Hard GCSE Spheres, cones and pyramids Questions

Challenging, exam-style GCSE Spheres, cones and pyramids questions with worked solutions. Stretch yourself on the hardest cone, volume and surface area formulas, exact answers in terms of pi, Pythagoras problems.

conevolume and surface area formulasexact answers in terms of piPythagorasslant heightsphere
GCSE Foundation34 questionsStep-by-step solutions
Question 1
5 markschallenging
A sphere of radius rr fits exactly inside a cylinder of radius rr and height 2r2r. The volume of the sphere is VsV_s and the volume of the cylinder is VyV_y. Which of these statements is correct?
Show worked solution

Worked solution

  1. Write down the volume of the cylinder.

    Vy=πr2×2rV_y = \pi r^2 \times 2r

    The cylinder has radius rr and height 2r2r, because the sphere must fit exactly inside it.

  2. Simplify the volume of the cylinder.

    Vy=2πr3V_y = 2\pi r^3

    Multiplying out gives 2πr32\pi r^3.

  3. Write down the volume of the sphere.

    Vs=43πr3V_s = \frac{4}{3}\pi r^3

    The sphere has radius rr, so its volume is 43πr3\frac{4}{3}\pi r^3.

  4. Divide one volume by the other.

    VsVy=43πr32πr3\frac{V_s}{V_y} = \frac{\frac{4}{3}\pi r^3}{2\pi r^3}

    Both volumes contain πr3\pi r^3, so the comparison will not depend on the radius at all.

  5. Simplify the fraction.

    VsVy=43÷2=23\frac{V_s}{V_y} = \frac{4}{3} \div 2 = \frac{2}{3}

    Everything cancels except the numbers, and 43÷2=23\frac{4}{3} \div 2 = \frac{2}{3}.

  6. Test the result with a radius of three.

    Vs=43π×27=36πV_s = \frac{4}{3}\pi \times 27 = 36\pi

    A sphere of radius 33 cm has volume 36π36\pi cm3\text{cm}^3.

  7. Work out the matching cylinder.

    Vy=π×9×6=54πV_y = \pi \times 9 \times 6 = 54\pi

    The cylinder round it has radius 33 cm and height 66 cm, so it holds 54π54\pi cm3\text{cm}^3.

  8. Compare the two numbers.

    36π54π=23\frac{36\pi}{54\pi} = \frac{2}{3}

    The sphere fills two thirds of the cylinder, exactly as the algebra said.

  9. Test the result with a radius of six.

    288π432π=23\frac{288\pi}{432\pi} = \frac{2}{3}

    A sphere of radius 66 cm holds 288π288\pi cm3\text{cm}^3 and its cylinder holds 432π432\pi cm3\text{cm}^3, and the ratio is the same.

  10. Rule out the half.

    12×54π=27π≠36π\frac{1}{2} \times 54\pi = 27\pi \ne 36\pi

    Half the cylinder would be 27π27\pi cm3\text{cm}^3, but the sphere holds 36π36\pi cm3\text{cm}^3, so it is more than half.

  11. Rule out the third.

    13×54π=18π≠36π\frac{1}{3} \times 54\pi = 18\pi \ne 36\pi

    One third of the cylinder is the volume of the CONE that fits inside it, not the sphere.

  12. Rule out the three quarters.

    34×54π=812π≠36π\frac{3}{4} \times 54\pi = \frac{81}{2}\pi \ne 36\pi

    Three quarters of the cylinder is too much: it would leave less empty space than there really is.

  13. Rule out the equality.

    Vs=Vy would leave no spaceV_s = V_y \text{ would leave no space}

    The sphere sits inside the cylinder with gaps at the sides, so it cannot have the same volume.

  14. State the general result.

    Vsphere=23VcylinderV_{\text{sphere}} = \frac{2}{3} V_{\text{cylinder}}

    This is Archimedes result: a sphere fills two thirds of the smallest cylinder that contains it, whatever its radius.

  15. Select the correct statement.

    Vs=23VyV_s = \frac{2}{3} V_y

    The sphere fills exactly two thirds of the cylinder that just contains it.

Answer
Vs=23VyV_s = \frac{2}{3} V_y
Question 2
5 markschallenging
A cone of base radius rr and perpendicular height hh has the same volume as a sphere of radius rr. Which of these statements is correct?
Show worked solution

Worked solution

  1. Write down the volume of the sphere.

    Vsphere=43πr3V_{\text{sphere}} = \frac{4}{3}\pi r^3

    The sphere has radius rr, so its volume is 43πr3\frac{4}{3}\pi r^3.

  2. Write down the volume of the cone.

    Vcone=13πr2hV_{\text{cone}} = \frac{1}{3}\pi r^2 h

    The cone has the same base radius rr and an unknown perpendicular height hh.

  3. Set the two volumes equal.

    13πr2h=43πr3\frac{1}{3}\pi r^2 h = \frac{4}{3}\pi r^3

    The two solids are said to have the same volume, so their formulas must give the same value.

  4. Multiply both sides by three.

    πr2h=4πr3\pi r^2 h = 4\pi r^3

    Clearing the thirds from both sides keeps the equation exact.

  5. Divide both sides by pi.

    r2h=4r3r^2 h = 4 r^3

    The factor of π\pi appears on both sides, so it cancels.

  6. Divide both sides by the square of the radius.

    h=4rh = 4r

    The radius is not zero, so dividing by r2r^2 is allowed, and it leaves h=4rh = 4r.

  7. Test the result with a radius of three.

    r=3:  43π×27=36πr = 3: \; \frac{4}{3}\pi \times 27 = 36\pi

    A sphere of radius 33 cm has volume 36π36\pi cm3\text{cm}^3.

  8. Check the matching cone.

    h=12:  13π×9×12=36πh = 12: \; \frac{1}{3}\pi \times 9 \times 12 = 36\pi

    A cone of radius 33 cm and height 1212 cm has volume 36π36\pi cm3\text{cm}^3 too, and 12=4×312 = 4 \times 3, as the rule says.

  9. Test the result with a radius of six.

    r=6:  43π×216=288π,13π×36×24=288πr = 6: \; \frac{4}{3}\pi \times 216 = 288\pi, \quad \frac{1}{3}\pi \times 36 \times 24 = 288\pi

    A sphere of radius 66 cm and a cone of radius 66 cm and height 2424 cm both hold 288π288\pi cm3\text{cm}^3, and 24=4×624 = 4 \times 6.

  10. Rule out the option with twice the radius.

    h=2r:  13πr2(2r)=23πr3h = 2r: \; \frac{1}{3}\pi r^2 (2r) = \frac{2}{3}\pi r^3

    That cone holds only 23πr3\frac{2}{3}\pi r^3, which is HALF the volume of the sphere.

  11. Rule out the option with the height equal to the radius.

    h=r:  13πr2(r)=13πr3h = r: \; \frac{1}{3}\pi r^2 (r) = \frac{1}{3}\pi r^3

    That cone holds a quarter of the volume of the sphere, so it is far too small.

  12. Rule out the four thirds option.

    h=43r:  13πr2(43r)=49πr3h = \frac{4}{3} r: \; \frac{1}{3}\pi r^2 \left(\frac{4}{3} r\right) = \frac{4}{9}\pi r^3

    That gives 49πr3\frac{4}{9}\pi r^3, which is only one third of the volume of the sphere. The 43\frac{4}{3} has been copied from the sphere formula without doing the algebra.

  13. Rule out the option with three times the radius.

    h=3r:  13πr2(3r)=πr3h = 3r: \; \frac{1}{3}\pi r^2 (3r) = \pi r^3

    That cone holds πr3\pi r^3, which is three quarters of the volume of the sphere, so it is still too small.

  14. Note that the answer does not depend on the radius.

    h=4r for every r>0h = 4r \text{ for every } r > 0

    The relationship is between the two lengths, so it holds for a cone and a sphere of any size.

  15. Select the correct statement.

    h=4rh = 4r

    A cone with the same radius as a sphere needs a perpendicular height of 4r4r to hold the same volume.

Answer
h=4rh = 4r
Question 3
5 markschallenging
A cone has base radius 55 cm and total surface area 90π90\pi cm2^2. Work out the perpendicular height of the cone.
Show worked solution

Worked solution

  1. Split the total surface area into the curved part and the base.

    A=πrl+πr2A = \pi r l + \pi r^2

    The total surface of a solid cone is the curved surface plus the flat circular base.

  2. Put the radius and the total area into the formula.

    π×5×l+π×52=90π\pi \times 5 \times l + \pi \times 5^2 = 90\pi

    The base radius is 55 cm and the total surface area is 90π90\pi cm2\text{cm}^2.

  3. Divide every term by pi.

    5l+52=905 l + 5^2 = 90

    Every term carries a factor of π\pi, so it cancels and the working stays exact.

  4. Work out the square of the radius.

    52=255^2 = 25

    The base circle contributes 25π25\pi cm2\text{cm}^2.

  5. Subtract the base from both sides.

    5l=90−25=655 l = 90 - 25 = 65

    What is left, 65π65\pi cm2\text{cm}^2, is the curved surface area.

  6. Divide by the radius to find the slant height.

    l=655=13l = \frac{65}{5} = 13

    The slant height is 1313 cm.

  7. Use Pythagoras to reach the perpendicular height.

    h2=l2−r2h^2 = l^2 - r^2

    The radius, the perpendicular height and the slant height form a right-angled triangle with the slant height as hypotenuse.

  8. Substitute the slant height and the radius.

    h2=132−52=169−25=144h^2 = 13^2 - 5^2 = 169 - 25 = 144

    So the square of the perpendicular height is 144144.

  9. Take the square root.

    h=144=12h = \sqrt{144} = 12

    The perpendicular height is 1212 cm.

  10. Check the slant height is longer than the perpendicular height.

    13>1213 > 12

    The hypotenuse is always the longest side of a right-angled triangle.

  11. Check by rebuilding the total surface area.

    π×5×13+π×25=90π\pi \times 5 \times 13 + \pi \times 25 = 90\pi

    The curved surface 65π65\pi plus the base 25π25\pi gives back the total 90π90\pi cm2\text{cm}^2.

  12. Check the units of the answer.

    cm33=cm\sqrt[3]{\text{cm}^3} = \text{cm}

    Undoing a volume with a cube root leaves a single length, so the answer is in centimetres.

  13. Find the volume of this cone as a bonus.

    V=13π×25×12=100πV = \frac{1}{3}\pi \times 25 \times 12 = 100\pi

    With the perpendicular height known, the volume is 100π100\pi cm3\text{cm}^3.

  14. Name the common error.

    905≠l\frac{90}{5} \ne l

    The base circle must be taken off BEFORE dividing by the radius, otherwise the slant height comes out too large.

  15. State the perpendicular height of the cone.

    h=169−25=12h = \sqrt{169 - 25} = 12

    The perpendicular height of the cone is 1212 cm.

Answer
h=12 cmh = 12 \text{ cm}
Question 4
6 markschallenging
A cone has base radius 66 cm and total surface area 96π96\pi cm2^2. Work out the perpendicular height of the cone.
Show worked solution

Worked solution

  1. Split the total surface area into the curved part and the base.

    A=πrl+πr2A = \pi r l + \pi r^2

    The total surface of a solid cone is the curved surface plus the flat circular base.

  2. Put the radius and the total area into the formula.

    π×6×l+π×62=96π\pi \times 6 \times l + \pi \times 6^2 = 96\pi

    The base radius is 66 cm and the total surface area is 96π96\pi cm2\text{cm}^2.

  3. Divide every term by pi.

    6l+62=966 l + 6^2 = 96

    Every term carries a factor of π\pi, so it cancels and the working stays exact.

  4. Work out the square of the radius.

    62=366^2 = 36

    The base circle contributes 36π36\pi cm2\text{cm}^2.

  5. Subtract the base from both sides.

    6l=96−36=606 l = 96 - 36 = 60

    What is left, 60π60\pi cm2\text{cm}^2, is the curved surface area.

  6. Divide by the radius to find the slant height.

    l=606=10l = \frac{60}{6} = 10

    The slant height is 1010 cm.

  7. Use Pythagoras to reach the perpendicular height.

    h2=l2−r2h^2 = l^2 - r^2

    The radius, the perpendicular height and the slant height form a right-angled triangle with the slant height as hypotenuse.

  8. Substitute the slant height and the radius.

    h2=102−62=100−36=64h^2 = 10^2 - 6^2 = 100 - 36 = 64

    So the square of the perpendicular height is 6464.

  9. Take the square root.

    h=64=8h = \sqrt{64} = 8

    The perpendicular height is 88 cm.

  10. Check the slant height is longer than the perpendicular height.

    10>810 > 8

    The hypotenuse is always the longest side of a right-angled triangle.

  11. Check by rebuilding the total surface area.

    π×6×10+π×36=96π\pi \times 6 \times 10 + \pi \times 36 = 96\pi

    The curved surface 60π60\pi plus the base 36π36\pi gives back the total 96π96\pi cm2\text{cm}^2.

  12. Check the units of the answer.

    cm33=cm\sqrt[3]{\text{cm}^3} = \text{cm}

    Undoing a volume with a cube root leaves a single length, so the answer is in centimetres.

  13. Find the volume of this cone as a bonus.

    V=13π×36×8=96πV = \frac{1}{3}\pi \times 36 \times 8 = 96\pi

    With the perpendicular height known, the volume is 96π96\pi cm3\text{cm}^3.

  14. Name the common error.

    966≠l\frac{96}{6} \ne l

    The base circle must be taken off BEFORE dividing by the radius, otherwise the slant height comes out too large.

  15. State the perpendicular height of the cone.

    h=100−36=8h = \sqrt{100 - 36} = 8

    The perpendicular height of the cone is 88 cm.

Answer
h=8 cmh = 8 \text{ cm}
Question 5
6 markschallenging
A pyramid has a square base of side 1010 cm and perpendicular height 1212 cm. Work out the total surface area of the pyramid.
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Worked solution

  1. Split the surface into the base and the four sloping faces.

    A=a2+4×12amA = a^2 + 4 \times \frac{1}{2} a m

    The base is a square, and the four sloping faces are congruent triangles. Here mm is the slant height of a face, measured from the apex to the middle of a base edge.

  2. Work out the area of the square base.

    a2=102=100a^2 = 10^2 = 100

    The base has area 100100 cm2\text{cm}^2.

  3. Set up the right-angled triangle for the slant height of a face.

    m2=h2+(a2)2m^2 = h^2 + \left(\frac{a}{2}\right)^2

    Drop the perpendicular height to the centre of the base, then travel to the midpoint of one base edge. That distance is half the side of the square, and the slant height of the face is the hypotenuse.

  4. Work out half the side of the base.

    a2=102=5\frac{a}{2} = \frac{10}{2} = 5

    Half the base edge is 55 cm.

  5. Substitute into Pythagoras.

    m2=122+52=144+25=169m^2 = 12^2 + 5^2 = 144 + 25 = 169

    So the square of the slant height of a face is 169169.

  6. Take the square root.

    m=169=13m = \sqrt{169} = 13

    Each sloping face has slant height 1313 cm.

  7. Work out the area of one triangular face.

    12×10×13=65\frac{1}{2} \times 10 \times 13 = 65

    A triangle of base 1010 cm and height 1313 cm has area 6565 cm2\text{cm}^2.

  8. Work out the area of all four faces.

    4×65=2604 \times 65 = 260

    The four sloping faces together have area 260260 cm2\text{cm}^2.

  9. Add the base to the sloping faces.

    A=100+260=360A = 100 + 260 = 360

    The total surface area is 360360 square centimetres.

  10. Check the units of the answer.

    cm×cm=cm2\text{cm} \times \text{cm} = \text{cm}^2

    An area multiplies two lengths together, so the answer is in square centimetres, written cm2\text{cm}^2.

  11. Check the slant height of a face is longer than the perpendicular height.

    13>1213 > 12

    The slant height of a face is the hypotenuse of a right-angled triangle with the perpendicular height as one leg, so it must be longer.

  12. Check the total is bigger than the base alone.

    360>100360 > 100

    The sloping faces add to the base, so the total surface area must exceed the area of the base.

  13. Distinguish the slant height of a face from the sloping edge.

    m=13≠122+52+52m = 13 \ne \sqrt{12^2 + 5^2 + 5^2}

    The sloping EDGE runs from the apex to a CORNER of the base and is longer than mm; the face area needs the distance to the MIDDLE of an edge.

  14. Name the common error.

    12×10×12≠area of a face\frac{1}{2} \times 10 \times 12 \ne \text{area of a face}

    Using the perpendicular height of the pyramid as the height of a triangular face is the standard mistake; that height goes up the middle of the solid, not up its surface.

  15. State the total surface area of the pyramid.

    A=100+260=360A = 100 + 260 = 360

    The total surface area of the pyramid is 360360 cm2\text{cm}^2.

Answer
A=360 cm2A = 360 \text{ cm}^2

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