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secondary 4 | E Maths
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Math question from BMAT
Date Posted: 4 years ago
Views: 306
Biggest circle - 2nd biggest + 2nd smallest - smallest
π(4d/2)² - π(3d/2)² + π(2d/2)² - π(d/2)²
= 4d²π - 9/4 d²π + d²π - ¼d²π
= πd²(4 - 9/4 + 1 - ¼)
= πd² (5/2)
= 5/2 πd²
Option A
π(4d/2)² - π(3d/2)² + π(2d/2)² - π(d/2)²
= 4d²π - 9/4 d²π + d²π - ¼d²π
= πd²(4 - 9/4 + 1 - ¼)
= πd² (5/2)
= 5/2 πd²
Option A
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Date Posted:
4 years ago
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