Question

A circle of having an area of 144 π sq.cm, is cut into two segments by a chord that is 6cm from the center of the circle.What is the area of the smaller segment?

Answer 1

the area would be 43

Answer 2

The area of the smaller segment is 88.38 squared centimeters.

The area of a circle is given in the question. Now, find the value of the radius of the circle with the help of its area.

$\rm{Area}=144 \pi\\\pi r^2=114 \pi\\r^2=144\\r=12\;\rm{cm}$

Now, the radius of the given circle is 12 cm. We need to determine the area of the smaller segment. The diagram is attached to the solution to give you a clear view of the solution.

Let us assume the angle AOD is $\theta$ and the length of AD be x units.

Now, in the triangle AOD.

$\begin{aligned}\cos \theta &= \dfrac{6}{12}\\&=\dfrac{1}{2}\\&=60^o \end{aligned}$

Similarly, $\angle BOD=60^o$

Therefore, $\angle AOB = 120^o$

Now,

The triangle AOD is a right triangle, apply the Pythagoras theorem to find AD.

Therefore,

$x^2+6^2=12^2\\x^2=144-36\\x^2=108\\x=10.39$

Similarly,

The triangle BOD is a right triangle, apply the Pythagoras theorem to find AD.

Therefore,

$x^2+6^2=12^2\\x^2=144-36\\x^2=108\\x=10.39$

Thus, the length of AB is 20.78 cm.

The area of smaller segment = Area of sector AOB - Area of triangle AOB

Now, calculate the area of sector AOB.

$Ar(AOB)=\dfrac{\pi \times 12^2 \times 120}{360}\\=48 \pi\\=150.72 \;\rm{cm^2}$

Now, calculate the area of triangle AOB.

$\begin{aligned}Ar(\triangle AOB)&=2 \times Ar(\triangle AOD)\\&=2 \times \dfrac{1}{2} \times AD \times OD\\&=10.39 \times 6\\&=62.34\;\rm{cm^2} \end{aligned}$

Thus, the area of the smaller segment is 88.38 squared centimeters.

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