More on the basic proportionality theorem is discussed in this exercise. A few problems to strengthen the key concepts are done by solving them in the right manner. To know how to present and make your solutions mistake-free, students can access the RD Sharma Solutions Class 10. Students aspiring to score high marks can download the RD Sharma Solutions for Class 10 Maths Chapter 4 Triangles Exercise 4.4 PDF provided below.
RD Sharma Solutions for Class 10 Maths Chapter 4 Triangles Exercise 4.4
Access RD Sharma Solutions for Class 10 Maths Chapter 4 Triangles Exercise 4.4
1. (i) In fig. 4.70, if AB∥CD, find the value of x.
Solution:
It’s given that AB∥CD.
Required to find the value of x.
We know that,
Diagonals of a parallelogram bisect each other.
So,
AO/ CO = BO/ DO
⇒ 4/ (4x – 2) = (x +1)/ (2x + 4)
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) In fig. 4.71, if AB∥CD, find the value of x.
Solution:
It’s given that AB∥CD.
Required to find the value of x.
We know that,
Diagonals of a parallelogram bisect each other
So,
AO/ CO = BO/ DO
⇒ (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)
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) In fig. 4.72, if AB ∥ CD. If OA = 3x – 19, OB = x – 4, OC = x- 3 and OD = 4, find x.
Solution:
It’s given that AB∥CD.
Required to find the value of x.
We know that,
Diagonals of a parallelogram bisect each other
So,
AO/ CO = BO/ DO
(3x – 19)/ (x – 3) = (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
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