Question

Find the area of the segment's shaded in figure ,if PQ=24cm,PR=7 cm and QR is the diameter of the circle with centre O
$(taken\pi = \frac{22}{7} )$

Answer 1

Hello Friend.....

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The answer of u r question is...

Ans:

Area of the segment's shaded =Area of the sector OQPR - Area of Triangle PQR...

Since,

QR is diameter ,<QPR =90° {Angle in a semicircle}

So,

using Pythagoras theorem..

In triangle PQR,

${qr}^{2} = \: {pq}^{2} + {pr}^{2}$

=24^2+7^2

=576+49

=625

QR = √625

=25cm...

Then radius
of the circle = 1/2 QR

=1/2(25)=25/2cm

Now,

area of semicircle OQPR=1/2 πr^2

=1/2×22/7×25/2×25/2

=245.53cm^2.....(1)

Area of right
angled triaQPR= 1/2×PR ×PQ

=1/2×7×24

=84cm^2....(2)

From 1 and 2

Area of shaded segments

=245.53-84

=161.53 cm^2

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Thank you...⭐️⭐️⭐️

Answer 2

Aloha user....."

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QUESTION:- Find the area of the segment's shaded in figure ,if PQ=24cm,PR=7 cm and QR is the diameter of the circle with centre O
(taken\pi = \frac{22}{7} )


ANSWER:- given

=> area of semi circle = 90 *

=> angle ROQ =90*

=> in triangle QPR, BY Pythagoras theorem

=> QR^2= PR^2 + PQ^2

=>. 7^2+ 24^2

=> 49+ 576

=> 625

=> QR^2= root of 625

=> QR= 25cm

=> here diameter = 25cm

=> radius = 25/2

=> Area of circle = pi R^2

=> semi circle = 1/2 Xpi R^2

=> 1/2 X 22/7 X (25/2)^2

=> 11 X 25 X 25 / 28

=> 6875/28cm

=> Area of triangle PQR

=> 1/2 X base X height

=> 1/2 X PQ X PR

=> 1/2 X 24 X 7

=> 12 X 7

=> 84 cm ^2

Now , area of the shaded region

=> area of semicircle --- area of triangle PQR

=> 6875/28 --- 84

=> 6875-84(28)/28

=> 6875-2352/28

=> 4523/28

=> 161.53cm^2

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