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# In a ΔABC, D and E are points on the sides AB and AC respectively such that DE || BC.If AD = 4x – 3, AE = 8x – 7, BD = 3x – 1, and CE = 5x – 3, find the value of x.

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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

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1. 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.

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