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secondary 4 | E Maths
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Someone please help me with part b thanks in advance
Date Posted: 4 years ago
Views: 637
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Recall Volume/capacity of regular shaped container = base area x height or cross section area x length.
So cross section area here
= capacity/volume ÷ length of container
= 2 000 000 cm³ ÷ 150 cm
= 13 333 ⅓ cm²
Now area of cross section = area of sector AOB
13 333 ⅓ cm² = πr²θ/360
13 333 ⅓ cm² = π(135/360)r²
r² = (13 333 ⅓ ÷ (135π/360) ) cm²
r = √(13 333 ⅓ ÷ (135π/360) ) cm
r ≈ 106.384608 cm
r = 106 cm (3s.f)
So cross section area here
= capacity/volume ÷ length of container
= 2 000 000 cm³ ÷ 150 cm
= 13 333 ⅓ cm²
Now area of cross section = area of sector AOB
13 333 ⅓ cm² = πr²θ/360
13 333 ⅓ cm² = π(135/360)r²
r² = (13 333 ⅓ ÷ (135π/360) ) cm²
r = √(13 333 ⅓ ÷ (135π/360) ) cm
r ≈ 106.384608 cm
r = 106 cm (3s.f)
done
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An alternate way to solve this question is to "Restore" the complete cylinder.
They have provided the volume of this cylindrical sector, and you have the angle of AOB.
Multiplying 2,000,000cm² by (360/135).
Dividing the result by 150cm (Height), you are left with the area of the base circle.
Just apply the Area of a Circle formula backwards and you can find the radius from there.
They have provided the volume of this cylindrical sector, and you have the angle of AOB.
Multiplying 2,000,000cm² by (360/135).
Dividing the result by 150cm (Height), you are left with the area of the base circle.
Just apply the Area of a Circle formula backwards and you can find the radius from there.
Date Posted:
4 years ago
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