Question

In the given figure $ABC$ is a triangle in which $CDEFG$ is a pentagon. Triangle $ADE$ and $BFG$ triangles each with side $2cm$and $EF=2cm$. Then the area of pentagon $CDEFG$ is?

• A. $8\sqrt{3}c{m}^2$
• B. $7\sqrt{3}c{m}^2$
• C. $15\sqrt{3}c{m}^2$
• D. $11,78c{m}^2$

Answer 1

The correct option is B $7\sqrt{3}c{m}^2$

$AB=6cm,\angle C={60}^{\circ },\angle B={60}^{\circ }$

$\therefore \angle A={60}^{\circ }$.

$\therefore △ABC$ is equilatiral triangle.

$\begin{array}{cc}\mathrm{Area}(△ABC{)}^{\prime }& =\frac{\sqrt{3}}{4}\times (6{)}^2\\ & =9\sqrt{3}{cm}^2\\ \mathrm{Area}(△ADE+\Delta BFG)& =2\times (\frac{\sqrt{3}}{4}\times (2{)}^2)\\ & =2\sqrt{3}{cm}^2\end{array}$

$\therefore$ Area of pentagon $=9\sqrt{3}-2\sqrt{3}$

$=7\sqrt{3}{cm}^2$

option B is correct.