Question

In the figure, AD = $12$ cm, AB=$20$ cm and AE = $10$ cm. Find the length of EC.

• A. $14$ cm
• B. $10$ cm
• C. $8$ cm
• D. $15$ cm

Answer 1

Answer: (A)

$14$ cm

$\angle DEA+\angle DEC={180}^{\circ }$ (Linear pair)

$\angle DEA={180}^{\circ }-{123}^{\circ }={57}^{\circ }$

Also, in $\Delta$ ADE.

$\angle ADE+\angle DAE+\angle DEA={180}^{\circ }$

$\angle ADE={180}^{\circ }-(80+57)={43}^{\circ }$

Similarly, in $\Delta$ ABC,

$\angle ABC+\angle BAC+\angle ACB={180}^{\circ }$

${57}^{\circ }+{80}^{\circ }+\angle ACB={180}^{\circ }$

$\angle ACB={43}^{\circ }$

So, $\Delta ADE\sim \Delta ACB$ (by AAA similarly)

$\Rightarrow \frac{AE}{AB}=\frac{AD}{DC}\Rightarrow \frac{10}{20}=\frac{12}{AC}$

$AC=24cm$

$EC=AC-AE$

$\Rightarrow EC=24-10=14cm$