Question

A chord of a circle of radius 15 cm subtends an angle of ${{60}^{\circ }}$ at the centre. Find the areas of the corresponding minor and major segment of the circle. $\left( \text{use }\pi =3.14\text{ and }\sqrt{3}=1.73 \right)$ .

Answer 1

Hint: Here, first we will draw a figure of a circle with ‘O’ as centre, then draw a chord AB and join it to O such that it forms a triangle ABO. Then we need to find the area of minor segment by subtracting area of a triangle from area of a sector, where area of a sector $\dfrac{\theta }{{{360}^{\circ }}}=\pi {{r}^{2}}$ & area of a triangle $=\dfrac{1}{2}\times \text{ base }\times \text{ height}$, then we will find the area of major segment by subtracting minor segment from the area of a circle i.e. $\pi {{r}^{2}}$ .

Complete step-by-step answer:

From the figure, we get to know that –
$\text{Radius OA=OB=15cm}$
$\text{AYB}=\text{ minor segment}$
$\angle \text{AOB}={{60}^{\circ }}$
$\text{OAYB}=\text{ sector}$
$\text{AOB }=\text{ major segment}$
Now, we will find the area of the minor segment AYB.
$=\text{Area of sector OAYB }-\text{ Area of }\Delta \text{OAB}$
We know that, area of the sector $\text{OAYB}=\dfrac{\theta }{{{360}^{\circ }}}\times \pi {{r}^{2}}$
$=\dfrac{{{60}^{\circ }}}{{{360}^{\circ }}}\times 3.14\times 15\times 15$
By simplifying the above equation, we get –
$117.75\text{ c}{{\text{m}}^{2}}$ ……………………………. (1)
For finding the area of $\Delta \text{OAB}$, we will draw $\text{OM}\bot \text{AB}$

Here, $OA=OB$. So, $\Delta AMO\cong \Delta BMO$
 And M is the midpoint of AB and $\angle \text{AOM}=\angle \text{BOM}$.
We know that, Area of the triangle $=\dfrac{1}{2}\times \text{ base }\times \text{ height}$.
Let us consider $OM=x\text{ cm}$
from $\Delta OMA,$ we will find the side of OM.
$\cos {{30}^{\circ }}=\dfrac{OM}{OA}$
We know that $\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}$ .
So, we get –
$\dfrac{x}{15}=\dfrac{\sqrt{3}}{2}$
By multiplying 15 on both sides, we get –
$x=\dfrac{15\sqrt{3}}{2}$
$\therefore OM=\dfrac{15\sqrt{3}}{2}cm$
Now, we will find the side of AM.
So, $\sin {{30}^{\circ }}=\dfrac{AM}{OA}$
We know that $\sin {{30}^{\circ }}=\dfrac{1}{2}$
So, $\dfrac{1}{2}=\dfrac{AM}{15}$
By multiplying 15 on both sides, we get –
$AM=\dfrac{15}{2}cm$
Therefore, $AB=2AM$ . So, we get –
$2\times \dfrac{15}{2}cm=15cm$
So, Area of the triangle $=\dfrac{1}{2}\times AB\times OM$
By applying the values of AB & OM, we get –
$=\dfrac{1}{2}\times 15\times \dfrac{15\sqrt{3}}{2}$
By simplifying the above equation and taking $\sqrt{3}=1.73$ we get –
$=97.3125c{{m}^{2}}$ …………………………. (2)
Therefore, Area of the minor segment AYB $=\text{Area of sector OAYB }-\text{ Area of }\Delta \text{OAB}$
By substituting equation (1) and (2) in the above equation, we get –
$=11775c{{m}^{2}}-97.3125c{{m}^{2}}$
$=20.4375c{{m}^{2}}$…………………………..(3)
Now, we will find the area of the major segment $AYB=\text{Area of a circle - Area of minor segment}$ .
We know that area of a circle $=\pi {{r}^{2}}$
$=3.14\times 15\times 15$
$=706.5c{{m}^{2}}$…………………………….. (4)
So, $AYB=\text{Area of a circle - Area of minor segment}$
By substituting equation (3) & (4) in the above equation, we get –
$\begin{align}
  & AYB=706.5-20.4375 \\
 & =686.0625c{{m}^{2}} \\
\end{align}$
Hence, Area of minor segment $=20.4375c{{m}^{2}}$ and Area of major segment $=686.0625c{{m}^{2}}$ .

Note: While solving this question students can be mistaken while finding the area of a triangle without bisecting OM & AB and taking radius as height of the triangle, which may lead to getting the wrong answer.
We can also check the answer is correct by equalizing RHS and LHS,
Area of a circle = area of a minor segment + area of major segment.
$\begin{align}
  & 706.5=20.4375+686.0625 \\
 & 706.5=706.5 \\
\end{align}$
Hence our answer is correct.