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Basic Proportionality Theorem

In the given ...

Question

In the given figure, $D E | | B C$ such that AD = x cm, DB = (3x + 4) cm, AE = (x + 3) cm and EC = (3x + 19) cm. Find the value of x.

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Solution

Triangle ABC, DE||BC, so by Basic Proportionality theorem,
AD /DB = AE /EC

substituting the values, we get,
x/3x+4 =x+3/3x+19

By cross multiplying, we get,

x(3x+19) =(x+3)(3x+4)

$3 x^{2} + 19 x = 3 x^{2} + 4 x + 9 x + 12$
19x-13x=12
6x=12
x=2

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Similar questions

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.

In the above figure ∆ABC~∆ADE. Given that AD=x cm, DB =x-3 cm, AE=4cm, EC=3 cm. What is the value of x?

∥

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

E C = (x - 1) c m,

x .

D E ∥ B C

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

E C = (x - 1) c m

x

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