Surface Area and Volume of 3D Solids

Spheres and Hemispheres

A basketball and a marble are perfectly round, but calculating exactly how much space they take up requires the mathematical constant $\pi$ . A sphere is a perfectly round 3D geometrical object where every point on the surface is an equal distance from the centre. This distance is the radius ( $r$ ). The distance all the way across, passing through the centre, is the diameter ( $d$ ).

For Edexcel GCSE Mathematics, you will often be given the formulas for a sphere in the exam, but you must know how to substitute into them accurately:

$V = \frac{4}{3}\pi r^3$

$SA = 4\pi r^2$

A hemisphere is exactly half of a sphere. Its volume is exactly half ( $\frac{2}{3}\pi r^3$ ), but its total surface area must include the curved surface ( $2\pi r^2$ ) plus the flat circular base ( $\pi r^2$ ), making the total surface area $3\pi r^2$ .

A sphere has a radius of $7\text{ cm}$ . Calculate its volume and surface area to 1 decimal place.

Step 1: Substitute the radius ( $r = 7$ ) into the volume formula.

$V = \frac{4}{3} \times \pi \times 7^3$

Step 2: Calculate the volume and state the units.

$V = 1436.755... \approx 1436.8\text{ cm}^3$

Step 3: Substitute the radius into the surface area formula.

$SA = 4 \times \pi \times 7^2$

Step 4: Calculate the surface area and state the units.

$SA = 615.752... \approx 615.8\text{ cm}^2$

The volume of a sphere is $12,000\text{ cm}^3$ . Calculate the radius $r$ to 1 decimal place.

Step 1: Set the volume formula equal to the given volume.

$12000 = \frac{4}{3}\pi r^3$

Step 2: Rearrange to isolate $r^3$ .

$r^3 = \frac{36000}{4\pi} = \frac{9000}{\pi}$

Step 3: Cube root to find $r$ .

$r = \sqrt[3]{\frac{9000}{\pi}} \approx 14.2\text{ cm}$

Pyramids

The Great Pyramid of Giza was built over 4,500 years ago, yet the mathematics used to find its volume remains identical today. A pyramid is a 3D shape with a polygon base and flat triangular faces that meet at a common point called the apex. The vertical distance from the centre of the base to the apex is the perpendicular height ( $h$ ).

You must memorise the volume of a pyramid, as it is often not provided in the exam:

$V = \frac{1}{3} \times \text{Area of Base} \times h$

A square-based pyramid has a base side length of $6\text{ cm}$ and a perpendicular height of $10\text{ cm}$ . Calculate its volume.

Step 1: Calculate the area of the square base.

$\text{Base Area} = 6 \times 6 = 36\text{ cm}^2$

Step 2: Substitute the base area and perpendicular height into the volume formula.

$V = \frac{1}{3} \times 36 \times 10$

Step 3: Calculate the final volume and state the units.

$V = 12 \times 10 = 120\text{ cm}^3$

To find the surface area, you calculate the area of the base and add the area of the triangular faces. To do this, you often need the slant height ( $l$ ), which is the distance from the edge of the base to the apex along the face of the triangle. You can use Pythagoras' Theorem ( $a^2 + b^2 = c^2$ ) to convert between perpendicular height and slant height.

A square-based pyramid has a base side length of $10\text{ cm}$ and a perpendicular height of $12\text{ cm}$ . Calculate its total surface area.

Step 1: Use Pythagoras' Theorem to find the slant height ( $l$ ). The base of the right-angled triangle is half the pyramid's base ( $5\text{ cm}$ ).

$l^2 = 12^2 + 5^2$
$l = \sqrt{144 + 25} = \sqrt{169} = 13\text{ cm}$

Step 2: Calculate the area of one triangular face.

$\text{Area of one triangle} = \frac{1}{2} \times \text{base} \times \text{slant height}$
$\text{Area} = \frac{1}{2} \times 10 \times 13 = 65\text{ cm}^2$

Step 3: Calculate the area of all four triangular faces (lateral area).

$\text{Lateral Area} = 4 \times 65 = 260\text{ cm}^2$

Step 4: Calculate the base area and add it to the lateral area for the total surface area.

$\text{Base Area} = 10 \times 10 = 100\text{ cm}^2$
$TSA = 100 + 260 = 360\text{ cm}^2$

Cones

You can think of a cone as a pyramid with a circular base, meaning its volume follows the exact same "one-third" rule. If the top portion of a cone or pyramid is cut off, the remaining shape is called a frustum.

The formulas for a cone are typically provided in the exam:

$V = \frac{1}{3}\pi r^2 h$

$CSA = \pi r l$

The Total Surface Area ( $TSA$ ) is the curved surface area plus the circular base ( $TSA = \pi r l + \pi r^2$ ). Just like with pyramids, the perpendicular height ( $h$ ), radius ( $r$ ), and slant height ( $l$ ) form a right-angled triangle, meaning $l^2 = r^2 + h^2$ .

A cone has a radius of $5.7\text{ cm}$ and a slant height of $12.6\text{ cm}$ . Calculate its volume to 3 significant figures.

Step 1: Use Pythagoras' Theorem to find the perpendicular height ( $h$ ).

$h = \sqrt{12.6^2 - 5.7^2}$
$h = \sqrt{158.76 - 32.49} = \sqrt{126.27} \approx 11.237\text{ cm}$

Step 2: Substitute into the volume formula.

$V = \frac{1}{3} \times \pi \times 5.7^2 \times 11.237$

Step 3: Calculate the final volume and round to 3 s.f.

$V = 382.3... \approx 382\text{ cm}^3$

Composite Solids

Most real-world objects aren't simple geometric shapes, but are built by combining multiple basic 3D shapes together to form a composite solid.

To find the volume of a composite solid, simply calculate the volume of each individual shape and add them together (or subtract if a piece has been removed). Finding the surface area is trickier: you must only calculate the exposed faces. Any internal faces where two shapes are joined together are hidden and must not be included in your final sum.

A composite solid is formed by joining a cone to a hemisphere. The radius of both shapes is $5\text{ cm}$ and the perpendicular height of the cone is $12\text{ cm}$ . Calculate the total volume and total surface area in terms of $\pi$ .

Step 1: Calculate the volume of the cone and the hemisphere separately.

$V_{\text{cone}} = \frac{1}{3} \times \pi \times 5^2 \times 12 = 100\pi\text{ cm}^3$
$V_{\text{hemisphere}} = \frac{2}{3} \times \pi \times 5^3 = \frac{250}{3}\pi\text{ cm}^3$

Step 2: Add the volumes together.

$V_{\text{total}} = 100\pi + \frac{250}{3}\pi = \frac{550}{3}\pi\text{ cm}^3$

Step 3: For surface area, first find the slant height ( $l$ ) of the cone.

$l = \sqrt{12^2 + 5^2} = \sqrt{169} = 13\text{ cm}$

Step 4: Calculate the exposed surface areas (ignore the flat circular base joining them).

$CSA_{\text{cone}} = \pi \times 5 \times 13 = 65\pi\text{ cm}^2$
$CSA_{\text{hemisphere}} = 2 \times \pi \times 5^2 = 50\pi\text{ cm}^2$

Step 5: Add the exposed surface areas together.

$SA_{\text{total}} = 65\pi + 50\pi = 115\pi\text{ cm}^2$

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Exam Tips

Common Mistake
Students often substitute the slant height (l) into the volume formula instead of the perpendicular height (h). This will score zero marks for that step.
2
Always check if a question gives you the diameter of a sphere or base; you must divide it by 2 to find the radius before substituting it into any formulas.
3
In 6-mark composite solid questions, examiners explicitly look for you to omit internal hidden faces from your total surface area calculations.
4
If a question asks for the answer "in terms of π" or as an "exact value", do not convert your final answer into a decimal.
5
Keep intermediate calculation steps to at least 4 significant figures to prevent compounding rounding errors in your final answer.

Key Terms(12)

Sphere: A perfectly round 3D geometrical object where every point on the surface is an equal distance from the centre.
Hemisphere: Exactly half of a sphere, possessing one curved surface and one flat circular face.
Radius: The distance from the centre of a circle or sphere to any point on its circumference or surface.
Diameter: The distance directly across a circle or sphere, passing through the centre, equal to twice the radius.
Pyramid: A 3D shape with a flat polygon base and triangular faces that meet at a common apex.
Apex: The top vertex of a pyramid or cone where all the lateral faces or curved surfaces meet.
Perpendicular height: The straight vertical distance from the centre of the base to the apex of a cone or pyramid.
Slant height: The distance from the edge of the base to the apex, measured along the lateral face or curved surface.
Cone: A 3D shape with a circular base that narrows smoothly to a pointed apex.
Frustum: The 3D shape remaining after the top portion of a cone or pyramid is removed by a plane cut parallel to the base.
Composite solid: A 3D shape created by joining two or more simpler geometric solids together.
Pythagoras' Theorem: A fundamental rule in geometry stating that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Previous NoteCircumference and Area of Circles and Composite 2D Shapes Next Subtopic: Arcs and SectorsArcs and Sectors

Back to Circles and Solids

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Key Terms(12)

Sphere: A perfectly round 3D geometrical object where every point on the surface is an equal distance from the centre.
Hemisphere: Exactly half of a sphere, possessing one curved surface and one flat circular face.
Radius: The distance from the centre of a circle or sphere to any point on its circumference or surface.
Diameter: The distance directly across a circle or sphere, passing through the centre, equal to twice the radius.
Pyramid: A 3D shape with a flat polygon base and triangular faces that meet at a common apex.
Apex: The top vertex of a pyramid or cone where all the lateral faces or curved surfaces meet.
Perpendicular height: The straight vertical distance from the centre of the base to the apex of a cone or pyramid.
Slant height: The distance from the edge of the base to the apex, measured along the lateral face or curved surface.
Cone: A 3D shape with a circular base that narrows smoothly to a pointed apex.
Frustum: The 3D shape remaining after the top portion of a cone or pyramid is removed by a plane cut parallel to the base.
Composite solid: A 3D shape created by joining two or more simpler geometric solids together.
Pythagoras' Theorem: A fundamental rule in geometry stating that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Surface Area and Volume of 3D Solids

Spheres and Hemispheres

For Edexcel GCSE Mathematics, you will often be given the formulas for a sphere in the exam, but you must know how to substitute into them accurately:

$V = \frac{4}{3}\pi r^3$

$SA = 4\pi r^2$

A sphere has a radius of $7\text{ cm}$ . Calculate its volume and surface area to 1 decimal place.

Step 1: Substitute the radius ( $r = 7$ ) into the volume formula.

$V = \frac{4}{3} \times \pi \times 7^3$

Step 2: Calculate the volume and state the units.

$V = 1436.755... \approx 1436.8\text{ cm}^3$

Step 3: Substitute the radius into the surface area formula.

$SA = 4 \times \pi \times 7^2$

Step 4: Calculate the surface area and state the units.

$SA = 615.752... \approx 615.8\text{ cm}^2$

The volume of a sphere is $12,000\text{ cm}^3$ . Calculate the radius $r$ to 1 decimal place.

Step 1: Set the volume formula equal to the given volume.

$12000 = \frac{4}{3}\pi r^3$

Step 2: Rearrange to isolate $r^3$ .

$r^3 = \frac{36000}{4\pi} = \frac{9000}{\pi}$

Step 3: Cube root to find $r$ .

$r = \sqrt[3]{\frac{9000}{\pi}} \approx 14.2\text{ cm}$

Pyramids

You must memorise the volume of a pyramid, as it is often not provided in the exam:

$V = \frac{1}{3} \times \text{Area of Base} \times h$

A square-based pyramid has a base side length of $6\text{ cm}$ and a perpendicular height of $10\text{ cm}$ . Calculate its volume.

Step 1: Calculate the area of the square base.

$\text{Base Area} = 6 \times 6 = 36\text{ cm}^2$

Step 2: Substitute the base area and perpendicular height into the volume formula.

$V = \frac{1}{3} \times 36 \times 10$

Step 3: Calculate the final volume and state the units.

$V = 12 \times 10 = 120\text{ cm}^3$

A square-based pyramid has a base side length of $10\text{ cm}$ and a perpendicular height of $12\text{ cm}$ . Calculate its total surface area.

Step 1: Use Pythagoras' Theorem to find the slant height ( $l$ ). The base of the right-angled triangle is half the pyramid's base ( $5\text{ cm}$ ).

$l^2 = 12^2 + 5^2$
$l = \sqrt{144 + 25} = \sqrt{169} = 13\text{ cm}$

Step 2: Calculate the area of one triangular face.

$\text{Area of one triangle} = \frac{1}{2} \times \text{base} \times \text{slant height}$
$\text{Area} = \frac{1}{2} \times 10 \times 13 = 65\text{ cm}^2$

Step 3: Calculate the area of all four triangular faces (lateral area).

$\text{Lateral Area} = 4 \times 65 = 260\text{ cm}^2$

Step 4: Calculate the base area and add it to the lateral area for the total surface area.

$\text{Base Area} = 10 \times 10 = 100\text{ cm}^2$
$TSA = 100 + 260 = 360\text{ cm}^2$

Cones

The formulas for a cone are typically provided in the exam:

$V = \frac{1}{3}\pi r^2 h$

$CSA = \pi r l$

A cone has a radius of $5.7\text{ cm}$ and a slant height of $12.6\text{ cm}$ . Calculate its volume to 3 significant figures.

Step 1: Use Pythagoras' Theorem to find the perpendicular height ( $h$ ).

$h = \sqrt{12.6^2 - 5.7^2}$
$h = \sqrt{158.76 - 32.49} = \sqrt{126.27} \approx 11.237\text{ cm}$

Step 2: Substitute into the volume formula.

$V = \frac{1}{3} \times \pi \times 5.7^2 \times 11.237$

Step 3: Calculate the final volume and round to 3 s.f.

$V = 382.3... \approx 382\text{ cm}^3$

Composite Solids

Most real-world objects aren't simple geometric shapes, but are built by combining multiple basic 3D shapes together to form a composite solid.

Step 1: Calculate the volume of the cone and the hemisphere separately.

$V_{\text{cone}} = \frac{1}{3} \times \pi \times 5^2 \times 12 = 100\pi\text{ cm}^3$
$V_{\text{hemisphere}} = \frac{2}{3} \times \pi \times 5^3 = \frac{250}{3}\pi\text{ cm}^3$

Step 2: Add the volumes together.

$V_{\text{total}} = 100\pi + \frac{250}{3}\pi = \frac{550}{3}\pi\text{ cm}^3$

Step 3: For surface area, first find the slant height ( $l$ ) of the cone.

$l = \sqrt{12^2 + 5^2} = \sqrt{169} = 13\text{ cm}$

Step 4: Calculate the exposed surface areas (ignore the flat circular base joining them).

$CSA_{\text{cone}} = \pi \times 5 \times 13 = 65\pi\text{ cm}^2$
$CSA_{\text{hemisphere}} = 2 \times \pi \times 5^2 = 50\pi\text{ cm}^2$

Step 5: Add the exposed surface areas together.

$SA_{\text{total}} = 65\pi + 50\pi = 115\pi\text{ cm}^2$

Sign up to continue reading

Get full access to revision notes, key terms, and exam tips for every subtopic.

Exam Tips

Common Mistake
Students often substitute the slant height (l) into the volume formula instead of the perpendicular height (h). This will score zero marks for that step.
2
Always check if a question gives you the diameter of a sphere or base; you must divide it by 2 to find the radius before substituting it into any formulas.
3
In 6-mark composite solid questions, examiners explicitly look for you to omit internal hidden faces from your total surface area calculations.
4
If a question asks for the answer "in terms of π" or as an "exact value", do not convert your final answer into a decimal.
5
Keep intermediate calculation steps to at least 4 significant figures to prevent compounding rounding errors in your final answer.

Key Terms(12)

Sphere: A perfectly round 3D geometrical object where every point on the surface is an equal distance from the centre.
Hemisphere: Exactly half of a sphere, possessing one curved surface and one flat circular face.
Radius: The distance from the centre of a circle or sphere to any point on its circumference or surface.
Diameter: The distance directly across a circle or sphere, passing through the centre, equal to twice the radius.
Pyramid: A 3D shape with a flat polygon base and triangular faces that meet at a common apex.
Apex: The top vertex of a pyramid or cone where all the lateral faces or curved surfaces meet.
Perpendicular height: The straight vertical distance from the centre of the base to the apex of a cone or pyramid.
Slant height: The distance from the edge of the base to the apex, measured along the lateral face or curved surface.
Cone: A 3D shape with a circular base that narrows smoothly to a pointed apex.
Frustum: The 3D shape remaining after the top portion of a cone or pyramid is removed by a plane cut parallel to the base.
Composite solid: A 3D shape created by joining two or more simpler geometric solids together.
Pythagoras' Theorem: A fundamental rule in geometry stating that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Previous NoteCircumference and Area of Circles and Composite 2D Shapes Next Subtopic: Arcs and SectorsArcs and Sectors

Back to Circles and Solids

Put your knowledge into practice — try past paper questions for Mathematics

Key Terms(12)

Sphere: A perfectly round 3D geometrical object where every point on the surface is an equal distance from the centre.
Hemisphere: Exactly half of a sphere, possessing one curved surface and one flat circular face.
Radius: The distance from the centre of a circle or sphere to any point on its circumference or surface.
Diameter: The distance directly across a circle or sphere, passing through the centre, equal to twice the radius.
Pyramid: A 3D shape with a flat polygon base and triangular faces that meet at a common apex.
Apex: The top vertex of a pyramid or cone where all the lateral faces or curved surfaces meet.
Perpendicular height: The straight vertical distance from the centre of the base to the apex of a cone or pyramid.
Slant height: The distance from the edge of the base to the apex, measured along the lateral face or curved surface.
Cone: A 3D shape with a circular base that narrows smoothly to a pointed apex.
Frustum: The 3D shape remaining after the top portion of a cone or pyramid is removed by a plane cut parallel to the base.
Composite solid: A 3D shape created by joining two or more simpler geometric solids together.
Pythagoras' Theorem: A fundamental rule in geometry stating that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.