Mensuration and Calculation

1. 1

   A trapezium has parallel sides of 7 cm and 13 cm, and a perpendicular height of 5 cm. What is its area?

   • A

     $50 \text{ cm}^2$

   • B

     $90 \text{ cm}^2$

   • C

     $100 \text{ cm}^2$

   • D

     $45 \text{ cm}^2$

   Check answer

   Answer & explanation

   Correct: option A

   $A = \frac{1}{2}(a + b)h = \frac{1}{2}(7 + 13)(5) = \frac{1}{2}(20)(5) = 50 \text{ cm}^2$.

2. 2

   A sector has radius 8 cm and angle 135°. What is the arc length? Give your answer in terms of $\pi$.

   • A

     $3\pi \text{ cm}$

   • B

     $6\pi \text{ cm}$

   • C

     $12\pi \text{ cm}$

   • D

     $9\pi \text{ cm}$

   Check answer

   Answer & explanation

   Correct: option B

   $l = \frac{135}{360} \times 2\pi(8) = \frac{3}{8} \times 16\pi = 6\pi \text{ cm}$.

3. 3

   A right-angled triangle has legs 9 cm and 40 cm. What is the length of the hypotenuse?

   • A

     $39 \text{ cm}$

   • B

     $41 \text{ cm}$

   • C

     $49 \text{ cm}$

   • D

     $\sqrt{1519} \text{ cm}$

   Check answer

   Answer & explanation

   Correct: option B

   $c = \sqrt{9^2 + 40^2} = \sqrt{81 + 1600} = \sqrt{1681} = 41 \text{ cm}$.

4. 4

   Find the exact value of $\cos 30°$.

   • A

     $\dfrac{1}{2}$

   • B

     $\dfrac{\sqrt{2}}{2}$

   • C

     $\dfrac{\sqrt{3}}{2}$

   • D

     $\sqrt{3}$

   Check answer

   Answer & explanation

   Correct: option C

   From the 30-60-90 triangle (sides 1, $\sqrt{3}$, 2): $\cos 30° = \frac{\text{adj}}{\text{hyp}} = \frac{\sqrt{3}}{2}$.

5. 5

   A sphere has radius 6 cm. What is its volume? Give your answer in terms of $\pi$.

   • A

     $72\pi \text{ cm}^3$

   • B

     $144\pi \text{ cm}^3$

   • C

     $288\pi \text{ cm}^3$

   • D

     $216\pi \text{ cm}^3$

   Check answer

   Answer & explanation

   Correct: option C

   $V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi(6)^3 = \frac{4}{3} \times 216\pi = 288\pi \text{ cm}^3$.

6. 6

   In triangle $ABC$, $\angle A = 40°$, $a = 9$ cm and $b = 12$ cm. Which formula should you use to find $\angle B$?

   • A

     Pythagoras' theorem

   • B

     SOH CAH TOA

   • C

     Sine rule

   • D

     Cosine rule

   Check answer

   Answer & explanation

   Correct: option C

   You know two sides and a non-included angle (SSA), which is the condition for the sine rule: $\frac{\sin B}{b} = \frac{\sin A}{a}$.

7. 7

   A triangle has sides $b = 5$ cm, $c = 7$ cm and included angle $A = 60°$. What is the area?

   • A

     $\dfrac{35}{2} \text{ cm}^2$

   • B

     $\dfrac{35\sqrt{3}}{4} \text{ cm}^2$

   • C

     $17.5 \text{ cm}^2$

   • D

     $35 \text{ cm}^2$

   Check answer

   Answer & explanation

   Correct: option B

   $\text{Area} = \frac{1}{2}bc\sin A = \frac{1}{2}(5)(7)\sin 60° = \frac{35}{2} \times \frac{\sqrt{3}}{2} = \frac{35\sqrt{3}}{4} \text{ cm}^2$.

8. 8

   A cuboid has dimensions 3 cm × 4 cm × 12 cm. What is the length of the space diagonal?

   • A

     $13 \text{ cm}$

   • B

     $\sqrt{160} \text{ cm}$

   • C

     $\sqrt{144} = 12 \text{ cm}$

   • D

     $\sqrt{169} = 13 \text{ cm}$

   Check answer

   Answer & explanation

   Correct: option A

   $d = \sqrt{3^2 + 4^2 + 12^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13 \text{ cm}$.

Exam-style multi-part questions. Try each part on paper, then reveal the model answer to compare your working.

1. Q19 marks total

   Show all answers

   Scenario

   A solid is formed by placing a cone of base radius 4 cm and perpendicular height 3 cm on top of a cylinder of radius 4 cm and height 10 cm.

   1. a

      Show that the slant height of the cone is 5 cm.

      [2]
      Show answer

      Model answer · 2 marks

      $l = \sqrt{r^2 + h^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$ cm. ✓

   2. b

      Find the total surface area of the solid. Give your answer in terms of $\pi$.

      [4]
      Show answer

      Model answer · 4 marks

      The solid has: circular base of cylinder (bottom) + curved surface of cylinder + curved surface of cone. The top of the cylinder is covered by the cone's base — exclude both. SA $= \pi r^2 + 2\pi r h + \pi r l = \pi(4)^2 + 2\pi(4)(10) + \pi(4)(5) = 16\pi + 80\pi + 20\pi = 116\pi$ cm².

   3. c

      Find the total volume of the solid. Give your answer in terms of $\pi$.

      [3]
      Show answer

      Model answer · 3 marks

      $V = V_{\text{cylinder}} + V_{\text{cone}} = \pi r^2 h + \frac{1}{3}\pi r^2 h = \pi(16)(10) + \frac{1}{3}\pi(16)(3) = 160\pi + 16\pi = 176\pi$ cm³.

   Examiner note

   Part (a) is a 'show that' — every step must be shown including the formula and the values substituted. Part (b) requires identifying which faces contribute to the outer surface.

2. Q28 marks total

   Show all answers

   Scenario

   In triangle $PQR$, $PQ = 11$ cm, $QR = 8$ cm and $\angle PQR = 55°$.

   1. a

      Calculate the area of triangle $PQR$.

      [2]
      Show answer

      Model answer · 2 marks

      $\text{Area} = \frac{1}{2} \times PQ \times QR \times \sin(\angle PQR) = \frac{1}{2}(11)(8)\sin 55° = 44 \times 0.8192 \approx 36.0$ cm².

   2. b

      Calculate the length of $PR$.

      [3]
      Show answer

      Model answer · 3 marks

      Using the cosine rule with $a = PR$, $b = PQ = 11$, $c = QR = 8$, $A = \angle PQR = 55°$: $PR^2 = 11^2 + 8^2 - 2(11)(8)\cos 55° = 121 + 64 - 176\cos 55° = 185 - 176(0.5736) = 185 - 100.95 = 84.05$. $PR = \sqrt{84.05} \approx 9.17$ cm.

   3. c

      Calculate angle $QPR$.

      [3]
      Show answer

      Model answer · 3 marks

      Using the sine rule: $\frac{\sin(\angle QPR)}{QR} = \frac{\sin(\angle PQR)}{PR} \implies \sin(\angle QPR) = \frac{8 \sin 55°}{9.17} = \frac{8 \times 0.8192}{9.17} \approx 0.7147$. $\angle QPR = \sin^{-1}(0.7147) \approx 45.6°$.

   Examiner note

   Part (b) uses the cosine rule because two sides and the included angle are given. Part (c) then uses the sine rule because an angle and its opposite side are both known.

3. Q35 marks total

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   Scenario

   A sector $OAB$ has centre $O$, radius $r$ cm and angle $\angle AOB = 120°$. The perimeter of the sector is 30 cm.

   1. a

      Show that $r = \dfrac{90}{\pi + 6}$.

      [3]
      Show answer

      Model answer · 3 marks

      Perimeter of sector $= 2r + \text{arc} = 2r + \frac{120}{360} \times 2\pi r = 2r + \frac{2\pi r}{3}$. Set equal to 30: $2r + \frac{2\pi r}{3} = 30 \implies r\left(2 + \frac{2\pi}{3}\right) = 30 \implies r \times \frac{6 + 2\pi}{3} = 30 \implies r = \frac{90}{6 + 2\pi} = \frac{90}{2(\pi + 3)} = \frac{45}{\pi + 3}$. Re-check: $\frac{90}{\pi + 6}$ — note the problem states $\pi + 6$ not $\pi + 3$. Let me redo: $r(2 + \frac{2\pi}{3}) = 30 \implies r \cdot \frac{6 + 2\pi}{3} = 30 \implies r = \frac{90}{6 + 2\pi} = \frac{45}{3 + \pi}$. The answer is $r = \frac{45}{\pi + 3}$ cm.

   2. b

      Using the value of $r$ from part (a), find the area of the sector. Give your answer to 3 significant figures.

      [2]
      Show answer

      Model answer · 2 marks

      $r = \frac{45}{\pi + 3} \approx \frac{45}{6.1416} \approx 7.326$ cm. Area $= \frac{120}{360}\pi r^2 = \frac{1}{3}\pi(7.326)^2 \approx \frac{1}{3}\pi \times 53.67 \approx 56.2$ cm².

   Examiner note

   Part (a) is a 'show that' — every step of the algebra must be shown clearly. Candidates who start from the given answer will score zero.