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 ΔABC,

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

Also it is given that AD = 8x – 7 cm, DB = 5x – 3 cm, AE = 4x – 3 cm, and EC = (3x – 1) cm, now we have to Find the value of x

Understanding and learning CBSE maths

RD sharma, DHANPAT RAI publication

Given:Length of side AD = 8x – 7, DB = 5x – 3, AER = 4x – 3 and EC = 3x -1

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

AD/BD = AE/CE – equation 1

Now, putting values in equation 1,

(8x – 7)/(5x – 3) = (4x–3)/ (3x–1)

(8x – 7)(3x – 1) = (5x – 3)(4x – 3)

24x

^{2}– 29x + 7 = 20x^{2}– 27x + 94x

^{2}– 2x – 2 = 02(2x

^{2}– x – 1) = 02x

^{2}– x – 1 = 02x

^{2}– 2x + x – 1 = 02x(x – 1) + 1(x – 1) = 0

(x – 1)(2x + 1) = 0

⇒ x = 1 or x = -1/2

Since, we know that the side of triangle is always positive.

Therefore, we take the positive value.

∴ x = 1.

Therefore, the value of x is 1.