6.1 – Volume of rectangular prisms, triangular prisms, cylinders, rectangular pyramids,cones, spheres,
Simple compound shapes
Curriculum Outcomes:
C36explore, determine, and apply relationships between perimeter and area, between surface area and volume
D1determine and apply formulas for perimeter, area, surface area, and volume
D7determine the accuracy and precision of a measurement (optional)
D13demonstrate an understanding of the concepts of surface area and volume
Volume
The volume of an object is the amount of space it occupies. The units for measuring volume are cubic units. Cubic centimetres (cm3) and cubic metres (m3) are examples of units used for measuring volume.
The amount of space inside a container is called capacity. The units for measuring capacity are millilitres (mL) and litres (L). Important conversions to remember are 1 mL – 1cm3 and therefore 1000 cm3 = 1 L.
Rectangular prisms, triangular prisms, trapezoidal prisms and cylinders all have a definite base that remains constant over its height. As a result, the product of the area of the base and height of the object produces its volume.
The volume of a pyramid or cone is calculated by using the formula V = Ah where A is the area of the circular base (A= πr2) and h is the perpendicular height.
The volume of a sphere, a ball-shaped round object, is found by the formula V = πr3.
There are many real-world objects that are prisms, pyramids, cones and spheres. Students should be shown these objects so they can become familiar with the names and properties of the various shapes. When they are able to distinguish between the shapes and to describe the properties of these shapes, they can proceed with mathematical calculations to determine the volume. Remind the students to include the correct units in their answers.
Examples
1.A beverage can has the following dimensions. What is its volume?
Solution 1
A = πr2 (Area of the Base)
A = (3.14) (8)2
A = 3.14 × 64
A = 200.96
A = 201 cm2
V = Ah
V = (201 cm2) (18 cm)
V = 3618 cm3
The volume of the beverage can is 3618 cm3.
2.A snow cone is sold in the following container. What volume of the snow cone would the container hold?
Solution 2
Diameter = 6.0 cm
Radius =
Radius =
Radius = 3.0 cm
V =
V = (28.3 cm2) (9cm)
V = 84.9 cm3
The volume of the snow cone is 84.9 cm3.
3.A chocolate bar is sold in the following box. Calculate the space inside the box.
Solution 3
V = Ah
V = (1600 mm2) (200 mm)
V = 320 000 mm3
The space inside the box is 320 000 mm3.
4.A company is making balls to send to all elementary schools. In order to make a container to store the balls, the company must determine the volume of the ball. Can you help the company?
Solution 4
The volume of the ball is 230 cm3.
5.Calculate the volume of metal in this length of copper pipe.
Solution 5
Step 1: All measurements should be in the same units.
2.5 m
7.0 cm = 0.07 m
5.0 mm = 0.005 m
Step 2: Find the circumference of the pipe. The circumference of the pipe is actually a dimension of the rectangle. It gives us the width. From here we can calculate the Area.
C = π × d
C = 3.14 × 0.07
C = 0.22 m
Step 3: Calculate the Area of the pipe.
Area = L × w
Area = 2.5 m × 0.22 m
Area = 0.55 m2
Step 4: The volume in the metal pipe is the area times the thickness (h).
Volume = Area × h
Volume = (0.55 m2) × (0.005 m)
Volume = (0.00275 m3)
Exercises: (Use 3.14 for π)
1.A rectangular bar of gold has dimensions 24 cm × 18 cm × 12 cm. The bar must be melted and recast into 6 cm cubes. How many gold cubes can be made from the rectangular bar?
2.Players from the provincial universities football teams are participating in a “build a snow fort” competition. Each team has the same amount of snow, the construction must be a team plan. The team from St. F.X. plans to build their fort with rectangular blocks of snow each measuring 20 cm × 15 cm × 10 cm. These blocks will be built with snow from their conical pile that is 6 m wide and 2.5 m high. How many snow blocks will the team have to build their fort?
3.Jack, Jim, and Tommy are participating in a “ball of wind” contest. Each contestant is given a deflated ball to inflate in a two-minute time period. The contestant who blows the most air into the ball will win the contest. After the time had elapsed, the judges measured each ball. Jack’s ball had a diameter of 18 cm, Jim’s ball 16 cm, and Tommy’s ball 21 cm. Show mathematically how much air each contestant had blown into his ball.
4.Toys-R-Us are selling plastic ice-cream cone drinking cups that have the following measurements:
How much juice will each cup hold?
5.For each of the following shapes:
a)name each and give a real-world example of the shape
b)explain how you could calculate the volume of shape ii
i) ii)
iii)
Compound Shapes
A compound shape consists of two or more objects combined to make a single object. To calculate the area, perimeter, surface area or volume of a compound shape it is necessary to determine the shapes that have been combined to make the new shape. All applicable formulas can be used on the individual shapes and then the results are combined to find the desired calculations of the compound shape.
Example 1:
A new basketball court has just been built in a school. It is necessary to paint the free throw areas (2) and the center circle dark red before the final lines are added. How much paint is required if it takes 1L for every 9.7m2. The dimensions of the court are in the diagrams below:
Solution 1:
The court is composed of two rectangles, a full circle and two half circles (or two full circles). The dimensions of the rectangle are 574cm by 365.8cm. Each circle or half circle has a radius of 189.9cm.
A conversion to meters is useful to determine how much paint is used.
574cm = 5.74m 365.8 = 3.66m 189.9 = 1.90m
Rectangle: A = L x w
A = 5.74 m x 3.66 m
A = 21.01m2
Circle:A = πr2
A = (3.14) (1.90)2
A = 11.34m2
2(21.01) + 2(11.34) = 64.70m2
64.7m2 ÷ 9.7m2 = 6.67 L
It takes about 6.7 L to paint the court.
Example 2:
- A breadbox has the following dimensions. What is its volume?
Solution 2:
The box is a compound shape made up of a rectangle and a quarter of a circle.
The dimensions of the rectangle are 40cm x 10cm x 35cm.
The quarter circle has a radius of 35cm.
Rectangle: V = L x wx h
V = 40 x 10 x 35
V = 14 000 cm3
The volume of the rectangle is 14 000 cm3
Circle: A = πr2
A = (3.14) (35)2
A = 3846.5 cm2
A = 3847 cm2
V = Ah
V = (3847) (35)
V = 134 645 cm3
The volume of the entire circle is 134 645cm3
One quarter of the volume is represented in the compound shape.
134 627 ÷ 4 = 33 661.25 cm3
V = 33 661 cm3
To find the volume of the compound shape, add the two shapes together.
14 000cm3+ 33 661cm3= 47 661 cm3
The volume of the breadbox is 47 661cm3
Exercise:
1. In a building construction a cement structure is to be poured to decorate the front of a courthouse building. The structure will be placed on either side of the front door. The specifications for that structure are shown below: (four round columns on a rectangular base, resting on a footing,supporting a top,). What is the volume of the part that is to be cement (shaded area)? Calculate one side only.
Answers:
The number of gold cubes that can be made is or 24.
All units should be the same for both shapes.
Area of Base = πr2
Area = (3.14) (300 cm)2
Area = 282 600 cm2
The number of snow blocks the team will have to build the fort is or 23 550.
The drinking glass will hold 1005 cm3 or 1005 mL of juice.
- a) The first shape is a rectangular prism. A real-world example would be a
box.
The second shape is a triangular prism. A real-world example would be a Toblerone bar.
The third shape is a cylinder. A real-world example would be a can.
b)To find the volume of the triangle prism, the area of one of the triangles (they are both the same) must be determined by using the formula A = ×b×h. The next step would be to multiply this area by the height of the shape.
Answer: (Compound Shapes)
1. There are four columns each with a radius of 30 cm and a height of 500 cm.
A = πr2V = A x h
A = (3.14) (30)2V = (2826 cm2) (500cm)
A = 2826 cm2V = 1 413 000 cm3
1 413 000 x 4 = 5 652 000cm3
The volume of cement in the columns is 5 652 000 cm3.
The base is a rectangle.
Length = 800 cm width = 150 cm height = 100 cm
V = L x w x h
V = 800 x 150 x 100
V = 12 000 000 cm3
The volume of cement in the base is 12 000 000 cm3.
The top support is a rectangle.
Length = 800 cm width = 150 cm height = 35cm
V = L x w x h
V = 800 x 150 x 35
V = 4 200 000 cm3
The volume of cement in the top support is 4 200 000 cm3.
The footing is a rectangle.
Length = 840 cm width = 190 cm height = 40cm
V = L x w x h
V = 840 x 190 x 40
V = 6 384 000 cm3
The volume of cement in the footing is 6 384 000 cm3.
Add thevalues of the four columns to the three rectangles (base, top, footing).
Columns base top footing
5 652 000 + 12 000 000 + 4 200 000 + 6 384 000 = 28 236 000cm3
The total amount of cement needed for the structure on one side of the door is 28 236 000 cm3or 28.24 m3