RD Sharma Solutions for Class 10 Chapter 4 Triangles Exercise 4.4

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

More on the basic proportionality theorem is discussed in this exercise. Few problems to strengthen the key concept is done by solving 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 Chapter 4 Triangles Exercise 4.4 Download PDF

 

rd sharma solutions for class 10 chapter 4 ex 4.4
rd sharma solutions for class 10 chapter 4 ex 4.4
rd sharma solutions for class 10 chapter 4 ex 4.4

 

Access RD Sharma Solutions for Class 10 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.

C:\Users\Tnluser\Downloads\extra-images-of-RD-Sharma-Chapter-4--ex-4.4-3.png

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

-4x+ 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.

C:\Users\Tnluser\Downloads\extra-images-of-RD-Sharma-Chapter-4--ex-4.4-1.png

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.

C:\Users\Tnluser\Downloads\extra-images-of-RD-Sharma-Chapter-4--ex-4.4-2.png

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

-x+ 19x – 88 = 0

x– 19x + 88 = 0

x2 – 11x – 8x + 88 = 0

x(x – 11) – 8(x – 11) = 0

x = 11 or x = 8


Also, access other exercise solutions of RD Sharma Class 10 Maths Chapter 4 Triangles

Exercise 4.1 Solutions

Exercise 4.2 Solutions

Exercise 4.3 Solutions

Exercise 4.5 Solutions

Exercise 4.6 Solutions

Exercise 4.7 Solutions

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