Question

Consider a circle with unit radius. There are seven adjacent sectors, ${{S}_{1}},{{S}_{2}},{{S}_{3}},\ldots ,{{S}_{7}}$, in the circle such that their total area is $\dfrac{1}{8}$ of the area of the circle. Further, the area of the ${{j}^{th}}$ sector is twice that of the ${{\left( j-1 \right)}^{th}}$ sector, for $j=2,\ldots ,7$. Find the area of the sector ${{S}_{1}}$.
(a) $\dfrac{3\pi }{1016}$
(b) $\dfrac{\pi }{508}$
(c) $\dfrac{\pi }{1016}$
(d) $\dfrac{\pi }{336}$

Answer 1

Hint: We will write the areas sectors ${{S}_{2}},\ldots ,{{S}_{7}}$ in terms of the area of sector ${{S}_{1}}$using the given relation between areas of two sectors. We will use the formula for the area of the circle. The area of the circle is given by $\pi {{r}^{2}}$ where $r$ is the radius of the circle. Then we will form an equation with the given condition that the total area of the seven sectors is $\dfrac{1}{8}$ of the area of the circle.

Complete step-by-step answer:
We have a circle with unit radius. We know that the area of a circle is given by $\pi {{r}^{2}}$ where $r$ is the radius of the circle. So, the area of the given circle is $\pi $ since the radius is 1.

Let us denote the area of a sector $S$ by $A(S)$.

Now, let us write the areas of sectors of the circle in terms of the area of sector ${{S}_{1}}$. We are given that the area of the ${{j}^{th}}$ sector is twice that of the ${{\left( j-1 \right)}^{th}}$ sector, for $j=2,\ldots ,7$. So, we have $A({{S}_{2}})=2A({{S}_{1}})$.
Next, we have $A({{S}_{3}})=2A({{S}_{2}})$.
Substituting $A({{S}_{2}})=2A({{S}_{1}})$ in the above equation, we get
$A({{S}_{3}})=2\times 2A({{S}_{1}})=4A({{S}_{1}})$
Similarly, we will get the following areas of sectors in terms of the area of sector ${{S}_{1}}$,
$A({{S}_{4}})=8A({{S}_{1}})$; $A({{S}_{5}})=16A({{S}_{1}})$; $A({{S}_{6}})=32A({{S}_{1}})$ and $A({{S}_{7}})=64A({{S}_{1}})$.
Now, we are given that the total area of the seven sectors is $\dfrac{1}{8}$ of the area of the circle. Hence we get the following equation,
$A({{S}_{1}})+A({{S}_{2}})+A({{S}_{3}})+A({{S}_{4}})+A({{S}_{5}})+A({{S}_{6}})+A({{S}_{7}})=\dfrac{1}{8}\times \pi $
We will substitute the values of the areas of sectors in terms of the area of sector ${{S}_{1}}$ in the above equation. We get the following equation,
$\begin{align}
  & A({{S}_{1}})+2A({{S}_{1}})+4A({{S}_{1}})+8A({{S}_{1}})+16A({{S}_{1}})+32A({{S}_{1}})+64A({{S}_{1}})=\dfrac{\pi }{8} \\
 & \Rightarrow A({{S}_{1}})\left( 1+2+4+8+16+32+64 \right)=\dfrac{\pi }{8} \\
 & \therefore A({{S}_{1}})\times 127=\dfrac{\pi }{8} \\
\end{align}$
We have to find the value of the area of sector ${{S}_{1}}$. Solving the above equation for $A({{S}_{1}})$, we get
$A({{S}_{1}})=\dfrac{\pi }{1016}$
Hence, the correct option is (c).

So, the correct answer is “Option (c)”.

Note: We should interpret the given information in the problem statement carefully. In this question, there is no need to use the formula for the area of the sector of a circle. The area of a sector of a circle cannot be calculated without knowing the central angle. It is important that we form the equations correctly, otherwise we will obtain an incorrect answer.