Question

In triangle ABC, DE parallel to bc, if BC=20cm ,DE=4cm and AE=3cm find AC

Answer 1

Answer:

It was good question. It has two basic complicated commands of triangle

Answer 2

In △ABC, DE∥BC

⇒  ∠B=∠D     (Corresponding angles )

⇒  ∠C=∠E    ( Corresponding angles )

⇒  ∠A=∠A    ( Common angle)

∴  4△ABC∼△ADE    ( By AAA criteria )

Let's consider CE = P

∴AC = 3 + P

From ΔAED & ΔABC

AE/AC = DE/BC

⇒ $\frac{3cm}{3cm+P}$ = $\frac{4}{20}$

⇒ $4 ( 3cm + P)$ = 3cm ×20

⇒12cm + 4P = 60cm

⇒4P= $60cm -12cm$ = 48cm

⇒ P =$\frac{48cm}{4cm}$= 12cm

Hence CE= 12cm,

∴ AC = 3 + P = (3 + 12 ) cm = 15cm

Ans :- AC will be 15cm.

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