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A circle has radius of 18 cm. Find the area of the smaller of the two regions determined by a chord with length of 18√2 cm . Hint: (Look for right triangles using Pythagorean Theorem Converse)

User Firephil
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2 Answers

1 vote

Answer:

81pi - 162

Explanation:

User Mridul Kashatria
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The area is 81*pi-162. Look at the triangle formed by the chord and the two radii connected to the endpoints of the chord. Since 18^2+18^2=(18*\sqrt2)^2, this is an isosceles right triangle by Pythagorean theorem. The area of the quarter circle containing the region we look for is pi*18^2/4=81*pi. The area of the right triangle is 18*18/2=162. The difference between these two areas, or 81*pi-162, is the area of the smaller region.
User DennisWelu
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