Question

In the figure $\Delta ABC,DE\parallel BC,A\left(\Delta ADE\right)=48$ sq.cm., $\frac{AD}{DB}=\frac{4}{5}$. Find the area of $\Delta BEC$.

• A. $60$ sq.cm
• B. $95$ sq.cm
• C. $108$sq.cm
• D. $135$ sq.cm

Answer 1

The correct option is D $135$ sq.cm

In $△ABC,DE\parallel BC,A\left(△ADE\right)=48$ sq. cm, $\frac{AD}{DB}=\frac{4}{5}$

$\frac{A\left(△AEB\right)}{A\left(△ADE\right)}=\frac{\frac{1}{2}\times DE\times AB}{\frac{1}{2}\times DE\times AD}$

$=\frac{AB}{AD}$

$=\frac{AD+DB}{AD}$

$=1+\frac{DB}{AD}=1+\frac{5}{4}$

$=\frac{9}{4}$

$⟹A\left(△AEB\right)=108$ sq. cm

In triangle $ABC$ and $ADE$, $\angle BAC=\angle DAE,\angle ABC=\angle ADE,\angle AED=\angle ACB$

$⟹△ABC\sim △ADE$

$⟹\frac{A(△ABC}{A\left(△ADE\right)}={\left(\frac{AB}{AD}\right)}^2$

$={\left(\frac{AD+DB}{AD}\right)}^2$

$={(1+\frac{DB}{AD})}^2$

$={\left(\frac{9}{4}\right)}^2=\frac{81}{16}$

$⟹A\left(△ABC\right)=\frac{81}{16}\times 48=243$

$A\left(△BEC\right)=A\left(△ABC\right)-A\left(△AEB\right)$

$=243-108=135$ sq. cm.

Hence, option D is correct.