Question

A triangle has m angel ABC = m angel AED , AE = 3 cm, EC = 2 cm, AD = 2 cm so what is the value of AB ?​

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}

Step-by-step explanation: