Volume of Shapes 4
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Question 1 of 6
1. Question
Find the volume of the Pyramid
Round your answer to one decimal place- `\text(Volume )=` (1326.6, 1324.9) `\text(cm)^3`
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Volume of a Pyramid
`\text(Volume )=1/3 times``\text(length)``times``\text(width)``times``\text(height)`Labelling the given lengths
`\text(length)=18``\text(width)=11``\text(height)=?``\text(b (right triangle))=9` `(``18``divide2)``\text(c (right triangle))=22`
First, we need to find the height perpendicular to its base.Label the sides of the right triangle formed within the pyramid
Use the Pythagorean Theorem Formula to solve for `a`, which is equal to the `\text(height)``a^2``+``b^2` `=` `c^2` Pythagoras’ Theorem Formula `\text(height)^2``+``9^2` `=` `22^2` Plug in the known lengths `\text(height)^2+81` `=` `484` Evaluate `9^2` and `22^2` `\text(height)^2` `=` `403` Subtract `81` from both sides `\text(height)` `=` `20.1 \text(cm)` Take the square root of both sides Next, find the area of the pyramid’s base, which is a rectangle`\text(Area)` `=` `\text(length)``times``\text(width)` Area of a Rectangle `=` `18``times``11` Plug in the known lengths `=` `198 \text(cm)^2` Finally, use the formula to find the volumeNote that `\text(area)``=``\text(length)``times``\text(width)``\text(Volume)` `=` `1/3 times``\text(length)``times``\text(width)``times``\text(height)` Volume of a Pyramid `=` `1/3 times``198``times``20.1` Plug in the known lengths `=` `1326.6 \text(cm)^3` The given measurements are in centimetres, so the volume is measured as centimetres cubed`\text(Volume)=1326.6 \text(cm)^3` -
Question 2 of 6
2. Question
What is the volume of this cylinder?
Round your answer to `2` decimal placesUse `pi~~3.14`- Volume`=` (5281.24)`mm^3`
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Volume of a Cylinder
`V=pi xx color(royalblue)(\text(radius)^2) xx color(darkviolet)(\text(height))`Labelling the given lengths
`color(darkviolet)(\text(height)=34)``color(royalblue)(\text(radius)=7)`Use the formula to find the volume`pi~~3.14``V` `=` `pi xx color(royalblue)(\text(radius)^2) xx color(darkviolet)(\text(height))` Volume of a cylinder formula `=` `3.14 xx color(royalblue)(7^2) xx color(darkviolet)(34)` Plug in the known lengths `=` `3.14 xx 49 xx 34` Simplify `=` `5,281.24 \ mm^3` Rounded to 2 decimal places The given measurements are in millimetres, so the volume is measured as millimetres cubedVolume`=5,281.24 \ mm^3` -
Question 3 of 6
3. Question
What is the volume of this cone?
Round your answer to `2` decimal placesUse `pi~~3.14`- Volume`=` (15113.87)`cm^3`
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Volume of a Cone
`V=1/3 xx pi xx color(royalblue)(\text(radius)^2) xx color(darkviolet)(\text(height))`Labelling the given lengths
`color(darkviolet)(\text(height)=40)``color(royalblue)(\text(radius)=19)`Use the formula to find the volume`pi~~3.14``V` `=` `1/3 xx pi xx color(royalblue)(\text(radius)^2) xx color(darkviolet)(\text(height))` Volume of a cone formula `=` `1/3 xx 3.14 xx color(royalblue)(19^2) xx color(darkviolet)(40)` Plug in the known lengths `=` `1/3 xx 3.14 xx 361 xx 40` Simplify `=` `15,113.86667` `=` `15,113.87 \ cm^3` Rounded to 2 decimal places The given measurements are in centimetres, so the volume is measured as centimetres cubedVolume`=15,113.87 \ cm^3` -
Question 4 of 6
4. Question
What is the volume of this Rectangular Pyramid?
- Volume`=` (50)`mm^3`
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Volume of a Rectangular Pyramid
`V=1/3 xx color(royalblue)(\text(length)) xx color(darkviolet)(\text(width))xx color(green)(\text(height))`Labelling the given lengths
`color(royalblue)(\text(length)=5)``color(darkviolet)(\text(width)=3)``color(green)(\text(height)=10)`Use the formula to find the volume`V` `=` `1/3 xx color(royalblue)(\text(length)) xx color(darkviolet)(\text(width))xx color(green)(\text(height))` Volume of a Rectangular Pyramid formula `=` `1/3 xx color(royalblue)(\text(5)) xx color(darkviolet)(\text(3))xx color(green)(\text(10))` Plug in the known lengths `=` `50` `=` `50 \ mm^3` The given measurements are in millimetres, so the volume is measured as millimetres cubedVolume`=50 \ mm^3` -
Question 5 of 6
5. Question
Find the volume of the Cone
Round your answer to `1` decimal placeUse `pi=3.141592654`- `\text(Volume )=` (5305.8, 5303.1, 5307.9) `\text(cm)^3`
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Volume of a Cone
`\text(Volume)=1/3 times pi times``\text(radius)^2``times``\text(height)`Labelling the given lengths
`\text(radius)=?``\text(diameter)=23``\text(height)=?``\text(c (right triangle))=40`
First, recall that the radius is equal to half of the diameter`\text(radius)` `=` `1/2 times ``23` `\text(radius)` `=` `11.5` Next, we need to find the height perpendicular to its base.Label the sides of the right triangle formed within the pyramid
Use the Pythagorean Theorem Formula to solve for `a`, which is equal to the `\text(height)``a^2``+``b^2` `=` `c^2` Pythagoras’ Theorem Formula `\text(height)^2``+``11.5^2` `=` `40^2` Plug in the known lengths `\text(height)^2+11.5^2` `=` `40^2` Evaluate `11.5^2` and `40^2` `\text(height)^2+132.25` `=` `1600` Subtract `132.25` from both sides `sqrt(\text(height)^2)` `=` `sqrt(1467.75)` Take the square root of both sides `\text(height)` `=` `38.3112255 \text(cm)` Finally, use the formula to find the volumeUse `pi=3.141592654` See `pi` explained`\text(Volume)` `=` `1/3 times pi times``\text(radius)^2``times``\text(height)` Volume of a Cone formula `=` `1/3 times 3.141592654 times``11.5^2``times``38.3112255` Plug in the known lengths `=` `1/3 times 3.141592654 times 132.25 times 38.3112255` Simplify `=` `5305.79349` `=` `5305.8 \text(cm)^3` Rounded to one decimal place The given measurements are in centimetres, so the volume is measured as centimetres cubed`\text(Volume)=5305.8 \text(cm)^3`The answer will depend on which `pi` you use.In this solution we used: `pi=3.141592654`.Using Answer `pi=3.141592654` `5305.8 cm^3` `pi=3.14` `5303.1 cm^3` `pi=(22)/(7)` `5307.9 cm^3` -
Question 6 of 6
6. Question
What is the volume of this hemisphere?
Round your answer to `2` decimal placesUse `pi~~3.14`- Volume`=` (1526.04)`cm^3`
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Volume of a Hemisphere
`V=2/3 xx pi xx color(royalblue)(\text(radius)^3)`Labelling the given lengths
`color(royalblue)(\text(radius)=9)`Use the formula to find the volume`pi~~3.14``V` `=` `2/3 xx pi xx color(royalblue)(\text(radius)^3)` Volume of a hemisphere formula `=` `2/3 xx 3.14 xx color(royalblue)(9^3)` Plug in the known lengths `=` `2/3 xx 3.14 xx 729` Simplify `=` `1,526.04 \ cm^3` Rounded to 2 decimal places The given measurements are in centimetres, so the volume is measured as centimetres cubedVolume`=1,526.04 \ cm^3`
Quizzes
- Volume of Shapes 1
- Volume of Shapes 2
- Volume of Shapes 3
- Volume of Shapes 4
- Volume of Composite Shapes 1
- Volume of Composite Shapes 2
- Surface Area of Shapes 1
- Surface Area of Shapes 2
- Surface Area of Shapes 3
- Surface Area and Volume Mixed Review 1
- Surface Area and Volume Mixed Review 2
- Surface Area and Volume Mixed Review 3
- Surface Area and Volume Mixed Review 4