Question

in the adjoining figure, angle ABC = angle AED given AD =3cm AE=5cmand EC=2cm calculate BD and area of triangle AED divided by area of triangle ABC

Answer 1

Answer:

$BD=\frac{26}{3}units$

$\frac{ar(AED)}{ar(ABC)}= (\frac{3}{7})^2=\frac{9}{49}$

Step-by-step explanation:

Given ∠ABC=∠AED and AD=3cm, AE=5cm, EC=2cm.

we have to find the length of BD and $\frac{ar(AED)}{ar(ABC)}$

In ΔADE and ΔACB

∠AED=∠ABC   (given)

∠A=∠A             (common)

By AA similarity rule, ΔAED~ΔABC

∴ the sides are in same proportion

$\frac{AD}{AC}=\frac{AE}{AB}$

⇒ $\frac{3}{5+2}=\frac{5}{3+BD}$

⇒ $\frac{3}{7}=\frac{5}{3+BD}$

⇒  $3+BD=\frac{7}{3}\times 5$

⇒ $BD=\frac{26}{3}units$

By the theorem, the ratio of the area of two similar triangles is equal to the square of sides.

⇒  $\frac{ar(AED)}{ar(ABC)}= (\frac{3}{7})^2=\frac{9}{49}$

Answer 2

Step-by-step explanation:

hope this will help you