Mathematics

Answer Key

Geometry

Pack A — Answers

# Question Answer
1 Find the area of a rectangle with length $8$ cm and width $5$ cm. 40 cm²
2 Find the area of a triangle with base $10$ cm and height $6$ cm. 30 cm²
3 Find the area of a circle with radius $5$ cm. Give your answer in terms of $\pi$. $25\pi$ cm²
4 In a right-angled triangle, the legs are $3$ and $4$ cm. Find the hypotenuse. 5 cm
5 A right-angled triangle has hypotenuse $10$ cm and one leg $6$ cm. Find the other leg. 8 cm
6 Find the circumference of a circle with radius $4$ cm, in terms of $\pi$. $8\pi$ cm
7 A ship sails due east. State its bearing. $090°$
8 Find the volume of a cuboid measuring $4$ cm × $5$ cm × $6$ cm. 120 cm³
9 In a right-angled triangle, the side opposite to angle $\theta$ has length $3$ cm and the hypotenuse is $5$ cm. Find $\sin\theta$. $\sin\theta = \dfrac{3}{5}$
10 Find the perimeter of a rectangle measuring $9$ cm by $4$ cm. 26 cm
11 Find the area of a trapezium with parallel sides $6$ cm and $10$ cm, height $4$ cm. 32 cm²
12 A shape consists of a rectangle $8$ × $5$ with a right triangle (legs $4$ and $3$) attached. Find the total area. 46 cm²
13 A right-angled triangle has hypotenuse $10$ cm and angle $\theta = 30°$. Find the length of the side opposite $\theta$ (to 2 d.p.). 5.00 cm
14 A right-angled triangle has hypotenuse $12$ cm and angle $\theta = 40°$. Find the side adjacent to $\theta$ (to 2 d.p.). 9.19 cm
15 A right-angled triangle has adjacent side $8$ cm and angle $\theta = 35°$. Find the opposite side (to 2 d.p.). 5.60 cm
16 In a right-angled triangle, the opposite side is $5$ cm and the hypotenuse is $10$ cm. Find $\theta$ (to 1 d.p.). $\theta = 30.0°$
17 A walker travels on a bearing of $50°$, then turns right through $90°$. What is the new bearing? $140°$
18 Find the volume of a cylinder with radius $4$ cm and height $10$ cm, in terms of $\pi$. $160\pi$ cm³
19 A right-angled triangle has legs $2$ and $3$ cm. Find the hypotenuse exactly. $\sqrt{13}$ cm
20 Find the area of a parallelogram with base $12$ cm and perpendicular height $7$ cm. 84 cm²
21 Find the area of a triangle with sides $a = 6$ cm, $b = 8$ cm, and included angle $C = 30°$ (to 2 d.p.). 12.00 cm²
22 A ship sails $50$ km on a bearing of $60°$. How far east and north does it travel? (1 d.p.) East: 43.3 km; North: 25.0 km
23 A tree of height $12$ m casts a shadow of length $5$ m. Find the angle of elevation of the sun (to 1 d.p.). $67.4°$
24 In a triangle, $A = 50°$, $B = 60°$, and the side opposite $A$ is $8$ cm. Find the side opposite $B$ (to 2 d.p.). 9.04 cm
25 In a triangle, two sides are $5$ cm and $7$ cm with included angle $60°$. Find the third side (to 2 d.p.). 6.24 cm
26 A ship sails from A on a bearing of $60°$ to B, $10$ km away. From B, it sails on bearing $150°$ to C, $12$ km away. (a) State the angle $\angle ABC$. (b) Use cosine rule to find $AC$ (2 d.p.). (a) $90°$ (b) $AC = \sqrt{244} \approx 15.62$ km
27 A right-angled triangle has hypotenuse $10$ cm and an angle of $30°$. Find its perimeter (to 2 d.p.). 23.66 cm
28 Find the length of the space diagonal of a cuboid measuring $3$ × $4$ × $12$ cm (to 2 d.p.). 13.00 cm
29 A pyramid has a square base of side $8$ cm and apex directly above the centre, $6$ cm above the base. Find the slant height from the apex to a midpoint of a base edge (to 2 d.p.). 7.21 cm
30 A triangle has sides $5, 7, 9$ cm. Find the largest angle (to 1 d.p.). $95.7°$
31 A ship at $A$ sails on bearing $60°$ for $30$ km to $B$. From $B$ it sails on bearing $150°$ for $40$ km to $C$. Find the distance $AC$ (to 2 d.p.). 50 km
32 Find the area of a sector of a circle with radius $10$ cm and angle $60°$ (to 2 d.p.). $\approx 52.36$ cm²
33 From a point $25$ m from the base of a tower the angle of elevation to the top is $38°$. Find the height of the tower (2 d.p.). $\approx 19.53$ m
34 In triangle $ABC$, $a = 8$ cm, $b = 10$ cm, $A = 40°$. Find angle $B$ (to 1 d.p.). $B \approx 53.5°$ (or $126.5°$)
35 A solid consists of a cylinder (radius $3$ cm, height $10$ cm) topped by a hemisphere (same radius). Find the volume in terms of $\pi$. $108\pi$ cm³
36 A triangle has sides $6, 8, 10$ cm. Use the cosine rule to find an angle, then compute the area (to 2 d.p.). 24 cm² (right triangle)
37 Two ships leave a port. Ship 1 sails on bearing $60°$ at $20$ km/h; Ship 2 on bearing $150°$ at $15$ km/h. How far apart are they after $2$ hours (to 2 d.p.)? 50 km
38 A ladder of length $5$ m leans against a vertical wall. The foot of the ladder is $3$ m from the wall. (a) How high up the wall does the ladder reach? (b) What angle does the ladder make with the ground (to 1 d.p.)? (a) 4 m (b) $\approx 53.1°$
39 In triangle $ABC$, $A = $50°$, $B = $70°$, and side $c = $10$ cm. Find side $a$ (to 2 d.p.). $\approx 8.85$ cm
40 A regular hexagon has side length $6$ cm. Find its area exactly (in surd form). $54\sqrt{3}$ cm²

Pack B — Answers

# Question Answer
1 Find the area of a rectangle with length $12$ cm and width $7$ cm. 84 cm²
2 Find the area of a triangle with base $14$ cm and height $8$ cm. 56 cm²
3 Find the area of a circle with radius $7$ cm. Give your answer in terms of $\pi$. $49\pi$ cm²
4 In a right-angled triangle, the legs are $5$ and $12$ cm. Find the hypotenuse. 13 cm
5 A right-angled triangle has hypotenuse $15$ cm and one leg $9$ cm. Find the other leg. 12 cm
6 Find the circumference of a circle with radius $6$ cm, in terms of $\pi$. $12\pi$ cm
7 A ship sails due east. State its bearing. $180°$
8 Find the volume of a cuboid measuring $3$ cm × $7$ cm × $10$ cm. 210 cm³
9 In a right-angled triangle, the side opposite to angle $\theta$ has length $5$ cm and the hypotenuse is $13$ cm. Find $\sin\theta$. $\sin\theta = \dfrac{5}{13}$
10 Find the perimeter of a rectangle measuring $15$ cm by $8$ cm. 46 cm
11 Find the area of a trapezium with parallel sides $5$ cm and $9$ cm, height $6$ cm. 42 cm²
12 A shape consists of a rectangle $10$ × $6$ with a right triangle (legs $6$ and $4$) attached. Find the total area. 72 cm²
13 A right-angled triangle has hypotenuse $15$ cm and angle $\theta = 50°$. Find the length of the side opposite $\theta$ (to 2 d.p.). 11.49 cm
14 A right-angled triangle has hypotenuse $20$ cm and angle $\theta = 60°$. Find the side adjacent to $\theta$ (to 2 d.p.). 10.00 cm
15 A right-angled triangle has adjacent side $10$ cm and angle $\theta = 45°$. Find the opposite side (to 2 d.p.). 10.00 cm
16 In a right-angled triangle, the opposite side is $7$ cm and the hypotenuse is $14$ cm. Find $\theta$ (to 1 d.p.). $\theta = 30.0°$
17 A walker travels on a bearing of $120°$, then turns right through $60°$. What is the new bearing? $180°$
18 Find the volume of a cylinder with radius $5$ cm and height $8$ cm, in terms of $\pi$. $200\pi$ cm³
19 A right-angled triangle has legs $4$ and $5$ cm. Find the hypotenuse exactly. $\sqrt{41}$ cm
20 Find the area of a parallelogram with base $9$ cm and perpendicular height $5$ cm. 45 cm²
21 Find the area of a triangle with sides $a = 10$ cm, $b = 12$ cm, and included angle $C = 45°$ (to 2 d.p.). 42.43 cm²
22 A ship sails $40$ km on a bearing of $45°$. How far east and north does it travel? (1 d.p.) East: 28.3 km; North: 28.3 km
23 A tree of height $8$ m casts a shadow of length $6$ m. Find the angle of elevation of the sun (to 1 d.p.). $53.1°$
24 In a triangle, $A = 40°$, $B = 70°$, and the side opposite $A$ is $10$ cm. Find the side opposite $B$ (to 2 d.p.). 14.62 cm
25 In a triangle, two sides are $8$ cm and $10$ cm with included angle $70°$. Find the third side (to 2 d.p.). 10.45 cm
26 A ship sails from A on a bearing of $45°$ to B, $8$ km away. From B, it sails on bearing $135°$ to C, $10$ km away. (a) State the angle $\angle ABC$. (b) Use cosine rule to find $AC$ (2 d.p.). (a) $90°$ (b) $AC = \sqrt{164} \approx 12.81$ km
27 A right-angled triangle has hypotenuse $13$ cm and an angle of $50°$. Find its perimeter (to 2 d.p.). 30.31 cm
28 Find the length of the space diagonal of a cuboid measuring $5$ × $12$ × $4$ cm (to 2 d.p.). 13.60 cm
29 A pyramid has a square base of side $10$ cm and apex directly above the centre, $12$ cm above the base. Find the slant height from the apex to a midpoint of a base edge (to 2 d.p.). 13.00 cm
30 A triangle has sides $6, 8, 11$ cm. Find the largest angle (to 1 d.p.). $102.6°$
31 A ship at $A$ sails on bearing $30°$ for $20$ km to $B$. From $B$ it sails on bearing $120°$ for $30$ km to $C$. Find the distance $AC$ (to 2 d.p.). $\sqrt{1300} \approx 36.06$ km
32 Find the area of a sector of a circle with radius $8$ cm and angle $45°$ (to 2 d.p.). $\approx 25.13$ cm²
33 From a point $40$ m from the base of a tower the angle of elevation to the top is $28°$. Find the height of the tower (2 d.p.). $\approx 21.27$ m
34 In triangle $ABC$, $a = 7$ cm, $b = 9$ cm, $A = 45°$. Find angle $B$ (to 1 d.p.). $B \approx 65.4°$ (or $114.6°$)
35 A solid consists of a cylinder (radius $5$ cm, height $8$ cm) topped by a hemisphere (same radius). Find the volume in terms of $\pi$. $\dfrac{850\pi}{3}$ cm³
36 A triangle has sides $5, 12, 13$ cm. Use the cosine rule to find an angle, then compute the area (to 2 d.p.). 30 cm² (right triangle)
37 Two ships leave a port. Ship 1 sails on bearing $30°$ at $25$ km/h; Ship 2 on bearing $120°$ at $20$ km/h. How far apart are they after $2$ hours (to 2 d.p.)? 64.03 km
38 A ladder of length $13$ m leans against a vertical wall. The foot of the ladder is $5$ m from the wall. (a) How high up the wall does the ladder reach? (b) What angle does the ladder make with the ground (to 1 d.p.)? (a) 12 m (b) $\approx 67.4°$
39 In triangle $ABC$, $A = $35°$, $B = $80°$, and side $c = $12$ cm. Find side $a$ (to 2 d.p.). $\approx 7.60$ cm
40 A regular hexagon has side length $4$ cm. Find its area exactly (in surd form). $24\sqrt{3}$ cm²

Problems — Worked Solutions

1

**Ladder safety.** A 6 m ladder leans against a wall with the foot 2 m from the base of the wall. (a) How high does the ladder reach (to 2 d.p.)? (b) What angle does the ladder make with the ground? (c) Safety advice: the angle should be between 70° and 80°. Is this ladder safely placed? If not, where should the foot be placed?

Answer

(a) 5.66 m (b) $\approx 70.5°$ (c) Yes — just within the safe range

(a) Height = $\sqrt{6^2 - 2^2} = \sqrt{32} = 4\sqrt{2} \approx 5.66$ m. (b) $\cos\theta = \frac{2}{6} = \frac{1}{3} \Rightarrow \theta = \cos^{-1}(1/3) \approx 70.5°$. (c) 70.5° is within the [70°, 80°] safe range — **just safe**. If the foot were moved closer to the wall (say 1 m away), $\cos\theta = \frac{1}{6}$, giving $\theta \approx 80.4°$ (a bit too steep). The safe range for this ladder corresponds to foot distance $6\cos 80° \leq d \leq 6\cos 70°$, i.e. $1.04 \leq d \leq 2.05$ m.
2

**Bearings triangle.** A walker sets out from camp $C$ on a bearing of $050°$ for 6 km to point $P$. At $P$, she changes to a bearing of $140°$ and walks 8 km to point $Q$. (a) Show that $\angle CPQ = 90°$. (b) Find the distance $CQ$. (c) Find the bearing of $Q$ from $C$ (to the nearest degree).

Answer

(a) See working (b) 10 km (c) $\approx 103°$

(a) At $P$, the walker turns from bearing $050°$ to $140°$. The interior angle at $P$ in the triangle $CPQ$ is $180° - (140° - 050°) = 180° - 90° = 90°$. (b) Right-angled triangle $CPQ$ with legs $CP = 6$ km and $PQ = 8$ km. $CQ = \sqrt{36 + 64} = 10$ km. (c) Bearing of $Q$ from $C$: angle clockwise from north. Component east: $6\sin 50° + 8\sin 140° = 4.596 + 5.142 = 9.74$ km. Component north: $6\cos 50° + 8\cos 140° = 3.857 - 6.128 = -2.27$ km. So $Q$ is south-east of $C$. Bearing = $180° - \tan^{-1}(9.74/2.27) = 180° - 76.9° \approx 103°$.
3

**Tower and shadow.** A tower of height $h$ m casts a shadow of length 15 m when the angle of elevation of the sun is $40°$. (a) Find $h$. (b) Later, the shadow has length 25 m. What is the new angle of elevation? (c) When is the sun's angle of elevation $30°$? (Find the shadow length.)

Answer

(a) $h \approx 12.59$ m (b) $\approx 26.7°$ (c) $\approx 21.81$ m

(a) $\tan 40° = \frac{h}{15}$, so $h = 15 \tan 40° \approx 15 \times 0.839 \approx 12.59$ m. (b) New angle: $\tan\theta = \frac{12.59}{25} = 0.504 \Rightarrow \theta \approx 26.7°$. (c) Shadow at 30°: $\tan 30° = \frac{12.59}{L} \Rightarrow L = \frac{12.59}{\tan 30°} = \frac{12.59}{0.5774} \approx 21.81$ m.
4

**Non-right-angled triangle.** Triangle $ABC$ has $a = 8$ cm, $b = 11$ cm, $C = 75°$. (a) Use the cosine rule to find $c$. (b) Find $\angle A$ using the sine rule. (c) Find the area of $\triangle ABC$.

Answer

(a) $c \approx 11.81$ cm (b) $A \approx 40.9°$ (c) $\approx 42.50$ cm²

(a) $c^2 = a^2 + b^2 - 2ab\cos C = 64 + 121 - 176\cos 75° \approx 185 - 45.55 \approx 139.45$. $c \approx \sqrt{139.45} \approx 11.81$ cm. (b) Sine rule: $\frac{\sin A}{a} = \frac{\sin C}{c}$. $\sin A = \frac{8 \sin 75°}{11.81} \approx \frac{7.73}{11.81} \approx 0.654$. $A \approx 40.9°$. (c) Area $= \frac{1}{2}ab \sin C = \frac{1}{2}(8)(11)\sin 75° = 44 \times 0.966 \approx 42.50$ cm².
5

**Compound area.** Find the area of an isosceles trapezium with parallel sides 10 cm and 16 cm, and slant sides of length 5 cm.

Answer

52 cm²

Drop perpendiculars from the ends of the shorter parallel side to the longer. The horizontal projection on each side: $\frac{16 - 10}{2} = 3$ cm. Height: $\sqrt{5^2 - 3^2} = \sqrt{16} = 4$ cm. Area $= \frac{1}{2}(10 + 16)(4) = \frac{1}{2}(26)(4) = \mathbf{52}$ cm².
6

**Cuboid diagonals.** A cuboid measures 4 cm × 6 cm × 12 cm. (a) Find the length of the space diagonal. (b) Find the angle the space diagonal makes with the base (i.e., the 4 × 6 face).

Answer

(a) 14 cm (b) $\approx 59.0°$

(a) Space diagonal $d = \sqrt{4^2 + 6^2 + 12^2} = \sqrt{16 + 36 + 144} = \sqrt{196} = 14$ cm. (b) Base diagonal: $\sqrt{16 + 36} = \sqrt{52} \approx 7.21$ cm. Angle: $\tan\theta = \frac{12}{\sqrt{52}}$. $\theta = \tan^{-1}(\frac{12}{\sqrt{52}}) \approx \tan^{-1}(1.664) \approx 59.0°$.
7

**Trig identities check.** For a right triangle with sides 3, 4, 5, take the angle $\theta$ opposite the side of length 3. (a) Find $\sin\theta$, $\cos\theta$, $\tan\theta$. (b) Verify $\sin^2\theta + \cos^2\theta = 1$. (c) Verify $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$.

Answer

(a) $\sin = 3/5, \cos = 4/5, \tan = 3/4$ (b) ✓ (c) ✓

(a) Opposite to $\theta$ is 3; adjacent is 4; hypotenuse is 5. - $\sin\theta = 3/5$ - $\cos\theta = 4/5$ - $\tan\theta = 3/4$ (b) $\sin^2\theta + \cos^2\theta = \frac{9}{25} + \frac{16}{25} = \frac{25}{25} = 1$ ✓. (c) $\frac{\sin\theta}{\cos\theta} = \frac{3/5}{4/5} = \frac{3}{4} = \tan\theta$ ✓.
8

**Two ships.** Two ships leave port at the same time. Ship $A$ sails on bearing $030°$ at 20 km/h; Ship $B$ on bearing $150°$ at 15 km/h. (a) Find the angle between the ships' paths. (b) After 2 hours, how far apart are they (2 d.p.)? (c) What is the bearing of $B$ from $A$ after 2 hours (nearest degree)?

Answer

(a) 120° (b) $\approx 60.83$ km (c) $\approx 185°$

(a) Angle between bearings $150° - 030° = 120°$. (b) After 2 hr: $|OA| = 40$ km, $|OB| = 30$ km, angle $AOB = 120°$. Cosine rule: $|AB|^2 = 40^2 + 30^2 - 2(40)(30)\cos 120° = 1600 + 900 - 2400(-0.5) = 2500 + 1200 = 3700$. $|AB| = \sqrt{3700} \approx 60.83$ km. (c) Position of $A$: $(40\sin 30°, 40\cos 30°) = (20, 34.64)$. Position of $B$: $(30\sin 150°, 30\cos 150°) = (15, -25.98)$. Vector from $A$ to $B$: $(-5, -60.62)$ — south and slightly west of $A$. Bearing (clockwise from north): pointing south is $180°$; the vector deviates slightly west of south by $\tan^{-1}(\frac{5}{60.62}) \approx 4.7°$. So bearing $= 180° + 4.7° \approx 185°$ (since vector points west-of-south).
9

**Pythagorean triples.** A "Pythagorean triple" is a set of three integers $(a, b, c)$ with $a^2 + b^2 = c^2$. (a) Check that $(3, 4, 5)$, $(5, 12, 13)$, and $(8, 15, 17)$ are all Pythagorean triples. (b) Show that if $(a, b, c)$ is a triple, then so is $(ka, kb, kc)$ for any positive integer $k$. (c) Find a triple where $c = 25$ (other than the trivial $(7, 24, 25)$).

Answer

(a) All check out (b) Scaling preserves the relation (c) $(15, 20, 25)$

(a) Checks: - $3^2 + 4^2 = 9 + 16 = 25 = 5^2$ ✓ - $5^2 + 12^2 = 25 + 144 = 169 = 13^2$ ✓ - $8^2 + 15^2 = 64 + 225 = 289 = 17^2$ ✓ (b) If $a^2 + b^2 = c^2$, then $(ka)^2 + (kb)^2 = k^2 a^2 + k^2 b^2 = k^2(a^2 + b^2) = k^2 c^2 = (kc)^2$ ✓. (c) Scale up $(3, 4, 5)$ by $k = 5$: $(15, 20, 25)$. Check: $225 + 400 = 625 = 25^2$ ✓.
10

**Hexagon.** A regular hexagon is inscribed in a circle of radius 10 cm. (a) Show that each side of the hexagon equals the radius. (b) Find the area of the hexagon (exact, in surd form). (c) Find the area of the circle (in terms of $\pi$) and compare.

Answer

(a) See working (b) $150\sqrt{3}$ cm² ≈ 259.81 (c) $100\pi \approx 314.16$ — hexagon ≈ 82.7% of circle

(a) Connect the centre to each vertex. This divides the hexagon into 6 congruent isoceles triangles, each with two radii of length $r$ and central angle $60°$ (i.e. $360°/6$). With two equal sides and a $60°$ included angle, the third side (a side of the hexagon) must also equal $r$ (forming equilateral triangles). (b) Six equilateral triangles of side 10. Each area: $\frac{\sqrt{3}}{4}(10)^2 = 25\sqrt{3}$. Total: $6 \times 25\sqrt{3} = 150\sqrt{3} \approx 259.81$ cm². (c) Circle area: $\pi r^2 = 100\pi \approx 314.16$ cm². Ratio: $\frac{150\sqrt{3}}{100\pi} \approx \frac{259.81}{314.16} \approx 82.7\%$. The hexagon fills about 83% of the circumscribing circle.
11

**Cone problem.** A cone has a slant height of 13 cm and a base radius of 5 cm. (a) Find the perpendicular height of the cone. (b) Find the volume of the cone in terms of $\pi$. (c) Find the total surface area in terms of $\pi$.

Answer

(a) 12 cm (b) $100\pi$ cm³ (c) $90\pi$ cm²

(a) Height $h = \sqrt{13^2 - 5^2} = \sqrt{144} = 12$ cm. (b) Volume = $\frac{1}{3}\pi r^2 h = \frac{1}{3}\pi(25)(12) = 100\pi$ cm³. (c) Total surface area = base + lateral = $\pi r^2 + \pi r \ell = 25\pi + \pi(5)(13) = 25\pi + 65\pi = 90\pi$ cm².
12

**Architecture.** A roof has a triangular cross-section. The base of the triangle is 8 m wide, and the two slopes meet at an apex 3 m above the base. (a) Find the length of each slope. (b) Find the angle each slope makes with the horizontal (to 1 d.p.). (c) If the roof is 12 m long, find the total area of the two rectangular sloped surfaces.

Answer

(a) 5 m each (b) $\approx 36.9°$ (c) 120 m²

(a) Each slope is the hypotenuse of a right triangle with horizontal $4$ m and vertical $3$ m. Slope length = $\sqrt{16 + 9} = 5$ m. (b) Angle: $\tan\theta = \frac{3}{4} \Rightarrow \theta = \tan^{-1}(0.75) \approx 36.9°$. (c) Each rectangular slope = $5 \times 12 = 60$ m². Two slopes total = $\mathbf{120}$ m².