Prism:
The volume of a prism measures the amount of space taken up by that prism. The general formula for calculating the volume of a prism is simple i.e. volume = A* h, where A represents area of the base and h the height. The volume of a triangular prism is equal to the product of its base and altitude. A triangular prism is a threedimensional solid with a base, 9 edges, 5 faces and 6 vertices. A triangular prism is basically half of a rectangular prism, composed of 3 rectangles and 2 triangles on each end.
An oblique prism is one, in which bases that are not aligned directly one above the other. Notice that the lateral faces of an oblique prism are parallelograms. In general the volume of an oblique prism is equal to that of a right prism having its base a right section of the oblique prism and its altitude a lateral edge of the oblique prism. You measure the volume of prism solids in cubic units of measurement.
If ‘A’ is the area of base and h is height of the prism, formula for volume = A * h.
(More text below video...)
(Continued from above)
Look at some of the examples to understand the concepts and use the applicable formula:
Example 1: given in the figure below (Fig: 2, NTS), find the volume of the triangular prism.
The formula for volume of prism is A = A * h
Area of triangular base (A) = 1/2 * (5 * 5) = 12.5 m^{2}
Volume of the prism = A * h =
12.5 * 13.5 = 168.75 m^{3}
Volume is 168.75 m^{3}, as the final answer.
Example 2: given in the figure above (Fig: 2, NTS), find
the volume of the rectangular prism.
The formula for volume of rectangular prism is A = A * h
Area of triangular base (A) = 1/2 * (5 * 8) = 20.0 in^{2}
Volume of the prism = A * h = 20 * 10.0 = 200 in^{3}
Volume is 200 in^{3}, as the final answer.
Example 3: Given in the Fig: 3 (NTS), there are two stacks of paper sheets with equal numbers. The shape of first stack is a right prism and the second an oblique prism. Also both he stacks have the same height, base and volume. Find out the volume of the oblique prism
Step 1: Find area of the base of right prism. The base of the right prism is a square with length of a side 8 in.
Step 2: The base area of the right prism equals A = 82 = 64 in^{2}.
Step 3: Volume of the right prism equals V = A * h =
64 * 8= 512 in^{3}
Step 4: Both the right prism and the oblique prism have the same volume (given).
Therefore, the volume of the oblique prism is 512 in^{3}.
Example 4: Given find out the volume of the prism shown in Fig: 4?
You may notice that the base of the prism is a trapezoid.
Step 1: The length of the two parallel sides of the trapezoid
is 5 cm and 7.5 cm respectively, and height of trapezoid is 6 cm
(given in figure).
Step 2: Height of trapezoidal prism is given as 12 cm.
Step 3: Area of base of the trapezoidal prism
= 1 / 2 * h × (sum of the lengths of the two parallel sides) = (1/2) * 6(5 + 7.5)
= 37.5 cm^{2}(you’ll get on simplification).
Step 4: Recall formula: Volume of trapezoidal prism
= area of the base * height
= 37.5 * 12
= 450 cm^{3}, as the final answer.
Example 5: Given Determine the volume of the solid shown in Fig: 4. Assume all are the right prisms and those seen with four sided bases are rectangular.
Step 1: Volume of a rectangular prism with base area A and height h
= A * h
Step 2: In the given situation, Volume = l * b * h =
15 * 12 * 5
= 900 ft^{3}
Step 3: As the upper prism has a triangular base, the base area is = 1/2 * 12 *
4 = 24 ft^{2}
Step 4: Volume of the right prism A* h =
24 *15
= 360 ft^{3}
Step 5: Add both the volumes, it equals to:
= (900
+ 360) = 1260 ft^{3}
The volume of solid is 1260 ft^{3} , as the final answer.
Cylinders:
You may recall from earlier learning of concepts and properties, cylinder is a prism with the circular base. If the height of cylinder is, h, i.e. the length of the line segment whose endpoints are the centers of the bases, and radius of the base is r:
Volume of Cylinder =r^{2}h
cubic units
E.g. volume of the cylinder, wit radius of the base 2 in and height 4 in, using this formula equals to 50.25 in^{3}.
Example 1: Given a hollow cylindrical tube (open at both ends) is made of steel 1 cm thick. If the external diameter is 15 cm and the length of tube is 75 cm. Find the volume of steel used in making the tube.
Step 1: The external diameter is 15 cm i.e. radius is 7.5 cm (r). Thickness being 1 cm, internal radius equals to = 7.5 1 = 6.5 cm.
Given height is 75 cm.
Step 2: Volume of steel used = (External volume – Internal volume)
=
(7.52  6.52) * h
= 3.14(56.25  42.25) * 75. Simplify it.
= 75 * 43.96 cm^{3}
= 3297 cm^{3}, as the final answer.
Example 2: Given A field is 150 m long and 75 m wide. A circular tank of radius 5 m and depth 25 m is dug in the field and the earth taken out of it is spread evenly over the field. How much will be the rise in height of the field?
Notice that the measurement units should be made same for all the dimensions.
Step 1: Volume of earth dug out =
r^{2}h = 3.14 * 5 * 5 * 25 = 1962.5 m^{3}
Step 2: Area of the field = (150 * 75) = 11250 m^{2}
Step 3: Base area of tank = r^{2} = 3.14
* 5 * 5 = 78.5 m^{2}
Step 4: Area, where earth can be spread =
11250 – 78.5 = 11175.5 m^{2}
Step 5: As the volume of earth removed is known and also available area is known, height can be determined by: volume/area =
1962.5 / 11175.5 = 0.175 m
Rise in height of the field = 0.175 m, as the final answer.
Example 3: Given A bucket cylindrical in shape has radius of 15 cm and height 20 cm, is filled with water full. You have to transfer this water into a rectangular shape tub 70 cm long and 30 cm wide. Find out to what eight the water will rise in the tub.
First check that all the dimensions are in the same units.
Step 1: Volume of water in the bucket =
r^{2}h
= 3.14 * 15 * 15 * 20 cm^{3} = 14130 cm^{3}
Step 2: Assume h will be the height to which water rises in the tub.
Step 3: Volume of water in the tub = 70 * 30 * h
cm^{3}
Step 4: As given, 70 * 30 * h = 14130 h
= 14130 / (70 * 30) = 6.7 cm
The water rises in the tub to a height of 6.7 cm.
The video above will explain more in detail about Volume  Prisms, Cylinders,
Cones, Pyramids and Spheres, with the help of several examples and done watching video.
Winpossible's online math courses and tutorials have gained rapidly popularity since
their launch in 2008. Over 100,000 students have benefited from Winpossible's courses...
these courses in conjunction with free unlimited homework help serve as a very effective
mathtutor for our students.

All of the Winpossible math tutorials have been designed by topnotch instructors
and offer a comprehensive and rigorous math review of that topic.

We guarantee that any student who studies with Winpossible, will get a firm grasp
of the associated problemsolving techniques. Each course has our instructors providing
stepbystep solutions to a wide variety of problems, completely demystifying the
problemsolving process!

Winpossible courses have been used by students for help with homework and by homeschoolers.

Several teachers use Winpossible courses at schools as a supplement for inclass
instruction. They also use our course structure to develop course worksheets.