Find the area of the segments shaded in figure, if PQ = 24 cm., PR = 7 cm. and QR is the diameter of the circle with centre O (Take π = 22/7 )
Answers
Given :
PQ= 24 cm ,PR = 7 cm
We know that any angle made by the diameter QR in the semicircle is 90°.
∠RPQ = 90°
In right angled ∆RPQ
RQ² = PQ² + PR²
[By pythagoras theorem]
RQ² = 24² + 7²
RQ² = 576 + 49
RQ² = 625
RQ = √625cm
RQ= 25 cm
radius of the circle (OQ)= 25 / 2 cm
Area of right ∆ RPQ= ½ × Base × height
Area of right ∆ RPQ= ½ × RP × PQ
Area of right ∆ RPQ = ½ × 7 × 24 = 7 × 12 = 84 cm²
Area of right ∆ RPQ = 84 cm²
Area of semicircle= πr²/2
= (22/7) × (25/2)² / 2
= (22 × 25 × 25)/ (7× 2 × 2 × 2)
= 11 × 625 /28 = 6875/28 cm²
Area of semicircle = 6875/28 cm²
Area of the shaded region = Area of semicircle - Area of right ∆ RPQ
= (6875/28 - 84 )cm²
= (6875 - 2532)/ 28
Area of the shaded region = 4523 / 28= 161.54 cm²
Hence, the area of the shaded region = 161.54 cm²
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Answer:
ANSWER
PQ=24cm ,PR =7 cm
We know that any angle made by the diameter QR in the semicircle is 90°.
∴∠RPQ=90°
In right angled ∆RPQ
RQ^2 =PQ^+ +PR ^2
[By pythagoras theorem]
RQ²=24²+7²
RQ²=576+49
RQ²=625
RQ=√625cm
RQ=25cm
radius of the circle (OQ)= RQ /2= 25/2cm
Area of right ∆RPQ= 1/2 ×Base×height
Area of right ∆RPQ= 1/2 ×RP×PQ
Area of right ∆RPQ= 1/2×7×24=7×12=84cm²
Area of right ∆RPQ=84cm²
Area of semicircle=πr²/2
=22/7×25/2×25/2×1/2= 11×25×25/28= 6875/28cm ^2
Area of the shaded region = Area of semicircle - Area of right ∆ RPQ 6875−2352/28
= 4523/28
=161.54 cm ^2
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