Question

A chord of a circle of radius 10cm subtends a right angle at the center. Find the area of the corresponding (i) Minor segment (ii) Major sector (use\[\pi =3.14\]).

Answer 1

Hint: Find the area of the minor segment by subtracting the area of the triangle formed by chord from the area of minor sector. Where\[\theta ={{90}^{\circ }}\] for the minor sector and for finding the area of the major sector, the angle becomes \[\left( 360-\theta \right)\].

Complete step by step answer:
Complete step-by-step answer:
(i) Minor segment
Given that the radius of the circle is 10cm.
It’s center is marked as O.
OA and OB are the radii of the triangle.

\[\therefore \]OA = OB = 10cm.
Here AB refers to the chord of the circle.
Given that the chord subtends a right angle at the center of the circle.
\[\therefore \theta ={{90}^{\circ }}\]
We need to find the area of the minor segment AOB.
Area is given by the formula, \[\dfrac{\theta }{360}\times \pi {{r}^{2}}\].
\[\therefore \] Area of minor sector OAPB\[=\dfrac{\theta }{360}\times \pi {{r}^{2}}=\dfrac{90}{360}\times 3.14\times {{\left( 10 \right)}^{2}}\]
\[\begin{align}
  & =\dfrac{1}{4}\times 3.14\times 100 \\
 & =\dfrac{314}{4}=78.5c{{m}^{2}} \\
\end{align}\]
Now we will find area of triangle AOB,
Area of triangle AOB = $\dfrac{1}{2} \times $ Base $ \times $ Height
Area of triangle AOB = $\dfrac{1}{2} \times $ OB $ \times $ AO
Area of triangle AOB = $\dfrac{1}{2} \times $ 10 $ \times $ 10
Area of triangle AOB = 50 $\text{cm}^2$
So now,
Area of minor segment = Area of Minor sector - Area of triangle AOB
Area of minor segment = 78.5 - 50
Area of minor segment = 28.5 $\text{cm}^2$

 (ii) Major sector

Angle =\[360-\theta =360-{{90}^{\circ }}\] of the shaded portion.
Here, radius = 10cm.
Area of major sector is given by formula,
\[\begin{align}
  & =\dfrac{360-\theta }{360}\times \pi {{r}^{2}} \\
 & =\dfrac{360-90}{360}\times 3.14\times {{10}^{2}}=\dfrac{270}{360}\times 314 \\
\end{align}\]
\[=\dfrac{3}{4}\times 314=235.5c{{m}^{2}}\].
\[\therefore \] Area of minor sector =28.5\[c{{m}^{2}}\].
Area of major sector = 235.5\[c{{m}^{2}}\].

So, Area of major sector = 235.5\[c{{m}^{2}}\].

Note: The sector and segment of a circle are entirely different.

The sector can be called a “pizza” slice and the segment which is cut from the circle by a “chord”.
Remember to take the value of \[\theta \] in the major sector as \[\left( 360-\theta \right)\] and not\[{{90}^{\circ }}\]. As we are finding the shaded region, we should that angle value.