Question

In the given figure, PQRSTV is a regular hexagon with each side of length 5 cm. A circle is drawn with its centre at V such that it passes through P. What is the area (in cm$^2$) of the shaded region? (The diagram is representative) \includegraphics[width=0.5\linewidth]{image37.png}

• A. $\dfrac{25\pi}{3}$
• B. $\dfrac{20\pi}{3}$
• C. $6\pi$
• D. $7\pi$

Hint:

Regular hexagon corner angle is $120^\circ$. If a circle is centred at a vertex and passes through an adjacent vertex, the sector between two adjacent sides is a $120^\circ$ sector with radius equal to the side length.

Answer 1

The Correct Option is A

Step 1: Identify the radius of the circle.
Since the circle is centred at $V$ and passes through $P$, its radius is $VP$. In a regular hexagon, all sides are equal and adjacent vertices are $5$ cm apart. Hence $VP=VT=5$ cm (adjacent sides of the hexagon). \(\Rightarrow r=5\) cm.

Step 2: Find the angle subtended at the centre $V$.
The interior angle at any vertex of a regular hexagon is $120^\circ$. The sector in question is formed by the two sides $VP$ and $VT$; therefore the central angle of the circular sector $\angle PVT=120^\circ$.

Step 3: Area of the shaded region.
From the diagram, the shaded part is exactly the sector of the circle between the radii $VP$ and $VT$ (no subtraction of the triangle is intended).
Area of a sector with angle $\theta$ and radius $r$: \(\displaystyle A_{\text{sector}}=\frac{\theta}{360^\circ}\pi r^2\).
Here, $\theta=120^\circ$, $r=5$ \(\Rightarrow\) \[ A_{\text{shaded}}=\frac{120^\circ}{360^\circ}\pi(5)^2 =\frac{1}{3}\cdot 25\pi =\boxed{\frac{25\pi}{3}}. \]