# Class 10 RD Sharma Solutions – Chapter 4 Triangles – Exercise 4.4

**Question 1. (i) In fig., if AB || CD, find the value of x.**

**Solution:**

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free classeswhich will definitely help them in making a wise career choice in the future.Given,

AB∥ CD.

To find the value of x.

Now,

AO/ CO = BO/ DO [Diagonals of a parallelogram bisect each other]

⇒ 4/ (4x – 2) = (x +1)/ (2x + 4)

4(2x + 4) = (4x – 2)(x +1)

8x + 16 = x(4x – 2) + 1(4x – 2)

8x + 16 = 4x

^{2}– 2x + 4x – 2-4x

^{2}+ 8x + 16 + 2 – 2x = 0-4x

^{2}+ 6x + 8 = 04x

^{2}– 6x – 18 = 04x

^{2}– 12x + 6x – 18 = 04x(x – 3) + 6(x – 3) = 0

(4x + 6) (x – 3) = 0

∴ x = – 6/4 or x = 3

**(ii) In fig., if AB || CD, find the value of x.**

**Solution:**

Given,

AB∥ CD.

To find the value of x.

Now,

AO/ CO = BO/ DO [Diagonals of a parallelogram bisect each other]

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

(6x – 5)(3x – 1) = (2x + 1)(5x – 3)

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

18x

^{2}– 10x^{2}– 21x + 5 + x +3 = 08x

^{2}– 20x + 8 = 08x

^{2}– 16x – 4x + 8 = 08x(x – 2) – 4(x – 2) = 0

(8x – 4)(x – 2) = 0

x = 4/8 = 1/2 or x = -2

∴ x= 1/2

**(iii) In fig., if AB || CD. If OA = 3x – 19, OB = x – 4, OC = x- 3 and OD = 4, find x.**

**Solution:**

Given,

AB∥ CD.

OA = 3x – 19, OB = x – 4, OC = x- 3 and OD = 4

To find the value of x.

Now,

AO/ CO = BO/ DO [Diagonals of a parallelogram bisect each other]

(3x – 19)/ (x – 3) = (x–4)/ 4

4(3x – 19) = (x – 3) (x – 4)

12x – 76 = x(x – 4) -3(x – 4)

12x – 76 = x

^{2}– 4x – 3x + 12-x

^{2}+ 7x – 12 + 12x -76 = 0-x

^{2}+ 19x – 88 = 0x

^{2}– 19x + 88 = 0x

^{2}– 11x – 8x + 88 = 0x(x – 11) – 8(x – 11) = 0

∴ x = 11 or x = 8