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Question

In the figure shown $D E ∥ B C$ and $A D = 3 x - 2, A E = 5 x - 4, B D = 7 x - 5$ and $C E = 5 x - 3$ . Therefore, the value of $x$ is

A

Only

1

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B

Only

\frac{7}{10}

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C

1

or

\frac{7}{10}

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D

\frac{10}{7}

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Solution

The correct option is C $1$ or $\frac{7}{10}$
$I n △ A B C D E ∥ B C a n d i n t e r s e c t s A B i n D a n d A C i n E . ∴ \frac{A D}{D B} = \frac{A E}{E C} (B a s i c P r o p o r t i o n a l i t y t h e o r m) A D = 3 x - 2, A E = 5 x - 4, B D = 7 x - 5 a n d C E = 5 x - 3. (G i v e n) \Rightarrow \frac{3 x - 2}{7 x - 5} = \frac{5 x - 4}{5 x - 3} \Rightarrow (3 x - 2) (5 x - 3) = (5 x - 4) (7 x - 5) \Rightarrow 15 x^{2} - 9 x - 10 x + 6 = 35 x^{2} - 25 x - 28 x + 20 \Rightarrow 20 x^{2} - 53 x + 19 x + 14 = 0 \Rightarrow 20 x^{2} - 34 x + 14 = 0 \Rightarrow 10 x^{2} - 17 x + 7 = 0 \Rightarrow 10 x^{2} - 10 x - 7 x + 70 \Rightarrow 10 x (x - 1) - 7 (x - 1) \Rightarrow x = 1 a n d x = \frac{7}{10}$

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Similar questions

Q. In the figure if

D E ∥ B C

A D = 3 x - 2

A E = 5 x - 4

B D = 7 X - 4

B D - 7 x - 5

C E = 5 x - 3

Q. In the figure given, if

D E ∥ B C

A B = 5 x - 4, B D = 7 x - 5, C E = 5 x - 3

A D = 3 x - 2

, find the value of

x

.

△

∥

A D = 4 x - 3

A E = 8 x - 7

B D = 3 x - 1

C E = 5 x - 3

then the value of

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△

ABC, D and E are points on the sides AB and AC respectively such that DE || BC. If AD =

4 x - 3

3 x - 1

8 x - 7

5 x - 3

x

Δ A B C

D

E

are points on the sides

A B

A C

respectively such that

D E | | B C

A D = 4 x - 3, A E = 8 x - 7, B D = 3 x - 1

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