Question

In the figure DE||BC, AD=P-3, BD=2P-2, AE/CE=1/4 then find the value of P.

Answer 1

If DE||BC, AD = P-3, BD = 2P-2, AE/CE = 1/4 then the value of P is 5.

Step-by-step explanation:

It is given that,

DE // BC

AD = P-3

BD = 2P-2

$\frac{AE}{CE} = \frac{1}{4}$

From the figure attached below, considering ΔADE and ΔABC, we have

∠A = ∠A ...... [common angle]

∠ADE = ∠ABC ...... [alternate angles, ∵ DE // BC (given)]

∴ By AA similarity, ΔADE ~ ΔABC

We know that the corresponding sides of two similar triangles are proportional to each other.

∴ $\frac{AD}{AB} = \frac{AE}{AC}$

⇒ $\frac{AD}{AD+DB} = \frac{AE}{AE+EC}$

⇒ $\frac{P - 3}{(P-3)+(2P - 2)} = \frac{1}{1+4}$

⇒ $\frac{P-3}{3P - 5} = \frac{1}{5}$

⇒ 5P - 15 = 3P - 5

⇒ 2P = 15 - 5

⇒ 2P = 10

⇒ P = 5

The value of P is 5.

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