Question

In the given figure, $ABCDEF$ is a regular hexagon. $AB,CD$ and $EF$ are the diameters of the semicircles. If $BC=7\text{cm}$, then the area of the shaded region is equal to

• A. $\frac{147}{8}(4\sqrt{3}-\pi ){\text{cm}}^2$
• B. $\frac{147}{4}(2\sqrt{3}-3\pi ){\text{cm}}^2$
• C. $147(\sqrt{3}-3\pi ){\text{cm}}^2$
• D. $\frac{147}{8}(2\sqrt{3}-3\pi ){\text{cm}}^2$

Answer 1

The correct option is A $\frac{147}{8}(4\sqrt{3}-\pi ){\text{cm}}^2$

Given that $ABCDEF$ is a regular hexagon of side $7\text{cm}$.
In a regular hexagon, each interior angle is of measure ${120}^{\circ }$.
The diagonals bisect interior angles and meet at a common point lets say, $O$.
$\therefore \angle OAB=\angle OBA={60}^{\circ }$
In $\Delta AOB$, by angle sum property
$\angle AOB={60}^{\circ }$
So, hexagon $ABCDEF$ is divided into six equilateral triangles of length $7\text{cm}$ each.
$\therefore$ Area of $\Delta AOB=\frac{\sqrt{3}}{4}\times {\text{Side}}^2$
$=\frac{\sqrt{3}}{4}\times 7^2=\frac{49}{4}\sqrt{3}{\text{cm}}^2$

Area of Hexagon $ABCDEF=6\times \frac{49}{4}\sqrt{3}{\text{cm}}^2$
$=\frac{147}{2}\sqrt{3}{\text{cm}}^2$

Area of semicircle with $AB$ as diameter $=\frac{1}{2}\times \pi \times {\left(\frac{AB}{2}\right)}^2$
$=\frac{\pi }{8}\times 7^2=\frac{49\pi }{8}{\text{cm}}^2$
$\Rightarrow$Area of semicircles with $AB,CD$ and $EF$ as diameters
$=3\times \frac{49}{8}\pi$
$=\frac{147\pi }{8}{\text{cm}}^2$

$\therefore$ Area of shaded region $=$ Area of hexagon $ABCDEF-$ Are of semicircles with $AB,CD$ and $EF$ as diameters
$=(\frac{147\sqrt{3}}{2}-\frac{147\pi }{8})$
$=\frac{147}{8}(4\sqrt{3}-\pi ){\text{cm}}^2$
Hence, the correct answer is option a.