In Figure, PQ is a diameter of the circle with centre O. PS and QR are perpendiculars to a line XY. PS intersects the circle at T. Prove that ST = QR.
Answers
Answer:
Given, Length (PQ)= 10cm
Given, Breadth (QR) = 7cm
Area of Rectangle PQRS will be = 10 * 7
= 70cm^2
Given that two semicircles of Diameter QR and PS are cut off, then
Area of remaining part will be Area of the rectangle - Area of the semicircle
Area of semicircle = pi * r * r/2
= (22/7 * 3.5 *3.5 )/2
= 38.5/2
= 19.25cm^2
Area of two semicircles = 19.25 + 19.25
= 38.50cm^2
Area of remaining portion = 70 - 38.50
= 31.50cm^2.
Hope this helps!
sanjayroyadv
Answer:
given , length (pq) = 10 CM
Given , breadth (qr) = 7 CM
Area of rectangle pqrs will be = 10×7 =70 cm^2
Given that two semicircles of diameter qr and PS are cut off then
Area of remaining part will be Area of the rectangle - Area of the semicircles
Area of semi circle = pi×r×r/2
22/7×3.5×3.5/2
38.5/2
=19.25 cm^2
Area of two semicircles = 19.25+19.25=38.50 cm^2
area of remaining portion = 70-38.50 = 31.50cm^2
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