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Area and Volume JCHL

Higher Level
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Area and Volume
Find the area of the rectangle.
π΄π‘Ÿπ‘’π‘Ž = πΏπ‘’π‘›π‘”π‘‘β„Ž × π‘€π‘–π‘‘π‘‘β„Ž
= 6 βˆ’ 11 6 + 11
= 6 6 + 11 βˆ’ 11 6 + 11
= 36 + 6 11 βˆ’ 6 11 βˆ’ 11
= 25
Circles
π΄π‘Ÿπ‘’π‘Ž = πœ‹π‘Ÿ 2
πΆπ‘–π‘Ÿπ‘π‘’π‘šπ‘“π‘’π‘Ÿπ‘’π‘›π‘π‘’ = 2πœ‹π‘Ÿ
πœƒ
360
πœƒ
πΏπ‘’π‘›π‘”π‘‘β„Ž π‘œπ‘“π΄π‘Ÿπ‘ = 2πœ‹π‘Ÿ ×
360
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘†π‘’π‘π‘‘π‘œπ‘Ÿ = πœ‹π‘Ÿ 2 ×
Cuboid
Cone
Cylinder
π‘‰π‘œπ‘™π‘’π‘šπ‘’ = 𝑙 × π‘ × β„Ž
π‘‰π‘œπ‘™π‘’π‘šπ‘’ =
1 2
πœ‹π‘Ÿ β„Ž
3
πΆπ‘’π‘Ÿπ‘£π‘’π‘‘ π‘†π‘’π‘Ÿπ‘“π‘Žπ‘π‘’ π΄π‘Ÿπ‘’π‘Ž = 2πœ‹π‘Ÿβ„Ž
πΆπ‘’π‘Ÿπ‘£π‘’π‘‘ π‘†π‘’π‘Ÿπ‘“π‘Žπ‘π‘’ π΄π‘Ÿπ‘’π‘Ž = πœ‹π‘Ÿπ‘™
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘†π‘’π‘Ÿπ‘“π‘Žπ‘π‘’ π΄π‘Ÿπ‘’π‘Ž = 2πœ‹π‘Ÿβ„Ž + 2πœ‹π‘Ÿ 2
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘†π‘’π‘Ÿπ‘“π‘Žπ‘π‘’ π΄π‘Ÿπ‘’π‘Ž = πœ‹π‘Ÿπ‘™ + πœ‹π‘Ÿ 2
2
𝑙 =β„Ž +π‘Ÿ
π‘†π‘’π‘Ÿπ‘“π‘Žπ‘π‘’ π΄π‘Ÿπ‘’π‘Ž = 2𝑙𝑏 + 2π‘β„Ž + 2π‘™β„Ž
Sphere
2
Hemisphere
4
π‘‰π‘œπ‘™π‘’π‘šπ‘’ = πœ‹π‘Ÿ 2 β„Ž
2
Prism
2
π‘‰π‘œπ‘™π‘’π‘šπ‘’ = 3 πœ‹π‘Ÿ 3
π‘‰π‘œπ‘™π‘’π‘šπ‘’ = 3 πœ‹π‘Ÿ 3
π‘†π‘’π‘Ÿπ‘“π‘Žπ‘π‘’ π΄π‘Ÿπ‘’π‘Ž = 4πœ‹π‘Ÿ 2
π‘†π‘’π‘Ÿπ‘“π‘Žπ‘π‘’ π΄π‘Ÿπ‘’π‘Ž = 2πœ‹π‘Ÿ 2
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘†π‘’π‘Ÿπ‘“π‘Žπ‘π‘’ π΄π‘Ÿπ‘’π‘Ž = 2πœ‹π‘Ÿ 2 + πœ‹π‘Ÿ 2
π‘‰π‘œπ‘™π‘’π‘šπ‘’ = π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘π‘Žπ‘ π‘’ × πΏπ‘’π‘›π‘”π‘‘β„Ž
A container in the shape of a cylinder has a capacity of
50 litres. The height of the cylinder is 0.7m. Find the
length of the diameter of the cylinder.
Give your answer correct to the nearest whole number.
A rectangular tank has a length of 0.6m, a
width of 0.35m and its height measures 15cm.
Find the capacity of the tank.
The diameter of
each sphere is
equal to the length
of each side of the
cube.
Find the volume
of the ornament.
0.6π‘š = 60π‘π‘š
0.35π‘š = 35π‘π‘š
π‘‰π‘œπ‘™π‘’π‘šπ‘’ = 𝑙 × π‘ × β„Ž
Change measurements:
1 π‘™π‘–π‘‘π‘Ÿπ‘’ = 1,000 π‘π‘š3
50 π‘™π‘–π‘‘π‘Ÿπ‘’ = 50,000 π‘π‘š3
2
π‘‰π‘œπ‘™π‘’π‘šπ‘’ = πœ‹π‘Ÿ β„Ž = 50,000
(3.14)π‘Ÿ 2 (70) = 50,000
219.8π‘Ÿ 2 = 50,000
50,000
π‘Ÿ2 =
219.8
π‘Ÿ 2 = 227.48
π‘Ÿ = 15.08
π‘‘π‘–π‘Žπ‘šπ‘’π‘‘π‘’π‘Ÿ = 2 × π‘Ÿπ‘Žπ‘‘π‘–π‘’π‘ 
= 60 × 35 × 15
1π‘š = 100π‘π‘š
0.7π‘š = 70π‘π‘š
= πŸ‘πŸπŸ“πŸŽπŸŽ π’„π’ŽπŸ‘
The rectangular tank is full of water. This is
then poured into the cylindrical container
above. Find the depth of water in the cylinder.
Length of Cube =
24
=6
4
Radius of Sphere = 3
π‘‰π‘œπ‘™π‘’π‘šπ‘’ = πœ‹π‘Ÿ 2 β„Ž
Volume = 2 Cubes + 2 Spheres
πœ‹π‘Ÿ 2 β„Ž = 31500
=2 𝑙×𝑏×β„Ž +2
4 3
πœ‹π‘Ÿ
3
= 2 6×6×6 +2
4
(3.14)(3)3
3
3.14 15.08 2 β„Ž = 31500
= 2 15.08
714.06β„Ž = 31500
= πŸ‘πŸŽ. πŸπŸ”π’„π’Ž or 𝟎. πŸ‘πŸŽπŸπŸ”π’Ž
31500
β„Ž=
714.06
𝒉 = πŸ’πŸ’. πŸπŸπ’„π’Ž
= 2 216 + 2 113.04
= 432 + 226.08
= πŸ”πŸ“πŸ–. πŸŽπŸ– π’„π’ŽπŸ‘
The dimensions of two solid cylinders are shown in the diagrams below.
A solid hemisphere has a radius of 12 cm. Calculate the volume of the
hemisphere in terms of 𝝅.
4
π‘‰π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ π‘†π‘β„Žπ‘’π‘Ÿπ‘’ = 3 πœ‹π‘Ÿ 3
2
π»π‘’π‘šπ‘–π‘ π‘β„Žπ‘’π‘Ÿπ‘’ = 3 πœ‹π‘Ÿ 3
2
= 3 πœ‹(12)3
= 1152πœ‹
A solid cone of radius 4cm and height 12cm is cut out from the
hemisphere. Calculate the volume of the cone in terms of 𝝅.
Calculate the ratio of the curved surface area of the smaller cylinder to the
curved surface area of the larger cylinder.
1
π‘‰π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ πΆπ‘œπ‘›π‘’ = πœ‹π‘Ÿ 2 β„Ž
3
1
= πœ‹(4)2 (12)
3
= 64πœ‹
πΆπ‘’π‘Ÿπ‘£π‘’π‘‘ π‘†π‘’π‘Ÿπ‘“π‘Žπ‘π‘’ π΄π‘Ÿπ‘’π‘Ž = 2πœ‹π‘Ÿβ„Ž
Smaller: Larger
2πœ‹π‘Ÿβ„Ž: 2πœ‹π‘Ÿβ„Ž
2πœ‹π‘Ÿβ„Ž: 2πœ‹(2π‘Ÿ)(2β„Ž)
2πœ‹π‘Ÿβ„Ž: 8πœ‹π‘Ÿβ„Ž
1: 4
The remaining metal in the hemisphere is melted down and recast into
cones of the same dimensions as the cone above. How many cones can be
formed from the remaining metal.
π‘…π‘’π‘šπ‘Žπ‘–π‘›π‘–π‘›π‘” π‘€π‘’π‘‘π‘Žπ‘™ = π»π‘’π‘šπ‘–π‘ π‘β„Žπ‘’π‘Ÿπ‘’ βˆ’ π‘π‘œπ‘›π‘’
= 1152πœ‹ βˆ’ 64πœ‹
= 1088πœ‹
1088πœ‹
= 17
64πœ‹
Area of the triangle is 10cm. Calculate 𝒙.
1
π‘β„Ž = 10
2
1
(2π‘₯)(π‘₯ + 3) = 10
2
(2π‘₯)(π‘₯ + 3) = 20
2π‘₯ 2 + 6π‘₯ = 20
2π‘₯ 2 + 6π‘₯ βˆ’ 20 = 0
2π‘₯ + 10 π‘₯ βˆ’ 2 = 0
2π‘₯ = βˆ’10
𝒙=𝟐
Not a
Solution

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