Sanjana's answer to ;-)'s Secondary 3 E Maths Singapore question.
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Perhaps there is a shorter approach but this would be mine:
Area enclosed = Area of trapezium - partial areas of circles
= [ (4p + p) x 4p x 0.5 ] - [(53.13/360) x pi x 16p^2] - {[(90-53.13 + 90)/360)] x pi x p^2 = 10p^2 - 7.4183p^2 - 1.1074p^2
= 1.4743 p^2
= 1.47 p^2
** the angle found using tangent should be left in 2 d.p or more as small differences here have very large difference in the working. Your answer will be off by +/- 0.01 dp or more depending on the d.p of the numbers in your working
Area enclosed = Area of trapezium - partial areas of circles
= [ (4p + p) x 4p x 0.5 ] - [(53.13/360) x pi x 16p^2] - {[(90-53.13 + 90)/360)] x pi x p^2 = 10p^2 - 7.4183p^2 - 1.1074p^2
= 1.4743 p^2
= 1.47 p^2
** the angle found using tangent should be left in 2 d.p or more as small differences here have very large difference in the working. Your answer will be off by +/- 0.01 dp or more depending on the d.p of the numbers in your working
Date Posted:
3 years ago
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