (7, 2)

AB = 25 = 5 units Fig. 6.20

We have area (∆ABD) = 1 × AB × h
2

⇒ 49 = 1 ×5× h ⇒ h = 49 × 2 = 49 = 9.8 units
2 2 2 5 5

9. 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). [NCERT]

Sol. Let the coordinates of A be (x, y).

Now, C is the centre of circle therefore, the coordinates of

C =  x + 1 , y+ 4 but coordinates of C are given as (2, – 3).
 2 2 

\ x + 1 = 2 ⇒ x + 1 = 4 \ x=3
2
Fig. 6.21

and y +4 = −3 ⇒ y + 4 = –6 \ y = –10
2

Hence, coordinates of A are (3, – 10). 3
7
10. If A and B are (– 2, – 2) and (2, – 4), respectively, find the coordinates of P such that AB = AB

and P lies on the line segment AB. [NCERT]
3
Sol. In Fig. 6.22, we have, AP = 7 AB

⇒ AP = 3 ⇒ AB = 7
AB 7 AP 3
Fig. 6.22
⇒ AP + PB 7 AP PB 7
AP = 3 ⇒ AP + AP = 3

1+ PB = 7 ⇒ PB = 7 − 1 = 4
⇒ AP 3 AP 3 3

⇒ AP = 3 ⇒ AP : PB = 3 : 4
PB 4

Let P (x, y) be the point which divides the joining of A(–2, –2) and B(2, –4) in the ratio 3 : 4.

\ x = 3× 2+ 4× −2 = 6−8 = −2 and y = 3× − 4 + 4 × − 2 = −12 − 8 = −20
3+ 4 7 7 3 + 4 7 7

Hence, the coordinates of the point P are  −2 , −20  .
 7 7 

146 Xam idea Mathematics–X

11. Find the coordinates of the points which divide the line segment joining A (–2, 2) and B (2, 8)

into four equal parts. [NCERT]

Sol. Let P, Q, R be the points that divide the line segment joining A(–2, 2) and B(2, 8) into four equal
parts. (Fig. 6.23)

Fig. 6.23

Since, Q divides the line segment AB into two equal parts, i.e., Q is the mid-point of AB.

\ Coordinates of Q are  −2 + 2, 2+ 8 i.e., (0, 5)
 2 2 

Now, P divides AQ into two equal parts i.e., P is the mid-point of AQ.

\ Coordinates of P are  −2 + 0 , 2+ 5 i.e.,  −1, 7
 2 2   2 

Again, R is the mid-point of QB.

Coordinates of R are  0 + 2 , 5+8 i.e., d1, 13 n.
 2 2  2

12. Find the area of a rhombus if its vertices (3, 0), (4, 5), (– 1, 4) and (– 2, – 1) are taken in order.

[NCERT]

Sol. Let A(3,0 ), B(4, 5), C(–1, 4) and D(–2, –1) be the vertices of a rhombus.

Therefore, its diagonals

AC = ( 1 3)2 (4 0)2 16 16 32 4 2

and BD = ( 2 4)2 ( 1 5)2 36 36 72 6 2

\ Area of rhombus ABCD = 1 × (Product of length of diagonals)
2

= 12 × AC × BD = 1 ×4 2 ×6 2 = 24 sq units.
2

13. If the points A (–2, 1), B (a, b) and C (4, –1) are collinear and a – b = 1, find the values of a and b.

[CBSE Delhi 2014]

Sol. Since the given points are collinear, then area of ∆ABC = 0.

⇒ 1 [x1 (y2 – y3) + x2(y3 – y1) + x3(y1 – y2)] = 0
2

Given, x1 = –2, y1 = 1, x2 = a, y2 = b, x3 = 4, y3 = –1
Putting the values, we get

1 [–2(b + 1) + a(–1 – 1) + 4(1 – b)] = 0
2

⇒ –2b – 2 – 2a + 4 – 4b = 0 ⇒ 2a + 6b = 2

⇒ a + 3b = 1 ...(i)

Given, a – b = 1 ...(ii)

Subtracting (i) from (ii), we have

– 4b = 0 ⇒ b = 0

Substituting the value of b in (ii), we have a = 1.

14. If the point P (k –1, 2) is equidistant from the points A (3, k) and B (k, 5), find the values of k.

[CBSE (AI) 2014]

Sol. Let the given line segment be divided by point P. Since P is equidistant from A and B,

AP = BP or AP2 = BP2

Coordinate Geometry 147

⇒ [3 – (k – 1)]2 + (k – 2)2 = [k – (k – 1)]2 + (5 – 2)2

⇒ (3 – k + 1)2 + (k – 2)2 = (k – k + 1)2 + (3)2

⇒ (4 – k)2 + (k – 2)2 = (1)2 + (3)2

⇒ 16 + k2 – 8k + k2 + 4 – 4k = 1 + 9

⇒ 2k2 – 12k + 20 = 10 ⇒ k2 – 6k + 10 = 5

⇒ k2 – 6k + 5 = 0 ⇒ k2 – 5k – k + 5 = 0

⇒ k (k – 5) –1(k – 5) = 0 ⇒ k = 1 or k = 5

15. Find the ratio in which the line segment joining the points A(3, –3) and B(–2, 7) is divided by

x-axis. Also find the coordinates of the point of division. [CBSE (AI) 2014]

Sol. Here, point Q is on x axis so its ordinate is 0.

Let ratio be k : 1 and coordinate of point Q be (x, 0).

So, Qy = (my2 + ny1 )
m + n

We are given that A (3, –3) and B (–2, 7)

\ 0 = k × 7 + 1 × (−3)
k +1

0 = 7k − 3 ⇒ 0 (k + 1) = 7k – 3
k +1
3
⇒ 7k = 3 ⇒ k= 7 Fig. 6.24

k:1 = 3:7 3 × (–2) + 1×3
7
Now, Qx = (mx2 + nx1 ) ⇒ Qx = 3 = 15 = 3
m + n 7 10 2
+ 1

∴ The co-ordinates of the point of division are d 3 , 0n .
2

16. Find the values of k if the points A(k + 1, 2k), B(3k, 2k + 3) and C(5k – 1, 5k) are collinear.

[NCERT Exemplar, CBSE (AI) 2017]

Sol.  Points A(k + 1, 2k), B (3k, 2k + 3) and C (5k – 1, 5k) are collinear.

\ Area of ∆ABC = 0

⇒ 1 |x1(y2 – y3) + x2(y3 – y1) + x3 (y1 – y2)| = 0
2

⇒ 1 |(k + 1) (2k + 3 – 5k) + 3k(5k – 2k) + (5k – 1){2k –(2k + 3)}| = 0
2

⇒ 1 |(k + 1) (–3k + 3) + 3k(3k) + (5k – 1)(2k – 2k – 3)| = 0
2

⇒ 1 |–3k2 + 3k –3k + 3 + 9k2 – 15k + 3| = 0
2

⇒ 1 |6k2 – 15k + 6| = 0 ⇒ 6k2 – 15k + 6 = 0
2 ⇒ 2k2 – 4k – k + 2 = 0

⇒ 2k2 – 5k + 2 = 0

⇒ (k – 2)(2k – 1) = 0 1
2
If k – 2 = 0, then k = 2 If 2k – 1 = 0, then k =

\ k = 2, 1
2

148 Xam idea Mathematics–X

17. In the given figure, ∆ABC is an equilateral triangle of side 3 units. Find the coordinates of the

other two vertices. [CBSE (F) 2017]

Sol. Since AB = BC = AC = 3 units

∴ Co-ordinates of B are (5, 0).

Let co-ordinates of C be (x, y).

AC2 = BC2 [ a ∆ABC is an equilateral triangle]

Using distance formula

(x – 2)2 + (y – 0)2 = (x – 5)2 + (y – 0)2 Y
C

⇒ x2 + 4 – 4x + y2 = x2 + 25 – 10x + y2

6x = 21
⇒
x = 21 = 7
6 2

Again (x – 2)2 + (y – 0)2 = 9 X' 0 X
Y'
7 2 A (2, 0) B
2
⇒ d – 2n + y2 = 9 Fig. 6.25

⇒ d 3 2 + y2 = 9
2
n

⇒ y2 = 9 – 9
4
33
⇒ y2 = 27 ⇒ y= 2 (+ve sign to be taken)
4

Hence, co-ordinate of Ce 7 , 3 3 o .
2 2

18. The area of a triangle is 5 sq units. Two of its vertices are (2, 1) and (3, –2). If the third vertex
7
is d 2 , yn , find the value of y. [CBSE Delhi 2017]

Sol. Given: ar(∆ABC) = 5 sq. units

⇒ 1 x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2) =5
⇒ 2
1 7
2 2 (–2 –y) + 3 (y – 1) + 2 (1 + 2) =5

⇒ |–4 – 2y + 3y – 3 + 7 + 7 |= 10
2

⇒ y+ 7 = 10 ⇒ y = 10 – 7 3 –2 y
⇒ 2 2 Fig. 6.26

y = 13
2

19. The coordinates of the points A, B and C are (6, 3), (–3, 5) and (4, –2) respectively. P(x, y) is any

point in the plane. Show that ar(∆PBC) = x + y − 2 . [CBSE (F) 2016]
ar(∆ABC) 7

Sol. Given points are P(x, y), B(–3, 5), C(4, –2), A(6, 3)

\ ar(∆PBC) = 12 |x(7) – 3(– 2– y) + 4(y – 5)| = 1 |7x + 7y – 14|
2

ar(∆ABC) = 1 |6 × 7 – 3(–5) + 4(– 2)| = 1 |42 + 15 – 8| = 49
2 2 2

Coordinate Geometry 149

ar(∆PBC) 1 7x + 7 y −14 x+ y−2 x+ y−2
ar(∆ABC) 2 7 7
LHS = = 1 = = = RHS
2
× 49

20. If a ≠ b ≠ 0, prove that the points (a, a2), (b, b2), (0, 0) will not be collinear. [CBSE Delhi 2017]

Sol. a We know that three points are collinear if area of triangle = 0.
∴ Area of triangle with vertices (a, a2), (b, b2) and (0, 0)

= 1 a (b2) + b (–a2) + 0 =0
2
1
<Using formula area of T = 2 x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2) = 0F

= 1 ab (b – a) ! 0 as a! b ! 0
2
a Area of T ! 0

∴ Given points are not collinear.

Long Answer Questions [5 marks]

1. 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 this area to the area of the given

triangle. [NCERT]

Sol. Let A (x1, y1) = (0, –1), B (x2, y2) = (2, 1), C (x3, y3) = (0, 3) be the vertices of ∆ABC.
Now, let P, Q, R be the mid-points of BC, CA and AB, respectively.

So, coordinates of P, Q, R are

P =  2 + 0 , 1 + 3  = (1, 2)
 2 2 

Q =  0 + 0 , 3 − 1 = (0, 1)
 2 2 
Fig. 6.27
2 + 0 ,1 1
R =  2 ar −  = (1, 0)
2  |1(1 –
1 1
Therefore, (∆PQR) = 2 0) + 0(0 – 2) + 1(2 – 1)| = 2 |1 + 1| = 1 sq. unit

Now, ar(∆ABC) = 1 |0(1 – 3) + 2 (3 + 1) + 0 (–1 –1)|
2

= 1 |0 + 8 + 0| = 8 = 4 sq. units
2 2

\ Ratio of ar (∆PQR) to the ar (∆ABC) = 1 : 4.

2. Find the area of the quadrilateral whose vertices, taken in order, are (– 4, – 2), (– 3, – 5), (3, – 2)

and (2, 3). [NCERT]

Sol. Let A(–4, –2), B(–3, –5), C(3, –2) and D(2, 3) be the vertices
of the quadrilateral ABCD.

Now, area of quadrilateral ABCD

= area of ∆ABC + area of ∆ADC

Fig. 6.28

= 1 |–4 (–5 + 2) – 3 (–2 + 2) + 3(–2 + 5)| + 1 |–4 (–2 –3) + 3(3 + 2) + 2 (–2 + 2)|
2 2
150 Xam idea Mathematics–X

= 1 |12 –0 + 9| + 1 |20 + 15 + 0|
2 2

= 1 |21 + 35| = 1 × 56 = 28 sq. units.
2 2

3. Find the ratio in which the point P (x, 2), divides the line segment joining the points A (12,5)

and B (4, –3). Also find the value of x. [CBSE Delhi 2014]

Sol.

Fig. 6.29
Let the ratio in which point P divides the line segment be k:1.

Then, coordinates of P :  4k + 12 , −3k + 5 
 k + 1 
 k +1 

Given, the coordinates of P as (x, 2).

\ 4k + 12 = x ...(i)

 and k + 1

−3k + 5 = 2 ...(ii)
k +1

–3k + 5 = 2k + 2 3
5
5k = 3 ⇒k=

Putting the value of k in (i), we have

4 × 3 + 12 x ⇒ 12 + 60 = x
5 3+5
3 =
5
+ 1

x= 72 ⇒ x=9
8
3
The ratio in which P divides the line segment is 5 , i.e., 3:5.

4. If A (4, 2), B (7, 6) and C (1, 4) are the vertices of a ∆ABC and AD is its median, prove that the

median AD divides into two triangles of equal areas. [CBSE (AI) 2014]

Sol. Given: AD is the median on BC.

⇒ BD = DC
The coordinates of midpoint D are given by

 x2 + x1 , y2 + y1     i.e.,  1 + 7 , 4 + 6 
 2 2  2 2 

Coordinates of D are (4, 5). Fig. 6.30
Now, area of triangle ABD = 12 x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)

= 1 4 (6 – 5) + 7 (5 – 2) + 4 (2 – 6) = 1 4 + 21 – 16 = 9 sq. units
2 2 2

Area of ∆ACD = 12 4 (4 – 5) + 1 (5 – 2) + 4 (2 – 4)

= 1 –4 + 3 –8 = 1 –9 = 9 sq. units
2 2 2

Hence, AD divides ∆ABC into two equal areas.

Coordinate Geometry 151

5. If the point A (2, – 4) is equidistant from P (3, 8) and Q (– 10, y), find the values of y. Also find

distance PQ. [NCERT Exemplar]

Sol. Given points are A(2, –4), P(3, 8) and Q(–10, y)

According to the question,

PA = QA

⇒ (2 − 3)2 + (−4 − 8)2 = (2 + 10)2 + (−4 − y)2

⇒ (−1)2 + (−12)2 = (12)2 + (4 + y)2

⇒ 1 + 144 = 144 + 16 + y2 + 8 y

⇒ 145 = 160 + y2 + 8 y

On squaring both sides, we get

145 = 160 + y2 + 8y ⇒ y2 + 8y + 160 – 145 = 0
⇒ y2 + 8y + 15 = 0
⇒ y2 + 5y + 3y + 15 = 0 ⇒ y (y + 5) + 3(y + 5) = 0
⇒ y + 5 = 0
⇒ (y + 5) (y + 3) = 0
⇒ y = –3
    ⇒   y = –5

and y + 3 = 0   
\ y = – 3, – 5

Now, PQ = (−10 − 3)2 + ( y − 8)2

For y = – 3 PQ = (−13)2 + (−3 − 8)2 = 169 + 121 = 290 units

and for y = – 5 PQ = (−13)2 + (−5 − 8)2 = 169 + 169 = 338 units

Hence, values of y are – 3 and – 5, PQ = 290 and 338 units.

6. The base BC of an equilateral triangle ABC lies on y-axis. The coordinates of point C are

(0, –3). The origin is the mid-point of the base. Find the coordinates of the points A and B. Also

find the coordinates of another point D such that BACD is a rhombus. [CBSE (F) 2015]

Sol. O is the mid-point of the base BC.

\ Coordinates of point B are (0, 3).

So, BC = 6 units

Let the coordinates of point A be (x, 0).

Using distance formula,

AB = (0 x)2 (3 0)2 x2 9

BC = (0 − 0)2 + (−3 − 3)2 = 36
Also,
AB = BC ( a ∆ABC is an equilateral triangle)

x2 + 9 = 36 Fig. 6.31

x2 + 9 = 36 ⇒ x2 – 27 = 0
x2 = 27    ⇒ (x + 3 3 ) (x – 3 3 ) = 0
⇒
x2 – (3 3)2 = 0   

x = – 3 3 or x = 3 3

x = ± 3 3

152 Xam idea Mathematics–X

\ C oordinates of point A = (x, 0) = ( 3 3 , 0)
Since BACD is a rhombus.
\ AB = AC = CD = DB
\ Coordinates of point D = (– 3 3 , 0)

7. In Fig. 6.32, the vertices of ∆ABC are A(4, 6), B(1, 5) and C(7, 2). A line-segment DE is drawn

to intersect the sides AB and AC at D and E respectively such that AA=DB AA=CE 1 . Calculate
3

the area of ∆ADE and compare it with area of ∆ABC. [CBSE (AI) 2016]

Sol. Since AA=DB AA=CE 1
3

⇒ D and E divide AB and AC respectively in the ratio 1 : 2.

Coordinates of D are  1(1) + 2(4)  , 1(5) + 2(6)  i.e., 3, 17 
 3  3  3 

Coordinates of E are § 1(7) 2(4) ,1(2) 2(6) · i.e.,§¨© 5,134 ·
¨© 3 3 ¹¸ ¹¸

Area of ∆ADE = 1 ;4c 17 – 14 m + 3c 14 – 6m + 5c6 – 17 mE
2 3 3 3 3
Fig. 6.32

= 1 4 + (– 4) + 5 = 5 sq. unit
2 3 6

Area of ∆ABC = 12 |4(3) + 1(–4) + 7(1)| = 15 sq. units
2

Area of ∆ADE: area of ∆ABC = 5 : 15 or 1:9
6 2

HOTS [Higher Order Thinking Skills]

1. The line joining the points (2, 1) and (5, –8) is trisected by the points P and Q. If the point P
lies on the line 2x – y + k = 0, find the value of k.

Sol. As line segment AB is trisected by the points P and Q. Therefore,

Case I: When AP : PB = 1 : 2.

Then, coordinates of P are

(1 # 5 + 2 # 2 , 1 # (–8) + 1 # 2 2 ⇒ P (3, –2)
1 + 2 1+2

Since the point P (3, – 2) lies on the line 2x – y + k = 0.

⇒ 2 × 3 – (–2) + k = 0

⇒ k = –8

Case II: When AP : PB = 2 : 1

Coordinates of point P are (–8) + 1
2 2+1 1
( 2 # 5 + 1 # 2 , # # 2 = {4, –5}
2 + 1
Fig. 6.33

Since the point P(4, –5) lies on the line 2x – y + k = 0

\ 2 × 4 –(–5) + k = 0 \ k = –13.

Coordinate Geometry 153

2. Prove that the diagonals of a rectangle bisect each other and are equal.

Sol. Let OACB be a rectangle such that OA is along x-axis and OB is along y-axis. Let OA = a and
OB = b.

Then, the coordinates of A and B are (a, 0) and (0, b) respectively.

Since, OACB is a rectangle. Therefore,

AC = OB ⇒ AC = b

Also, OA = a ⇒ BC = a

So, the coordinates of C are (a, b).

The coordinates of the mid-points of OC are

 a +2 0 , b + 0  =  a , b 
2   2 2 

Also, the coordinates of the mid-points of AB =  a + 0 , 0 + b  =  a , b  Fig. 6.34
 2 2   2 2 

Clearly, coordinates of the mid-point of OC and AB are same.

Hence, OC and AB bisect each other.

Also, OC = a2 + b2 and BA = (a − 0)2 + (0 − b)2 = a2 + b2

\ OC = AB. .

3. In what ratio does the y-axis divide the line segment joining the point P (–4, 5) and Q (3, – 7)?
Also, find the coordinates of the point of intersection.

Sol. Suppose y-axis divides PQ in the ratio k : 1. Then, the coordinates
of the point of division are

R  3k − 4 , −7k + 5 
 k +1 k +1 

 R lies on y-axis and x-coordinate of every point on y-axis is
zero.

\ 3k − 4 =0
k +1

⇒ 3k – 4 = 0 ⇒ k = 4 Fig. 6.35
3

Hence, tkh=e re43quinirtehdercaotioordisin43ate: s1of i.e., 4 : 3. its coordinates are  0, −13  .
Putting R, we find that  7 

4. Find the centre of a circle passing through the points (6, –6), (3, –7) and (3, 3).

Sol. Let O(x, y) be the centre of circle. Given points are A(6, – 6), B(3, –7) and C(3, 3).

Then, OA = (x − 6)2 + ( y + 6)2; OB = (x − 3)2 + ( y + 7)2

and OC = (x − 3)2 + ( y − 3)2

Since, each point on the circle is equidistant from centre.
\ OA = OB = OC = Radius
Since, OA = OB

⇒ (x − 6)2 + ( y + 6)2 = (x − 3)2 + ( y + 7)2

154 Xam idea Mathematics–X

Squaring both sides, we get,

(x – 6)2 + (y + 6)2 = (x – 3)2 + (y + 7)2

⇒ x2 – 12x + 36 + y2 + 12y + 36 = x2 – 6x + 9 + y2 + 14y + 49

or –6x – 2y = – 14

or 3x + y = 7 ... (i)
Similarly, OB = OC

⇒ (x − 3)2 + ( y + 7)2 = (x − 3)2 + ( y − 3)2

Squaring both sides, we get,

(x – 3)2 + (y + 7)2 = (x – 3)2 + (y – 3)2 ⇒ (y + 7)2 = (y – 3)2

⇒ y2 + 14y + 49 = y2 – 6y + 9 or 20y = – 40

⇒ y = –2

Putting y = –2 in (i), we get

3x – 2 = 7 ⇒ x = 3

Hence, the coordinates of the centre of circle are (3, –2).

5. If the coordinates of the mid-points of the sides of a triangle
are (1, 1), (2, – 3) and (3, 4). Find its centroid.

Sol. Let P(1, 1), Q(2, –3), R(3, 4) be the mid-points of sides AB, BC

and CA respectively, of triangle ABC. Let A(x1, y1), B(x2, y2) and
C(x3, y3) be the vertices of triangle ABC. Then, P is the mid-point
of AB.
x1 + x2 y1 + y2 Fig. 6.36
2 = 1, 2 =1
⇒

⇒ x1 + x2 = 2 and y1 + y2 = 2 ...(i)
Q is the mid-point of BC

⇒ x2 + x3 = 2, y2 + y3 = −3      ⇒ x2 + x3 = 4 and y2 + y3 = –6 ...(ii)
2 2

R is the mid-point of AC

⇒ x1 + x3 = 3 and y1 + y3 =4   ⇒ x1 + x3 = 6 and y1 + y3 = 8 ...(iii)
2 2

From (i), (ii) and (iii), we get

x1 + x2 + x2 + x3 + x1 + x3 = 2 + 4 + 6

and y1 + y2 + y2 + y3 + y1 + y3 = 2 – 6 + 8

⇒ x1 + x2 + x3 = 6 and y1 + y2 + y3 = 2 ...(iv)

The coordinates of the centroid of ∆ABC are

 x1 + x2 + x3 , y1 + y2 + y3  =  6 , 2 =  2, 2 [Using (iv)]
 3 3   3 3   3 
 
AP 1
6. Point P divides the line segment joining the points A(2,1) and B(5,–8) such that PB = 3 . If P

lies on the line 2x – y + k = 0, find the value of k. [CBSE 2019 (30/5/1)]

Sol. Let the point P be (x, y) which divides AB such that

AP = 1 & AB = 3 & AB – 1= 3 –1
AB 3 AP 1 AP 1

⇒ AB – AP = 2 & PB = 2
AP 1 AP 1

Coordinate Geometry 155

⇒ AP = 1 & AP : BP =1: 2
BP 2

∴ Using section formula, we have

x = 1× 5 + 2 ×2 = 9 =3
1 + 2 3

y = 1× (– 8) + 2 ×1 = – 8+2 = –6 = – 2
1+2 3 3

∴ Co-ordinate of P be (3, –2).

Now, P(3, – 2) lies on the line 2x – y + k = 0.

∴ 2 × 3 – (–2) + k = 0

⇒ 6 + 2 + k = 0 ⇒ k = – 8

PROFICIENCY EXERCISE

QQ Objective Type Questions: [1 mark each]

1. Choose and write the correct option in each of the following questions.

(i) The distance of the point P(3, 4) from the origin is

(a) 7 units (b) 5 units (c) 4 units (d) 3 units

(ii) The mid point of the line segment joining the points (– 5, 7) and (– 1, 3) is

(a) (– 3, 7) (b) (– 3, 5) (c) (– 1, 5) (d) (5, – 3)

(iii) If the point C (x, 3) divides the line joining points A(2, 6) and B(5, 2) in the ratio 2 : 1 then

the value of x is

(a) 4 (b) 8 (c) 6 (d) 3

(iv) The distance of the point (– 8, – 7) from y-axis is

(a) 5 units (b) 1 units (c) 8 units (d) 7 units

(v) The ratio in which x-axis divides the line segment joining A(2, –3) and B (5, 6) is

(a) 3 : 5 (b) 1 : 2 (c) 2 : 1 (d) 2 : 3

2. Fill in the blanks.
(i) The point on x-axis which is equidistant from points A (– 1, 0) and B(5, 0) is ___________ .

(ii) The distance between the points (– 5, 7) and (– 1, 3) is ___________ .
(iii) The point (0, 4) lies on ___________ axis.

(iv) Centroid of a triangle with vertices A(x1, y1); B(x2, y2); C (x3, y3) is given by ___________.
(v) The area of the triangle with vertices (– 5, – 1), (3, – 5), (5, 2) is ______________.

QQ Very Short Answer Questions: [1 mark each]

3. Find the distance of a point P(x, y) from the origin. [CBSE 2018 (30/1)]

4. A(5, 1); B(1, 5) and C(– 3, –1) are the vertices of ∆ABC. Find the length of median AD.

[CBSE 2018 (C) (30/1)]

5. Find the coordinates of a point A, where AB is diameter of a circle whose centre is (2, – 3) and B

is the point (1, 4). [CBSE 2019 (30/1/1)]

156 Xam idea Mathematics–X

6. Write the coordinates of a point P on x-axis which is equidistant from the points A(– 2, 0) and
B(6, 0). [CBSE 2019 (30/2/1)]

7. Find the value(s) of x, if the distance between the points A(0, 0) and B(x, – 4) is 5 units.

[CBSE 2019 (30/3/1)]

8. Find the distance between the points (a, b) and (–a, –b). [CBSE 2019 (30/4/2)]

9. Find the value of ‘a’ so that the point (3, a) lies on the line represented by 2x – 3y = 5.

[CBSE 2019 (30/5/1)]

10. Find the values of x for which the distance between the points A(x, 2) and B(9, 8) is 10 units.

[CBSE 2019 (30/5/3)]

11. In what ratio is the line segment joining the points P(3, – 6) and Q(5, 3) divided by x-axis?

[CBSE 2019 (30/1/1)]

12. The point which divides the line segment joining the points (7, –6) and (3, 4) in ratio 1 : 2

internally lies in which quadrant? [NCERT Exemplar]

13. If P  a , 4  is the mid-point of the line segment joining the points Q (–6, 5) and R (–2, 3), then
 3 
find the value of a. [NCERT Exemplar]

14. A line intersects the y-axis and x-axis at the points P and Q respectively. If (2, –5) is the mid-point

of PQ, then find the coordinates of P and Q. [NCERT Exemplar, CBSE (AI) 2017]

15. Where does the perpendicular bisector of the line segment joining the points A (1, 5) and

B (4, 6) cut the y-axis? [NCERT Exemplar]

16. If the points (k, 2k), (3k, 3k) and (3, 1) are collinear, then find the value of k.

17. If A (5, 2), B (2, –2) and C (–2, t) are the vertices of a right angled triangle with ∠B = 90°, then

find the value of t. [CBSE Delhi 2015]

18. If points (a, 0), (0, b) and (1, 1) are collinear, then find the value of 1 + 1 .
a b

19. If the centroid of a triangle formed by the points (a, b), (b, c) and (c, a) is at the origin, then find

the value of a3 + b3 + c3.

20. What is the ratio in which the point P  −2 , 6  divides the line joining A (– 4, 3) and B (2, 8)?
QQ Short Answer Questions–I:  5  [2 marks each]

21. Find the ratio in which P(4, m) divides the line segment joining the points A(2, 3) and B(6, –3).

Hence find m. [CBSE 2018 (30/1)]

22. Find the linear relation between x and y such that P(x, y) is equidistant from the points A(1, 4)

and B(–1, 2). [CBSE 2018 (C) (30/1)]

23. Find the ratio in which the segment joining the points (1, – 3) and (4, 5) is divided by x-axis? Also
find the coordinates of this point on x-axis. [CBSE 2019 (30/1/1)]

24. Find a relation between x and y if the points A(x, y); B(– 4, 6) and C(– 2, 3) are collinear.

[CBSE 2019 (30/2/1)]

25. Find the area of a triangle whose vertices are given as (1, – 1); (– 4, 6) [CBSE 2019 (30/2/1)]

26. Points A(3, 1), B(5, 1), C(a, b) and D(4, 3) are vertices of a parallelogram ABCD. Find the values
of a and b. [CBSE 2019 (30/3/1)]

27. Points P and Q trisect the line segment joining the points A(– 2, 0) and B(0, 8) such that P is near

to A. Find the coordinates of points P and Q. [CBSE 2019 (30/3/1)]

Coordinate Geometry 157

28. The point R divides the line segment AB, where A(– 4, 0) and B(0, 6) such that AR = 3 AB. Find
4
the coordinates of R. [CBSE 2019 (30/4/2)]

29. Find the relation between x and y such that the points (x, y), (1, 2) and (7, 0) are collinear.

[CBSE 2019 (C) (30/1/1)]

30. If A(–2, 2), B(5, 2) and C(k, 8) are the vertices of a right-angled triangle ABC with ∠B = 90°, then
find the value of k. [CBSE 2019 (C) (30/1/2)]

31. If A (2, 2), B (– 4, 4) and C (5, –8) are the vertices of a triangle, then find the length of the median
through vertex C.

32. What is the distance between the points A (5cos q, 0) and B(0, 5sin q)?

33. If the distance between the points (x, 0) and (0, 3) is 5, what are the values of x?

34. What is the area of the triangle formed by the points O (0, 0), A (4, 0) and B (0, 3)?
35. If the centroid of the triangle formed by points A (a, b), B (b, c) and C (c, a) is at the origin, what

is the value of a2 + b2 + c2 ?
bc ca ab

36. If the mid-point of a segment joining A x , y +1  and B(x + 1, y – 3) is C(5, –2), find x, y.
2 2 

37. Prove that the points (3, 0), (6, 4) and (–1, 3) are the vertices of a right angled isosceles triangle.

[CBSE (AI) 2016]

QQ Short Answer Questions–II: [3 marks each]

38. If A(– 2, 1), B(a, 0), C( 4, b) and D(1, 2) are vertices of a parallelogram ABCD, find the values of

a and b. Hence find the lengths of its sides. [CBSE 2018 (30/1)]

39. If A(– 5, 7), B(– 4, – 5), C( – 1, – 6) and D(4, 5) are vertices of a quadrilateral, find the area of the
quadrilateral ABCD. [CBSE 2018 (30/1)]

40. If coordinates of two adjacent vertices of a parallelogram are (3, 2), (1, 0) and diagonals bisect

each other at (2, – 5), find coordinates of the other two vertices. [CBSE 2018 (C) (30/1)]

41. If the area of triangle with vertices (x, 3), (4, 4) and (3, 5) is 4 square units, find x.

[CBSE 2018 (C) (30/1)]

42. Find the point on y-axis which is equidistant from the points (5, – 2) and (– 3, 2).

[CBSE 2019 (30/1/1)]

43. The line segment joining the points A(2, 1) and B(5, – 8) is trisected at the points P and Q such
that P is nearer to A. If P also lies on the line given by 2x – y + k = 0, find the value of k.

[CBSE 2019 (30/1/1)]

44. Find the ratio in which the line x – 3y = 0 divides the line segment joining the points (– 2, – 5)

and (6, 3). Find the coordinates of the point of intersection.

[CBSE 2019 (30/2/1)]

45. Point A lies on the line segment XY joining X(6, – 6) and Y(– 4, – 1) in such a way that XA = 2 .
XY 5
If point A also lies on the line 3x + k (y + 1) = 0, find the value of k. [CBSE 2019 (30/2/2)]

46. Find the area of the triangle formed by joining the mid-points of the sides of the triangle ABC,

whose vertices are A(0, – 1), B(2, 1) and C(0, 3). [CBSE 2019 (30/3/1)]

47. Find the value of k so that the area of triangle ABC with A(k + 1, 1), B(4, – 3) and C(7, – k) is 6
square units. [CBSE 2019 (30/3/3)]

158 Xam idea Mathematics–X

48. In what ratio does the point P(– 4, y) divide the line segment joining the point A(– 6, 10) and
B(3, – 8)? Hence find the value of y. [CBSE 2019 (30/4/2)]

49. Find the value of p for which the points (– 5, 1), (1, p) and (4, – 2) are collinear.
[CBSE 2019 (30/4/2)]

50. The line segment joining the points A(2, 1) and B(5, – 8) is trisected by the points P and Q, where
P is nearer to A. If the point P also lies on the line 2x – y + k = 0, find the value of k.
[CBSE 2019 (C) (30/1/1)]

51. Show that (a, a), (– a, – a), and (– 3 a, 3 a) are vertices of an equilateral triangle.

[CBSE 2019 (C) (30/1/1)]
52. Find a point which is equidistant from the points P(–5, 4) and Q (–1, 6). How many such points

are there?

53. Find the points on the x-axis which are at a distance of 2 5 from the point (7, 4). How many
such points are there?

54. What type of a quadrilateral do the points A (2, –2), B (7, 3), C (11, –1) and D (6, – 6) taken in that
order, will form?

55. Find the coordinates of the point Q on the x-axis which lies on the perpendicular bisector of the
line segment joining the points A (–5, 2) and B (4, –2). Name the type of triangle formed by the
points Q, A and B.

56. Name the type of triangle ABC formed by the points A( 2, 2), B(− 2, − 2) and C(− 6, 6) .

57. The line segment joining the points A(3, – 4) and B(1, 2) is trisected at the points P and Q, where

P is nearer to A. Find the coordinates of point P. [CBSE Delhi 2017 (C)]

58. Point P divides the line segment joining the points A (–1, 3) and B (9, 8) such that AP = k . If P
lies on the line x – y + 2 = 0, find the value of k. PB 1

59. Point A lies on the line segment PQ joining P(6, – 6) and Q(– 4, –1) in such a way that PA = 2 .
PQ 5

If point P also lies on the line 3x + k (y + 1) = 0, find the value of k. [CBSE (F) 2015]

60. If R (x, y) is a point on the line segment joining the points P (a, b) and Q (b, a), then prove that
x + y = a + b.

61. If the points (p, q), (m, n) and (p – m, q – n) are collinear, show that pn = qm.

62. If the mid-point of the line segment joining the points A (3, 4) and B (k, 6) is P (x, y) and P lies on
the line x + y – 10 = 0, find the value of k.

63. If P (9a – 2, – b) divides the line segment joining A (3a + 1, –3) and B (8a, 5) in the ratio 3 : 1,
find the values of a and b.

64. In what ratio does the x-axis divide the line segment joining the points (– 4,–6) and (–1, 7)? Find
the coordinates of the point of division.

65. If D  − 1 , 5  , E (7, 3) and F  7 , 7  are the mid-points of sides of ∆ABC, find the area of the
 2 2   2 2 

∆ABC.

66. If (a, b) is the mid-point of the line segment joining the points A (10, –6) and B (k, 4) and

a – 2b = 18, find the value of k and the distance AB.

Coordinate Geometry 159

67. The centre of a circle is (2a, a – 7). Find the values of a if the circle passes through the point
(11, –9) and has diameter 10 2 units.

68. Find the relation between x and y, if the points (x, y), (1, 1) and (–2, –2) are collinear.

69. Find the coordinates of the point R on the line segment joining the points P(–1, 3) and Q(2, 5)

such that PR = 3 PQ.
5

70. Find the third vertex of a triangle, if two of its vertices are at (–3, 1) and (0, –2) and the centroid

is at the origin.

71. The vertices of a triangle are (a, b – c), (b, c – a) and (c, a – b). Prove that its centroid lies on x-axis.

72. The base AB of two equilateral triangles ABC and ABC′ with side 2a lies along the x-axis, such that
the mid point of AB is at the origin. Find coordinates of vertices C and C′ of the triangles.

73. If the points (a, b), (– a′, – b′) and [(a – a′), (b – b′)] are collinear, show that ab′ = a′b.

74. The co-ordinates of A, B, C are (6, 3), (–3, 5) and (4, –2) respectively and P is a point (x, y). Show

that ar(∆PBC) = x + y − 2 .
ar(∆ABC) 7

75. Show that (1, –1) is the centre of the circle circumscribing the triangle whose vertices are (4, 3),

(–2, 3) and (6, –1).

76. If the points P (–3, 9), Q (a, b) and R (4, –5) are collinear and a + b = 1, find the values of a

and b. [CBSE Delhi 2014]

77. Find the value(s) of k for which the points (3k – 1, k – 2), (k, k – 7) and (k – 1, –k – 2) are collinear.

[CBSE (F) 2014]

78. Points P, Q, R and S divide the line segment joining the points A (1, 2) and B (6, 7) in 5 equal

parts. Find the coordinates of the points P, Q and R. [CBSE (F) 2014]

79. Find the coordinates of the point on the y-axis which is equidistant from the points A(5, 3) and

B(1, –5). [CBSE Delhi 2017 (C)]

QQ Long Answer Questions: [5 marks each]

80. A(6, 1), B(8, 2) and C(9, 4) are three vertices of a parallelogram ABCD. If E is the mid-point of
DC, find the area of ∆ADE.

81. The mid-points D, E, F of the sides of a triangle ABC are (3, 4), (8, 9) and (6, 7). Find the
coordinates of the vertices of the triangle.

82. The points A(x1, y1), B(x2, y2) and C(x3, y3) are 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 point 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 are the coordinates of the centroid of the triangle ABC?  [NCERT Exemplar]

83. The mid-point P of the line segment joining the points A (–10, 4) and B (–2, 0) lies on the line

segment joining the points C (–9, –4) and D (–4, y). Find the ratio in which P divides CD. Also find

the value of y. [CBSE (F) 2014]

84. If A(–4, 8), B(–3, –4), C(0, –5) and D(5, 6) are the vertices of a quadrilateral ABCD, find its area.

[CBSE Delhi 2015]

160 Xam idea Mathematics–X

85. Find the area of quadrilateral ABCD, the co-ordinates of whose vertices are A(1, 2), B(6, 2),

C(5, 3) and D(3, 4). [CBSE (F) 2016]

86. Find the area of a quadrilateral PQRS whose vertices are P(4, 3), Q(10, –1), R(15, 4) and

S(10, 23). [CBSE Delhi 2017 (C)]

Answers

1. (i) (b) (ii) (b) (iii) (a) (iv) (c) (v) (b)

2. (i) (2, 0) (ii) 4 2 units (iii) y-axis (iv) x1 + x2 + x3 , y1 + y2 + y3 (v) 32 sq units
3 3

3. x2 + y2 4. 37 5. A(3, – 10) 6. (2, 0) 7. x =  3 8. 2 a2 + b2
13. a = –12
9. a = 1 10. x = 17, x = 1 11. 2 : 1 12. IV 18. 1
3 17. t = 1
1
14. (0, –10) and (4, 0) 15. (0, 13) 16. k = – 3

19. 3abc 20. 3 : 2 21. 1 : 1; m = 0 22. x + y – 3 = 0 23. 3 : 5, c 17 , 0 m
8

24. 3x + 2y = 0 25. 24 sq. units2 6. a = 6, b = 3 27. Pc –4 , 8 m, Q c –2 , 16 m
3 3 3 3

28. R c –1, 9 m 29. x + 3y – 7 = 0 30. k = 5 31. 157 units 32. 5 units
2

33. ±4 34. 6 square units 35. 3 36. x = 6, y = –1

38. a = 1, b = 1; üüü 39. 72 sq. units 40. C(1, – 12); D(3, – 10) 41. x = – 3
42. (0, – 2) 43. k = – 8 1 sq. unit
44. 13 : 3, c 9 , 3 m 45. k = 2 46.
2 2

47. k = 3 48. 2 : 7; y = 6 49. p = – 1 50. k = – 8 52. (–3, 5), infinitely many

53. (9, 0), (5, 0); 2 points 54. Rectangle 55.  −1 , 0  , isosceles triangle 56. Equilateral
 2 
7 2
57.  3 , −2 58. k = 3 59. k = 2 62. k = 7 63. a=1, b=–3


64. 6 : 7;  −34 , 0  65. 11 sq. units.6 6. k = 22, 2 61 67. a = 5, 3 68. x = y
 13 

69.  4 , 21  70. (3, 1) 72. C(0, 3 a), C'(0, 3 a) 76. a = 2, b = – 1
 5 5 

77. k = 0 or 3 78. P(2, 3), Q(3, 4), R(4, 5) 79. 0, 1 80. 3 sq. unit
2 4

81. A (1, 2), B (5, 6), C (11, 12) 82. (i) § x2 x3 , y2 y3 · (ii) § x1 x2 x3 , y1 y2 y3 ·
¨© 2 2 ¸¹ ¨© 3 3 ¸¹

(iii) Q  x1 + x2 + x3 , y1 + y2 + y3  , R  x1 + x2 + x3 , y1 + y2 + y3  (iv) § x1 x2 x3 , y1 y2 y3 ·
 3 3   3 3  ¨© 3 3 ¹¸

83. 3 : 2, y = 6 84. 72 sq. units 85. 11 sq. units 86. 132 sq. units
2

Coordinate Geometry 161

SELF-ASSESSMENT TEST

Time allowed: 1 hour Max. marks: 40

Section A

1. Choose and write the correct option in the following questions. (4 × 1 = 4)

(i) The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is

(a) 5 units (b) 12 units (c) 11 units (d) 12 units

(ii) Point on y-axis has coordinate

(a) (–a, b) (b) (a, 0) (c) (0, b) (d) (– a, – b)

(iii) I(fa)t he52 p oints (1, x), (5(,b2) )–a25nd (9, 5) are co(lcli)n –e1a r then value of x is
(d) 1

(iv) The point which lies on the perpendicular bisector of the line segment joining the

points A (–2, –5) and B (2, 5) is

(a) (0, 0) (b) (0, 2) (c) (2, 0) (d) (–2, 0)

2. Fill in the blanks. (3 × 1 = 3)

(i) The point P (3, 2) is at distance 3 units from y-axis and _______________ units from x-axis.

(ii) In isosceles triangle, we prove that any _______________ sides are equal.

(iii) If rhombus is not a square, we prove that its all sides are equal but diagonals are
______________ equal.

3. Solve the following questions. (3 × 1 = 3)

(i) In which quadrant does the point, dividing line segment joining the points (7, – 6) and

(3, 4) in ratio 1 : 2, internally lies?

(ii) If (–4, 0); (4, 0) and (0, 3) are vertices of a triangle. Which type of triangle is it?

(iii) What is the equation of a line parallel to x-axis at a distance of 5 units below x-axis?

Section B

QQ Solve the following questions. (3 × 2 = 6)

4. In what ratio does the point P(–4, 6) divide the line segment joining the points. A(– 6, 10) and

B(3, –8) ? [CBSE Delhi 2017 (C)]

5. Prove that the points (2, –2), (–2, 1) and (5, 2) are the vertices of a right angled triangle. Also find
the area of this triangle. [CBSE (F) 2016]

6. Find the value of k for which the points (–5, 1), (1, k) and (4, –2) are collinear. [CBSE Delhi 2017 (C)]

QQ Solve the following questions. (3 × 3 = 9)

7. In given figure ABC is a triangle coordinates of whose vertex A are (0, –1). D and E respectively
are the mid-points of the sides AB and AC and their coordinates are (1, 0) and (0, 1) respectively.
If F is the mid-point of BC, find the areas of ∆ABC and ∆DEF. [CBSE Delhi 2016]

162 Xam idea Mathematics–X

Fig. 6.37

8. Points A(–1, y) and B(5, 7) lie on a circle with centre O(2, –3y). Find the values of y. Hence find the

radius of the circle. [CBSE Delhi 2014]

9. If A(–3, 5), B(–2, –7), C(1, –8) and D(6, 3) are the vertices of a quadrilateral ABCD, find its area.

[CBSE (AI) 2014]

QQ Solve the following questions. (3 × 5 = 15)

10. A(4, –6), B(3, –2) and C(5, 2) are the vertices of a ∆ABC and AD is its median. Prove that the

median AD divides ∆ABC into two triangles of equal areas. [CBSE (AI) 2014]

11. The base QR of an equilateral triangle PQR lies on x-axis. The coordinates of point Q are (– 4, 0)

and the origin is the mid-point of the base. Find the coordinates of the points P and R.

[CBSE (F) 2015]

12. Find the values of k so that the area of the triangle with vertices (1, –1), (–4, 2k), and (–k, –5) is

24 sq. units. [CBSE (AI) 2015]

OR

If the distance of P(x, y) from A(5, 1) and B(–1, 5) are equal, prove that 3x = 2y. [CBSE (AI) 2017]

Answers

1. (i) (b) (ii) (c) (iii) (c) (iv) (a)

2. (i) 2 (ii) two (iii) not

3. (i) IV quadrant (ii) isosceles triangle (iii) y = –5
7. 4 sq. units, 1 sq. unit
4. 2:7 5. 25 sq. units 6. k=–1
2

8. y = – 1, radius =5 units or y = 7, radius = 28.1 units

9. 72 sq. units 11. P (0, 4 3) or P (0, – 4 3) ; R (4, 0) 12. k = 3, –9
2
zzz

Coordinate Geometry 163

7 Triangles

BASIC CONCEPTS – A FLOW CHART

164 Xam idea Mathematics–X

MORE POINTS TO REMEMBMERORE POINTS TO REMEMBER

Bisector Theorems
 The internal bisector of an angle of a triangle divides the

opposite side internally in the ratio of the sides containing
the angle.

For example, If ABC is a triangle and AD is the bisector of ∠A,

then AB = BD Fig. 7.1
AC CD
BD AB
 In a triangle ABC, if D is a point on BC such that DC = AC , then AD is the bisector of ∠A.

 The external bisector of an angle of a triangle divides the
opposite side externally in the ratio of the sides containing
the angle.

For example, If ABC be a triangle and AD is the external

bisector of ∠A, then AB = BD Fig. 7.2
AC CD
BD AB
 In a triangle ABC, if D is a point on BC produced such that CD = AC , then AD bisects ∠A
externally, i.e., ∠XAD = ∠CAD.

Fig. 7.3 [1 mark]

Multiple Choice Questions [NCERT Exemplar]

Choose and write the correct option in the following questions.
1. In Fig. 7.4, ∠BAC = 90° and AD ⊥ BC. Then,

(a) BD . CD = BC2 Fig. 7.4
(c) BD . CD = AD2
(b) AB . AC = BC2
(d) AB . AC = AD2

2. The length of each side of a rhombus whose diagonals are of lengths 10 cm and 24 cm is

(a) 25 cm (b) 13 cm (c) 26 cm (d) 34 cm

Triangles 165

3. If DABC ~ DEDF and DABC is not similar to DDEF, then which of the following is not true?

[NCERT Exemplar]

(a) BC . EF = AC . FD (b) AB . EF = AC . DE

(c) BC . DE = AB . EF (d) BC . DE = AB . FD

4. If DABC ~ DDEF, ar (DABC) = 9 , BC = 21 cm, then EF is equal to
ar (DDEF) 25

(a) 9 cm (b) 6 cm (c) 35 cm (d) 25 cm

5. D and E are respectively the points on the sides AB and AC of a triangle ABC such that
AD = 3 cm, BD = 5 cm, BC = 12.8 cm and DE ||BC. Then length of DE (in cm) is

(a) 4.8 cm (b) 7.6 cm (c) 19.2 cm (d) 2.5 cm

6. If in two triangles ABC and PQR, AB = BC = CA , then
QR PR PQ

(a) DPQR ~ DCAB (b) DPQR ~ DABC

(c) DCBA ~ DPQR (d) DBCA ~ DPQR

7. If DPRQ ~ DXYZ, then

(a) PR = RQ (b) PQ = PR
XZ YZ XY XZ
PQ QR QR
(c) XZ = YZ (d) XZ = PR
XY

8. A square and a rhombus are always

(a) similar (b) congruent
(d) neither similar nor congruent
(c) similar but not congruent

9. In triangles ABC and DEF, ∠B = ∠E, ∠F = ∠C and AB = 3DE. Then, the two triangles are

[NCERT Exemplar]

(a) congruent but not similar (b) similar but not congruent

(c) neither congruent nor similar (d) congruent as well as similar

10. It is given that DABC ~ DDFE, ∠A = 30°, ∠C = 50°, AB = 5 cm, AC = 8 cm and DF = 7.5 cm.

Then, the following is true: [NCERT Exemplar]

(a) DE = 12 cm, ∠F = 50° (b) DE = 12 cm, ∠F = 100°

(c) EF = 12 cm, ∠D = 100° (d) EF = 12 cm, ∠D = 30°

11. Two circles are always (b) neither similar nor congruent
(d) none of these
(a) congruent
(c) similar but may not be congruent

12. In Fig. 7.5, two line segments AC and BD intersect each other at the point P such that

PA= 6 cm, PB = 3 cm, PC = 2.5 cm, PD = 5 cm, ∠APB = 50° and ∠CDP = 30°. Then, ∠PBA

is equal to [NCERT Exemplar]

A 6 cm D
5 cm 30°
(b) 30°
P

50°

B 3 cm 2.5 cm C

Fig. 7.5

(a) 50° (c) 60° (d) 100°

166 Xam idea Mathematics–X

13. If DABC ~ DQ RP, aarr ((DDPAQBCR)) = 9 , AB = 18 cm and BC = 15 cm, then PR is equal to
4 [NCERT Exemplar]

(a) 10 cm (b) 12 cm (c) 20 cm (d) 8 cm
3

14. In DLMN and DPQR, ∠L = ∠P, ∠N = ∠R and MN = 2QR. Then the two triangles are

(a) Congruent but not similar (b) Similar but not congruent

(c) neither congruent nor similar (d) Congrunt as well as similar

15. In DABC and DRPQ, AB = 4.5 cm, BC = 5 cm, CA = 6 2 cm, PR = 12 2 cm, PQ = 10 cm,

QR = 9 cm. If ∠A = 75° and ∠B = 55°, then ∠P is equal to

(a) 75° (b) 55° (c) 50° (d) 130°

16. If in triangles ABC and DEF, AB = AC , then they will be similar when
EF DE

(a) ∠A = ∠D (b) ∠A = ∠E (c) ∠B = ∠E (d) ∠C = ∠F

17. If DPQR ~ DXYZ and PQ = 5 , then ar (DXYZ) is equal to
XY 2 ar (DPQR)

(a) 4 (b) 2 (c) 25 (d) 5
25 5 4 2

18. It is given that ar(DABC) = 81 square units and ar(DDEF) = 64 square units. If DABC ~ DDEF,

then AB2
DE2
(a) AB = 81 (b) = 9
DE 64 8

(c) AB = 9 (d) AB = 81 units, DE = 64 units
DE 8

19. DABC and DBDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the
area of triangles ABC and BDE is

(a) 2 : 1 (b) 1 : 2 (c) 1 : 4 (d) 4 : 1

20. In triangle ABC, if AB = 6 3 cm, AC = 12 cm and BC = 6 cm, then ∠B is

(a) 120° (b) 60° (c) 90° (d) 45°

Answers

1. (c) 2. (b) 3. (c) 4. (c) 5. (a) 6. (a)
7. (c) 8. (d) 9. (b) 10. (b) 11. (c) 12. (d)
13. (a) 14. (b) 15. (c) 16. (b) 17. (a) 18. (c)
19. (d) 20. (c)

Fill in the Blanks [1 mark]

Complete the following statements with appropriate word(s) in the blank space(s).

1. Two figures having the same shape and size are said to be _______________ .
2. Two figures are said to be _______________ if they have same shape but not necessarily the same

size.
3. All similar figures need not be _______________ .
4. If two polygons are similar then the same ratio of the corresponding sides is referred to as the

_______________ .
5. Two triangles are said to be _______________ if corresponding angles of two triangles are equal.

Triangles 167

6. _______________ theorem states that if a line is drawn parallel to one side of a triangle to intersect
the other two sides in distinct points, the other two sides are divided in the same ratio.

7. _______________ theorem states that in a right triangle, the square of the hypotenuse is equal to
the sum of the squares of the other two sides.

8. If a line divides any two sides of a triangle in the same ratio, then the line is _______________ to
the third side.

9. The ratio of the areas of two similar triangles is equal to the square of the ratio of their
________________ .

10. All circles are _______________ .

11. If DDEF  DQRP then ∠D = _______________ and ∠E = _______________ .

12. In a DABC right angled at B, AC = 26 cm, AB = 24 cm then BC = _______________ cm

13. In DDEF  DPQR ar(DDEF) = 9 sq.units and ar(DPQR) : ar(DDEF) = 4 : 3, then ar(DPQR) is equal
to _______________ .

14. If DABC is an isosceles triangle right angled at C then AB2 = _______________ times AC2.

15. The _______________ bisector of an angle of a triangle divides the opposite side externally in the
ratio of the sides containing the angle.

Answers

1. congruent 2. similar 3. congruent 4. scale factor 5. equiangular

6. Basic proportionality/Thales 7. Pythagoras 8. parallel 9. corresponding sides

10. similar 11. ∠Q; ∠R 12. 10 13. 16 sq. units 14. two 15. external

Very Short Answer Questions [1 mark]

1. Two sides and the perimeter of one triangle are respectively three times the corresponding
sides and the perimeter of the other triangle. Are the two triangles similar? Why?

Sol. Since the perimeters and two sides are proportional

\ The third side is proportional to the corresponding third side.

i.e., The two triangles will be similar by SSS criterion.

2. A and B are respectively the points on the sides PQ and PR of a ∆PQR such that PQ = 12.5 cm,
PA = 5 cm, BR = 6 cm, and PB = 4 cm. Is AB || QR ? Give reason.

Sol. Yes, PA = 5 5 = 5 = 2
AQ 12.5 − 7.5 3

BPRB= 64= 2
Since 3
\
AP=QA BP=RB 2
3

AB || QR Fig. 7.6

3. If ∆ABC ~ ∆QRP, ar (TABC) = 8 , AB = 15 cm and BC = 12 cm, then find the length of PR.
ar (TPQR) 6

Sol. Area of ∆ABC = BC2 ⇒ 8 = ]12g2
Area of ∆QRP RP2 6 RP2

\ RP2 = 144× 6 = 864 =108 ⇒ RP = 3 12 cm
8 8

168 Xam idea Mathematics–X

4. If it is given that ∆ABC ~ ∆PQR with BC = 1 , then find ar ( ∆PQR ) . [NCERT Exemplar]
QR 3 ar ( ∆ABC )
BC 1
Sol. QR = 3 (Given)

aarr((∆∆PAQBCR)) = (QR)2 [ Ratio of area of similar triangles
(BC)2 is equal to the ratio of square of its
corresponding sides]
 QR 2 3 2 9
=  BC  =  1  = 1 =9:1


5. ∆DEF ~ ∆ABC, if DE : AB = 2:3 and ar(∆DEF) is equal to 44 square units. Find the area (∆ABC).

Sol. Since ∆DEF ~ ∆ABC [ Ratio of area of similar triangles

aarr((∆∆DABECF)) = (DE)2 is equal to the ratio of square of its
( AB)2 corresponding sides]

ar(∆4A4BC) =  2 2 ⇒ ar(∆ABC) = 44 ×9
 3  4

So, ar(∆ABC) = 99 cm2

6. The area of two similar triangles are 25 sq. cm and 121 sq. cm. Find the ratio of their

corresponding sides. [CBSE 2019 (30/5/1)]

Sol. Ratio of corresponding sides of two similar triangles.

= 25 = 5
121 11

Required ratio is 5 : 11.

Short Answer Questions-I [2 marks]

1. In triangles PQR and TSM, ∠P = 55°, ∠Q = 25°, ∠M = 100°, and ∠S = 25°. Is ∆QPR ~ ∆TSM?

Why?

Sol. Since, ∠R = 180° – (∠P + ∠Q)

= 180° – (55° + 25°) = 100° = ∠M

∠Q = ∠S = 25° (Given)

∆QPR ~ ∆STM (By AA similarity)

i.e., ∆QPR is not similar to ∆TSM.

2. If ABC and DEF are similar triangles such that ∠A = 47° and ∠E = 63°, then what is the

measures of ∠C.

Sol. Since ∆ABC ~ ∆DEF (Given)

\ ∠A = ∠D = 47° and ∠B = ∠E = 63°

\ ∠C = 180° – (∠A + ∠B) = 180° – (47° + 63°) = 70°

3. Let ∆ABC ~ ∆DEF and their areas be respectively 64 cm2 and 121 cm2. If EF = 15.4 cm, find

BC. [NCERT]
area of ∆ABC BC2
Sol. We have, area of ∆DEF = EF 2   (as ∆ABC ~ ∆DEF)

⇒ 16241 = BC2   ⇒ 64 = BC2
EF 2 121 (15.4)2

⇒ 1B5C.4 = 8
11

\ BC = 8 × 15.4 = 11.2 cm Fig. 7.7
11

Triangles 169

4. ABC is an isosceles triangle right-angled at C. Prove that AB2 = 2AC2. [NCERT]

Sol. ∆ABC is right-angled at C.

\ AB2 = AC2 + BC2 (By Pythagoras theorem)

⇒ AB2 = AC2 + AC2 ( AC = BC)

⇒ AB2 = 2AC2
5. Sides of triangle 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 [NCERT]

Sol. (i) Let a = 7 cm, b = 24 cm and c = 25 cm.

Here, largest side, c = 25 cm
We have, a2 + b2 = (7)2 + (24)2 = 49 + 576 = 625 = c2 ( c = 25)

So, the triangle is a right triangle.

Hence, c is the hypotenuse of right triangle.

(ii) Let a = 3 cm, b = 8 cm and c = 6 cm

Here, largest side, b = 8 cm
We have, a2 + c2 = (3)2 + (6)2 = 9 + 36 = 45 ≠ b2

So, the triangle is not a right triangle.

6. If triangle ABC is similar to triangle DEF such that
2AB = DE and BC = 8 cm. Then find the length of EF.

Sol.  ∆ABC ~ ∆DEF (Given)
\ ( DE = 2AB)
AB = BC
DE EF

\ AB = 8 Fig. 7.8
2 AB EF

1 = 8
2 EF
EF = 16 cm

7. If the ratio of the perimeter of two similar triangles is 4:25, then find the ratio of the areas of
the similar triangles.

Sol.  Ratio of perimeter of 2 ∆’s = 4 : 25

\ Ratio of corresponding sides of the two ∆’s = 4 : 25

Now, the ratio of area of 2 ∆’s = Ratio of square of its corresponding sides.

= (4)2 = 16
(25)2 625
8. In an isosceles ∆ABC, if AC = BC and AB2 = 2AC2, then find ∠C.

Sol. AB2 = 2AC2 (Given)

AB2 = AC2 + AC2

AB2 = AC2 + BC2 ( AC = BC)

Hence AB is the hypotenuse and ∆ABC is a right angle ∆.

So, ∠C = 90° Fig. 7.9

9. The length of the diagonals of a rhombus are 16 cm and 12 cm. Find the
length of side of the rhombus.

Sol.  The diagonals of rhombus bisect each other at 90°.

170 Xam idea Mathematics–X

\ In the right angle ∆BOC

BO = 8 cm and CO = 6 cm

\ By Pythagoras Theorem
BC2 = BO2 + CO2 = 64 + 36
BC2 = 100

BC = 10 cm Fig. 7.10

10. A man goes 24 m towards West and then 10 m towards North. How far is he from the starting
point?

Sol. By Pythagoras Theorem in DABC
AC2 = AB2 + BC2 = (24)2 + (10)2
AC2 = 676

AC = 26 m

\ The man is 26 m away from the starting point. Fig. 7.11

11. ∆ABC ~ ∆DEF such that AB = 9.1 cm and DE = 6.5 cm. If the perimeter of ∆DEF is 25 cm, what
is the perimeter of ∆ABC?

Sol. Since ∆ABC ~ ∆DEF

PPeerriimmeetteerr of ∆DEF = DE
of ∆ABC AB

25 = 6.5
9.1
Perimeter of ∆ABC

Perimeter of ∆ABC = 25 × 91 = 35 cm
65

Short Answer Questions-II [3 marks]

1. In Fig. 7.12, DE  BC. If AD = x, DB = x – 2, AE = x + 2 and EC = x – 1, find the value of x.

Sol. In ∆ABC, we have DE  BC

\ AD = AE [By Basic Proportionality Theorem]
DB EC
⇒ x(x – 1) = (x – 2) (x + 2)
⇒ x x = x+2 ⇒ x = 4
–2 x –1

⇒ x2 – x = x2 – 4 Fig. 7.12

2. E and F are points on the sides PQ and PR respectively of a ∆PQR. Show that EF ||QR

if PQ = 1.28 cm, PR= 2.56 cm, PE = 0.18 cm and PF = 0.36 cm. [NCERT]

Sol. We have, PQ = 1.28 cm, PR = 2.56 cm

PE = 0.18 cm, PF = 0.36 cm

Now, EQ = PQ – PE = 1.28 – 0.18 = 1.10 cm

and FR = PR – PF = 2.56 – 0.36 = 2.20 cm

Now, PE = 0.18 = 18 = 9
EQ 1.10 110 55

and, =FPRF 02=..3260 36 = 9 \ PE = PF Fig. 7.13
Therefore, 220 55 EQ FR

EF  QR [By the converse of Basic Proportionality Theorem]

Triangles 171

3. 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. [NCERT]

Sol. Let AB be a vertical pole of length 6 m and BC be its shadow and DE be tower and EF be its

shadow. Join AC and DF.

Now, in ∆ABC and ∆DEF, we have

∠B = ∠E = 90°

∠C = ∠F (Angle of elevation of the Sun)

\ ∆ABC ~ ∆DEF (By AA criterion of similarity)

Thus, AB = BC
DE EF
Fig. 7.14
6 4
⇒ h = 28 (Let DE = h)

⇒ 6 = 1 ⇒ h = 42
h 7

Hence, height of tower, DE = 42 m AM AN
AB AD
4. In Fig. 7.15, if LM || CB and LN  CD , prove that = . [NCERT]
Sol. Firstly, in ∆ABC, we have

LM ||CB (Given)

Therefore, by Basic Proportionality Theorem, we have

AAMB = AL ...(i)
AC

Again, in ∆ACD, we have

LN ||CD (Given)

\ By Basic Proportionality Theorem, we have Fig. 7.15
[NCERT]
AAND = AL ...(ii)
AC
AM AN
Now, from (i) and (ii), we have AB = AD .

5. In Fig. 7.16, DE ||OQ and DF ||OR , Show that EF ||QR .

Sol. In ∆POQ, we have

DE||OQ (Given)

\ By Basic Proportionality Theorem, we have

EPQE = PD ...(i)
DO

Similarly, in ∆POR, we have

DF ||OR (Given)

PD PF ...(ii) Fig. 7.16
DO FR
=

Now, from (i) and (ii), we have

EPQE = PF ⇒ EF ||QR
FR

[Applying the converse of Basic Proportionality Theorem in ∆PQR]

6. 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. [NCERT]

Sol. Given: ∆ABC in which D and E are the mid-points of sides AB and AC respectively.

172 Xam idea Mathematics–X

To prove: DE|| BC

Proof: Since, D and E are the mid-points of AB and AC respectively

\ AD = DB and AE = EC
AD
⇒ DB = 1 and AE = 1
EC
⇒
AD = AE Fig. 7.17
DB EC

Therefore, DE|| BC (By the converse of Basic Proportionality Theorem)

7. In ∆ABC, ∠B = 90° and D is the mid point of BC. Prove that AC2 = AD2 + 3CD2.

[CBSE 2019 (30/5/1)]

Sol. We have ∆ABC is a right triangle right angled at B.

Also, D is the mid point of BC. A

\ BD = CD = BC
2
In right ∆ABC, we have

AC2 = AB2 + BC2

⇒ AC2 = AB2 + (2CD)2

⇒ AC2 = AB2 + 4CD2 … (i) BD C
Fig. 7.18
In ∆ABD, we have

AD2 = AB2 + BD2

⇒ AB2 = AD2 – BD2 = AD2 – CD2

Putting the value in (i), we have

AC2 = AD2 – CD2 + 4CD2 = AD2 + 3CD2

⇒ AC2 = AD2 + 3CD2 Hence proved.

8. In Fig. 7.19, OA=CO OB=OD 1 and AB = 5cm. Find the value of DC.
2

Sol. In ∆AOB and ∆COD, we have

∠AOB = ∠COD (Vertically opposite angles)

and AO = BO (Given)
OC OD
Fig. 7.19
So, by SAS criterion of similarity, we have

∆AOB ~ ∆COD

⇒ OA=CO OB=DO AB ⇒ 1 = 5 [ AB = 5 cm]
⇒ DC 2 DC

DC = 10 cm.

9. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show

that ∆ABE ~ ∆CFB. [NCERT]

Fig. 7.20

Sol. In ∆ABE and ∆CFB, we have

Triangles 173

∠AEB = ∠CBF (Alternate angles)

∠A = ∠C (Opposite angles of a parallelogram)

\ ∆ABE ~ ∆CFB (By AA criterion of similarity)

10. S and T are points on sides PR and QR of ∆PQR such that ∠P = ∠RTS.

Show that ∆RPQ ~ ∆RTS. [NCERT]

Sol. In ∆RPQ and ∆RTS, we have

∠RPQ = ∠RTS (Given)

∠PRQ = ∠TRS = ∠R (Common)

\ ∆RPQ ~ ∆RTS (By AA criterion of similarity) Fig. 7.21

11. In Fig. 7.22, ABC and AMP are two right triangles right-angled at B and M respectively. Prove

that: CA BC
PA MP
(i) ∆ABC ~ ∆AMP (ii) =   [NCERT]

Sol. (i) In ∆ABC and ∆AMP, we have

∠ABC = ∠AMP = 90° (Given)

And, ∠BAC = ∠MAP (Common angle)

\ ∆ABC ~ ∆AMP (By AA criterion of similarity) Fig. 7.22
(ii) As ∆ABC ~ ∆AMP (Proved above)

\ CA = BC (Sides of similar triangles are proportional)
PA MP
12. D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that
CA2 = CB.CD.
[NCERT]

Sol. In ∆ABC and ∆DAC, we have

∠BAC = ∠ADC (Given)

and ∠C = ∠C (Common)

\ ∆ABC ~ ∆DAC (By AA criterion of similarity)
⇒
DA=BA BA=CC AC
DC

⇒ CB = CA ⇒ CA2 = CB × CD Fig. 7.23
CA CD [NCERT]

13. ABC is an equilateral triangle of side 2a. Find each of its altitudes. Fig. 7.24

Sol. Let ABC be an equilateral triangle of side 2a units.

We draw AD ⊥ BC. Then D is the mid-point of BC.

⇒ BD = BC = 2a =a
2 2
Now, ABD is a right triangle right-angled at D.

\ AB2 = AD2 + BD2 [By Pythagoras Theorem]

⇒ (2a)2 = AD2 + a2
⇒ AD2 = 4a2 – a2 = 3a2 ⇒ AD = 3a

Hence, each altitude = 3a unit.

14. 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 1 hours? [NCERT]
2

174 Xam idea Mathematics–X

Sol. Let the first aeroplane starts from O and goes upto A towards north where

OA = 1000 × 3  km = 1500 km (Distance = Speed × Time)
2 

Again let second aeroplane starts from O at the same time

and goes upto B towards west where
3
OB = 1200 × 2 = 1800 km

Now, we have to find AB.

In right angled ∆ABO, we have

AB2 = OA2 + OB2 [By using Pythagoras Theorem] Fig. 7.25

⇒ AB2 = (1500)2 + (1800)2
⇒ AB2 = 2250000 + 3240000 ⇒ AB2 = 5490000

\ AB = 100 549 = 100 × 23.4307 = 2343.07 km.

15. In the given Fig. 7.26, ∆ABC and ∆DBC are on the same base BC. If AD intersects BC at O. Prove

that ar(∆ABC) = AO .
ar(∆DBC) DO

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

To Prove: ar(∆ABC) = AO
ar(∆DBC) DO

Construction: Draw AL ⊥ BC and DM ⊥ BC

Proof: In ∆ALO and ∆DMO, we have

∠ALO = ∠DMO = 90° and

∠AOL = ∠DOM (Vertically opposite angles)

\ ∆ALO ~ ∆DMO (By AA similarity)

⇒ AL = AO ....(i) Fig. 7.26
DM DO

ar(∆ABC) 1 BC × AL AL AO
ar(∆DBC) 2 DM DO
\ = 1 = = (Using (i))
2
BC × DM

Hence, ar(∆ABC) = AO
ar(∆DBC) DO
1 1 1
16. In Fig. 7.27, AB|| PQ || CD , AB = x units, CD = y units and PQ = z units. Prove that x + y = z .

Sol. In ∆ADB and ∆PDQ,

Since AB||PQ
∠ABQ = ∠PQD (Corresponding angles)

∠ADB = ∠PDQ (Common)

∆ADB ~ ∆PDQ (By AA similarity)

\ DQ = PQ ⇒ DQ = z ...(i)
DB AB DB x

Similarly, ∆PBQ ~ ∆CBD Fig. 7.27

and BQ = z ...(ii)
DB y

Adding (i) and (ii), we get

Triangles 175

xz + z = DQ + BQ = BD
y DB BD

z + z =1 ⇒ 1 + 1 = 1
x y x y z

17. In Fig. 7.28, if ∆ABC ~ ∆DEF and their sides are of lengths (in cm) as marked along them, then
find the lengths of the sides of each triangle.

Sol. ∆ABC ~ ∆DEF (Given)

therefore, DA=EB EB=CF CA
FD

So, 2x −1 = 2x + 2 = 3x
18 3x + 9 6x
Fig. 7.28
Now, taking 2x −1 3x , we have
18 = 6x

2x −1 = 1
18 2

⇒ 4x – 2 = 18 ⇒ x=5

\ AB = 2 × 5 – 1 = 9, BC = 2 × 5 + 2 = 12

CA = 3 × 5 = 15, DE = 18, EF = 3 × 5 + 9 = 24 and FD = 6 × 5 = 30

Hence, AB = 9 cm, BC = 12 cm, CA = 15 cm

DE = 18 cm, EF = 24 cm, FD = 30 cm

18. In ∆ABC, it is given that AB = BD . If ∠B = 70° and ∠C = 50° then find ∠BAD.
Sol. In ∆ABC AC DC

 ∠A + ∠B + ∠C = 180° (Angle sum property)

∠A + 70° + 50° = 180°

⇒ ∠A = 180° – 120° ⇒ ∠A = 60°

\ AB = BD (Given)
AC DC
Fig. 7.29
∠1 = ∠2 ...(i)

[Because if a line through one vertex of a triangle divides the opposite sides in the ratio of other
two sides, then the line bisects the angle at the vertex.]

But ∠1 + ∠2 = 60° ...(ii)

From (i) and (ii) we get, 60°
2
2∠1 = 60° ⇒ ∠1 = = 30°

Hence, ∠BAD = 30°

19. If the diagonals of a quadrilateral divide each other proportionally, prove that it is a trapezium.

Sol. AO = CO (Given)
BO DO

⇒ AO = BO ...(i)
CO DO

In ∆ABD, EO||AB (Construction)

\ AE = BO (By BPT) ...(ii) Fig. 7.30
ED DO
176 Xam idea Mathematics–X

From equations (i) and (ii)

AE = AO ⇒ EO||DC (Converse of BPT)
ED CO (Construction)

But EO||AB

\ AB||DC

⇒ In quad ABCD since AB||DC  ⇒  ABCD is a trapezium.

20. In the given Fig. 7.31, PS = PT and ∠PST = ∠PRQ. Prove that PQR is an isosceles triangle.
SQ TR

Sol. Given: PS = PT and ∠PST = ∠PRQ
SQ TR

To Prove: PQR is isosceles triangle.

Proof: PS = PT
SQ TR
Fig. 7.31
By converse of BPT, we get

ST||QR

\ ∠PST = ∠PQR (Corresponding angles) ...(i)

But, ∠PST = ∠PRQ (Given) ...(ii)

From equation (i) and (ii)

∠PQR = ∠PRQ ⇒ PR = PQ

So, ∆PQR is an isosceles triangle.
21. The diagonals of a trapezium ABCD in which AB||DC, intersect at O. If AB = 2CD, then find

the ratio of areas of triangles AOB and COD.

Sol. In ∆AOB and ∆COD

∠COD = ∠AOB (Vertically opposite angles)

∠CAB = ∠DCA (Alternate angles)

\ ∆AOB ~ ∆COD (By AA similarity)

By area theorem Fig. 7.32

ar (∆AOB) = AB2 ⇒ ar (∆AOB) = (2CD)2 = 4
ar (∆COD) DC2 ar (∆COD) 1
CD2

Hence, ar(∆AOB) : ar(∆COD) = 4 : 1.
22. In the given Fig. 7.33, find the value of x in terms of a, b and c.

Sol. In ∆LMK and ∆PNK
We have, ∠M = ∠N = 50° and ∠K = ∠K (Common)
∴ ∆LMK ~ ∆PNK (By AA similarity)

⇒ LPMN = KM
KN

⇒ a = b + c ⇒ x = ac Fig. 7.33
x c b+c

23. In the given Fig. 7.34, CD|| LA and DE||AC. Find the length of CL, if BE = 4 cm and EC = 2 cm.

Sol. In ∆ABC, DE||AC, (Given)

⇒ BD = BE (By BPT) …(i)
DA EC

Triangles 177

In ∆ABL DC||AL

⇒ BD = BC (By BPT) ...(ii)
DA CL

From (i) and (ii) we get

EBEC = BC ⇒ 4 = 6    ⇒ CL = 3 cm Fig. 7.34
CL 2 CL

24. In the given Fig. 7.35, AB = AC. E is a point on CB produced. If AD is perpendicular to BC and

EF perpendicular to AC, prove that ∆ ABD is similar to ∆ECF. [CBSE 2019 (30/5/1)]

Sol. AB = AC (Given)

⇒ ∠ABC = ∠ACB (Equal sides have equal opposite angles)

Now, in ∆ABD and ∆ECF

∠ABD = ∠ECF (Proved above)

∠ADB = ∠EFC (Each 90°)

So, ∆ABD ~ ∆ECF (By AA similarity) Fig. 7.35

Long Answer Questions [5 marks]

1. 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. [NCERT]

Sol. Given: A ∆ABC in which D is the mid-point of AB and DE is drawn parallel to BC, which meets
AC at E.

To prove: AE = EC

Proof: In ∆ABC, DE||BC

\ By Basic Proportionality Theorem, we have

AD = AE ...(i)
DB EC

Now, since D is the mid-point of AB

⇒ AD = BD ...(ii)

From (i) and (ii), we have Fig. 7.36

BD = AE ⇒ 1= AE
BD EC EC

⇒ AE = EC

Hence, E is the mid-point of AC.

2. ABCD is a trapezium in which AB||DC and its diagonals intersect each other at the point O.

Show that AO = CO . [NCERT]
BO DO
Sol. Given: ABCD is a trapezium, in which AB||DC and its diagonals intersect each other at the point O.

To prove: AO = CO
BO DO
Construction: Through O, draw OE||AB i.e., OE||DC.

Proof: In ∆ADC, we have OE||DC (Construction)

\ By Basic Proportionality Theorem, we have

AE = AO ...(i) Fig. 7.37
ED CO

178 Xam idea Mathematics–X

Now, in ∆ABD, we have OE||AB ( Construction)

\ By Basic Proportionality Theorem, we have

ED = DO ⇒ AE = BO ...(ii)
AE BO ED DO

From (i) and (ii), we have

AO = BO ⇒ AO = CO
CO DO BO DO

3. If AD and PM are medians of triangles ABC and PQR respectively, where ∆ABC ~ ∆PQR, prove

that AB = AD . [NCERT]
PQ PM

Sol. In ∆ABD and ∆PQM we have

∠B = ∠Q ( ∆ABC ~ ∆PQR) …(i)

PAQB = BC ( ∆ABC ~ ∆PQR)
QR

AB 1 BC Fig. 7.38
PQ 2 QR
⇒ = 1
2

(Since AD and PM are the medians of ∆ABC and ∆PQR respectively)
AB BD
⇒ PQ = QM …(ii)

From (i) and (ii), it is proved that ∆ABD ~ ∆PQM (By SAS criterion of similarity)

⇒ AB = BD = AD ⇒ AB = AD
PQ QM PM PQ PM

4. In Fig. 7.39, ABCD is a trapezium with AB||DC. If ∆AED is similar to ∆BEC, prove that AD = BC.

Sol. In ∆EDC and ∆EBA we have
∠1 = ∠2 (Alternate angles)
∠3 = ∠4 (Alternate angles)
and ∠CED = ∠AEB (Vertically opposite angles)

\ ∆EDC ~ ∆EBA (By AA criterion of similarity)

⇒ ED = EC ⇒ ED = EB …(i)
EB EA EC EA

It is given that ∆AED ~ ∆BEC Fig. 7.39

\ EE=DC EE=BA AD ...(ii)
BC ⇒ EB = EA

From (i) and (ii), we get

EEBA = EA ⇒ (EB)2 = (EA)2
EB

Substituting EB = EA in (ii), we get

EEAA = AD ⇒ AD = 1 ⇒ AD = BC
BC BC

5. Prove that the area of an equilateral triangle described on a side of a right-angled isosceles

triangle is half the area of the equilateral triangle described on its hypotenuse.

Sol. Given: A ∆ABC in which ∠ABC = 90° and AB = BC.

∆ABD and ∆CAE are equilateral triangles.

Triangles 179

To Prove: ar(∆ABD) = 1 × ar(∆CAE)
2

Proof: Let AB = BC = x units.

\ hyp. CA = x2 + x2 = x 2 units.

Each of the ∆ABD and ∆CAE being equilateral, has each
angle equal to 60°.

\ ∆ABD ~ ∆CAE (By AA similarity)

But, the ratio of the areas of two similar triangles is equal Fig. 7.40

to the ratio of the squares of their corresponding sides.
ar^DABDh AB2 x2 x2
\ ar^DCAEh = CA2 = ^x 2 h2 = 2x2 = 1
2

Hence, ar(∆ABD) = 1 × ar(∆CAE)
2

6. If the areas of two similar triangles are equal, prove that they are congruent. [NCERT]

Sol. Given: Two triangles ABC and DEF, such that

∆ABC ~ ∆DEF and area (∆ABC) = area (∆DEF)

To prove: ∆ABC  ∆DEF
Proof: ∆ABC ~ ∆DEF

⇒ ∠A = ∠D, ∠B = ∠E, ∠C = ∠F

and DA=EB BE=CF AC
DF
Now, ar(∆ABC) = ar(∆DEF) (Given)
Fig. 7.41

\ ar (∆ABC) = 1 …(i)
ar (∆DEF)

and AB2 = BC2 = AC2 = ar (∆ABC) ( ∆ABC ~ ∆DEF) …(ii)
DE2 EF 2 DF 2 ar (∆DEF)

From (i) and (ii), we have

AB2 = BC2 = AC2 =1 ⇒ AB = BC = AC =1
DE2 EF 2 DF 2 DE EF DF

⇒ AB = DE, BC = EF, AC = DF

Hence, ∆ABC  ∆DEF (By SSS criterion of congruency)

7. Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of

their corresponding medians. [NCERT]

Sol. Let ∆ABC and ∆PQR be two similar triangles. AD and PM are the medians of ∆ABC and ∆PQR

respectively.
ar (∆ABC)
To prove: ar (∆PQR) AD2

Proof: Since ∆ABC ~ ∆= PPQMR2 aarr ((∆∆PAQBCR)) = AB2 …(i)
PQ 2

In ∆ABD and ∆PQM 1
2
 1 BC 
 AB BC 2 QR 
PAQB = BD  PQ = QR = 
QM 

and ∠B = ∠Q ( ∆ABC ~ ∆PQR)

Fig. 7.42

180 Xam idea Mathematics–X

Hence, ∆ABD ~ ∆PQM (By SAS similarity criterion)

AB = AD …(ii)
PQ PM

From (i) and (ii), we have

ar (∆ABC) = AD2
ar (∆PQR) PM2

8. In Fig. 7.43, 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 [NCERT]

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

Sol. Join OA, OB and OC.

(i) In right ∆’s OFA, ODB and OEC, we have ...(i)
OA2 = AF2 + OF2 ...(ii)
OB2 = BD2 + OD2 ...(iii)
and OC2 = CE2 + OE2

Adding (i), (ii) and (iii), we have Fig. 7.43
OA2 + OB2 + OC2 = AF2 + BD2 + CE2 + OF2 + OD2 + OE2
⇒ OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2
(ii) We have, OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2
⇒ (OA2 – OE2) + (OB2 – OF2) + (OC2 – OD2 ) = AF2 + BD2 + CE2
⇒ AE2 + CD2 + BF2 = AF2 + BD2 + CE2 

  (Using Pythagoras Theorem in ∆AOE, ∆BOF and ∆COD)

9. The perpendicular from A on side BC of a ∆ABC intersects BC at D such that DB = 3CD (see
Fig. 7.44). Prove that 2AB2 = 2AC2 + BC2.
[NCERT]

Sol. We have, DB = 3CD

Now, BC = BD + CD

⇒ BC = 3 CD + CD = 4CD (Given DB = 3CD)
1
\ CD = 4 BC

and DB = 3CD = 3 BC
4

Now, in right-angled triangle ABD using Pythagoras Theorem we have Fig. 7.44

AB2 = AD2 + DB2 …(i)

Again, in right-angled triangle ∆ADC, we have
AC2 = AD2 + CD2 …(ii)


Subtracting (ii) from (i), we have

AB2 – AC2 = DB2 – CD2

⇒ AB2 – AC2 = 3 BC 2 − 1 BC 2 = 9 − 1 BC2 = 8 BC2
 4   4   16 16  16
⇒
\ AB2 – AC2 = 1 BC2
2
2AB2 – 2AC2 = BC2 ⇒   2AB2 = 2AC2 + BC2

10. In an equilateral triangle, prove that three times the square of one side is equal to four times

the square of one of its altitudes. [NCERT]

Sol. Let ABC be an equilateral triangle and let AD ⊥ BC.

Triangles 181

\ BD = DC

Now, in right-angled triangle ADB, we have

AB2 = AD2 + BD2 (Using Pythagoras Theorem)

⇒ AB2 = AD2 + 1 BC 2 ⇒ AB2 = AD2 + 1 BC2
 2  4

⇒ AB2 = AD2 + AB2 (AB = BC) Fig. 7.45
⇒ 4
AB2 3AB2
AB2 – 4 = AD2 ⇒ 4 = AD2

⇒ 3AB2 = 4AD2

11. Prove that, if a line is drawn parallel to one side of a triangle to intersect the other two sides in

distinct points, the other two sides are divided in the same ratio.

Using the above result, do the following:

In Fig. 7.46 DE||BC and BD = CE. Prove that ∆ABC is an isosceles triangle.

Sol. Given: A triangle ABC in which a line parallel to side BC intersects other two

sides AB and AC at D and E respectively.

To Prove: AD = AE Fig. 7.46
DB EC

Construction: Join BE and CD and then draw DM ⊥ AC and EN ⊥ AB.

Proof: Area of ∆ADE =  12 base × height 


So, ar(∆ADE) = 12 (AD × EN)

and ar(∆BDE) = 12 (DB × EN)

Similarly, ar(∆ADE) = 12 (AE × DM) Fig. 7.47

and ar(∆DEC) = 12 (EC × DM)

Therefore, ar^DADEh = 1 AD × EN = AD …(i)
ar^DBDEh 2 DB × EN DB
1
2

ar^DADEh 1 AE × DM AE
ar^DDECh EC × DM EC
and = 2 = …(ii)
1

2

Now, ∆BDE and ∆DEC are on the same base DE and between the same parallel lines BC and DE.

So, ar(∆BDE) = ar(∆DEC) ...(iii)
AD AE
Therefore, from (i), (ii) and (iii) we have, DB = EC

Second part :

As DE||BC

\ AD = AE ⇒ AD +1 = AE +1
DB EC DB EC

182 Xam idea Mathematics–X

⇒ AD + DB = AE + EC ⇒ AB = AC
DB EC DB EC

⇒ AB = AC (As DB = EC)

\ ∆ABC is an isosceles triangle.
12. In Fig. 7.48, ABD is a triangle right-angled at A and AC⊥BD Show that

(i) AB2 = BC . BD (ii) AD2 = BD . CD    (iii) AC2 = BC . DC [NCERT]

Sol. Given: ABD is a triangle right-angled at A and AC⊥BD.
(i) AB2 = BC . BD
To prove: (ii) AD2 = BD . CD

(iii) AC2 = BC . DC



Proof: (i) In ∆ACB and ∆DAB, we have

∠ACB = ∠DAB = 90°

∠ABC = ∠DBA = ∠B (Common) Fig. 7.48

\ ∆ACB ~ ∆DAB (By AA criterion of similarity)

\ BC = AB   ⇒  AB2 = BC . BD
AB DB

(ii) In ∆ACD and ∆BAD, we have

∠ACD = ∠BAD = 90°

∠CDA = ∠BDA = ∠D (Common)

\ ∆ACD ~ ∆BAD (By AA criterion of similarity)

\ AD = CD   ⇒  AD2 = BD . CD
BD AD

(iii) We have ∆ACB ~ ∆DAB

⇒ ∆BCA ~ ∆BAD …(i)

and ∆ACD ~ ∆BAD …(ii)

From (i) and (ii), we have

∆BCA ~ ∆ACD

⇒ BC = AC ⇒  AC2 = BC . DC
AC DC

13. Prove that ratio of areas of two similar triangles is equal to the ratio of the squares of their

corresponding sides.

Using the above result do the following:

Diagonals of a trapezium ABCD with AB||DC intersect each other at the point O. If AB = 2CD,
find the ratio of the areas of triangles AOB and COD.

Sol. Given: Two triangles ABC and PQR such that ∆ABC ~ ∆PQR

To Prove: ar (∆ABC) =  AB 2 =  BC 2 =  CA 2
ar (∆PQR)  PQ     RP 
   QR 

Construction: Draw AM ⊥ BC and PN ⊥ QR.

Proof: ar(∆ABC) = 12 × BC × AM

and ar(∆PQR) = 12 × QR × PN Fig. 7.49

So, ar (∆ABC) = 1 × BC × AM = BC × AM ...(i)
ar (∆PQR) 2 × QR × PN QR × PN
1
2 Triangles 183

Now, in ∆ABM and ∆PQN,

∠B = ∠Q (As ∆ABC ~ ∆PQR)

and ∠M = ∠N (Each 90°)

So, ∆ABM ~ ∆PQN (By AA similarity criterion)

Therefore, AM = AB …(ii)
PN PQ

Also, ∆ABC ~ ∆PQR (Given)

So, =PAQB =QBRC CA …(iii)
RP

Therefore, ar (∆ABC) = AB × AM (From (i) and (iii))
ar (∆PQR) PQ PN

= AB × AB =  AB 2 (From (ii))
PQ PQ  
 PQ 

Now using (iii), we get ar (∆ABC) =  AB 2 =  BC 2 =  CA 2
ar (∆PQR)  PQ     RP 
   QR 

Second part:

In ∆AOB and ∆COD we have

∠AOB = ∠COD (Vertically opposite angles)

and ∠OAB = ∠OCD (Alternate angles)

\ ∆AOB ~ ∆COD (By AA criterion of similarity]

⇒ area of ∆AOB = AB2 Fig. 7.50
area of ∆COD DC2

⇒ area of ∆AOB = (2DC)2 = 4
area of ∆COD DC2 1

Hence, the ratio of areas of ∆AOB and ∆COD = 4 : 1.

14. Prove that, in a right triangle, the square of the hypotenuse is equal to the sum of squares of the
other two sides.

Using the above, do the following:
Prove that, in a ∆ABC if AD is perpendicular to BC, then AB2 + CD2 = AC2 + BD2.

Sol. Given: A right triangle ABC right-angled at B.
To Prove: AC2 = AB2 + BC2

Construction: Draw BD ⊥ AC

Proof: In ∆ADB and ∆ABC

∠A = ∠A (Common)

∠ADB = ∠ABC (Both 90°) Fig. 7.51

\ ∆ADB ~ ∆ABC (By AA similarity criterion)

So, AD = AB (Sides are proportional)
AB AC

or AD . AC = AB2 …(i)

184 Xam idea Mathematics–X

In ∆BDC and ∆ABC

∠C = ∠C (Common)

∠BDC = ∠ABC (Each 90°)

∆BDC ~ ∆ABC (By AA similarity)

So, CD = BC or, CD . AC = BC2 …(ii)
BC AC

Adding (i) and (ii), we get
AD . AC + CD . AC = AB2 + BC2

or, AC (AD + CD) = AB2 + BC2 Fig. 7.52

or, AC . AC = AB2 + BC2 Fig. 7.53
Fig. 7.54
or, AC2 = AB2 + BC2 (From fig. 7.51) Fig. 7.55

Second part : Triangles 185

In Fig. 7.52, As AD ⊥ BC

Therefore, ∠ADB = ∠ADC = 90°

By Pythagoras Theorem, we have
AB2 = AD2 + BD2 …(i)
AC2 = AD2 + DC2 …(ii)

Subtracting (ii) from (i), we get

AB2 – AC2 = AD2 + BD2 – (AD2 + DC2)

AB2 – AC2 = BD2 – DC2

⇒ AB2 + DC2 = BD2 + AC2

15. In a triangle, if the square on one side is equal to the sum of the

squares on the other two sides, prove that the angle opposite to

the first side is a right angle. [CBSE 2019 (30/5/1)]

Use the above theorem to find the measure of ∠PKR in Fig. 7.53.

Sol. Given: A triangle ABC in which AC2 = AB2 + BC2.

To Prove: ∠B = 90°.

Construction: We construct a ∆PQR right-angled at Q such that
PQ = AB and QR = BC

Proof: In ∆PQR, we have,

PR2 = PQ2 + QR2 (Pythagoras Theorem)

or, PR2 = AB2 + BC2 (By construction) ...(i)

But AC2 = AB2 + BC2 (Given) …(ii)

So, AC2 = PR2 (From (i) and (ii))

\ AC = PR ...(iii)

Now, in ∆ABC and ∆PQR,

AB = PQ (By construction)

BC = QR (By construction)

AC = PR (Proved in (iii))

So, ∆ABC  ∆PQR (By SSS congruency)
Therefore,

∠B = ∠Q (CPCT)
But
∠Q = 90° (By construction)

So, ∠B = 90°

Second part:

In ∆PQR, (Fig. 7.55)

By Pythagoras Theorem, we have

PR2 = (26)2 – (24)2 ⇒ PR2 = 676 – 576

PR = 100 = 10 cm

Now, in ∆PKR we have

PK2 + KR2 = (8)2 + (6)2 = 64 + 36 = 100 = PR2

Hence, ∠PKR = 90° (By Converse of Pythagoras Theorem)

16. ABC is a triangle in which AB = AC and D is a point on AC such that BC2 = AC × CD. Prove that

BD = BC.

Sol. Given: ∆ABC in which AB = AC and D is a point on the side AC such that BC2 = AC × CD

To prove: BD = BC

Construction: Join BD

Proof: We have, BC AC
CD BC
BC2 = AC × CD ⇒ = …(i)

Thus, in ∆ABC and ∆BDC, we have

AC = BC (From (i)) Fig. 7.56
BC CD

and ∠C = ∠C (Common)

\ ∆ABC ~ ∆BDC (By SAS similarity criterion)

⇒ AB = BC …(ii)
BD CD

From (i) and (ii), we get

AC = AB \ BD = BC ( AB = AC)
BC BD

17. Prove that the area of an equilateral triangle described on one side of a square is equal to half

the area of the equilateral triangle described on one of its diagonals. [NCERT]

Sol. Let ABCD be a square and ∆BCE and ∆ACF have been drawn on side BC and the diagonal AC

respectively. 1
2
To prove: area (∆BCE) = area (∆ACF)

Proof: Since ∆BCE and ∆ACF are equilateral triangles

\ ∆BCE ~ ∆ACF (By AA similarity criterion)

⇒ area (∆BCE) = BC2 Fig. 7.57
area (∆ACF ) AC2

⇒ ( )area(∆BCE) = BC2 (  Diagonal = 2 side, AC = 2 BC)
2BC 2
⇒ area (∆ACF )

area (∆BCE) = 1 ⇒ area (∆BCE) = 1 area (∆ACF)
area (∆ACF ) 2 2

18. 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.
[NCERT]

186 Xam idea Mathematics–X

OR

If P and Q are the points on side CA and CB respectively of ∆ ABC, right angled at C, prove that
(AQ2 + BP2 ) = (AB2 + PQ2).
[CBSE 2019]

Sol. In right angled ∆ACE and ∆DCB, we have

AE2 = AC2 + CE2 (Pythagoras Theorem) …(i)

and BD2 = DC2 + BC2 …(ii)

Adding (i) and (ii), we have

⇒ AE2 + BD2 = AC2 + CE2 + DC2 + BC2

⇒ AE2 + BD2 = (AC2 + BC2) + (DC2 + CE2)

⇒ AE2 + BD2 = AB2 + DE2 Fig. 7.58

( AC2 + BC2 = AB2 and DC2 + EC2 = DE2 in right-angled triangle ABC and CDE respectively.)

OR

[D replace by P and E replace by Q only and solution is similar]

HOTS [Higher Order Thinking Skills]

1. In Fig. 7.59, ∆FEC  ∆GDB and ∠1 = ∠ 2. Prove that ∆ADE ~ ∆ABC.

Sol. Since, ∆FEC  ∆GDB

⇒ EC = BD ...(i)

It is given that

∠1 = ∠2 Sides opposite to equal 

⇒ AE = AD  angles are equal  ...(ii)
 
Dividing (ii) by (i), we have
AE AD
EC = BD Fig. 7.59

⇒ DE||BC (By the converse of basic proportionality theorem)
⇒ ∠1 = ∠3 and ∠2 = ∠4 (Corresponding angles]

Thus, in ∆’ s ADE and ABC, we have

∠A = ∠A (Common)
∠2 = ∠4 (Proved above)

⇒ ∆ADE ~ ∆ABC (By AA similarity)

2. Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ
and PR and median PM of another triangle PQR. Show that ∆ABC ~ ∆PQR.

Sol. Given: In ∆ABC and ∆PQR, AD and PM are their medians respectively

such that =PAQB P=AMD AC ... (i)
PR

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, RN.
Proof: Q uadrilateral ABEC and PQNR are ||gm because their diagonals bisect each other at D and

M respectively.

⇒ BE = AC and QN = PR

⇒ BE = AC ⇒ BE = AB (From (i))
QN PR QN PQ

Triangles 187

i.e., AB = BE …(ii)
PQ QN

From (i) =PAQB P=AMD 2 AD = AE
2PM PN

⇒ AB = AE …(iii)
PQ PN

From (ii) and (iii) Fig. 7.60

P=AQB QB=NE AE
PN

⇒ ∆ABE ~ ∆PQN (By SSS similarity criterion)
…(iv)
⇒ ∠1 = ∠2
⇒ ∠3 = ∠4 …(v)
Similarly, we can prove
⇒ ∠A = ∠P
∆ACE ~ ∆PRN (Given)

Adding (iv) and (v), we get

∠1 + ∠3 = ∠2 + ∠4

and AB = AC
PQ PR

\ ∆ABC ~ ∆PQR (By SAS criterion of similarity)

3. In Fig. 7.61, P is the mid-point of BC and Q is the mid-point of AP. If BQ when produced meets

AC at R, prove that RA = 1 CA.
3
Sol. Given: In ∆ABC, P is the mid-point of BC, Q is the mid-point of AP such that BQ produced meets

AC at R. 1
3
To prove: RA = CA

Construction: Draw PS|| BR , meeting AC at S.

Proof: In ∆BCR, P is the mid-point of BC and PS|| BR

\ S is the mid-point of CR.

⇒ CS = SR …(i) Fig. 7.61
In ∆APS, Q is the mid-point of AP and QR  PS

\ R is the mid-point of AS.

⇒ AR = RS …(ii)

From (i) and (ii), we get

AR = RS = SC 1 1
3 3
⇒ AC = AR + RS + SC = 3 AR ⇒ AR = AC = CA

4. In Fig. 7.62, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show

that ar (∆ABC) = AO .
ar (∆DBC) DO

Sol. Given: Two triangles ∆ABC and ∆DBC which stand on the same base

but on opposite sides of BC.

To prove: ar (∆ABC) = AO Fig. 7.62
ar (∆DBC) DO

Construction: We draw AE ⊥ BC and DF ⊥ BC.
Proof: In ∆AOE and ∆DOF, we have

188 Xam idea Mathematics–X

∠AEO = ∠DFO = 90°
∠AOE = ∠DOF (Vertically opposite angles)
\ ∆AOE ~ ∆DOF (By AA criterion of similarity)

AE = AO …(i)
Now, DF DO
⇒ 1
ar (∆ABC) 2 × BC × AE
ar (∆DBC) BC × DF
= 1
2
×

ar (∆ABC) = AE …(ii)
ar (∆DBC) DF
Fig. 7.63

From (i) and (ii), we have

ar (∆ABC) = AO
ar (∆DBC) DO

5. Two poles of height a metres and b metres are p metres apart. Prove that the height of the point
of intersection of the lines joining the top of each pole to the foot of the opposite pole is given

by ab metres.
a+b
Sol. Let AB and CD be two poles of height a and b metres respectively such that the poles are p metres

apart i.e., AC = p metres. Suppose the lines AD and BC meet at O such that OL = h metres.

Let CL = x and LA = y. Then, x + y = p.

In ∆ABC and ∆LOC, we have

∠CAB = ∠CLO (Each equal to 90°)

∠C = ∠C (Common)

\ ∆ABC ~ ∆LOC (By AA criterion of similarity)

⇒ CA = AB ⇒ p = a
CL LO x h

⇒ x= ph …(i)
a

In ∆ALO and ∆ACD, we have

∠ALO = ∠ACD (Each equal to 90°)

∠A = ∠A (Common)

\ ∆ALO ~ ∆ACD (By AA criterion of similarity) Fig. 7.64

⇒ AL = OL ⇒ y = h
AC DC p b

⇒ y= ph …(ii)
b
From (i) and (ii), we have

x+y= ph + ph ⇒ p = ph 1 + 1  [ x + y = p]
a b  a b 

⇒ 1=h  a+b  ⇒ h= ab metres.
 ab  a+b

Hence, the height of the intersection of the lines joining the top of each pole to the foot of the

opposite pole is ab metres.
a+b

Triangles 189

6. In an equilateral triangle ABC, D is a point on side BC such that BD = 1 BC. Prove that
3
9AD2 = 7AB2. [NCERT, CBSE 2018]

Sol. Given: An equilateral triangle ABC and D be a point on BC such that BD = 1 BC.
To Prove: 9AD2 = 7AB2 3

Construction: Draw AE ⊥ BC. Join AD.
Proof: ∆ABC is an equilateral triangle and AE ⊥ BC so BE = EC .

Thus, we have

BD = 1 BC and DC = 2 BC and BE = EC = 1 BC
3 3 2
In ∆ AEB Fig. 7.65

AE2 + BE2 = AB2 (Using Pythagoras Theorem)
AE2 = AB2 – BE2
AD2 – DE2 = AB2 – BE2 ( In ∆AED, AD2 = AE2 + DE2)



AD2 = AB2 – BE2 + DE2
2
AD2 = AB2 –  1 BC  + (BE – BD)2
 2

AD2 = AB2 – 1 BC2 +  1 BC − 1 BC 2
4  2 3 

AD2 = AB2 – 1 BC2 +  BC 2 ⇒ AD2 = AB2 – BC2 1 − 1
4  6   4 36 

9AADD2 2==A9BA2B–2B–C22AB3286  ⇒ (9AADB2 = 9AB2 – 2BC2
= BC)

9AD2 = 7AB2

7. Through the mid-point M of the side CD of a parallelogram ABCD, the line BM is drawn

intersecting AC at L and AD produced to E. Prove that EL = 2BL.

Sol. In ∆BMC and ∆EMD, we have

MC = MD ( M is the mid-point of CD)

∠CMB = ∠DME (Vertically opposite angles)

and ∠MBC = ∠MED (Alternate angles)

So, by AAS criterion of congruence, we have

∆BMC  ∆EMD

⇒ BC = DE (CPCT)

Also, BC = AD ( ABCD is a parallelogram)

Now, in ∆AEL and ∆CBL, we have

∠ALE = ∠CLB (Vertically opposite angles)

∠EAL = ∠BCL (Alternate angles)

∴ ∆AEL ~ ∆CBL (By AA similarity) Fig. 7.66

⇒ EL = AE ⇒ EL = 2BC ( AE = AD + DE = BC + BC = 2BC)
BL CB BL BC



⇒ EL = 2 ⇒ EL = 2BL
BL

190 Xam idea Mathematics–X

PROFICIENCY EXERCISE [1 mark each]

QQ Objective Type Questions:
1. Choose and write the correct option in each of the following questions.
(i) In DABC, DE  BC Fig. 7.67, then BD is equal to

A

4 cm 6 cm

DE
3 cm

B C

(b) 8 cm Fig. 7.67

(a) 2 cm (c) 3 cm (d) 9 cm

(ii) If Manish goes 3 km towards east and then 4 km towards North. His distance from starting
point is

(a) 3 km (b) 4 km (c) 5 km (d) 2 km

(iii) In a triangle ABC if AB = 13 cm, BC = 12 cm and AC = 5 cm, then the triangle is right
angled at

(a) A (b) B (c) C (d) can't say
ar (DABC)
(iv) If DABC ~ DPQR and ar (DPQR) = 9 , also AB = 4 cm, then PQ is equal to
4

(a) 8 units (b) 8 units (c) 3 units (d) 8 units
3 2 2

(v) If in DABC and DDEF, AB = DE , then they will be similar when
BC FD

(a) ∠B = ∠E (b) ∠A = ∠D (c) ∠B = ∠D (d) ∠A = ∠F

2. Fill in the blanks.

(i) The areas of two similar triangles are in the ratio 16 : 25, then the sides of these triangles
are in the ratio _____________ .

(ii) If a line divides any two sides of a triangle in the same ratio then the line is _____________
to the third side.

(iii) In DABC if AB = 6 3 cm, AC = 12 cm, BC = 6 cm, then ∠B is _____________ .
(iv) All equilateral triangles are _____________ .

(v) In a right angled triangle _____________ is the longest side

QQ Very Short Answer Questions : [1 mark each]

3. Given DABC ~ DPQR, if AB = 1 , then find arDABC . [CBSE 2018 (30/1)]
PQ 3 arDPQR

4. If DABC ~ DQRP, ar (DABC) = 9 and BC =15 cm, then find PR [CBSE 2018 (C) (30/1)]
ar (DQRP) 4

5. In Fig. 7.68, DE  BC, AD = 1 cm and BD = 2 cm. What is the ratio of the ar(ABC) to the

ar(DADE)? [CBSE 2019 (30/1/1)]

Triangles 191

Fig. 7.68

6. In Fig. 7.69, ABC is an isosceles triangle right angled at C with AC = 4 cm. Find the length of AB.
[CBSE 2019 (30/2/1)]

Fig. 7.69

7. In Fig. 7.70, DEBC. Find the length of side AD, given that AE = 1.8 cm, BD = 7.2 cm and

CE = 5.4 cm. [CBSE 2019 (30/2/1)]

1.8 cm

7.2 cm 5.4 cm

Fig. 7.70

8. Let DABC ~ DDEF and their areas be respectively, 64 cm2 and 121 cm2. If EF = 15.4 cm, find BC

[CBSE 2019 (30/4/2)]

9. In Fig. 7.71, GCBD and GEBF. If AC = 3 cm and CD = 7 cm, then find the value of AE .
AF
D [CBSE 2019(C) (30/1/1)]

7 cm B
C
3 cm

AG

E

F

Fig. 7.71

10. If ∆ABC and ∆DEF are two triangles such that EA=FB FB=DC CA = 3 , then find ar(∆DEF) : ar(∆ABC).
DE 4

11. If ∆ABC ~ ∆PQR, ar(∆ABC) = 16 , AB = 2 cm and AC = 12 cm, then find the value of PR.
ar(∆PQR) 9

12. In ∆ABC, AB = 24 cm, BC = 10 cm and AC = 26 cm. Is this triangle right triangle?
[NCERT Exemplar]

192 Xam idea Mathematics–X