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:

Given,

ABâˆ¥ CD.

To find the value of x.

Now,

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

â‡’ 4/ (4x â€“ 2) = (x +1)/ (2x + 4)

4(2x + 4) = (4x â€“ 2)(x +1)

8x + 16 = x(4x â€“ 2) + 1(4x â€“ 2)

8x + 16 = 4x2 â€“ 2x + 4x â€“ 2

-4x2 + 8x + 16 + 2 â€“ 2x = 0

-4x2 + 6x + 8 = 0

4x2 â€“ 6x â€“ 18 = 0

4x2 â€“ 12x + 6x â€“ 18 = 0

4x(x â€“ 3) + 6(x â€“ 3) = 0

(4x + 6) (x â€“ 3) = 0

âˆ´ x = â€“ 6/4 or x = 3

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

Solution:

Given,

ABâˆ¥ CD.

To find the value of x.

Now,

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

â‡’ (6x â€“ 5)/ (2x + 1) = (5x â€“ 3)/ (3x â€“ 1)

(6x â€“ 5)(3x â€“ 1) = (2x + 1)(5x â€“ 3)

3x(6x â€“ 5) â€“ 1(6x â€“ 5) = 2x(5x â€“ 3) + 1(5x â€“ 3)

18x2 â€“ 10x2 â€“ 21x + 5 + x +3 = 0

8x2 â€“ 20x + 8 = 0

8x2 â€“ 16x â€“ 4x + 8 = 0

8x(x â€“ 2) â€“ 4(x â€“ 2) = 0

(8x â€“ 4)(x â€“ 2) = 0

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

âˆ´ x= 1/2

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

Solution:

Given,

ABâˆ¥ CD.

OA = 3x â€“ 19, OB = x â€“ 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 â€“ 19)/ (x â€“ 3) = (xâ€“4)/ 4

4(3x â€“ 19) = (x â€“ 3) (x â€“ 4)

12x â€“ 76 = x(x â€“ 4) -3(x â€“ 4)

12x â€“ 76 = x2 â€“ 4x â€“ 3x + 12

-x2 + 7x â€“ 12 + 12x -76 = 0

-x2 + 19x â€“ 88 = 0

x2 â€“ 19x + 88 = 0

x2 â€“ 11x â€“ 8x + 88 = 0

x(x â€“ 11) â€“ 8(x â€“ 11) = 0

âˆ´ x = 11 or x = 8

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