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Tagged: Measurement, volume
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Cube

$$V=s^3$$
Rectangular Prism

$$V=LWH$$
Triangular Prism

$$V=\frac{1}{2}bhL$$
Pyramid

$$V=\frac{1}{3}LWH$$
Cylinder

$$V=\pi r^2h$$
Cone

$$V=\frac{1}{3}\pi r^2h$$
Sphere

$$V=\frac{4}{3}\pi r^3$$
Truncated Cone

$$V=\frac13\pi R^2\,H-\frac13\pi r^2\,h $$
or
$$V=\frac13\pi(R^2\,H- r^2\,h)$$
The diagram shows a cone inscribed in a hemisphere. What is the ratio of the volume of the cone to the volume of the hemisphere?
a. 1:3
b. 2:3
c. 1:2
d. More information neededVolume of hemisphere = $$\frac12\times\frac43\times\pi\times r^3$$
$$=\frac23\pi r^3$$
Volume of cone = $$\frac13\times\pi\times r^2\times h$$
since h = rV =$$\frac13\times\pi\times r^2\times r$$
$$=\frac13\pi r^3$$
Ratio of volume of Cone to volume of Hemisphere
$$\frac13\pi r^3:\frac23\pi r^3$$
$$\frac13:\frac23$$ × both sides by 3
C 1:2
A cylinder has a radius of 10cm and a height of 15 cm.
Find the volume.
##V=\pi\times10^2\times20##
= 6283.19 cm2
V = $$\pi r^2h$$
r = 10, h = 20
A Christmas tree decoration in the shape of a sphere is 4.5cm in diameter.
a. Find the volume of the decoration. V =4/3πr3.
b. The 12 decorations are to fit into a rectangular box which is 10cm wide and 5cm high. What is the minimum length that the box can be if it is to hold all 12 decorations?a. V = 4/3 × π × 2.253
= 47.7cm3
Use the volume formula for a sphere: $$V=\frac{4}{3}\pi r^3$$ and calculate, where r = ½ of the diameter
radius = ½ × 4.5 = 2.25b. 2 × 1 = 2 If it is 10cm wide, that means it can fit 2 balls across its width. 2 × 1 × 6 = 12 balls If 5cm high, can only fit one ball. so if we can fit two per row, and we need 12, so it has to be able to fit six rows along its length, 2 × 1 × 6 = 12 balls 6 × 4.5 = 27cm now if each ball is 4.5cm in diameter, then 6 balls long to fit the 12, so 6 × 4.5 -
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