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secondary 3 | E Maths
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Celine
Celine

secondary 3 chevron_right E Maths chevron_right Singapore

Pls help, thanks!

Date Posted: 2 years ago
Views: 278
Eric Nicholas K
Eric Nicholas K
2 years ago
We take a 144° chunk out of the full 360° of a circle (i.e. a fraction 144/360 or 2/5 of a circle), so we take only 2/5 of the area of the full circle to which each sector comes from.

[That's what the θ/360° in the area formula θ/360° x πr² or the arc length formula θ/360° x 2πr means]

For the big sector, its area is 144°/360° times π(5 + x)², or simply put, 2/5 times π(5 + x)².

For the small sector, its area is 144°/360° times π(5)², or simply put, 2/5 times π(5)²,

Their difference in areas give the shaded region of area 30π.

So, we have
2/5 times π(5 + x)² - 2/5 times π(5)² = 30π

Dividing all sides by 2/5,
π(5 + x)² - π(5)² = 30π / (2/5)
π(25 + 5x + 5x + x²) - 25π = 75π
π(25 + 10x + x²) = 100π
25 + 10x + x² = 100
x² + 10x - 75 = 0
(x + 15) (x - 5) = 0
x + 15 = 0 or x - 5 = 0
x = -15 or x = 5

Of course, x is a length which much be positive in value, so x = 5.

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arnold
Arnold's answer
58 answers (Tutor Details)
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Hope this helps and also take note not to reject for this question
Celine
Celine
2 years ago
Clear, neat, and good explanation.