Question taken from RD sharma

Class 10th

Chapter no. 4

Chapter name:- Triangles

Exercise :- 4.2

This is very basic and important questions.

In this question we have been given that ΔABC,

In which D and E are points on the sides AB and AC respectively such that DE || BC.

Also it is given that AD = 4x – 3, AE = 8x – 7, BD = 3x – 1, and CE = 5x – 3,

Now we have to find the value of x.

Understanding and learning CBSE maths

RD sharma, DHANPAT RAI publication

Given:Length of side AD = 4x – 3, BD = 3x – 1, AE = 8x – 7 and EC = 5x – 3

To find : Value of xBy using Thales Theorem, we get

AD/BD = AE/CE – equation 1

Now, putting values in equation 1,

We get,

(4x – 3)/(3x – 1) = (8x – 7)/(5x – 3)

(4x – 3)(5x – 3) = (3x – 1)(8x – 7)

4x(5x – 3) -3(5x – 3) = 3x(8x – 7) -1(8x – 7)

20x

^{2}– 12x – 15x + 9 = 24x^{2}– 29x + 720x

^{2}-27x + 9 = 24x^{2}– 29x + 7⇒ -4x

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

^{2}– 2x – 2 = 04x

^{2}– 4x + 2x – 2 = 04x(x – 1) + 2(x – 1) = 0

(4x + 2)(x – 1) = 0

⇒ x = 1 or x = -2/4

We know that the side of triangle is always positive

Therefore, we only take the positive value.

∴ x = 1

Therefore, the value of x is 1.