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Question

A chord 10 cm long is drawn in a circle whose radius is 52 cm. Find the areas of both the segments.

[Takeπ=3.14.]

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Solution

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Given, a chord AB of length 10 cm and radius = OA = OB = 52 cm.

Construction: Draw OC perpendicular to AB.

Now, AC = BC = 102 = 5 cm [The perpendicular drawn from the centre of a circle to a chord always bisect the chord.]

In OAC,

sin x=ACOAsin x=552sin x=12sin x=sin 45ox=45o

Similarly, BOC=45oAOB=AOC+BOC=45o+45o=90o

We know that, area of sector =θ360×πr2

Therefore, area of sector OAB=θ360×πr2=90360×π(52)2=50π4=252×227=2757=39.28 cm2

Again, in OAC,

cos x=OCOAcos 45o=OC5212=OC52OC=5 cm

Therefore, area of OAB

=12×OC×AB=12×5×10=25 cm2

Now, area of minor segment = Area of sector OAB - Area of OAB

= 39.28 cm2 - 25 cm2​​​​​​​

= 14.28 cm2​​​​​​​

Similarly, area of major segment = Area of circle - Area of minor segment

⇒ Area of major segment

=πr214.28=227×(52)214.28=157.142814.28=142.86 cm2


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