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 x
By 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)
24x2 – 29x + 7 = 20x2– 27x + 9
4x2 – 2x – 2 = 0
2(2x2 – x – 1) = 0
2x2 – x – 1 = 0
2x2 – 2x + x – 1 = 0
2x(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.