Question

Find the area of the minor segment of a circle of radius 42 cm, if the length of the corresponding arc is 44 cm.

Answer 1

Hint: In this sum, we will first draw figure for better understanding. After that, we will find angles subtended by an arc of the given length. For finding the area of the segment, we need to find areas of sector and area of a triangle formed. We can find the area of the sector by using angles found. For finding the area of the triangle we will use basic formula but for finding base and height we will use congruency, right-angled triangle properties.

Complete step-by-step solution
Formulas which we will use are:
I) $\text{Length of arc}={\displaystyle \frac{\theta }{{360}^{\circ }}}\times 2\pi r$ where r is radius of circle.
II) Area of segment = Area of the sector - Area of a triangle.
III) $\text{Area of sector}={\displaystyle \frac{\theta }{{360}^{\circ }}}\times \pi r^2$ where r is radius of circle.
IV) $\text{Area of triangle}={\displaystyle \frac{1}{2}}\times \text{base}\times \text{height}$
Let us draw the figure first:

Now, let us find angle subtended by arc with help of length of arc 44 cm, radius 42 cm. As we know,
$$\text{Length of arc}={\displaystyle \frac{\theta }{{360}^{\circ }}}\times 2\pi r$$
Putting the values we get:
$$\begin{array}{rl}& \text{44}={\displaystyle \frac{\theta }{{360}^{\circ }}}\times 2\pi \times 42\\ & 44={\displaystyle \frac{\theta }{{360}^{\circ }}}\times 2\times {\displaystyle \frac{22}{7}}\times 42\\ & {\displaystyle \frac{44\times {360}^{\circ }\times 7}{2\times 22\times 42}}=\theta \\ & \theta ={60}^{\circ }\cdots \cdots \cdots \left(1\right)\end{array}$$
As we can see from diagram, area of segment APB = Area of sector OAPB - Area of triangle OAB . . . . . . . . . (2) so, let us first find area of sector OAPB and area of triangle OAB. Given radius r = 42 cm.
$$\begin{array}{rl}& \text{Area of sector OAPB}={\displaystyle \frac{\theta }{{360}^{\circ }}}\times \pi r^2\\ & \Rightarrow {\displaystyle \frac{\theta }{{360}^{\circ }}}\times {\displaystyle \frac{22}{7}}\times {\left(42\right)}^2\end{array}$$
Putting value of $\theta$ from (1), we get:
$$\begin{array}{rl}& \Rightarrow {\displaystyle \frac{{60}^{\circ }}{{360}^{\circ }}}\times {\displaystyle \frac{22}{7}}\times 42\times 42\\ & \Rightarrow 924c{m}^2\end{array}$$
$$\text{Area of sector OAPB}=924c{m}^2$$
$$\text{Area of triangle AOB}={\displaystyle \frac{1}{2}}\times \text{base}\times \text{height}$$
But we don't know values of base or height. So let us find that,
Draw $OM\mathrm{\perp }AB$
Therefore, $\mathrm{\angle }OMB=\mathrm{\angle }OMA={90}^{\circ }$

In $\mathrm{\Delta }OMA\text{ and }\mathrm{\Delta }OMB$ we can see that $\mathrm{\angle }OMB=\mathrm{\angle }OMA={90}^{\circ }$
OA = OB as both is the radius of the same circle.
OM = OM which is common in both triangles.
Therefore, by RHS congruency criterion,
$$\mathrm{\Delta }OMA\widetilde{=}\mathrm{\Delta }OMB$$
By using CPCT (corresponding parts of congruent triangle) we can say,
$$\mathrm{\angle }AOM=\mathrm{\angle }BOM\text{ and }BM=AM$$
Therefore, $$\mathrm{\angle }AOM=\mathrm{\angle }BOM={\displaystyle \frac{1}{2}}\mathrm{\angle }BO{A}^{\circ }$$
As we have find $\theta =\mathrm{\angle }BOA={60}^{\circ }$
Therefore, $$\mathrm{\angle }AOM=\mathrm{\angle }BOM={\displaystyle \frac{1}{2}}\times {60}^{\circ }={30}^{\circ }$$
Also, $$BM=AM={\displaystyle \frac{1}{2}}AB\cdots \cdots \cdots \left(3\right)$$
Now from $\mathrm{\Delta }OMA$
$$\begin{array}{rl}& \sin {30}^{\circ }={\displaystyle \frac{\text{side opposite of angle 3}{\text{0}}^{\circ }}{\text{hypotenuse}}}\\ & \Rightarrow {\displaystyle \frac{1}{2}}={\displaystyle \frac{AM}{AO}}\\ & \Rightarrow {\displaystyle \frac{1}{2}}={\displaystyle \frac{AM}{42}}\\ & \Rightarrow AM=21\end{array}$$
Also,
$$\begin{array}{rl}& \cos {30}^{\circ }={\displaystyle \frac{\text{side adjacent of angle 3}{\text{0}}^{\circ }}{\text{hypotenuse}}}\\ & \Rightarrow {\displaystyle \frac{\sqrt{3}}{2}}={\displaystyle \frac{OM}{AO}}\\ & \Rightarrow {\displaystyle \frac{\sqrt{3}}{2}}={\displaystyle \frac{OM}{42}}\\ & \Rightarrow OM=21\sqrt{3}\end{array}$$
From (3) AB = 2AM
Therefore, putting value of AM, we get:
AB = 42 cm
Also, $OM=21\sqrt{3}cm$
So, we have found length of base and perpendicular for triangle AOB.
$$\begin{array}{rl}& \text{Area of triangle AOB}={\displaystyle \frac{1}{2}}\times \text{AB}\times \text{OM}\\ & \text{Area of triangle AOB}={\displaystyle \frac{1}{2}}\times \text{42}\times \text{21}\sqrt{\text{3}}\\ & \text{Area of triangle AOB}=441\sqrt{3}c{m}^2\end{array}$$
From (2) as we know, area of segment APB = area of sector OAPB - area of triangle OAB.
We have, $$\text{Area of segment}=(924-441\sqrt{3})c{m}^2$$

Note: Students should not get confused with length of arc or length of chord such that, by two points. Students should always draw diagrams for better understanding. Also, don't get confused with formulas of $\sin \theta \text{ and }\cos \theta$. Don't forget to put units after finding a value for area use squared units and length as linear units.