# 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\[\pi =3.14\]).

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**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.

Last updated date: 21st Sep 2023

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