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 x
By 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)
20x2 – 12x – 15x + 9 = 24x2 – 29x + 7
20x2 -27x + 9 = 24x2 – 29x + 7
⇒ -4x2+ 2x + 2 = 0
4x2 – 2x – 2 = 0
4x2 – 4x + 2x – 2 = 0
4x(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.