Find the area of the segment shaded in pQ24cm pr=7cm and Qr is the diameter of the circle with center O
Answers
Answer:
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