# RD Sharma Solutions Class 10 Triangles Exercise 4.4

## RD Sharma Solutions Class 10 Chapter 4 Exercise 4.4

### RD Sharma Class 10 Solutions Chapter 4 Ex 4.4 PDF Free Download

#### Exercise 4.4

Q1) In fig. (i)  if $AB \parallel CD$, find the value of x.

(ii)  In fig. if $AB \parallel CD$, find the value of x.

(iii) in fig. if $AB \parallel CD$ .and OA = 3x – 19, OB = x – 4, OC = x- 3 and OD = 4, find x.

Sol:

(i) it is given that $AB \parallel CD$

We have to find the value of x.

Diagonals of the parallelogram,

As we know, $\frac{DO}{OA} = \frac{CO}{OB}$

$\frac{4x – 2}{4} = \frac{2x + 4}{x + 1}$

4(2x + 4) = (4x – 2)(x +1)

8x + 16 = x(4x – 2) + 1(4x – 2)

8x + 16 = 4x2 – 2x + 4x – 2

-4x2 + 8x + 16 + 2 – 2x = 0

-4x2 + 6x + 8 = 0

4x2 – 6x – 18 = 0

4x2 – 12x + 6x – 18 = 0

4x(x – 3) + 6(x – 3) = 0

(4x + 6) (x – 3) = 0

X = – 6/4 or x = 3

(ii) it is given that $AB \parallel CD$

We need to find the value of x.

Now, $\frac{DO}{OA} = \frac{CO}{OB}$

$\frac{6x – 5}{2x + 1} = \frac{5x – 3}{3x – 1}$

(6x – 5)(3x – 1) = (2x + 1)(5x – 3)

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

18x2 – 10x2 – 21x + 5 + x +3 = 0

8x2 – 16x – 4x + 8 = 0

8x(x – 2) – 4(x – 2) = 0

(8x – 4)(x – 2) = 0

X = 4/8 = 1/2   or  x = -2

X= 1/2

(iii) it is given that $AB \parallel CD$

And OA = 3x – 19 OB = x – 4 OC = x – 3 and OD = 4

We need to find the value of x,

Now, Now, $\frac{AO}{OC} = \frac{BO}{OD}$

$\frac{3x – 19}{x – 3} = \frac{x – 4}{4}$

4(3x – 19) = (x – 3) (x – 4)

12x – 76 = x(x – 4) -3(x – 4)

12x – 76 = x2 – 4x – 3x + 12

-x2 + 7x – 12 + 12x -76 = 0

-x2 + 19x – 88 = 0

X2 – 19x + 88 = 0

X2 – 11x – 8x + 88 = 0

X(x – 11) – 8(x – 11) = 0

X = 11 or x = 8