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Pythagoras, mensuration and 3D

Pythagoras in 2D and 3D, areas of composite shapes, volumes and surface areas of prisms, cylinders, cones, spheres and pyramids, trig ratios in right-angled triangles.

Spec M5.7, M5.11-M5.13, M5.15-M5.17Section 1 Part 2: Higher GCSE mathematics17 questions3 real past questions
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Click an option to answer. A correct first attempt marks the question secure; a wrong one marks it needs work and opens the worked solution. Keys A-H answer the question nearest the top of the screen.

1Pythagoras, mensuration and 3D · spec M5.7 · Paper 1 style · ★☆☆

A cuboid measures \(3\) cm by \(4\) cm by \(12\) cm. Find the length of its longest diagonal.

  1. A\(\sqrt{153}\) cm
  2. B\(13\) cm
  3. C\(15\) cm
  4. D\(5\) cm
  5. E\(\sqrt{119}\) cm
  6. F\(12.5\) cm
Worked solution

\(\sqrt{9 + 16 + 144} = \sqrt{169} = 13\).

Full method — how and why
The space diagonal of a cuboid: Pythagoras applied twice, which gives the three-dimensional formula.
  1. First find the diagonal of the base: \(\sqrt{3^2 + 4^2} = 5\).
    That is a 3-4-5 triangle - the base diagonal lies in a horizontal plane.
  2. Now use that with the height: \(\sqrt{5^2 + 12^2} = \sqrt{169} = 13\).
    The space diagonal, the base diagonal and the vertical edge form a second right-angled triangle. This is a 5-12-13 triple.
  3. Equivalently, \(\sqrt{a^2+b^2+c^2} = \sqrt{9 + 16 + 144} = 13\) cm.
    The two applications of Pythagoras combine into one formula.
Where it goes wrong
  • Using only two of the three dimensions, giving \(5\) or \(\sqrt{153}\).
  • Adding the lengths: \(3 + 4 + 12 = 19\).
Takeaway Space diagonal \(= \sqrt{a^2+b^2+c^2}\) - Pythagoras twice in one step.
Answer B · Pythagoras, mensuration and 3D (M5.7) · question MEN-01
2Pythagoras, mensuration and 3D · spec M5.16 · Paper 1 style · ★☆☆

A trapezium has parallel sides of length \(8\) cm and \(14\) cm and area \(77\) cm\(^2\). Find the perpendicular distance between the parallel sides.

  1. A\(14\) cm
  2. B\(6\) cm
  3. C\(7\) cm
  4. D\(11\) cm
  5. E\(3.5\) cm
  6. F\(5.5\) cm
Worked solution

\(\frac12(8 + 14)h = 11h = 77 \Rightarrow h = 7\).

Full method — how and why
Trapezium area formula, rearranged for the height.
  1. Write the formula: area \(= \frac12(a + b)h\) where \(a, b\) are the parallel sides.
    The height is the perpendicular distance between the parallel sides.
  2. Substitute: \(\frac12(8 + 14)h = 11h\).
    \(\frac12 \times 22 = 11\).
  3. Solve: \(11h = 77\), so \(h = 7\) cm.
    Divide both sides by \(11\).
Where it goes wrong
  • Forgetting the \(\frac12\) and answering \(3.5\) - an offered option.
  • Using the difference of the parallel sides instead of the sum.
Takeaway Trapezium: average the parallel sides, then multiply by the perpendicular height.
Answer C · Pythagoras, mensuration and 3D (M5.16) · question MEN-03
3Pythagoras, mensuration and 3D · spec M5.16 · Paper 1 style · ★☆☆

A cone has base radius \(6\) cm and slant height \(10\) cm. Find its volume. (Volume of a cone \(= \frac{1}{3}\pi r^2 h\).)

  1. A\(60\pi\) cm\(^3\)
  2. B\(72\pi\) cm\(^3\)
  3. C\(288\pi\) cm\(^3\)
  4. D\(48\pi\) cm\(^3\)
  5. E\(96\pi\) cm\(^3\)
  6. F\(120\pi\) cm\(^3\)
Worked solution

Height \(= \sqrt{100 - 36} = 8\). Volume \(= \frac13\pi\cdot36\cdot8 = 96\pi\).

Full method — how and why
The cone's slant height, radius and vertical height form a right-angled triangle. Find the height, then use the volume formula.
  1. Use Pythagoras: \(h = \sqrt{l^2 - r^2} = \sqrt{100 - 36} = 8\) cm.
    The slant height is the hypotenuse, so subtract; this is another 6-8-10 triangle.
  2. Apply the given formula: \(V = \frac13\pi r^2 h = \frac13\pi(36)(8)\).
    The formula is provided in the question, as the specification promises.
  3. Compute: \(\frac13 \times 288 = 96\), so \(V = 96\pi\) cm\(^3\).
    \(36 \times 8 = 288\).
Where it goes wrong
  • Using the slant height as the vertical height, giving \(120\pi\) - an offered option.
  • Forgetting the \(\frac13\) and answering \(288\pi\).
Takeaway Cone: slant\(^2 = r^2 + h^2\); and the volume carries a factor of \(\frac13\) compared with a cylinder.
Answer E · Pythagoras, mensuration and 3D (M5.16) · question MEN-04
4Pythagoras, mensuration and 3D · spec M5.17 · Paper 1 style · ★☆☆

A sphere and a cube have the same volume. The cube has side length \(a\). Find the radius of the sphere. (Volume of a sphere \(= \frac{4}{3}\pi r^3\).)

  1. A\(\dfrac{a}{\sqrt[3]{\pi}}\)
  2. B\(a\sqrt[3]{\dfrac{4\pi}{3}}\)
  3. C\(\dfrac{a}{2}\)
  4. D\(\dfrac{3a}{4\pi}\)
  5. E\(a\sqrt{\dfrac{3}{4\pi}}\)
  6. F\(a\sqrt[3]{\dfrac{3}{4\pi}}\)
Worked solution

\(\frac43\pi r^3 = a^3 \Rightarrow r^3 = \frac{3a^3}{4\pi} \Rightarrow r = a\sqrt[3]{\frac{3}{4\pi}}\).

Full method — how and why
Equate the two volumes and solve for \(r\) - an algebraic rearrangement with a cube root.
  1. Set the volumes equal: \(\frac43\pi r^3 = a^3\).
    The sphere's formula is given; the cube's is \(a^3\).
  2. Isolate \(r^3\): \(r^3 = \dfrac{3a^3}{4\pi}\).
    Multiply both sides by \(\frac{3}{4\pi}\).
  3. Take the cube root: \(r = a\sqrt[3]{\dfrac{3}{4\pi}}\).
    The cube root of \(a^3\) is \(a\), and the constant keeps its cube root.
Where it goes wrong
  • Inverting the fraction, giving \(a\sqrt[3]{\frac{4\pi}{3}}\) - an offered option.
  • Taking a square root instead of a cube root.
Takeaway Equate volumes, isolate \(r^3\), then cube-root everything including the constant.
Answer F · Pythagoras, mensuration and 3D (M5.17) · question MEN-05
5Pythagoras, mensuration and 3D · spec M5.16 · Paper 1 style · ★☆☆

A square of side \(10\) has a quarter-circle of radius \(10\) drawn inside it, centred at one corner. Find the area of the square not covered by the quarter-circle.

  1. A\(100 - 50\pi\)
  2. B\(100 - 25\pi\)
  3. C\(100 - 100\pi\)
  4. D\(100\pi - 100\)
  5. E\(50 - 25\pi\)
  6. F\(25\pi\)
Worked solution

\(100 - \frac14\pi\cdot100 = 100 - 25\pi\).

Full method — how and why
A composite area: subtract the quarter-circle from the square.
  1. Square area: \(10^2 = 100\).
    Side squared.
  2. Quarter-circle area: \(\frac14\pi r^2 = \frac14\pi(100) = 25\pi\).
    The radius equals the side, and a quarter of a circle is \(\frac14\) of \(\pi r^2\).
  3. Subtract: \(100 - 25\pi\).
    Leave \(\pi\) exact - no calculator is available anyway.
Where it goes wrong
  • Forgetting the quarter and subtracting \(100\pi\), which would give a negative area.
  • Using \(\frac14 \pi r\) (circumference) instead of area.
Takeaway Composite areas: subtract the parts, and check the answer is positive (\(25\pi \approx 78.5 < 100\) ✓).
Answer B · Pythagoras, mensuration and 3D (M5.16) · question MEN-06
6Pythagoras, mensuration and 3D · spec M5.7 · Paper 1 style · ★☆☆

A ladder of length \(6.5\) m leans against a vertical wall with its foot \(2.5\) m from the wall on horizontal ground. How far up the wall does it reach?

  1. A\(7\) m
  2. B\(5\) m
  3. C\(4\) m
  4. D\(5.5\) m
  5. E\(6\) m
  6. F\(\sqrt{48.5}\) m
Worked solution

\(\sqrt{6.5^2 - 2.5^2} = \sqrt{42.25 - 6.25} = \sqrt{36} = 6\).

Full method — how and why
A ladder against a wall is a right-angled triangle with the ladder as hypotenuse.
  1. Identify the hypotenuse: the ladder, \(6.5\) m - the longest side, opposite the right angle at the base of the wall.
    The wall and ground are perpendicular.
  2. Apply Pythagoras: height \(= \sqrt{6.5^2 - 2.5^2} = \sqrt{42.25 - 6.25}\).
    Subtract, because we want a shorter side.
  3. Compute: \(\sqrt{36} = 6\) m.
    Decimals chosen to give a whole number - a sign the working is right.
Where it goes wrong
  • Adding the squares, giving \(\sqrt{48.5}\) - an offered option.
  • Arithmetic: \(6.5^2 = 42.25\), not \(36.25\).
Takeaway Hypotenuse known \(\Rightarrow\) subtract squares; hypotenuse unknown \(\Rightarrow\) add them.
Answer E · Pythagoras, mensuration and 3D (M5.7) · question MEN-07
7Pythagoras, mensuration and 3D · spec M5.18 · Paper 1 style · ★☆☆

In a right-angled triangle the hypotenuse is \(10\) and one angle is \(30^\circ\). Find the area of the triangle.

  1. A\(\dfrac{25}{2}\)
  2. B\(25\)
  3. C\(25\sqrt3\)
  4. D\(\dfrac{25\sqrt{3}}{2}\)
  5. E\(50\sqrt3\)
  6. F\(\dfrac{25\sqrt3}{4}\)
Worked solution

Sides \(10\sin30^\circ = 5\) and \(10\cos30^\circ = 5\sqrt3\). Area \(= \frac12\cdot5\cdot5\sqrt3 = \frac{25\sqrt3}{2}\).

Full method — how and why
Right-angled trigonometry to get the two legs, then the standard triangle area.
  1. The side opposite the \(30^\circ\) angle: \(10\sin 30^\circ = 10 \times \frac12 = 5\).
    \(\sin = \frac{\text{opposite}}{\text{hypotenuse}}\), so opposite \(=\) hypotenuse \(\times \sin\).
  2. The adjacent side: \(10\cos 30^\circ = 10 \times \frac{\sqrt3}{2} = 5\sqrt3\).
    Exact values must be recalled: \(\cos 30^\circ = \frac{\sqrt3}{2}\).
  3. Area \(= \frac12 \times\) base \(\times\) height \(= \frac12(5)(5\sqrt3) = \dfrac{25\sqrt3}{2}\).
    The two legs are perpendicular, so they serve as base and height.
Where it goes wrong
  • Forgetting the \(\frac12\), giving \(25\sqrt3\) - an offered option.
  • Using the hypotenuse as one of the perpendicular sides.
Takeaway In a right-angled triangle the two legs are base and height; get them from \(\sin\) and \(\cos\) of the hypotenuse.
Answer D · Pythagoras, mensuration and 3D (M5.18) · question MEN-08
8Pythagoras, mensuration and 3D · spec M5.16 · Paper 1 style · ★☆☆

The volume of a cube is numerically equal to its surface area. Find the length of its side.

  1. A\(\sqrt6\)
  2. B\(3\)
  3. C\(2\)
  4. D\(12\)
  5. E\(1\)
  6. F\(6\)
Worked solution

\(a^3 = 6a^2 \Rightarrow a = 6\).

Full method — how and why
Set the two expressions equal and solve. The equation is cubic but factorises immediately.
  1. Write both: volume \(= a^3\), surface area \(= 6a^2\) (six square faces).
    A cube has six identical faces of area \(a^2\).
  2. Set them equal: \(a^3 = 6a^2\).
    'Numerically equal' means the numbers match, though the units differ.
  3. Factorise rather than divide: \(a^2(a - 6) = 0\), so \(a = 0\) or \(a = 6\). A cube must have \(a > 0\), so \(a = 6\).
    Dividing by \(a^2\) is fine here (a side length is non-zero), but factorising is the safer habit.
Where it goes wrong
  • Taking a square root somewhere and answering \(\sqrt6\) - an offered option.
  • Using \(4a^2\) or \(a^2\) for the surface area.
Takeaway Cube: \(V = a^3\), \(S = 6a^2\); equate and factorise rather than dividing by a variable.
Answer F · Pythagoras, mensuration and 3D (M5.16) · question MEN-09
9Pythagoras, mensuration and 3D · spec M5.16 · Paper 2 style · ★☆☆

Which of the following statements is/are true?

  1. IDoubling the radius of a circle doubles its circumference.
  2. IIDoubling the radius of a sphere multiplies its volume by \(8\).
  3. IIIDoubling the radius of a cylinder while halving its height leaves its volume unchanged.
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Worked solution

\(C = 2\pi r\) is linear in \(r\) (I). \(V \propto r^3\) (II). \(\pi(2r)^2\cdot\frac h2 = 2\pi r^2 h\): doubled, not unchanged (III false).

Full method — how and why
Three scaling claims. Each is checked by substituting the new dimensions into the formula.
  1. I: \(C = 2\pi r\) is linear in \(r\), so doubling \(r\) doubles \(C\). True.
    Circumference scales with the first power of the radius.
  2. II: \(V = \frac43\pi r^3\), so doubling \(r\) multiplies the volume by \(2^3 = 8\). True.
    Volume scales with the cube.
  3. III: new volume \(= \pi(2r)^2\left(\frac h2\right) = \pi(4r^2)\left(\frac h2\right) = 2\pi r^2 h\) - double, not unchanged. False.
    The radius is squared, so doubling it quadruples that factor; halving the height only removes a factor of \(2\).
  4. I and II only.
    Match to the option list.
Where it goes wrong
  • Assuming the doubling and halving in III cancel - the radius is squared, so they do not.
  • Confusing area and volume scaling in II.
Takeaway Substitute the new dimensions into the formula; squared quantities change faster than linear ones.
Answer E · Pythagoras, mensuration and 3D (M5.16) · question MEN-11
10Pythagoras, mensuration and 3D · spec M5.16 · Paper 1 style · ★☆☆

A running track consists of two straights of length \(100\) m and two semicircular ends of radius \(r\). The total length of the track is \(400\) m. Find \(r\).

  1. A\(100\pi\)
  2. B\(\dfrac{100}{\pi}\)
  3. C\(\dfrac{200}{\pi}\)
  4. D\(\dfrac{400}{\pi}\)
  5. E\(\dfrac{300}{\pi}\)
  6. F\(\dfrac{50}{\pi}\)
Worked solution

\(200 + 2\pi r = 400 \Rightarrow r = \frac{100}{\pi}\).

Full method — how and why
The two semicircular ends together make one full circle. Write the perimeter and solve.
  1. Total length \(=\) two straights \(+\) two semicircles \(= 2(100) + 2\left(\pi r\right)\).
    A semicircular arc has length \(\pi r\); two of them make a full circumference \(2\pi r\).
  2. Set up: \(200 + 2\pi r = 400\).
    Substitute the given total.
  3. Solve: \(2\pi r = 200\), so \(r = \dfrac{100}{\pi}\).
    Divide by \(2\pi\).
Where it goes wrong
  • Using \(\pi r\) for both ends together, giving \(\frac{200}{\pi}\) - an offered option.
  • Using the area formula instead of arc length.
Takeaway Two semicircles of the same radius make one circle: total arc \(= 2\pi r\).
Answer B · Pythagoras, mensuration and 3D (M5.16) · question MEN-12
11Pythagoras, mensuration and 3D · spec M5.16 · Paper 1 style · ★☆☆

A cylindrical tin of radius \(4\) cm and height \(10\) cm is full of water. The water is poured into an empty cylindrical jug of radius \(5\) cm. How deep is the water in the jug?

  1. A\(6\) cm
  2. B\(12.5\) cm
  3. C\(8\) cm
  4. D\(7.2\) cm
  5. E\(5\) cm
  6. F\(6.4\) cm
Worked solution

\(\pi\cdot16\cdot10 = \pi\cdot25\cdot h \Rightarrow h = \frac{160}{25} = 6.4\).

Full method — how and why
Conservation of volume: the water keeps its volume, so equate the two cylinder volumes.
  1. Volume in the tin: \(\pi r^2 h = \pi(16)(10) = 160\pi\).
    \(r = 4\) so \(r^2 = 16\).
  2. In the jug the same volume occupies \(\pi(25)h\).
    Radius \(5\), unknown depth \(h\); the \(\pi\) will cancel.
  3. Equate and solve: \(160\pi = 25\pi h \Rightarrow h = \dfrac{160}{25} = 6.4\) cm.
    The wider jug means a shallower depth - a good sanity check.
Where it goes wrong
  • Using the radii rather than their squares, giving \(8\) cm - an offered option.
  • Expecting the depth to increase because the jug is 'bigger'.
Takeaway Pouring conserves volume: equate \(\pi r_1^2h_1 = \pi r_2^2h_2\); a wider container gives a shallower depth.
Answer F · Pythagoras, mensuration and 3D (M5.16) · question MEN-14
12Pythagoras, mensuration and 3D · spec M5.17 · Paper 1 style · ★☆☆

A solid consists of a hemisphere of radius \(3\) on top of a cylinder of radius \(3\) and height \(4\). Find its volume. (Sphere volume \(\frac{4}{3}\pi r^3\).)

  1. A\(90\pi\)
  2. B\(45\pi\)
  3. C\(63\pi\)
  4. D\(36\pi\)
  5. E\(72\pi\)
  6. F\(54\pi\)
Worked solution

Cylinder \(\pi\cdot9\cdot4 = 36\pi\); hemisphere \(\frac12\cdot\frac43\pi\cdot27 = 18\pi\). Total \(54\pi\).

Full method — how and why
A composite solid: add the two volumes, halving the sphere formula for the hemisphere.
  1. Cylinder: \(\pi r^2 h = \pi(9)(4) = 36\pi\).
    \(r = 3\), \(h = 4\).
  2. Hemisphere: half a sphere, \(\frac12 \times \frac43\pi r^3 = \frac23\pi(27) = 18\pi\).
    \(\frac23 \times 27 = 18\); the sphere formula is given if needed.
  3. Add: \(36\pi + 18\pi = 54\pi\).
    The solids sit on top of each other, so volumes simply add.
Where it goes wrong
  • Using the full sphere, giving \(72\pi\) - an offered option.
  • Computing \(r^3\) as \(9\) instead of \(27\).
Takeaway Composite solids: compute each piece separately and add; a hemisphere is exactly half a sphere.
Answer F · Pythagoras, mensuration and 3D (M5.17) · question MEN-15
13Pythagoras, mensuration and 3D · spec M5.18 · Paper 1 style · ★☆☆

From the top of a cliff \(40\) m high, the angle of depression of a boat is \(30^\circ\). How far is the boat from the foot of the cliff?

  1. A\(20\) m
  2. B\(20\sqrt3\) m
  3. C\(40\sqrt{3}\) m
  4. D\(40\) m
  5. E\(80\) m
  6. F\(\dfrac{40}{\sqrt3}\) m
Worked solution

\(\tan30^\circ = \frac{40}{d} \Rightarrow d = \frac{40}{1/\sqrt3} = 40\sqrt3\).

Full method — how and why
An angle of depression from the top equals the angle of elevation from the boat. Then it is right-angled trigonometry.
  1. Draw the triangle: the cliff is vertical (\(40\) m), the sea horizontal, and the angle at the boat is \(30^\circ\).
    The angle of depression from the cliff top equals the angle of elevation from the boat (alternate angles with the horizontal).
  2. Relative to the \(30^\circ\) angle, the cliff is opposite and the distance \(d\) is adjacent: \(\tan 30^\circ = \dfrac{40}{d}\).
    \(\tan = \frac{\text{opposite}}{\text{adjacent}}\).
  3. Solve: \(d = \dfrac{40}{\tan 30^\circ} = \dfrac{40}{1/\sqrt3} = 40\sqrt3\) m.
    Dividing by \(\frac{1}{\sqrt3}\) multiplies by \(\sqrt3\).
Where it goes wrong
  • Multiplying by \(\tan 30^\circ\) instead of dividing, giving \(\frac{40}{\sqrt3}\) - an offered option.
  • Placing the \(30^\circ\) angle at the top of the cliff inside the triangle, which would swap opposite and adjacent.
Takeaway Depression from the top \(=\) elevation from the bottom; a shallow angle means a large horizontal distance.
Answer C · Pythagoras, mensuration and 3D (M5.18) · question MEN-16
14Pythagoras, mensuration and 3D · spec M5.16 · Paper 2 style · ★☆☆

\(P\): 'a solid cylinder and a solid cube have equal volumes'. \(Q\): 'the cylinder and the cube have equal surface areas'. Which one of the following is true?

  1. AP is necessary and sufficient for Q.
  2. BP is necessary but not sufficient for Q.
  3. CP is sufficient but not necessary for Q.
  4. DP is neither necessary nor sufficient for Q.
Worked solution

Equal volumes say nothing about surface areas (a long thin cylinder has a large surface area), and equal surface areas say nothing about volumes. Neither implies the other.

Full method — how and why
Volume and surface area are independent quantities. Neither condition constrains the other, so test both directions.
  1. Not sufficient: take a cube of side \(1\) (volume \(1\), surface \(6\)) and a cylinder of volume \(1\) that is long and thin - say radius \(0.1\), height \(\frac{1}{0.01\pi} \approx 31.8\), whose surface area is far larger than \(6\).
    Equal volumes with wildly different shapes give wildly different surface areas.
  2. Not necessary: two solids can share a surface area while having different volumes, by the same flexibility of shape.
    Stretching a cylinder thin at fixed surface area shrinks its volume.
  3. So \(P\) is neither necessary nor sufficient for \(Q\).
    Match to the list.
Where it goes wrong
  • Assuming 'bigger volume means bigger surface area' - among different shapes, no such rule holds.
  • Testing only one direction and then guessing the other.
Takeaway Volume and surface area are independent across shapes; a sphere minimises surface area for a given volume, and everything else can be made worse without limit.
Answer D · Pythagoras, mensuration and 3D (M5.16) · question X2-29
15Pythagoras, mensuration and 3D · spec M5.16 · Paper 1 style · ★★☆

A cylinder has radius \(r\) and height \(h\). Its volume is \(54\pi\) and \(h = 2r\). Find its total surface area.

  1. A\(27\pi\)
  2. B\(45\pi\)
  3. C\(108\pi\)
  4. D\(54\pi\)
  5. E\(72\pi\)
  6. F\(36\pi\)
Worked solution

\(\pi r^2 \cdot 2r = 2\pi r^3 = 54\pi \Rightarrow r = 3\), \(h = 6\). Surface area \(= 2\pi r^2 + 2\pi rh = 18\pi + 36\pi = 54\pi\).

Full method — how and why
Use the volume condition to find \(r\), then substitute into the surface-area formula. Both formulas must be recalled.
  1. Write the volume with \(h = 2r\): \(V = \pi r^2 h = \pi r^2 (2r) = 2\pi r^3\).
    Substituting the relationship immediately reduces to one unknown.
  2. Solve \(2\pi r^3 = 54\pi\): \(r^3 = 27\), so \(r = 3\) and \(h = 6\).
    The \(\pi\) cancels, which is a good sign.
  3. Total surface area \(= 2\pi r^2\) (two circular ends) \(+\ 2\pi r h\) (curved surface).
    'Total' means include both ends; 'curved surface area' would be just \(2\pi rh\).
  4. Substitute: \(2\pi(9) + 2\pi(3)(6) = 18\pi + 36\pi = 54\pi\).
    Coincidentally equal to the volume's numerical value - a deliberate touch.
Where it goes wrong
  • Giving only the curved surface \(36\pi\) - an offered option.
  • Forgetting that there are two circular ends.
Takeaway Cylinder: \(V = \pi r^2h\), curved surface \(2\pi rh\), total surface \(2\pi r^2 + 2\pi rh\).
Answer D · Pythagoras, mensuration and 3D (M5.16) · question MEN-02
16Pythagoras, mensuration and 3D · spec M5.7 · Paper 1 style · ★★☆

A square-based pyramid has base side \(6\) and all slant edges of length \(9\). Find its height.

  1. A\(\sqrt{81 - 36}\)
  2. B\(\sqrt{45}\)
  3. C\(6\)
  4. D\(\sqrt{63}\)
  5. E\(\sqrt{72}\)
  6. F\(3\sqrt{6}\)
Worked solution

Half the base diagonal is \(3\sqrt2\), so \(h^2 = 81 - 18 = 63\).

Full method — how and why
The height of a pyramid, a slant edge and half the base diagonal form a right-angled triangle.
  1. Find the base diagonal: a square of side \(6\) has diagonal \(6\sqrt2\).
    Pythagoras on the square base: \(\sqrt{36 + 36} = 6\sqrt2\).
  2. Half of it is \(3\sqrt2\) - the horizontal distance from the centre to a corner.
    The apex sits above the centre of the base, so this is the horizontal leg.
  3. Apply Pythagoras with the slant edge as hypotenuse: \(h^2 = 9^2 - (3\sqrt2)^2 = 81 - 18 = 63\).
    \((3\sqrt2)^2 = 9 \times 2 = 18\) - square the coefficient too.
  4. So \(h = \sqrt{63}\).
    \(= 3\sqrt7\), if you prefer it simplified.
Where it goes wrong
  • Using half the side (\(3\)) instead of half the diagonal, giving \(\sqrt{72}\) - an offered option.
  • Computing \((3\sqrt2)^2\) as \(6\) or \(9\).
Takeaway For a slant edge, use half the base diagonal; for a slant face height, use half the base side.
Answer D · Pythagoras, mensuration and 3D (M5.7) · question MEN-10
17Pythagoras, mensuration and 3D · spec M5.7 · Paper 1 style · ★★☆

The points \(A(1, 2)\), \(B(4, 6)\) and \(C(k, 2)\) with \(k > 1\) form a triangle with a right angle at \(B\). Find \(k\).

  1. A\(\dfrac{16}{3}\)
  2. B\(7\)
  3. C\(\dfrac{28}{3}\)
  4. D\(\dfrac{25}{3}\)
  5. E\(5\)
  6. F\(8\)
Worked solution

Gradient \(AB = \frac43\), so \(BC\) must have gradient \(-\frac34\): \(\dfrac{2 - 6}{k - 4} = -\dfrac34 \Rightarrow 16 = 3(k - 4) \Rightarrow k = \dfrac{28}{3}\).

Full method — how and why
A right angle in coordinate geometry means perpendicular gradients - the vertex of the right angle is the shared point.
  1. Compute the gradient of \(AB\): \(\dfrac{6-2}{4-1} = \dfrac43\).
    The right angle is at \(B\), so the two sides meeting there are \(BA\) and \(BC\).
  2. Perpendicular gradient for \(BC\): \(-\dfrac34\).
    Negative reciprocal.
  3. Write the gradient of \(BC\) using \(C(k,2)\): \(\dfrac{2 - 6}{k - 4} = \dfrac{-4}{k-4}\), and set it equal to \(-\frac34\).
    Keep the subtraction order consistent: from \(B\) to \(C\) in both coordinates.
  4. Solve: \(\dfrac{-4}{k-4} = -\dfrac34 \Rightarrow 16 = 3(k-4) \Rightarrow k = 4 + \dfrac{16}{3} = \dfrac{28}{3}\).
    Cross-multiply; both sides are negative so the signs cancel.
Where it goes wrong
  • Assuming the right angle is at \(A\) or \(C\).
  • Dropping the \(+4\) at the end and answering \(\frac{16}{3}\) - an offered option.
Takeaway Right angle at a named vertex: use the two sides through that vertex and set the gradient product to \(-1\).
Answer C · Pythagoras, mensuration and 3D (M5.7) · question MEN-13

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