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Triangles

In the given ...

Question

In the given figure, $D E ∥ B C$ . If $A D = x c m, D B = (x - 2) c m, A E = (x + 2) c m$ and $E C = (x - 1) c m$ , find the value of $x$ .

x = 10

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x = 9

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x = 6

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x = 4

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Solution

The correct option is C $x = 4$
Given: AD = 3.6 cm, AB = 9 cm and AE = 2.4 cm $D E ∥ B C$
Since, $D E ∥ B C$ , by Equal Intercept Theorem,
$\frac{A D}{B D} = \frac{A E}{E C}$
$\frac{x}{x - 2} = \frac{x + 2}{x - 1}$
$x (x - 1) = (x - 2) (x + 2)$
$x^{2} - x = x^{2} - 4$
$x = 4$

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A D = 3 \cdot 6 c m, A B = 9 c m

A E = 2 \cdot 4 c m,

E C

\frac{A D}{D B} = \frac{3}{5}

A C = 5 \cdot 6 c m,

A E

A D = x c m, D B = (x - 2) c m, A E = (x + 2) c m

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x .

D and E are points on the sides AB and AC respectively of a $Δ A B C$ such that DE || BC
Find the value of x, when

(i) AD = x cm, DB = (x - 2) cm,

AE = (x + 2) cm and EC = (x - 1) cm.

(ii) AD = 4 cm, DB = (x - 4) cm, AE = 8 cm and EC = (3x - 19) cm.

(iii) AD = (7x - 4) cm, AE = (5x - 2) cm, DB = (3x + 4 ) cm and EC = 3x cm.

D E ∥ B C .

A D = x, D B = x - 2

A E = x + 2, E C = x - 1

x ?

∥

D E | | B C

A D = x, D B = x - 2, A E = x + 2

E C = x - 1,

x

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