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In the given figure, $D E ‖ B C$ and $A D = 4 x − 3$ , $A E = 8 x − 7$ , $B D = 3 x − 1$ and $C E = 5 x − 3$ . Find the value of x.

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\frac{1}{2}

, 1

\frac{1}{2}

, 6

\frac{2}{3}

, 3

\frac{2}{5}

, 2

answer is A.

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Detailed Solution

It is given that

D E ‖ B C

A D = 4 x − 3

A E = 8 x − 7

B D = 3 x − 1

and

C E = 5 x − 3

According to Thales theorem we can write,

A D B D = A E C E

Substituting the values in the above equation,

4 x − 3 3 x − 1 = 8 x − 7 5 x − 3 4 x − 3 5 x − 3 = 3 x − 1 8 x − 7 20 x 2 − 12 x − 15 x + 9 = 24 x 2 − 21 x − 8 x + 7 4 x 2 − 2 x − 2 = 0

2 x 2 − x − 1 = 0 2 x 2 − 2 x − x + 1 2 x x − 1 + 1 x − 1

2 x + 1 x − 1 = 0 x = − 12 & x = 1

The value of

x = − 12 & x = 1

.
Hence, the correct option is 1.

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