Sec 2 Maths · Pyramids, cones and spheres

Pyramids questions with answers

Practice questions on pyramids, from the basics up to exam level, each with a full worked solution. Try them on paper first, then open the solution to check your working.

Worked example

The diagram shows a paperweight in the shape of a pyramid with a rectangular base $6$ cm by $10$ cm. Its apex is $12$ cm vertically above the centre of the base. Find the volume of the pyramid.

Step 1 of 2

Area of the base

Base area

= 6 \times 10

= 60

cm²

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Questions

Question 1Volume of a pyramid
The diagram shows a glass building in the shape of a pyramid with a square base of side $12$ m. Its apex is $6$ m vertically above the centre of the base. Find the volume of the pyramid.
Show the worked solution
1. Step 1: Area of the baseBase area $= 12 \times 12$ $= 144$ m²
2. Step 2: Volume of a pyramid = ⅓ × base area × height $V = \dfrac{1}{3} \times 144 \times 6$ $= 288$ m³
Question 2Height from the volume
The diagram shows a paperweight in the shape of a pyramid with a square base of side $12$ cm. Its volume is $720$ cm³. Find its height, $h$ cm.
Show the worked solution
1. Step 1: Area of the baseBase area $= 12 \times 12$ $= 144$ cm²
2. Step 2: Volume = ⅓ × base area × height: write an equation in h and solve it $\dfrac{1}{3} \times 144 \times h = 720$ $h = \dfrac{720 \times 3}{144}$ $= 15$ cm
Question 3Total surface area (slant heights given)
The diagram shows a solid pyramid with a square base of side $12$ cm. The slant height of each triangular face is $10$ cm. Find the total surface area of the pyramid.
Show the worked solution
1. Step 1: Area of the baseBase area $= 12 \times 12$ $= 144$ cm²
2. Step 2: Each triangular face: ½ × base × slant heightArea of one face $= \dfrac{1}{2} \times 12 \times 10$ $= 60$ cm²
3. Step 3: Total surface area: the base and all four facesTotal surface area $= 144 + 4 \times 60$ $= 384$ cm²
Question 4Slant height first, then surface area
The diagram shows a paperweight in the shape of a pyramid with a square base of side $20$ cm. Its apex is $8$ cm vertically above the centre of the base. Find the slant height, $l$ cm, of a triangular face, and hence the total surface area of the pyramid, correct to 3 significant figures.
Show the worked solution
1. Step 1: Slant height: the height, half the base and the slant height make a right-angled triangle $l^2 = 8^2 + 10^2$ $= 164$ $l = \sqrt{164}$ $= 12.8062\ldots$
2. Step 2: Area of the baseBase area $= 20 \times 20$ $= 400$ cm²
3. Step 3: Each triangular face: ½ × base × slant heightArea of one face $= \dfrac{1}{2} \times 20 \times 12.8062\ldots$ $= 128.0624\ldots$
4. Step 4: Total surface area: the base and all four facesTotal surface area $= 400 + 4 \times 128.0624\ldots$ $= 912.2499\ldots$ $= 912$ cm² (3 s.f.)
Question 5Height from a slant height or edge, then the volume
The diagram shows a pyramid with a square base of side $6$ cm. Its apex $V$ is vertically above the centre $O$ of the base, and each of its four slant edges is $12$ cm long. Find the height of the pyramid, and hence its volume, correct to 3 significant figures.
Show the worked solution
1. Step 1: Half the diagonal of the base: O is the centre, so OB is half of BD (Pythagoras on the base) $OB^2 = 3^2 + 3^2$ $= 18$
2. Step 2: The height: triangle VOB has a right angle at O, with the edge VB as hypotenuse $h^2 = 12^2 - 18$ $= 126$ $h = \sqrt{126}$ $= 11.2249\ldots$ cm
3. Step 3: Volume of a pyramid = ⅓ × base area × height (keep the full value of h) $V = \dfrac{1}{3} \times 6^2 \times 11.2249\ldots$ $= 134.6996\ldots$ $= 135$ cm³ (3 s.f.)
Question 6The base side from the volume
The diagram shows a pyramid with a square base of side $x$ cm. Its apex is $3$ cm vertically above the centre of the base, and its volume is $36$ cm³. Find $x$ .
Show the worked solution
1. Step 1: Volume = ⅓ × base area × height: write an equation in x $\dfrac{1}{3} \times x^2 \times 3 = 36$
2. Step 2: Make x² the subject: multiply by 3, divide by the height $x^2 = \dfrac{36 \times 3}{3}$ $\phantom{x^2} = 36$
3. Step 3: Take the square root (a length is positive) $x = \sqrt{36}$ $\phantom{x} = 6$
Question 7Hexagonal pyramids: surface area, mass and density
The diagram shows a glass paperweight in the shape of a pyramid. Its base is a regular hexagon of side $4$ cm and area $41.6$ cm², and the slant height of each triangular face is $12.5$ cm. Find the total surface area of the pyramid, correct to 3 significant figures.
Show the worked solution
1. Step 1: Each of the six triangular faces: ½ × base × slant heightArea of one face $= \dfrac{1}{2} \times 4 \times 12.5$ $= 25$ cm²
2. Step 2: Total surface area: the hexagonal base and all six facesTotal surface area $= 41.6 + 6 \times 25$ $= 191.6000\ldots$ $= 192$ cm² (3 s.f.)

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Worked example

The diagram shows a paperweight in the shape of a pyramid with a rectangular base

6

cm by

10

cm. Its apex is

12

cm vertically above the centre of the base. Find the volume of the pyramid.

Step 1 of 2

Area of the base

Base area

= 6 \times 10

= 60

cm²

Questions

Question 1Volume of a pyramid

The diagram shows a glass building in the shape of a pyramid with a square base of side $12$ m. Its apex is $6$ m vertically above the centre of the base. Find the volume of the pyramid.

Show the worked solution

Step 1: Area of the baseBase area $= 12 \times 12$ $= 144$ m²
Step 2: Volume of a pyramid = ⅓ × base area × height $V = \dfrac{1}{3} \times 144 \times 6$ $= 288$ m³

Question 2Height from the volume

The diagram shows a paperweight in the shape of a pyramid with a square base of side $12$ cm. Its volume is $720$ cm³. Find its height, $h$ cm.

Show the worked solution

Step 1: Area of the baseBase area $= 12 \times 12$ $= 144$ cm²
Step 2: Volume = ⅓ × base area × height: write an equation in h and solve it $\dfrac{1}{3} \times 144 \times h = 720$ $h = \dfrac{720 \times 3}{144}$ $= 15$ cm

Question 3Total surface area (slant heights given)

The diagram shows a solid pyramid with a square base of side $12$ cm. The slant height of each triangular face is $10$ cm. Find the total surface area of the pyramid.

Show the worked solution

Step 1: Area of the baseBase area $= 12 \times 12$ $= 144$ cm²
Step 2: Each triangular face: ½ × base × slant heightArea of one face $= \dfrac{1}{2} \times 12 \times 10$ $= 60$ cm²
Step 3: Total surface area: the base and all four facesTotal surface area $= 144 + 4 \times 60$ $= 384$ cm²

Question 4Slant height first, then surface area

The diagram shows a paperweight in the shape of a pyramid with a square base of side $20$ cm. Its apex is $8$ cm vertically above the centre of the base. Find the slant height, $l$ cm, of a triangular face, and hence the total surface area of the pyramid, correct to 3 significant figures.

Show the worked solution

Step 1: Slant height: the height, half the base and the slant height make a right-angled triangle $l^2 = 8^2 + 10^2$ $= 164$ $l = \sqrt{164}$ $= 12.8062\ldots$
Step 2: Area of the baseBase area $= 20 \times 20$ $= 400$ cm²
Step 3: Each triangular face: ½ × base × slant heightArea of one face $= \dfrac{1}{2} \times 20 \times 12.8062\ldots$ $= 128.0624\ldots$
Step 4: Total surface area: the base and all four facesTotal surface area $= 400 + 4 \times 128.0624\ldots$ $= 912.2499\ldots$ $= 912$ cm² (3 s.f.)

Question 5Height from a slant height or edge, then the volume

The diagram shows a pyramid with a square base of side $6$ cm. Its apex $V$ is vertically above the centre $O$ of the base, and each of its four slant edges is $12$ cm long. Find the height of the pyramid, and hence its volume, correct to 3 significant figures.

Show the worked solution

Step 1: Half the diagonal of the base: O is the centre, so OB is half of BD (Pythagoras on the base) $OB^2 = 3^2 + 3^2$ $= 18$
Step 2: The height: triangle VOB has a right angle at O, with the edge VB as hypotenuse $h^2 = 12^2 - 18$ $= 126$ $h = \sqrt{126}$ $= 11.2249\ldots$ cm
Step 3: Volume of a pyramid = ⅓ × base area × height (keep the full value of h) $V = \dfrac{1}{3} \times 6^2 \times 11.2249\ldots$ $= 134.6996\ldots$ $= 135$ cm³ (3 s.f.)

Question 6The base side from the volume

The diagram shows a pyramid with a square base of side $x$ cm. Its apex is $3$ cm vertically above the centre of the base, and its volume is $36$ cm³. Find $x$ .

Show the worked solution

Step 1: Volume = ⅓ × base area × height: write an equation in x $\dfrac{1}{3} \times x^2 \times 3 = 36$
Step 2: Make x² the subject: multiply by 3, divide by the height $x^2 = \dfrac{36 \times 3}{3}$ $\phantom{x^2} = 36$
Step 3: Take the square root (a length is positive) $x = \sqrt{36}$ $\phantom{x} = 6$

Question 7Hexagonal pyramids: surface area, mass and density

The diagram shows a glass paperweight in the shape of a pyramid. Its base is a regular hexagon of side $4$ cm and area $41.6$ cm², and the slant height of each triangular face is $12.5$ cm. Find the total surface area of the pyramid, correct to 3 significant figures.

Show the worked solution

Step 1: Each of the six triangular faces: ½ × base × slant heightArea of one face $= \dfrac{1}{2} \times 4 \times 12.5$ $= 25$ cm²
Step 2: Total surface area: the hexagonal base and all six facesTotal surface area $= 41.6 + 6 \times 25$ $= 191.6000\ldots$ $= 192$ cm² (3 s.f.)