Contents:
Volume: Cube, Cuboid, Cylinder, Cone, Sphere, Prism, Pyramid.
The volume of
a cube and a cuboid is heigth x width x depth; for rectangles: (h x w x d),
for a cube where all sides have identical lengths: (h
).
Example: Find the volume of a cube of
height: 5mm, width 5mm, depth 5mm.
Answer: 5 x 5 x 5 = 125mm. (5= 125mm).
Example: Find the volume of a cuboid of
height 2mm, width 5mm, depth 2mm.
Answer: 2 x 5 x 2 = 20mm.
The surface area of a cuboid may be measured
by opening up the shape into a mesh. It is seen that the top,
bottom, front and back have an area of 5 x 2 and the two sides
have an area of 2 x 2. This gives: 4 x 5 x 2 + 2 x 2 x 2. Total
surface area: 48mm
.
The volume of a cylinder is the area of the
base (circle) x height; (
rh).
The volume of a cone is the area of the base (circle) x height
3; (rh3).
Example: Find the volume of a cylinder radius
2mm, height 5mm.
Answer = 3
14 x 4 x 5
= 628mm.
Example: Find the volume of a cone radius 2mm, height 5mm.
Answer = 314 x 4 x 5 3 = 209mm.
The surface area of a cylinder is found
by first calculating the areas of the top and bottom circle. Then
by opening up the wall. The length of the wall is the circumference
of the circle, this is multiplied with the height of the wall.
Area = (2 x r) + (2 xr x h).
Example: Find the surface area of a cylinder radius 2mm, height 5mm.
Answer = (2 x 314
x 4) + (2 x 314 x 2 x 5) = (2512) + (628) = 879mm.
The volume of a sphere
is (4 xr)3.
The surface area of a sphere is 4 xr.
Example: Find the volume of a sphere of radius
4mm.
Answer: (4 x 314
x 64)3 = 267.95mm.
Example: Find the radius of a sphere volume
268mm.
Answer: formula 268 = 
xr, rearranged:
r = 
(268 x 3)
4 x , = 4mm.
Example: Find the surface area of a sphere of radius 4mm.
Answer: formula 4 x 314
x 16 = 201mm.
The volume of
the triangular prism is the area of the triangle x length or depth; (h x b x
l 2).
Example: Find the volume of a triangular
prism height 4mm, base 5mm, length 8mm.
Answer: 4 x 5 x 82
= 80mm.
The surface area is the area of the end triangles plus the 3 rectangles. The areas of the rectangles depend on the dimentions of the end triangle. (If the triangle was equilateral then the three rectangle areas would be identical.) The length of the sloping edge of the triangle otherwise has to be known, but can be calculated by Pythagaros theorem (see Pyramid).
Example: Find the surface area of a triangular prism height 4mm, base 5mm, length 8mm, sloping side 7mm.
Answer: triangle area: 4 x 25
x 2 = 20mm; rectangle area 7 x
8 x 4 = 224mm; total area = 244mm.
The volume of a pyramid
is similar to the cuboid. The area of the base is calculated x height 3; (l x w x h3).
Example: Find the volume of a pyramid
with a base area of 15mm, and a height
of 6mm.
Answer: 15 x 63
= 30mm.
The surface area of a pyramid involves knowing the length of the sloping edge and using Pythagoras' theorem. Using one of the faces drop a perpendicular line from apex to base, this creates a right angle triangle. This line is calculated by taking the square of the sloping edge subtracting the square of half of the base and square rooting the result. This measurement is multiplied by half of the base measurement to give the area for one face. Times this by 4 and add the floor area for the total surface area. (This is for a pyramid of square base, if rectangular the differing sides need to be calculated).
Example: Find the surface area of a pyramid
with a square base of 8mm and a vertical slope of 64mm.
Answer: length of perpendicular = 64- 4= 25, 
25 = 5mm,
area: 5 x 4 = 20mm. Floor area
= 16mm, + 4 sides of 20mm = 96mm.
Exercise in calculating volumes:
