Question

In the given figure, ABC is a triangle in which CDEFG is a pentagon. Triangles ADE and BFG are equilateral triangles each with side 2 cm and EF = 2 cm. Find the area of the pentagon:

• A. $8\sqrt{3}c{m}^2$
• B. $7\sqrt{3}c{m}^2$
• C. $15\sqrt{3}c{m}^2$
• D. $11.28c{m}^2$

Answer 1

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

$AB=6cm,\angle C={60}^{\circ }$
and $\angle A=\angle B={60}^{\circ }$
$\therefore \Delta ABC$ is an equilateral triangle
$\therefore$ Area of $\Delta ABC=\frac{\sqrt{3}}{4}\times (6{)}^2=9\sqrt{3}$
Area of $(\Delta ADE+\Delta BFG)=2\times \frac{\sqrt{3}}{4}\times (2{)}^2=2\sqrt{3}$
$\therefore$ Area of pentagon $=9\sqrt{3}-2\sqrt{3}=7\sqrt{3}c{m}^2$

Alternatively,
Since all angles of triangle ABC are 60 degrees each, it is an equilateral triangle.
Now, ABC has (1 + 3 + 5) = 9 equilateral triangles of side 2 cm each.
Area of pentagon = $7\times \frac{\sqrt{3}}{4}\times (2{)}^2=7\sqrt{3}c{m}^2$