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Pyramid Problems

Surface area and volume of pyramid problems along with with detailed solutions are presented.

Lesson Solved problems on volume of pyramids

Volume of a Pyramid Practice Questions

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• Worksheet on Volume of a Pyramid

Math worksheet on volume of a pyramid will help us to practice the different questions for finding volume of a pyramid.

1.  The base of a right pyramid 10 √7 feet high is a triangle whose sides are 9 feet, 11feet and 16 feet. Find the volume of the pyramid. 

2.  The base of right pyramid is a triangle whose sides are 28 cm, 25cm and 17 cm. If the volume of the pyramid be 11120 cubic cm, find its height. 

3.  The base of a right pyramid is a square of 40 cm and its slant height is 25 cm. If the value of cube is equal to the volume of the pyramid find the length of a side of the cube. 

4.  The base of a right pyramid is a square of side 40 cm and length of an edge through the vertex is 5√41 cm. If the volume of a cube is equal to the volume of the pyramid, then find length of the side of the cube. 

5.  The base of a right pyramid is a square of side 12 cm. If its slant height is 6√5 cm, find its volume. 

6.  The base of a right pyramid is a rectangle whose length and breadth are 18 cm and 15 cm respectively. If its height be 24 cm, find its volume. 

7. The base of a right pyramid is a rectangle whose length and breadth are 24 cm and 18 cm respectively. If its length of its slant edge is 17 cm, find the height and volume of the pyramid.

8. The base of a right pyramid, 5√3 cm height, is a regular hexagon of side 6 cm. Find its volume. 

9. OA, OB, OC are there mutually perpendicular straight lines in space. If OA = a, OB = b, and OC = c, prove that the volume of the pyramid OABC is (1/6) abc.

10. The base of a right pyramid 15 cm height cm is a regular octagon. If the volume of the pyramid 160(√2 + 1) cubic cm, find the length of a side of the octagon.

11. The base of a right pyramid is a square of side 12 cm and the dihedral angle between its base and a lateral face is 60°. Find its height and volume.

12. The base of right pyramid, 15 cm. High, is a square of side 16 cm. Its upper part is cut off by a plane parallel to the base and through the middle of its height. Find the volume of the frustum of the pyramid formed.

13. The lower and upper faces of the frustum of a right pyramid are squares of sides 16 cm and 9 cm respectively. If the height of the pyramid be 12 cm, then find its volume.

14. The upper and lower faces of the frustum of a right pyramid are a regular of hexagons of sides 8 cm and 12 cm respectively. If the height of the frustum be 2√3 cm, find its volume.

15. The base of right pyramid, h cm height is a square. It is divided into two parts by a plane parallel to the base so that volumes of the two parts are equal. Show that the distance of the plane form the vertex is h/(3√2) cm.

Answers for the worksheet on volume of a pyramid are given below to check the exact answers of the above questions.

1. 420 cu. ft. 2. 16 cm. 3. 20 cm. 4. 20 cm. 5. 576 cu. cm. 6. 2160 cu. cm. 7. 8 cm and 1152 cu. cm. 8. 270 cu. cm. 10. 4 cm. 11. 6√3 cm. and 72√3 cu. cm. 12. 1120 cu. cm. 13. 1924 cu. cm. 14. 2736 cu. cm.

●  Mensuration

• Formulas for 3D Shapes
• Volume and Surface Area of the Prism
• Worksheet on Volume and Surface Area of Prism
• Volume and Whole Surface Area of Right Pyramid
• Volume and Whole Surface Area of Tetrahedron
• Volume of a Pyramid
• Volume and Surface Area of a Pyramid
• Problems on Pyramid
• Worksheet on Volume and Surface Area of a Pyramid

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In Praise of the Open Middle with Pyramid Puzzles

Sometimes you hear the perfect word to describe something you were already doing, but didn’t quite realize you were doing. My most favorite recent case of this came from openmiddle.com .

We’ve all heard of open-ended problems, the genuine, intriguing problems that can take us on journeys to unexpected answers. A lesson based around an open-ended problem can be a beautiful thing, and done right, there’s hardly a more exciting educational experience you can have.

The trouble is, it doesn’t always go so well. Guiding a class through a discussion and exploration of an open-ended problem requires tremendous insight and understanding on the part of the teacher, and even the most experienced teachers sometimes have them go awry. Will this student’s comments lead in a productive direction, or just muddle the issue, and leave us all frustrated? Is this problem so hard that we can’t hope for more than a partial understanding of it, or is there some key idea that we’re all missing? When you have to make assessments on the fly, things don’t always go well.

One trick to coax these explorations in productive directions is to game them just a little bit. The teacher can go in with a plan, or an arc of where they expect the lesson to go. So they can “discover” something like the fact that the midpoints of quadrilaterals seem to form parallelograms and ask, “that doesn’t always happen, does it?”, all the while knowing that it is, in fact, true, and with a couple of good ideas about how to prove it. This little bit of lesson planning can greatly improve the chances of an individual lesson feeling like a success, even if it gives up a little bit of the total free of open ended lessons.

Enter the Open Middle. The idea of Open Middle problems is that the beginning (the problem) and the end (the answer) are both clearly well-defined, but the middle of the problem—the part where you look for the solution—is wide open. The middle is where you cast around for ideas, structures, tools, or just blind guesses about what’s going on. Because Open Middle problems have such a well-defined structure, there’s much less of a chance of them going off the rails. This makes them a fantastic tool for teachers who might be taking their first steps into less regimented math teaching, or who don’t feel like taking a chance on an exploration yielding a broad discussion that doesn’t go anywhere. (Read more about Open Middle problems here .)

I produced a number of these Pyramid Puzzles recently, and I think they’re great examples of Open Middle problems. it’s relatively easy to adjust their difficulty as well.

Pyramid Puzzle 1.

Each number must be the sum of the two directly below it in the pyramid. Fill in the blanks.

This is a puzzle that seems easy, but the solution is just out of reach. We can put a 7 above the 2 and 5, but what then? There’s no obvious next step, and our best guess is to take a stab at it, and see what happens. Some people like to work from top to bottom, and try taking a guess of what two numbers might go below the 30 (say, 12 and 18?), while some prefer to work up from the bottom, putting a number in the blank and going from there.

Suppose I put a 10 in the bottom blank. Then I’d have this:

A 10 on the bottom leads to a 51 at the top, which is a problem. But it’s also a useful mistake, since now I know the number in the bottom must be smaller than 10. One wild guess gives me traction, and it’s only a matter of time before I solve this problem.

Pyramid Puzzle 2.

Each number must be the sum of the two directly below it in the pyramid AND no number can appear more than once. Fill in the blanks with positive integers so that the top of the Pyramid is 20.

Bonus: Could the 20 at the top of the pyramid be replaced with a smaller number and the pyramid still be solved? Show how if it’s possible, or show why it’s impossible.

This is a much subtler puzzle, and solving the bonus problem in particular requires you to delve into the workings of these puzzles. Is there some kind of theory that can tell us what the maximum number at the top of the pyramid can be, given the rules that we’re filling them only with positive integers, and never the same one twice? What about differently-sized pyramids?

This is clearly the smallest number that can be atop a pyramid 2 stories high.

And this seems to be the smallest number that can be on top of a pyramid three stories high.

So what’s the smallest for four stories high, or more?

We’re in the territory of open-ended problems now, which is where I always seem to end up. But that’s the exciting thing about mathematical exploration: there are always questions pointing in directions you’ve never gone before.

Find more of our Pyramid Puzzles here, at our lessons page .

Check out openmiddle.com for more examples of Open Middle problems.

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Volume of Pyramids

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• Mixed Pyramids

Volume of a Pyramid Worksheets

Work out the volume of the pyramids with rectangular, triangular and polygonal base faces. Calculate the volume by plugging in the measures expressed as integers and decimals in the appropriate formulas. The 8th grade and high school printable worksheets are classified into two levels. Level 1 comprises polygons with 3 or 4-sided base faces, while level 2 includes polygonal base faces. Practice finding the missing measures as well. Explore some of these worksheets for free!

Finding Volume using Base Area | Integers

Jump-start your learning with this batch of PDF worksheets for 8th grade and 9th grade students on mixed pyramids featuring rectangular and triangular pyramids. Apply apt formulas to find the volume using the base area measure expressed as integers.

• Download the set

Finding Volume using Base Area | Decimals

Advance to the next level consisting of pyramids whose base faces are rectangles or triangles, and the base area is indicated as decimals. Multiply the base area with the height and divide by 3 to compute the volume.

Volume of Rectangular Pyramids

Exclusively dealing with rectangular pyramids, these PDF worksheets for 9th grade and 10th grade students are a must-have for thorough knowledge and practice in finding the volume of rectangular pyramids offering varied levels of difficulty.

(24 Worksheets)

Volume of Triangular Pyramids

Make strides in your practice with these printable worksheets on finding the volume of triangular pyramids or tetrahedrons. Try the easy, moderate and challenging levels with integer and decimal dimensions.

Finding Volume of Polygonal Pyramids using Apothem | Level 1

Another essential step is to determine the volume of pyramids with polygonal base faces. Plug in the apothem(a), number of sides(n), side length(s) and height(h) in the formula V = 1/6 * ansh and compute the volume.

Volume of Polygonal Pyramids using Side length or Perimeter | Level 2

The side length or perimeter and height are provided. First, find the apothem (apothem = side / 2 tan (180/n)), and then assign the known values in the volume formula to solve for volume of polygonal pyramids.

Volume of Pyramids | Level 1 - Integers - Easy

The easy level contains pyramids with integer dimensions ≤ 20. One-third of the base area of the square, rectangle or a triangle multiplied by the height results in the volume of the pyramid. Recommended for grade 8 and grade 9 children.

Volume of Pyramids | Level 1 - Integers - Moderate

Polish up your skills in finding the volume of pyramids with this section of high school geometry worksheets presenting pyramids as 3D shapes and in the word format with 3 or 4-sided base faces involving integer dimensions ≥ 20.

Volume of Pyramids | Level 1 - Decimals

Scale new heights as you practice solving for volume with these pdf worksheets posing a challenge with decimal dimensions. Apply relevant formula, substitute and calculate the volume of each pyramid.

Volume of Pyramids | Level 2

With the sides of the base face, increases the level of difficulty. Figure out the volume of the triangular, rectangular and polygonal pyramids applying suitable formulas and bolster skills.

Volume of Pyramids | Missing Measure

Motivate learners to give these printable worksheets a quick try. All they have to do is simply rearrange the formula, making the missing measure (x) the subject, plug in the known values and solve!

Related Worksheets

» Volume of Prisms

» Volume of Mixed Shapes

» Volume of Composite Shapes

» Volume of Cones

» Volume of Spheres

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A pyramid is a 3-dimensional geometric solid . It consists of a base that is a polygon and a point not on the plane of the polygon, called the vertex . The edges of the pyramid are the sides of the polygonal base together with line segments which join the vertex of the pyramid to each vertex of the polygon.

Some well-known pyramids include the tetrahedron , which has a triangle for its base. (A regular tetrahedron has all edges of equal length, and is one of the Platonic solids ). Another is the regular square pyramid. Two of these with their bases joined form an octahedron , which is another Platonic solid.

Introductory

Intermediate

• Corners are sliced off a unit cube so that the six faces each become regular octagons . What is the total volume of the removed tetrahedra? ( 2007 AMC 12A Problems/Problem 20 )
• In a regular tetrahedron the centers of the four faces are the vertices of a smaller tetrahedron. Find the ratio of the volume of the smaller tetrahedron to that of the larger. ( 2003 AIME II, #4 )
• A square pyramid with base ABCD and vertex E has eight edges of length 4. A plane passes through the midpoints of AE, BC, and CD. Find the area of the plane's intersection with the pyramid. ( 2007 AIME I, #13 )
• Prove that pyramid volume is 1/3 bh using parallelipiped and dimensional analysis.

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• Volume of Pyramid – Explanation & Examples

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Volume of a pyramid formula

Volume of a square pyramid, volume of a trapezoidal pyramid, volume of a triangular pyramid, practice questions, volume of pyramid – explanation & examples.

Pyramids are named after the shape of their bases. For instance, a rectangular pyramid has a rectangular base, a triangular pyramid has a triangular base, a pentagonal pyramid has a pentagonal base, etc.

How to Find the Volume of a Pyramid?

In this article, we discuss how to find the volume of pyramids with different types of bases and solve word problems involving a pyramid’s volume.

The volume of a pyramid is defined as the number of cubic units occupied by the pyramid. As stated before, the name of a pyramid is derived from the shape of its base. Therefore, the volume of a pyramid also depends on the shape of the base.

To find the pyramid’s volume, you only need the dimensions of the base and the height.

The general volume of a pyramid formula is given as:

Volume of a pyramid = 1/3 x base area x height.

V= 1/3 A b h

Where A b = area of the polygonal base and h = height of the pyramid.

Calculate the volume of a rectangular pyramid whose base is 8 cm by 6 cm and the height is 10 cm.

For a rectangular pyramid, the base is a rectangle.

Area of a rectangle = l x w

= 48 cm 2 .

And by the volume of a pyramid formula, we have,

Volume of a pyramid = 1/3A b h

= 1/3 x 48 cm 2 x 10 cm

= 160 cm 3 .

The volume of a pyramid is 80 mm 3 . If the pyramid’s base is a rectangle that is 8 mm long and 6 mm wide, find the pyramid’s height.

⇒ 80 = 1/3 x (8 x 6) x h

⇒ 80 = 15.9h

By dividing both sides by 15.9, we get,

Thus, the height of the pyramid is 5 mm.

To obtain the formula for the volume of a square pyramid, we substitute the base area (A b ) with the area of a square (Area of a square = a 2 )

Therefore, the volume of a square pyramid is given as:

Volume of a square pyramid = 1/3 x a 2 x h

V = 1/3 a 2  h

Where a = side length of the base (a square) and h = height of the pyramid.

A square pyramid has a base length of 13 cm and a height of 20 cm. Find the volume of the pyramid.

Length of the base, a = 13 cm

height = 20 cm

Volume of a square pyramid = 1/3 a 2  h

By substitution, we have,

Volume = 1/3 x 13 x 13 x 20

= 1126.7 cm 3

The volume of a square pyramid is 625 cubic feet. If the height of the pyramid is 10 feet, what are the dimensions of the pyramid’s base?

Volume = 625 cubic feet.

height = 10 feet

By the volume of a square formula,

⇒ 625 = 1/3 a 2  h

⇒ 625 = 1/3 x a 2 x 10

⇒ 625 = 3.3a 2

⇒ a 2 =187.5

⇒ a = = √187.5

a =13.7 feet

So, the dimensions of the base will be 13.7 feet by 13.7 feet.

The base length of a square pyramid is twice the height of the pyramid. Find the dimensions of the pyramid if it has a volume of 48 cubic yards.

Let the height of the pyramid = x

the length = 3x

volume = 48 cubic yards

But, the volume of a square pyramid = 1/3 a 2  h

Substitute.

⇒ 48 = 1/3 (3x) 2 (x)

⇒ 48 = 1/3 (9x 3 )

⇒ 48 = 3x 3

Divide both sides by 3 to get,

⇒ x = 3 √16

Therefore, the height of the pyramid = x ⇒2.53 yards,

and each side of the base is 7.56 yards

A trapezoidal pyramid is a pyramid whose base is a trapezium or a trapezoid.

Since we know, area of a trapezoid = h 1 (b 1 + b 2 )/2

Where h = height of the trapezoid

b 1 and b 2 are the lengths of the two parallel sides of a trapezoid.

Given the general formula for the volume of a pyramid, we can derive the formula for the volume of a trapezoidal pyramid as:

Volume of a trapezoidal pyramid = 1/6 [h 1 (b 1 + b 2 )] H

Note: When using this formula, always remember that h is the height of the trapezoidal base and H is the height of the pyramid.

The base of a pyramid is a trapezoid with parallel sides of length 5 m and 8 m and a height of 6 m. If the pyramid has a height of 15 m, find the volume of the pyramid.

h = 6 m, H = 15 m, b 1 =5 m and b 2 = 8 m

Volume of a trapezoidal pyramid = 1/6 [h 1 (b 1 + b 2 )] h

= 1/6 x 6 x 15 (5 + 8)

As we know, the area of a triangle;

Area of a triangle = 1/2 b h

Volume of a triangular pyramid = 1/3 (1/2 b h) H

Where b and h are the base length and height of the triangle. H is the height of the pyramid.

Find the area of a triangular pyramid whose base area is 144 in 2 and the height is 18 in.

Base area = 144 in 2

= 1/3 x 144 x 18

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Volume Of Square Based Pyramid

Here we will learn about the volume of a square based pyramid, including how to calculate the volume of a square based pyramid and how to solve problems involving the volume.

There are also square based pyramid worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

What is volume of a square based pyramid?

The volume of a square based pyramid is how much space there is inside a square based pyramid.

A square based pyramid is a three dimensional shape made up of flat faces. It has a square base and triangular faces which meet at a point, called the apex . The vertical height is the length from the square base to the apex and is perpendicular to the base of the pyramid.

To calculate the volume of a square based pyramid we use the formula

\text{Volume}=\cfrac{1}{3}\times \text{area of base} \times \text{height} .

The pyramid height should be perpendicular to its base.

The formula can also be written as

• V represents the volume of the pyramid,
• B represents the area of its base,
• h represents the perpendicular height of the pyramid.

If we are not given the perpendicular height, we can use Pythagoras’ theorem to find it from the given slant height or the lateral edge length.

There are lots of different types of pyramids because the base of a pyramid can be any polygon.

For example, a rectangular pyramid has a rectangular base and a triangular pyramid (often known as a tetrahedron) has a triangular base.

The square pyramid formula for volume can be used for any pyramid.

How to calculate the volume of a square based pyramid

In order to calculate the volume of a square based pyramid:

Calculate the area of the base.

Substitute values into the formula and solve.

Write the answer, including the units.

Explain how to calculate the volume of a square based pyramid

Volume and surface area of a pyramid worksheet (includes volume of square based pyramid)

Get your free volume of square based pyramid worksheet of 20+ volume and surface area of a pyramid questions and answers. Includes reasoning and applied questions.

Volume of square based pyramid examples

Example 1: calculating the volume, given the height, of a square pyramid.

Find the volume of the square based pyramid.

The area of the base can be found by squaring the base edge.

2 Substitute values into the formula and solve.

The volume of a square based pyramid can be found by using the formula.

3 Write the answer, including the units.

The volume is 351 \ m^{3}.

Example 2: calculating the volume, given the height, of a square pyramid

The volume of a square based pyramid can be found by using the formula. But the height is given in a different unit of measurement to the side of the base, 2.5 \ m is 250 \ cm.

\begin{aligned} V&=\cfrac{1}{3}Bh \\\\ V&=\cfrac{1}{3}\times 6400\times 250 \\\\ V&=533 \ 333.33… \end{aligned}

The volume is 533 \ 000 \ cm^3 (to 3 sf). This can be also written as 0.533 \ m^{3}.

Example 3: calculating the volume of a square pyramid not given perpendicular height

Find the volume of the square based pyramid. The apex of the pyramid is directly above the centre of the square base.

The volume of a square based pyramid can be found by using the formula. But the perpendicular height is not given. It can be found using Pythagoras’ theorem. The slant height of the pyramid 8 \ cm is the hypotenuse.

h=\sqrt{8^2-3^2}=\sqrt{55} \\

\begin{aligned} V&=\cfrac{1}{3}Bh \\\\ V&=\cfrac{1}{3}\times 36\times \sqrt{55} \\\\ V&=12\sqrt{55} \\\\ V&=88.9943… \end{aligned}

The volume is 89.0 \ cm^3 (to 3 sf).

Example 4: calculating the volume of a square pyramid not given perpendicular height

The volume of a square based pyramid can be found by using the formula. But we do not know the perpendicular height. We have been given the lateral edge length. We will need to apply Pythagoras’ theorem twice to find the perpendicular height.

The value 7 comes from using the midpoint of the base edge to make right angled triangles.

First we find the slant length from the lateral edge length.

x=\sqrt{18^2-7^2}=5\sqrt{11}

Then we can find the perpendicular height of the square based pyramid.

h=\sqrt{(5\sqrt{11})^2-7^2}=\sqrt{226} \\

\begin{aligned} V&=\cfrac{1}{3}Bh \\\\ V&=\cfrac{1}{3}\times 196\times \sqrt{226} \\\\ V&=982.175… \end{aligned}

The volume is 982 \ m^3 (to 3 sf).

Example 5: calculating the height given the volume

The volume of the square based pyramid is 87.6 \ cm^{3}. Find the perpendicular height, h. Give your answer to 3 significant figures.

6.5^2=42.25

The formula for the volume of the square based pyramid can be used to find the perpendicular height, h.

\begin{aligned} V&=\cfrac{1}{3}Bh \\\\ 87.6&=\cfrac{1}{3}\times 42.25\times h \\\\ h&=\cfrac{87.6\times 3}{42.25} \\\\ h&=6.2201… \end{aligned}

The perpendicular height h is 6.22 \ cm (to 3 sf).

Example 6: calculating the length of the base given the volume

The volume of the square based pyramid is 1200 \ m^{3}. Find the base edge. Give your answer correct to 1 decimal place.

The area of the base can not be found yet. But, when we find it we can square root it to find the length of the base (the base edge).

The formula for the volume of the square based pyramid can be used to find the base area.

\begin{aligned} V&=\cfrac{1}{3}Bh \\\\ 1200&=\cfrac{1}{3}\times B\times 24.7 \\\\ h&=\cfrac{1200\times 3}{24.7}\\\\ h&=145.7489… \end{aligned}

We can then square root the base area to find the length of the base.

\sqrt{145.7489…}=12.072…

The length of the base is 12.1 \ m (to 1 dp).

Common misconceptions

• The height is the perpendicular height

The height of a square pyramid needs to be the perpendicular height. This is the height that is at a right-angle to the base.

• Use the correct units

Remember to use cubic units for volume such as cm^{3} or m^{3}. Square units such as cm^{2} or m^{2} are for areas.

• Use the correct volume formula

There are many different volume formulas in maths. Make sure that you use the correct one to work out the volume of a pyramid. \text{Volume}=\cfrac{1}{3}\times \text{area of base} \times \text{height} Or V=\cfrac{1}{3}Bh

• Be accurate

When there are two or more steps in your workings, do not round your workings. Only round at the end of your solution so that your answer is accurate.

• Take care with rounding

At the end of the question, make sure you round your answer to the correct number of decimal places or significant figures.

Related lessons

Volume of a pyramid is part of our series of lessons to support revision on pyramids. You may find it helpful to start with the main pyramid lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:

• Surface area of a pyramid
• Triangular based pyramid
• Square based pyramid
• Volume of a pyramid

Practice volume of square based pyramid questions

1. Work out the volume of this square based pyramid.

We can work out the volume of a pyramid by using the formula

The base area can be found by squaring the side length of the base square.

We can substitute the values we have been given into the formula and work out the volume.

So the volume is 2484 \ m^{3}.

2. Work out the volume of this square based pyramid. Give your answer in cm^{3}.

The base area can be found by squaring the side length. The side length is in mm so needs converting the cm.

The height of 6.7 \ m needs to be converted to 670 \ cm .

So the volume is 4522.5 \ cm^{3} .

3. Work out the volume of this square based pyramid. Give your answer in cm^{3} correct to three significant figures.

The base area can be found by squaring the side length.

We have been given the slant length. We can use this to find the perpendicular height of the pyramid.

So the volume is 11 \ 100 \ cm^{3} (to 3 sf).

4. Work out the volume of this square based pyramid. V is directly above B and VC is 10 \ cm. Give your answer in cm^{3}.

We have been given a slant length. We can use this to find the perpendicular height of the pyramid.

So the volume is 128 \ cm^{3} .

5. The volume of this square based pyramid is 460 \ cm^{3}. Find the perpendicular height of the pyramid. Give your answer correct to 3 significant figures.

We can substitute the values we are given into volume of a pyramid formula and rearrange it to find the height, h.

So the perpendicular height is 8.17 \ cm^{3} (to 3 sf).

6. The volume of this square based pyramid is 8900 \ mm^{3}. Find the side length of the pyramid. Give your answer correct to 3 significant figures.

We can substitute the values we are given into the volume of a pyramid formula and rearrange it to find the base area.

The perpendicular height is given in cm, so needs converting to mm.

The area of the base needs to be square rooted to find the side length.

So the perpendicular height is 18.5 \ mm (to 3 sf).

Volume of square based pyramid GCSE questions

1. The diagram shows a square based pyramid.

Calculate the volume of the square based pyramid.

Give your answer correct to 3 significant figures

Give the correct units for your answer.

Area of the base is 120^{2}=14400 .

2. The diagram shows a square based pyramid.

Area of the base is 32^{2}=1024 .

3. The diagram shows a square based pyramid.

It is based on one of the pyramids of Giza.

It has a volume of 2.6 million m^{3}.

It has a height of 147 \ m.

Find the side length, x of its square base.

Give your answer to 3 significant figures

2.6 million = 2 \ 600 \ 000

4. The diagram shows a truncated square pyramid.

Find the volume of the truncated square based pyramid.

Give your answer as an exact value.

For using similar triangles to find the missing height.

For the height of large pyramid or the height of small pyramid, 15 \ cm and 5 \ cm.

For the volume of large pyramid or the volume of small pyramid, 720 or 26\cfrac{2}{3}.

Learning checklist

You have now learned how to:

• Recognise a square based pyramid and its features
• Calculate the volume of a square based pyramid
• Solve problems involving volume
• Find the total surface area of a square based pyramid

The next lessons are

• Pythagoras theorem
• Trigonometry

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Volume Problem Solving

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To solve problems on this page, you should be familiar with the following: Volume - Cuboid Volume - Sphere Volume - Cylinder Volume - Pyramid

This wiki includes several problems motivated to enhance problem-solving skills. Before getting started, recall the following formulas:

• Volume of sphere with radius \(r:\) \( \frac43 \pi r^3 \)
• Volume of cube with side length \(L:\) \( L^3 \)
• Volume of cone with radius \(r\) and height \(h:\) \( \frac13\pi r^2h \)
• Volume of cylinder with radius \(r\) and height \(h:\) \( \pi r^2h\)
• Volume of a cuboid with length \(l\), breadth \(b\), and height \(h:\) \(lbh\)

Volume Problem Solving - Basic

Volume - problem solving - intermediate, volume problem solving - advanced.

This section revolves around the basic understanding of volume and using the formulas for finding the volume. A couple of examples are followed by several problems to try.

Find the volume of a cube of side length \(10\text{ cm}\). \[\begin{align} (\text {Volume of a cube}) & = {(\text {Side length}})^{3}\\ & = {10}^{3}\\ & = 1000 ~\big(\text{cm}^{3}\big).\ _\square \end{align}\]

Find the volume of a cuboid of length \(10\text{ cm}\), breadth \(8\text{ cm}\). and height \(6\text{ cm}\). \[\begin{align} (\text {Area of a cuboid}) & = l × b × h\\ & = 10 × 8 × 6\\ & = 480 ~\big(\text{cm}^{3}\big).\ _\square \end{align}\]

I made a large ice cream cone of a composite shape of a cone and a hemisphere. If the height of the cone is 10 and the diameter of both the cone and the hemisphere is 6, what is the volume of this ice cream cone? The volume of the composite figure is the sum of the volume of the cone and the volume of the hemisphere. Recall the formulas for the following two volumes: \( V_{\text{cone}} = \frac13 \pi r^2 h\) and \( V_{\text{sphere}} =\frac43 \pi r^3 \). Since the volume of a hemisphere is half the volume of a a sphere of the same radius, the total volume for this problem is \[\frac13 \pi r^2 h + \frac12 \cdot \frac43 \pi r^3. \] With height \(h =10\), and diameter \(d = 6\) or radius \(r = \frac d2 = 3 \), the total volume is \(48\pi. \ _\square \)

Find the volume of a cone having slant height \(17\text{ cm}\) and radius of the base \(15\text{ cm}\). Let \(h\) denote the height of the cone, then \[\begin{align} (\text{slant height}) &=\sqrt {h^2 + r^2}\\ 17&= \sqrt {h^2 + 15^2}\\ 289&= h^2 + 225\\ h^2&=64\\ h& = 8. \end{align}\] Since the formula for the volume of a cone is \(\dfrac {1}{3} ×\pi ×r^2×h\), the volume of the cone is \[ \frac {1}{3}×3.14× 225 × 8= 1884 ~\big(\text{cm}^{2}\big). \ _\square\]

Find the volume of the following figure which depicts a cone and an hemisphere, up to \(2\) decimal places. In this figure, the shape of the base of the cone is circular and the whole flat part of the hemisphere exactly coincides with the base of the cone (in other words, the base of the cone and the flat part of the hemisphere are the same). Use \(\pi=\frac{22}{7}.\) \[\begin{align} (\text{Volume of cone}) & = \dfrac {1}{3} \pi r^2 h\\ & = \dfrac {1 × 22 × 36 × 8}{3 × 7}\\ & = \dfrac {6336}{21} = 301.71 \\\\ (\text{Volume of hemisphere}) & = \dfrac {2}{3} \pi r^3\\ & = \dfrac {2 × 22 × 216}{3 × 7}\\ & = \dfrac {9504}{21} = 452.57 \\\\ (\text{Total volume of figure}) & = (301.71 + 452.57) \\ & = 754.28.\ _\square \end{align} \]

Try the following problems.

Find the volume (in \(\text{cm}^3\)) of a cube of side length \(5\text{ cm} \).

A spherical balloon is inflated until its volume becomes 27 times its original volume. Which of the following is true?

Bob has a pipe with a diameter of \(\frac { 6 }{ \sqrt { \pi } }\text{ cm} \) and a length of \(3\text{ m}\). How much water could be in this pipe at any one time, in \(\text{cm}^3?\)

What is the volume of the octahedron inside this \(8 \text{ in}^3\) cube?

A sector with radius \(10\text{ cm}\) and central angle \(45^\circ\) is to be made into a right circular cone. Find the volume of the cone.

\[\] Details and Assumptions:

• The arc length of the sector is equal to the circumference of the base of the cone.

Three identical tanks are shown above. The spheres in a given tank are the same size and packed wall-to-wall. If the tanks are filled to the top with water, then which tank would contain the most water?

A chocolate shop sells its products in 3 different shapes: a cylindrical bar, a spherical ball, and a cone. These 3 shapes are of the same height and radius, as shown in the picture. Which of these choices would give you the most chocolate?

\[\text{ I. A full cylindrical bar } \hspace{.4cm} \text{ or } \hspace{.45cm} \text{ II. A ball plus a cone }\]

How many cubes measuring 2 units on one side must be added to a cube measuring 8 units on one side to form a cube measuring 12 units on one side?

This section involves a deeper understanding of volume and the formulas to find the volume. Here are a couple of worked out examples followed by several "Try It Yourself" problems:

\(12\) spheres of the same size are made from melting a solid cylinder of \(16\text{ cm}\) diameter and \(2\text{ cm}\) height. Find the diameter of each sphere. Use \(\pi=\frac{22}{7}.\) The volume of the cylinder is \[\pi× r^2 × h = \frac {22×8^2×2}{7}= \frac {2816}{7}.\] Let the radius of each sphere be \(r\text{ cm}.\) Then the volume of each sphere in \(\text{cm}^3\) is \[\dfrac {4×22×r^3}{3×7} = \dfrac{88×r^3}{21}.\] Since the number of spheres is \(\frac {\text{Volume of cylinder}}{\text {Volume of 1 sphere}},\) \[\begin{align} 12 &= \dfrac{2816×21}{7×88×r^3}\\ &= \dfrac {96}{r^3}\\ r^3 &= \dfrac {96}{12}\\ &= 8\\ \Rightarrow r &= 2. \end{align}\] Therefore, the diameter of each sphere is \[2\times r = 2\times 2 = 4 ~(\text{cm}). \ _\square\]

Find the volume of a hemispherical shell whose outer radius is \(7\text{ cm}\) and inner radius is \(3\text{ cm}\), up to \(2\) decimal places. We have \[\begin{align} (\text {Volume of inner hemisphere}) & = \dfrac{1}{2} × \dfrac{4}{3} × \pi × R^3\\ & = \dfrac {1 × 4 × 22 × 27}{2 × 3 × 7}\\ & = \dfrac {396}{7}\\ & = 56.57 ~\big(\text{cm}^{3}\big) \\\\ (\text {Volume of outer hemisphere}) & = \dfrac{1}{2} × \dfrac{4}{3} × \pi × r^3\\ & = \dfrac {1 × 4 × 22 × 343}{2 × 3 × 7}\\ & = \dfrac {2156}{7}\\ & = 718.66 ~\big(\text{cm}^{3}\big) \\\\ (\text{Volume of hemispherical shell}) & = (\text{V. of outer hemisphere}) - (\text{V. of inner hemisphere})\\ & = 718.66 - 56.57 \\ & = 662.09 ~\big(\text{cm}^{3}\big).\ _\square \end{align}\]

A student did an experiment using a cone, a sphere, and a cylinder each having the same radius and height. He started with the cylinder full of liquid and then poured it into the cone until the cone was full. Then, he began pouring the remaining liquid from the cylinder into the sphere. What was the result which he observed?

There are two identical right circular cones each of height \(2\text{ cm}.\) They are placed vertically, with their apex pointing downwards, and one cone is vertically above the other. At the start, the upper cone is full of water and the lower cone is empty.

Water drips down through a hole in the apex of the upper cone into the lower cone. When the height of water in the upper cone is \(1\text{ cm},\) what is the height of water in the lower cone (in \(\text{cm}\))?

On each face of a cuboid, the sum of its perimeter and its area is written. The numbers recorded this way are 16, 24, and 31, each written on a pair of opposite sides of the cuboid. The volume of the cuboid lies between \(\text{__________}.\)

A cube rests inside a sphere such that each vertex touches the sphere. The radius of the sphere is \(6 \text{ cm}.\) Determine the volume of the cube.

If the volume of the cube can be expressed in the form of \(a\sqrt{3} \text{ cm}^{3}\), find the value of \(a\).

A sphere has volume \(x \text{ m}^3 \) and surface area \(x \text{ m}^2 \). Keeping its diameter as body diagonal, a cube is made which has volume \(a \text{ m}^3 \) and surface area \(b \text{ m}^2 \). What is the ratio \(a:b?\)

Consider a glass in the shape of an inverted truncated right cone (i.e. frustrum). The radius of the base is 4, the radius of the top is 9, and the height is 7. There is enough water in the glass such that when it is tilted the water reaches from the tip of the base to the edge of the top. The proportion of the water in the cup as a ratio of the cup's volume can be expressed as the fraction \( \frac{m}{n} \), for relatively prime integers \(m\) and \(n\). Compute \(m+n\).

The square-based pyramid A is inscribed within a cube while the tetrahedral pyramid B has its sides equal to the square's diagonal (red) as shown.

Which pyramid has more volume?

Please remember this section contains highly advanced problems of volume. Here it goes:

Cube \(ABCDEFGH\), labeled as shown above, has edge length \(1\) and is cut by a plane passing through vertex \(D\) and the midpoints \(M\) and \(N\) of \(\overline{AB}\) and \(\overline{CG}\) respectively. The plane divides the cube into two solids. The volume of the larger of the two solids can be written in the form \(\frac{p}{q}\), where \(p\) and \(q\) are relatively prime positive integers. Find \(p+q\).

If the American NFL regulation football

has a tip-to-tip length of \(11\) inches and a largest round circumference of \(22\) in the middle, then the volume of the American football is \(\text{____________}.\)

Note: The American NFL regulation football is not an ellipsoid. The long cross-section consists of two circular arcs meeting at the tips. Don't use the volume formula for an ellipsoid.

Answer is in cubic inches.

Consider a solid formed by the intersection of three orthogonal cylinders, each of diameter \( D = 10 \).

What is the volume of this solid?

Consider a tetrahedron with side lengths \(2, 3, 3, 4, 5, 5\). The largest possible volume of this tetrahedron has the form \( \frac {a \sqrt{b}}{c}\), where \(b\) is an integer that's not divisible by the square of any prime, \(a\) and \(c\) are positive, coprime integers. What is the value of \(a+b+c\)?

Let there be a solid characterized by the equation \[{ \left( \frac { x }{ a } \right) }^{ 2.5 }+{ \left( \frac { y }{ b } \right) }^{ 2.5 } + { \left( \frac { z }{ c } \right) }^{ 2.5 }<1.\]

Calculate the volume of this solid if \(a = b =2\) and \(c = 3\).

• Surface Area

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How To : Solve Word Problems Involving Volume of a Pyramid

Learn how to solve geometry word problems.

For example, how would you solve the following problem?

A prism has a square base a prism has a square base with a side of 3 the volume is 90 a square pyramid has the same width, base and height as the pyramid, what is the volume?

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