GCSE Maths: Pyramids and cones (Higher)

Key words. The apex is the top point of a pyramid. The vertical height is measured straight down from the apex to the base. The slant height is the height of a sloping triangular face.

Method. Volume of a pyramid = one third × area of the base × vertical height.

1. Write down the formula. A pyramid takes up exactly one third of the prism with the same base and the same height.
   Volume = ⅓ × base area × vertical height
2. Find the area of the base. The base is a square 6 cm by 6 cm.
   6 × 6 = 36 cm²
3. Multiply by the vertical height. Use the vertical height, from the top point straight down to the base (not the slanting edge).
   36 × 4 = 144
4. Divide by 3. Taking one third means dividing by 3.
   144 ÷ 3 = 48
5. Write the unit.
   V = 48 cm³

Check. The prism with the same base and height has volume 144. Three pyramids fill it: 3 × 48 = 144. ✓

Watch out. Do not forget the ⅓. Use the vertical (perpendicular) height, not the sloping edge or slant height.

Key words. The apex is the top point of a pyramid. The vertical height is measured straight down from the apex to the base. The slant height is the height of a sloping triangular face.

Method. Volume of a pyramid = one third × area of the base × vertical height.

1. Write down the formula. A pyramid takes up exactly one third of the prism with the same base and the same height.
   Volume = ⅓ × base area × vertical height
2. Find the area of the base. The base is a rectangle 8 cm by 5 cm.
   8 × 5 = 40 cm²
3. Multiply by the vertical height. Use the vertical height, from the top point straight down to the base (not the slanting edge).
   40 × 9 = 360
4. Divide by 3. Taking one third means dividing by 3.
   360 ÷ 3 = 120
5. Write the unit.
   V = 120 cm³

Check. The prism with the same base and height has volume 360. Three pyramids fill it: 3 × 120 = 360. ✓

Watch out. Do not forget the ⅓. Use the vertical (perpendicular) height, not the sloping edge or slant height.

Key words. The apex is the top point of a pyramid. The vertical height is measured straight down from the apex to the base. The slant height is the height of a sloping triangular face.

Method. Volume of a pyramid = one third × area of the base × vertical height.

1. Write down the formula. A pyramid takes up exactly one third of the prism with the same base and the same height.
   Volume = ⅓ × base area × vertical height
2. Find the area of the base. The base is a square 10 cm by 10 cm.
   10 × 10 = 100 cm²
3. Multiply by the vertical height. Use the vertical height, from the top point straight down to the base (not the slanting edge).
   100 × 12 = 1200
4. Divide by 3. Taking one third means dividing by 3.
   1200 ÷ 3 = 400
5. Write the unit.
   V = 400 cm³

Check. The prism with the same base and height has volume 1200. Three pyramids fill it: 3 × 400 = 1200. ✓

Watch out. Do not forget the ⅓. Use the vertical (perpendicular) height, not the sloping edge or slant height.

Key words. The apex is the top point of a pyramid. The vertical height is measured straight down from the apex to the base. The slant height is the height of a sloping triangular face.

Method. Volume of a pyramid = one third × area of the base × vertical height.

1. Write down the formula. A pyramid takes up exactly one third of the prism with the same base and the same height.
   Volume = ⅓ × base area × vertical height
2. Find the area of the base. The base is a rectangle 9 cm by 4 cm.
   9 × 4 = 36 cm²
3. Multiply by the vertical height. Use the vertical height, from the top point straight down to the base (not the slanting edge).
   36 × 5 = 180
4. Divide by 3. Taking one third means dividing by 3.
   180 ÷ 3 = 60
5. Write the unit.
   V = 60 cm³

Check. The prism with the same base and height has volume 180. Three pyramids fill it: 3 × 60 = 180. ✓

Watch out. Do not forget the ⅓. Use the vertical (perpendicular) height, not the sloping edge or slant height.

Key words. The surface area of a solid is the total area of all its faces added together. It is measured in square units (cm²), because it is an area.

Method. Surface area = area of the square base + the area of the 4 triangular faces (each uses the slant height).

1. Find the slant height first. The sloping faces are triangles. Their height is the slant height, which is the hypotenuse of a right-angled triangle made from the vertical height and half of the base.
2. Draw the right-angled triangle. The slant height of a sloping face l is the longest side (the hypotenuse) of a right-angled triangle with shorter sides 4 cm and 3 cm.
3. Use Pythagoras: a² + b² = c². The square on the hypotenuse equals the sum of the squares on the other two sides.
   l² = 4² + 3² = 16 + 9 = 25
4. Square root. Undo the squaring.
   l = √25 = 5 cm
5. Area of the square base.
   6 × 6 = 36
6. Area of one triangular face. Area of a triangle = ½ × base × height. Here the height of the triangle is the slant height.
   ½ × 6 × 5 = 15
7. Four triangular faces. A square-based pyramid has 4 identical sloping faces.
   4 × 15 = 60
8. Add the base to the triangles. Surface area = base + 4 triangles.
   36 + 60 = 96
9. Write the unit.
   Surface area = 96 cm²

Check. Quick check: 4 triangles = 2 × base × slant = 2 × 6 × 5 = 60; adding the base 36 gives 96. ✓

Watch out. The triangles use the slant height, not the vertical height of the pyramid.

Key words. The surface area of a solid is the total area of all its faces added together. It is measured in square units (cm²), because it is an area.

Method. Surface area = area of the square base + the area of the 4 triangular faces (each uses the slant height).

1. Find the slant height first. The sloping faces are triangles. Their height is the slant height, which is the hypotenuse of a right-angled triangle made from the vertical height and half of the base.
2. Draw the right-angled triangle. The slant height of a sloping face l is the longest side (the hypotenuse) of a right-angled triangle with shorter sides 12 cm and 5 cm.
3. Use Pythagoras: a² + b² = c². The square on the hypotenuse equals the sum of the squares on the other two sides.
   l² = 12² + 5² = 144 + 25 = 169
4. Square root. Undo the squaring.
   l = √169 = 13 cm
5. Area of the square base.
   10 × 10 = 100
6. Area of one triangular face. Area of a triangle = ½ × base × height. Here the height of the triangle is the slant height.
   ½ × 10 × 13 = 65
7. Four triangular faces. A square-based pyramid has 4 identical sloping faces.
   4 × 65 = 260
8. Add the base to the triangles. Surface area = base + 4 triangles.
   100 + 260 = 360
9. Write the unit.
   Surface area = 360 cm²

Check. Quick check: 4 triangles = 2 × base × slant = 2 × 10 × 13 = 260; adding the base 100 gives 360. ✓

Watch out. The triangles use the slant height, not the vertical height of the pyramid.

Key words. The surface area of a solid is the total area of all its faces added together. It is measured in square units (cm²), because it is an area.

Method. Surface area = area of the square base + the area of the 4 triangular faces (each uses the slant height).

1. Find the slant height first. The sloping faces are triangles. Their height is the slant height, which is the hypotenuse of a right-angled triangle made from the vertical height and half of the base.
2. Draw the right-angled triangle. The slant height of a sloping face l is the longest side (the hypotenuse) of a right-angled triangle with shorter sides 15 cm and 8 cm.
3. Use Pythagoras: a² + b² = c². The square on the hypotenuse equals the sum of the squares on the other two sides.
   l² = 15² + 8² = 225 + 64 = 289
4. Square root. Undo the squaring.
   l = √289 = 17 cm
5. Area of the square base.
   16 × 16 = 256
6. Area of one triangular face. Area of a triangle = ½ × base × height. Here the height of the triangle is the slant height.
   ½ × 16 × 17 = 136
7. Four triangular faces. A square-based pyramid has 4 identical sloping faces.
   4 × 136 = 544
8. Add the base to the triangles. Surface area = base + 4 triangles.
   256 + 544 = 800
9. Write the unit.
   Surface area = 800 cm²

Check. Quick check: 4 triangles = 2 × base × slant = 2 × 16 × 17 = 544; adding the base 256 gives 800. ✓

Watch out. The triangles use the slant height, not the vertical height of the pyramid.

Key words. The surface area of a solid is the total area of all its faces added together. It is measured in square units (cm²), because it is an area.

Method. Surface area = area of the square base + the area of the 4 triangular faces (each uses the slant height).

1. Area of the square base.
   8 × 8 = 64
2. Area of one triangular face. Area of a triangle = ½ × base × height. Here the height of the triangle is the slant height.
   ½ × 8 × 5 = 20
3. Four triangular faces. A square-based pyramid has 4 identical sloping faces.
   4 × 20 = 80
4. Add the base to the triangles. Surface area = base + 4 triangles.
   64 + 80 = 144
5. Write the unit.
   Surface area = 144 cm²

Check. Quick check: 4 triangles = 2 × base × slant = 2 × 8 × 5 = 80; adding the base 64 gives 144. ✓

Watch out. The triangles use the slant height, not the vertical height of the pyramid.

Key words. The slant height (l) of a cone is the distance from the point (apex) down the sloping side to the edge of the base. The vertical height (h) goes straight down from the apex to the centre of the base. They are different lengths.

Method. Volume of a cone = one third of πr²h, where h is the vertical height.

1. Write down the formula. A cone is one third of the cylinder with the same base and height.
   V = ⅓πr²h
2. Substitute.
   V = ⅓ × π × 3² × 4
3. Work out r² and multiply by h. Powers first, then multiply.
   3² = 9
   9 × 4 = 36
4. Take one third. Divide by 3.
   36 ÷ 3 = 12, so V = 12π
5. Write the answer in terms of π. Leave the answer in terms of π: write the number in front of π and do not multiply by 3.14.
   12π cm³

Check. The cylinder with the same r and h has volume 36π; the cone is one third of it: 12π. ✓

Watch out. Use the vertical height h in V = ⅓πr²h. If you are given the slant height, use Pythagoras to find h first.

Key words. The slant height (l) of a cone is the distance from the point (apex) down the sloping side to the edge of the base. The vertical height (h) goes straight down from the apex to the centre of the base. They are different lengths.

Method. Volume of a cone = one third of πr²h, where h is the vertical height.

1. Write down the formula. A cone is one third of the cylinder with the same base and height.
   V = ⅓πr²h
2. Substitute.
   V = ⅓ × π × 6² × 5
3. Work out r² and multiply by h. Powers first, then multiply.
   6² = 36
   36 × 5 = 180
4. Take one third. Divide by 3.
   180 ÷ 3 = 60, so V = 60π
5. Write the answer in terms of π. Leave the answer in terms of π: write the number in front of π and do not multiply by 3.14.
   60π cm³

Check. The cylinder with the same r and h has volume 180π; the cone is one third of it: 60π. ✓

Watch out. Use the vertical height h in V = ⅓πr²h. If you are given the slant height, use Pythagoras to find h first.

Key words. The slant height (l) of a cone is the distance from the point (apex) down the sloping side to the edge of the base. The vertical height (h) goes straight down from the apex to the centre of the base. They are different lengths.

Method. Volume of a cone = one third of πr²h, where h is the vertical height.

1. Find the vertical height first. The volume formula needs the vertical height, but we were given the slant height.
2. Draw the right-angled triangle. The vertical height h is one of the shorter sides. The slant side 13 cm is the hypotenuse and the other short side is 5 cm.
3. Use Pythagoras with the unknown on a short side. a² + b² = c², so b² = c² − a².
   h² = 13² − 5² = 169 − 25 = 144
4. Square root. Undo the squaring.
   h = √144 = 12 cm
5. Write down the formula. A cone is one third of the cylinder with the same base and height.
   V = ⅓πr²h
6. Substitute.
   V = ⅓ × π × 5² × 12
7. Work out r² and multiply by h. Powers first, then multiply.
   5² = 25
   25 × 12 = 300
8. Take one third. Divide by 3.
   300 ÷ 3 = 100, so V = 100π
9. Write the answer in terms of π. Leave the answer in terms of π: write the number in front of π and do not multiply by 3.14.
   100π cm³

Check. The cylinder with the same r and h has volume 300π; the cone is one third of it: 100π. ✓

Watch out. Use the vertical height h in V = ⅓πr²h. If you are given the slant height, use Pythagoras to find h first.

Key words. The slant height (l) of a cone is the distance from the point (apex) down the sloping side to the edge of the base. The vertical height (h) goes straight down from the apex to the centre of the base. They are different lengths.

Method. Volume of a cone = one third of πr²h, where h is the vertical height.

1. Find the radius. The formulas use the radius, which is half the diameter.
   r = 10 ÷ 2 = 5
2. Write down the formula. A cone is one third of the cylinder with the same base and height.
   V = ⅓πr²h
3. Substitute.
   V = ⅓ × π × 5² × 12
4. Work out r² and multiply by h. Powers first, then multiply.
   5² = 25
   25 × 12 = 300
5. Take one third. Divide by 3.
   300 ÷ 3 = 100, so V = 100π
6. Write the answer in terms of π. Leave the answer in terms of π: write the number in front of π and do not multiply by 3.14.
   100π cm³

Check. The cylinder with the same r and h has volume 300π; the cone is one third of it: 100π. ✓

Watch out. Use the vertical height h in V = ⅓πr²h. If you are given the slant height, use Pythagoras to find h first.

Key words. The slant height (l) of a cone is the distance from the point (apex) down the sloping side to the edge of the base. The vertical height (h) goes straight down from the apex to the centre of the base. They are different lengths.

Method. Curved surface area of a cone = πrl, where l is the slant height.

1. Find the slant height first. The surface area formulas use the slant height l, but we were given the vertical height.
2. Draw the right-angled triangle. The slant height l is the longest side (the hypotenuse) of a right-angled triangle with shorter sides 3 cm and 4 cm.
3. Use Pythagoras: a² + b² = c². The square on the hypotenuse equals the sum of the squares on the other two sides.
   l² = 3² + 4² = 9 + 16 = 25
4. Square root. Undo the squaring.
   l = √25 = 5 cm
5. Write down the formula. The curved surface of a cone opens out into part of a circle (a sector).
   Curved surface area = πrl
6. Substitute and multiply. r is the radius of the base; l is the slant height.
   π × 3 × 5 = 15π
7. Write the answer in terms of π. Leave the answer in terms of π: write the number in front of π and do not multiply by 3.14.
   15π cm²

Check. Sector check: arc length = circumference of base = 2π × 3. The sector area is ½ × l × arc = ½ × 5 × 6π = 15π. ✓

Watch out. The curved surface area uses the slant height l. Add the base circle only if the question asks for the total surface area.

Key words. The slant height (l) of a cone is the distance from the point (apex) down the sloping side to the edge of the base. The vertical height (h) goes straight down from the apex to the centre of the base. They are different lengths.

Method. Curved surface area of a cone = πrl, where l is the slant height.

1. Find the slant height first. The surface area formulas use the slant height l, but we were given the vertical height.
2. Draw the right-angled triangle. The slant height l is the longest side (the hypotenuse) of a right-angled triangle with shorter sides 5 cm and 12 cm.
3. Use Pythagoras: a² + b² = c². The square on the hypotenuse equals the sum of the squares on the other two sides.
   l² = 5² + 12² = 25 + 144 = 169
4. Square root. Undo the squaring.
   l = √169 = 13 cm
5. Write down the formula. The curved surface of a cone opens out into part of a circle (a sector).
   Curved surface area = πrl
6. Substitute and multiply. r is the radius of the base; l is the slant height.
   π × 5 × 13 = 65π
7. Write the answer in terms of π. Leave the answer in terms of π: write the number in front of π and do not multiply by 3.14.
   65π cm²

Check. Sector check: arc length = circumference of base = 2π × 5. The sector area is ½ × l × arc = ½ × 13 × 10π = 65π. ✓

Watch out. The curved surface area uses the slant height l. Add the base circle only if the question asks for the total surface area.

Key words. The slant height (l) of a cone is the distance from the point (apex) down the sloping side to the edge of the base. The vertical height (h) goes straight down from the apex to the centre of the base. They are different lengths.

Method. Curved surface area of a cone = πrl, where l is the slant height.

1. Write down the formula. The curved surface of a cone opens out into part of a circle (a sector).
   Curved surface area = πrl
2. Substitute and multiply. r is the radius of the base; l is the slant height.
   π × 8 × 17 = 136π
3. Write the answer in terms of π. Leave the answer in terms of π: write the number in front of π and do not multiply by 3.14.
   136π cm²

Check. Sector check: arc length = circumference of base = 2π × 8. The sector area is ½ × l × arc = ½ × 17 × 16π = 136π. ✓

Watch out. The curved surface area uses the slant height l. Add the base circle only if the question asks for the total surface area.

Key words. The slant height (l) of a cone is the distance from the point (apex) down the sloping side to the edge of the base. The vertical height (h) goes straight down from the apex to the centre of the base. They are different lengths.

Method. Total surface area of a cone = curved surface (πrl) + circular base (πr²).

1. Write down the formula. Total surface area = the curved surface + the circular base.
   Total surface area = πrl + πr²
2. Curved surface.
   π × 5 × 13 = 65π
3. Circular base.
   π × 5² = 25π
4. Add them. Add the numbers in front of π.
   65π + 25π = 90π
5. Write the answer in terms of π. Leave the answer in terms of π: write the number in front of π and do not multiply by 3.14.
   90π cm²

Check. Factorise: πr(l + r) = π × 5 × 18 = 90π. ✓

Watch out. The curved surface area uses the slant height l. Add the base circle only if the question asks for the total surface area.

Key words. The slant height (l) of a cone is the distance from the point (apex) down the sloping side to the edge of the base. The vertical height (h) goes straight down from the apex to the centre of the base. They are different lengths.

Method. Total surface area of a cone = curved surface (πrl) + circular base (πr²).

1. Find the slant height first. The surface area formulas use the slant height l, but we were given the vertical height.
2. Draw the right-angled triangle. The slant height l is the longest side (the hypotenuse) of a right-angled triangle with shorter sides 3 cm and 4 cm.
3. Use Pythagoras: a² + b² = c². The square on the hypotenuse equals the sum of the squares on the other two sides.
   l² = 3² + 4² = 9 + 16 = 25
4. Square root. Undo the squaring.
   l = √25 = 5 cm
5. Write down the formula. Total surface area = the curved surface + the circular base.
   Total surface area = πrl + πr²
6. Curved surface.
   π × 3 × 5 = 15π
7. Circular base.
   π × 3² = 9π
8. Add them. Add the numbers in front of π.
   15π + 9π = 24π
9. Write the answer in terms of π. Leave the answer in terms of π: write the number in front of π and do not multiply by 3.14.
   24π cm²

Check. Factorise: πr(l + r) = π × 3 × 8 = 24π. ✓

Watch out. The curved surface area uses the slant height l. Add the base circle only if the question asks for the total surface area.

Key words. The slant height (l) of a cone is the distance from the point (apex) down the sloping side to the edge of the base. The vertical height (h) goes straight down from the apex to the centre of the base. They are different lengths.

Method. Total surface area of a cone = curved surface (πrl) + circular base (πr²).

1. Write down the formula. Total surface area = the curved surface + the circular base.
   Total surface area = πrl + πr²
2. Curved surface.
   π × 6 × 10 = 60π
3. Circular base.
   π × 6² = 36π
4. Add them. Add the numbers in front of π.
   60π + 36π = 96π
5. Write the answer in terms of π. Leave the answer in terms of π: write the number in front of π and do not multiply by 3.14.
   96π cm²

Check. Factorise: πr(l + r) = π × 6 × 16 = 96π. ✓

Watch out. The curved surface area uses the slant height l. Add the base circle only if the question asks for the total surface area.

Key words. The slant height (l) of a cone is the distance from the point (apex) down the sloping side to the edge of the base. The vertical height (h) goes straight down from the apex to the centre of the base. They are different lengths.

Method. Volume of a cone = one third of πr²h, where h is the vertical height.

1. Write down the formula. A cone is one third of the cylinder with the same base and height.
   V = ⅓πr²h
2. Substitute.
   V = ⅓ × π × 2.5² × 6
3. Work out r² and multiply by h. Powers first, then multiply.
   2.5² = 6.25
   6.25 × 6 = 37.5
4. Take one third. Divide by 3.
   37.5 ÷ 3 = 12.5, so V = 12.5π
5. Use your calculator. Type the calculation in using the π button on your calculator (not 3.14). Write down the whole display before you round.
   ⅓ × π × 2.5² × 6 = 39.2699…
6. Round the answer. The question wants 3 significant figures. Look at the digit just after the last one you are keeping: 5 or more rounds up, 4 or less leaves it.
   39.2699… = 39.3 cm³

Check. The cylinder with the same r and h has volume 37.5π; the cone is one third of it: 12.5π. ✓

Watch out. Use the vertical height h in V = ⅓πr²h. If you are given the slant height, use Pythagoras to find h first.

Key words. The slant height (l) of a cone is the distance from the point (apex) down the sloping side to the edge of the base. The vertical height (h) goes straight down from the apex to the centre of the base. They are different lengths.

Method. Volume of a cone = one third of πr²h, where h is the vertical height.

1. Find the vertical height first. The volume formula needs the vertical height, but we were given the slant height.
2. Draw the right-angled triangle. The vertical height h is one of the shorter sides. The slant side 25 cm is the hypotenuse and the other short side is 7 cm.
3. Use Pythagoras with the unknown on a short side. a² + b² = c², so b² = c² − a².
   h² = 25² − 7² = 625 − 49 = 576
4. Square root. Undo the squaring.
   h = √576 = 24 cm
5. Write down the formula. A cone is one third of the cylinder with the same base and height.
   V = ⅓πr²h
6. Substitute.
   V = ⅓ × π × 7² × 24
7. Work out r² and multiply by h. Powers first, then multiply.
   7² = 49
   49 × 24 = 1176
8. Take one third. Divide by 3.
   1176 ÷ 3 = 392, so V = 392π
9. Use your calculator. Type the calculation in using the π button on your calculator (not 3.14). Write down the whole display before you round.
   ⅓ × π × 7² × 24 = 1231.5043…
10. Round the answer. The question wants 3 significant figures. Look at the digit just after the last one you are keeping: 5 or more rounds up, 4 or less leaves it.
    1231.5043… = 1230 cm³

Check. The cylinder with the same r and h has volume 1176π; the cone is one third of it: 392π. ✓

Watch out. Use the vertical height h in V = ⅓πr²h. If you are given the slant height, use Pythagoras to find h first.

Method. Write the volume formula with the numbers, then undo it step by step to find the height.

1. Area of the base. The base is 6 cm by 6 cm.
   6 × 6 = 36 cm²
2. Write the formula with the numbers. V = ⅓ × base area × height.
   96 = ⅓ × 36 × h
3. Multiply both sides by 3. This removes the ⅓.
   288 = 36 × h
4. Divide by the base area.
   h = 288 ÷ 36 = 8 cm

Check. ⅓ × 36 × 8 = 96. ✓

Watch out. Multiply the volume by 3 (do not divide by 3) to undo the ⅓.

Method. Write the volume formula with the numbers, then undo it step by step to find the height.

1. Area of the base. The base is 5 cm by 8 cm.
   5 × 8 = 40 cm²
2. Write the formula with the numbers. V = ⅓ × base area × height.
   200 = ⅓ × 40 × h
3. Multiply both sides by 3. This removes the ⅓.
   600 = 40 × h
4. Divide by the base area.
   h = 600 ÷ 40 = 15 cm

Check. ⅓ × 40 × 15 = 200. ✓

Watch out. Multiply the volume by 3 (do not divide by 3) to undo the ⅓.

Method. Put the numbers into V = ⅓πr²h, cancel π, then undo the ⅓ and the r².

1. Write the formula with the numbers.
   144π = ⅓ × π × 4² × h
2. Divide both sides by π and work out r². The π cancels.
   144 = ⅓ × 16 × h
3. Multiply both sides by 3. This undoes the ⅓.
   432 = 16 × h
4. Divide by the number in front of h.
   h = 432 ÷ 16 = 27 cm

Check. ⅓ × π × 16 × 27 = 144π. ✓

Watch out. Multiply by 3 (not divide) to undo the ⅓.

Method. Put the numbers into V = ⅓πr²h, cancel π, then undo the ⅓ and the r².

1. Write the formula with the numbers.
   100π = ⅓ × π × 5² × h
2. Divide both sides by π and work out r². The π cancels.
   100 = ⅓ × 25 × h
3. Multiply both sides by 3. This undoes the ⅓.
   300 = 25 × h
4. Divide by the number in front of h.
   h = 300 ÷ 25 = 12 cm

Check. ⅓ × π × 25 × 12 = 100π. ✓

Watch out. Multiply by 3 (not divide) to undo the ⅓.