In the ABC, DE BC and AD = 3x ? 2, AE = 5x ? 4, BD = 7x ? 5, CE = 5x ? 3, then find the value of x

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Solution

Q.
In the $∆$ ABC, DE $| |$ BC and AD = 3x − 2, AE = 5x − 4, BD = 7x − 5, CE = 5x − 3, then find the value of x

1 1

2 7/10

3 both (a) & (b)

4 none of these

Ans.
(3)

Given that, AD = 3x − 2, AE = 5x − 4, BD = 7x − 5, CE = 5x − 3

By Basic Proportionality theorem, we have $\frac{A D}{B D} = \frac{A E}{E C}$

$\Rightarrow \frac{3 x - 2}{7 x - 5} = \frac{5 x - 4}{5 x - 3} \Rightarrow (3 x - 2) (5 x - 3) = (5 x - 4) (7 x - 5)$

$\Rightarrow 15 x^{2} - 19 x + 6 = 35 x^{2} - 53 x + 20$

$\Rightarrow 20 x^{2} - 34 x + 14 = 0 \Rightarrow 10 x^{2} - 17 x + 7 = 0$

$\Rightarrow (x - 1) (10 x - 7) = 0 \Rightarrow x = 1, x = \frac{7}{10}$

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