A chord of a circle of radius 20, cm subtends | vedclass

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Question

Let $A, B$ be the chord of a circle of radius $10\, cm$, with $O$ as the centre of the circle (see $Fig.$)

Here, $\angle A$ O $B =90^{\circ}$ and w

Vedclass · Accepted Answer

$285.5$

Vedclass · Answer

Let $A, B$ be the chord of a circle of radius $10\, cm$, with $O$ as the centre of the circle (see $Fig.$)

Here, $\angle A$ O $B =90^{\circ}$ and we have to find the area of the major segment (which is shaded). As $\angle AOB =90^{\circ},$ therefore angle of the major sector $=360^{\circ}-90^{\circ}=270^{\circ}$

So, area of the major sector $=\frac{270}{360} 	imes \pi 	imes(10)^{2} \,cm ^{2}$

$=\frac{3}{4} 	imes 3.14 	imes 100\, cm ^{2}$

$=75 	imes 3.14 cm ^{2}=235.5\, cm ^{2}$

Now, to find the area of $	riangle$ $OAB,$ draw $OM$ $\perp$ $AB.$

$So , AM =\frac{1}{2} A B$ and $\angle AOM =\frac{1}{2} 	imes 90^{\circ}=45^{\circ}$

Now, $\frac{	ext { AM }}{	ext { OA }}=\sin 45^{\circ}=\frac{1}{\sqrt{2}}$

So, $\quad A M=10 	imes \frac{1}{\sqrt{2}}\, cm$

Therefore, $\quad A B=10 \sqrt{2} c m$ and $O M=O A \cos 45^{\circ}=10 	imes \frac{1}{\sqrt{2}} cm =5 \sqrt{2}\, cm$

So, area of $	riangle$ $OAB$ $=\frac{1}{2}$ base $	imes$ height

$=\frac{1}{2} 10 \sqrt{2} 	imes 5 \sqrt{2}\, cm ^{2}=50 \,cm ^{2}$

Therefore, the area of the required major segment

$=235.5 \,cm ^{2}+50 \,cm ^{2}=285.5 \,cm ^{2}$

A chord of a circle of radius $20\, cm$ subtends an angle of $90^{\circ}$ at the centre. Find the area of the corresponding major segment of the circle. (Use $\pi=3.14)$ (in $cm^2$)

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