Volume of Shapes 4

Topics

Geometry

Surface Area and Volume

Volume of Shapes

Volume of Shapes 4

Try VividMath Premium to unlock full access

Start Free Trial

Time limit: 0

Quiz summary

0 of 6 questions completed

Questions:

Information

–

You have already completed the quiz before. Hence you can not start it again.

Quiz is loading...

You must sign in or sign up to start the quiz.

You have to finish following quiz, to start this quiz:

Lets see your results!

Points

Score

Time

Retry Quiz

Answered
Review

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) $\text(cm)^3$

Hint

Help Video

Correct

Great Work!

Incorrect

Need Text

Play

Current Time 0:00

Duration Time 0:00

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

Video Acceleration:

Off

00:00

Mute

Playback Rate

2x
1.5x
1.25x
1x
0.75x
0.5x

Subtitles

subtitles off

Captions

captions off
English

Chapters

Chapters

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 volume

Note 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 places

Use

$pi~~3.14$

Volume $=$ (5281.24) $mm^3$

Correct

Keep Going!

Incorrect

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 cubed

Volume

$=5,281.24 \ mm^3$

Question 3 of 6

3. Question

What is the volume of this cone?

Round your answer to

$2$ decimal places

Use

$pi~~3.14$

Volume $=$ (15113.87) $cm^3$

Correct

Keep Going!

Incorrect

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 cubed

Volume

$=15,113.87 \ cm^3$

Question 4 of 6

4. Question

What is the volume of this Rectangular Pyramid?

Volume $=$ (50) $mm^3$

Correct

Keep Going!

Incorrect

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 cubed

Volume

$=50 \ mm^3$

Question 5 of 6

5. Question

Find the volume of the Cone

Round your answer to

$1$ decimal place

Use

$pi=3.141592654$

$\text(Volume )=$ (5305.8, 5303.1, 5307.9) $\text(cm)^3$

Hint

Help Video

Correct

Exceptional!

Incorrect

Need Text

Play

Current Time 0:00

Duration Time 0:00

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

Video Acceleration:

Off

00:00

Mute

Playback Rate

2x
1.5x
1.25x
1x
0.75x
0.5x

Subtitles

subtitles off

Captions

captions off
English

Chapters

Chapters

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 volume

Use

$pi=3.141592654$ 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 places

Use

$pi~~3.14$

Volume $=$ (1526.04) $cm^3$

Correct

Keep Going!

Incorrect

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 cubed

Volume

$=1,526.04 \ cm^3$

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