Question

$ABC$ is an equilateral triangle of side $16\text{cm}$. Taking $A,B$ and $C$ as centres, three sectors are drawn from each vertex of radii $4\text{cm},6\text{cm}$ and $8\text{cm}$ as shown in the given figure. The area of the shaded region is

• A. $(64\sqrt{3}-32\pi ){\text{cm}}^2$
• B. $(64\sqrt{3}-22\pi ){\text{cm}}^2$
• C. $(121+128\sqrt{3}){\text{cm}}^2$
• D. $(32\sqrt{3}-44\pi ){\text{cm}}^2$

Answer 1

The correct option is B $(64\sqrt{3}-22\pi ){\text{cm}}^2$


Side of $\Delta ABC=16\text{cm}$
Area of equilateral $\Delta ABC=\frac{\sqrt{3}}{4}\times {\text{side}}^2$

$=\frac{\sqrt{3}}{4}\times {16}^2=64\sqrt{3}{\text{cm}}^2$

Area of sector $AFJ=\frac{\theta }{{360}^{\circ }}\times \pi \times (AJ{)}^2$

$=\frac{{60}^{\circ }}{{360}^{\circ }}\times \pi \times 8^2$ ($\therefore$Interior angle of an equilateral triangle is ${60}^{\circ }$)

$=\frac{32}{3}\pi {\text{cm}}^2$
$\therefore$ Area of region $FIJ=$ Area of $\Delta ABC-3\times$Area of sector $AFJ$
$=64\sqrt{3}-3\times \frac{32\pi }{3}$
$=(64\sqrt{3}-32\pi ){\text{cm}}^2$ ... (i)

Area of region $DHGE=$ Area of sector $AGE-$ Area of sector $AHD$
$=\frac{{60}^{\circ }}{{360}^{\circ }}\times \pi (A{G}^2-A{H}^2)$
$=\frac{\pi }{6}\times (6^2-4^2)$
$=\frac{10\pi }{3}{\text{cm}}^2$

$\therefore$ Area of shaded region
$=$ Area of region $FIJ+3\times$ Area of region $DHGE$
$=(64\sqrt{3}-32\pi )+3\times \frac{10\pi }{3}$
$=64\sqrt{3}-32\pi +10\pi$
$=(64\sqrt{3}-22\pi ){\text{cm}}^2$
Hence, the correct answer is option b.