Answer Key
9.1 Surds and Pythagoras
Pack A — Answers
| # | Question | Answer |
|---|---|---|
| 1 | A right-angled triangle has the two shorter sides of length 3 cm and 4 cm. Find the hypotenuse. | 5 cm |
| 2 | Simplify $\sqrt{50}$ fully. | $5\sqrt{2}$ |
| 3 | Simplify $\sqrt{20}$ fully. | $2\sqrt{5}$ |
| 4 | A triangle has sides 5, 12, 13 cm. Use Pythagoras' converse to decide whether it is right-angled. | Yes — right-angled |
| 5 | Simplify $\sqrt{3} \times \sqrt{12}$. | 6 |
| 6 | Identify which numbers are rational vs irrational: $5, \sqrt{2}, \pi, 0.\overline{3}, \sqrt{4}$. | Rational: 5, $0.\overline{3}$, $\sqrt{4} = 2$. Irrational: $\sqrt{2}, \pi$. |
| 7 | Estimate $\sqrt{50}$ to 1 d.p. | ≈ 7.1 |
| 8 | Find the hypotenuse of a right-angled triangle with legs 9 and 12 cm. | 15 cm |
| 9 | Add: $\sqrt{50} + \sqrt{18}$ (give simplified form). | $8\sqrt{2}$ |
| 10 | A right-angled triangle has hypotenuse 10 cm and one shorter side 6 cm. Find the other side. | 8 cm |
| 11 | Pythagoras with surds: legs $2\sqrt{3}$ and 3. Find the hypotenuse. | $\sqrt{21}$ |
| 12 | Find the distance between $(1, 2)$ and $(4, 6)$. Give answer as a surd in simplest form. | 5 |
| 13 | Simplify $\sqrt{8} + \sqrt{32} - \sqrt{18}$. | $3\sqrt{2}$ |
| 14 | Find the hypotenuse of a right-angled triangle with legs 4 and 6 cm. Give answer as a simplified surd. | $2\sqrt{13}$ |
| 15 | Simplify $\sqrt{75} \div \sqrt{3}$. | 5 |
| 16 | A right-angled triangle has hypotenuse $\sqrt{50}$ and one leg $\sqrt{18}$. Find the other leg. | $\sqrt{32} = 4\sqrt{2}$ |
| 17 | Find the distance between $(-2, 1)$ and $(3, 13)$. | 13 |
| 18 | Find the perimeter of a right-angled triangle with legs 6 and 8 cm. | 24 cm |
| 19 | Square the surd: $(2 + \sqrt{3})^2$. Expand. | $7 + 4\sqrt{3}$ |
| 20 | Rationalise the denominator: $\dfrac{6}{\sqrt{3}}$. | $2\sqrt{3}$ |
| 21 | A right-angled triangle has shorter sides 4 and 6. Hypotenuse as a simplified surd. | $2\sqrt{13}$ |
| 22 | Find the length of the space diagonal of a cuboid with sides 6, 4, 3. Use 3D Pythagoras. | $\sqrt{61}$ ≈ 7.81 cm |
| 23 | Pythagoras in 3D: cuboid 12 × 9 × 6. Find the space diagonal. | ≈ 16.16 |
| 24 | A square has diagonal 10 cm. Find its side length and area. | Side $5\sqrt{2}$; area 50 |
| 25 | A ladder of length 10 m leans against a wall. Foot 3 m from the base. Find the height reached. | $\sqrt{91} \approx 9.54$ m |
| 26 | Simplify $(3 + \sqrt{2})(2 - \sqrt{2})$. | $4 - \sqrt{2}$ |
| 27 | The area of a square is 12 cm². Find the side as a simplified surd. | $2\sqrt{3}$ |
| 28 | Find the length of the diagonal of a rectangle 6 cm by 8 cm. | 10 cm |
| 29 | Show that the triangle with sides 5, 5, $5\sqrt{2}$ is right-angled. | $5^2 + 5^2 = 50 = (5\sqrt{2})^2$ ✓ |
| 30 | A ladder of length 8 m makes a 60° angle with the ground. Find the height up the wall (using trig). | $4\sqrt{3}$ m ≈ 6.93 |
| 31 | A cuboid has length 8 cm, width 6 cm, height 4 cm. Find the angle between the space diagonal and the base, to 1 d.p. | ≈ 21.8° |
| 32 | Make $a$ the subject of $c = \sqrt{a^2 + b^2}$. | $a = \sqrt{c^2 - b^2}$ |
| 33 | A cliff of height $h$ m stands on level ground. From point A 28 m from the base, the line of sight to the top is 52 m. From B further away, sight to top is 60 m. Find $h$ and B's distance. | $h \approx 43.82$, B ≈ 40.99 m |
| 34 | Apply Pythagoras to find a diagonal of a square pyramid: base side 8, apex height 12 above the centre. Find the slant distance from a base vertex to the apex. | $\sqrt{144 + 32} = \sqrt{176} \approx 13.27$ |
| 35 | Rationalise: $\dfrac{1}{1 + \sqrt{2}}$. | $\sqrt{2} - 1$ |
| 36 | An isosceles triangle has equal sides 13 cm and base 10 cm. Find (a) the perpendicular height, (b) the area. | (a) 12 cm (b) 60 cm² |
| 37 | Find the area of an equilateral triangle of side $s = 10$ cm. | $25\sqrt{3} \approx 43.3$ cm² |
| 38 | A right-angled triangle has legs $a$ and $b$ with $a + b = 14$ and $a^2 + b^2 = 100$. Find $a$ and $b$. | $a = 6, b = 8$ (or swap) |
| 39 | Two ladders lean against opposite walls of an alley. One reaches 6 m up the left wall; the other 8 m up the right wall. The alley is 5 m wide. They cross at some height $h$. Find $h$ (use similar triangles). | $h = \dfrac{6 \cdot 8}{6 + 8} = \dfrac{48}{14} \approx 3.43$ m |
| 40 | A triangle has sides $\sqrt{8}$, $\sqrt{18}$, $\sqrt{32}$. Show it is right-angled. | Yes — $8 + 18 = 26$? No: $8 + 18 = 26 \neq 32$. So not right-angled this way. Try: longest side $\sqrt{32}$ → squared 32. Other squares 8 + 18 = 26. Not equal. So NOT right-angled. |
Pack B — Answers
| # | Question | Answer |
|---|---|---|
| 1 | A right-angled triangle has the two shorter sides of length 6 cm and 8 cm. Find the hypotenuse. | 10 cm |
| 2 | Simplify $\sqrt{72}$ fully. | $6\sqrt{2}$ |
| 3 | Simplify $\sqrt{45}$ fully. | $3\sqrt{5}$ |
| 4 | A triangle has sides 7, 24, 25 cm. Use Pythagoras' converse to decide whether it is right-angled. | Yes |
| 5 | Simplify $\sqrt{5} \times \sqrt{20}$. | 10 |
| 6 | Identify which numbers are rational vs irrational: $5, \sqrt{2}, \pi, 0.\overline{3}, \sqrt{4}$. | Same. |
| 7 | Estimate $\sqrt{50}$ to 1 d.p. | ≈ 5.5 |
| 8 | Find the hypotenuse of a right-angled triangle with legs 9 and 12 cm. | 17 cm |
| 9 | Add: $\sqrt{50} + \sqrt{18}$ (give simplified form). | $5\sqrt{3}$ |
| 10 | A right-angled triangle has hypotenuse 13 cm and one shorter side 5 cm. Find the other side. | 12 cm |
| 11 | Pythagoras with surds: legs $2\sqrt{3}$ and 3. Find the hypotenuse. | $\sqrt{34}$ |
| 12 | Find the distance between $(1, 2)$ and $(4, 6)$. Give answer as a surd in simplest form. | $5\sqrt{2}$ |
| 13 | Simplify $\sqrt{8} + \sqrt{32} - \sqrt{18}$. | $7\sqrt{3}$ |
| 14 | Find the hypotenuse of a right-angled triangle with legs 5 and 10 cm. Give answer as a simplified surd. | $5\sqrt{5}$ |
| 15 | Simplify $\sqrt{75} \div \sqrt{3}$. | 10 |
| 16 | A right-angled triangle has hypotenuse $\sqrt{50}$ and one leg $\sqrt{18}$. Find the other leg. | $\sqrt{40} = 2\sqrt{10}$ |
| 17 | Find the distance between $(-2, 1)$ and $(3, 13)$. | 10 |
| 18 | Find the perimeter of a right-angled triangle with legs 6 and 8 cm. | 30 cm |
| 19 | Square the surd: $(2 + \sqrt{3})^2$. Expand. | $11 - 6\sqrt{2}$ |
| 20 | Rationalise the denominator: $\dfrac{6}{\sqrt{3}}$. | $2\sqrt{5}$ |
| 21 | A right-angled triangle has shorter sides 5 and 10. Hypotenuse as a simplified surd. | $5\sqrt{5}$ |
| 22 | Find the length of the space diagonal of a cuboid with sides 8, 6, 4. Use 3D Pythagoras. | $\sqrt{116}$ ≈ 10.77 cm |
| 23 | Pythagoras in 3D: cuboid 12 × 9 × 6. Find the space diagonal. | ≈ 18.38 |
| 24 | A square has diagonal 10 cm. Find its side length and area. | Side $7\sqrt{2}$; area 98 |
| 25 | A ladder of length 10 m leans against a wall. Foot 3 m from the base. Find the height reached. | 12 m |
| 26 | Simplify $(3 + \sqrt{2})(2 - \sqrt{2})$. | $7 - 3\sqrt{3}$ |
| 27 | The area of a square is 12 cm². Find the side as a simplified surd. | $4\sqrt{3}$ |
| 28 | Find the length of the diagonal of a rectangle 6 cm by 8 cm. | 13 cm |
| 29 | Show that the triangle with sides 5, 5, $5\sqrt{2}$ is right-angled. | Same approach. |
| 30 | A ladder of length 8 m makes a 60° angle with the ground. Find the height up the wall (using trig). | 4 m |
| 31 | A cuboid has length 8 cm, width 6 cm, height 4 cm. Find the angle between the space diagonal and the base, to 1 d.p. | ≈ 21.8° |
| 32 | Make $a$ the subject of $c = \sqrt{a^2 + b^2}$. | Same. |
| 33 | A cliff of height $h$ m stands on level ground. From point A 28 m from the base, the line of sight to the top is 52 m. From B further away, sight to top is 60 m. Find $h$ and B's distance. | $h = 45$, B ≈ 46.90 |
| 34 | Apply Pythagoras to find a diagonal of a square pyramid: base side 8, apex height 12 above the centre. Find the slant distance from a base vertex to the apex. | $\sqrt{144 + 50} = \sqrt{194} \approx 13.93$ |
| 35 | Rationalise: $\dfrac{1}{1 + \sqrt{2}}$. | $\sqrt{5} + 2$ |
| 36 | An isosceles triangle has equal sides 13 cm and base 10 cm. Find (a) the perpendicular height, (b) the area. | (a) 24 cm (b) 168 cm² |
| 37 | Find the area of an equilateral triangle of side $s = 10$ cm. | $9\sqrt{3} \approx 15.59$ cm² |
| 38 | A right-angled triangle has legs $a$ and $b$ with $a + b = 14$ and $a^2 + b^2 = 100$. Find $a$ and $b$. | $a = 5, b = 12$ |
| 39 | Two ladders lean against opposite walls of an alley. One reaches 6 m up the left wall; the other 8 m up the right wall. The alley is 5 m wide. They cross at some height $h$. Find $h$ (use similar triangles). | Same. |
| 40 | A triangle has sides $\sqrt{8}$, $\sqrt{18}$, $\sqrt{32}$. Show it is right-angled. | Check. |
Problems — Worked Solutions
**Leaning ladder.** A ladder of length 10 m leans against a vertical wall. The foot of the ladder is 3 m from the base. (a) Sketch the situation. (b) Find the height reached, to 2 d.p. (c) If the wall is 12 m tall, find the distance from the top of the ladder to the top of the wall.
(b) ≈ 9.54 m (c) ≈ 2.46 m
**Pythagorean triples.** Pythagorean triples are sets of three positive integers $(a, b, c)$ satisfying $a^2 + b^2 = c^2$. (a) Verify $(3, 4, 5)$, $(5, 12, 13)$, $(8, 15, 17)$ are triples. (b) Show $(7, 24, 25)$ is also a triple. (c) Multiply $(3, 4, 5)$ by 7 to find another triple. (d) Show that if $(a, b, c)$ is a triple, then so is $(ka, kb, kc)$ for any positive integer $k$.
(a) All ✓ (b) ✓ (c) (21, 28, 35) (d) Multiply each by $k$
**3D Pythagoras.** A cuboid has length 4, width 3, height 2 m. (a) Find the diagonal of the base. (b) Find the space diagonal (corner to opposite corner). (c) Find the angle between the space diagonal and the base.
(a) 5 m (b) $\sqrt{29} \approx 5.39$ m (c) ≈ 21.8°
**Surds simplification.** Simplify each: (a) $\sqrt{72}$ (b) $\sqrt{12} + \sqrt{27}$ (c) $\sqrt{50} \times \sqrt{2}$ (d) $\dfrac{8}{\sqrt{2}}$ (rationalised)
(a) $6\sqrt{2}$ (b) $5\sqrt{3}$ (c) 10 (d) $4\sqrt{2}$
**Diagonal of a cube.** A cube has side length $a$. (a) Find the face diagonal in terms of $a$. (b) Find the space diagonal. (c) For $a = 5$ cm, compute both diagonals.
(a) $a\sqrt{2}$ (b) $a\sqrt{3}$ (c) Face $5\sqrt{2} \approx 7.07$; space $5\sqrt{3} \approx 8.66$
**The fish tank.** A fish tank is 60 × 40 × 30 cm. A diagonal frame is built from corner to opposite corner. (a) Find the length of the frame. (b) Find the angle of the frame above the bottom. (c) Find the lengths needed for two diagonals across each face.
(a) $\sqrt{6100} \approx 78.1$ cm (b) ≈ 22.4° (c) Front face: $\sqrt{4500} \approx 67.1$; side face: $\sqrt{2500} = 50$; top: $\sqrt{5200} \approx 72.1$
**Triangle perimeter with surds.** A triangle has vertices $A(0, 0), B(3, 4), C(0, 4)$. (a) Find each side. (b) Show one side is a Pythagorean triple-like length. (c) Find the perimeter.
(a) AB = 5, BC = 3, AC = 4 (b) 3-4-5 triple (c) 12
**Diagonal of a rectangular field.** A rectangular field 60 m × 80 m has a diagonal path. (a) Find the path length. (b) Find the angle the path makes with the longer side. (c) The field is enlarged by a scale factor of 1.5. Find the new diagonal length.
(a) 100 m (b) ≈ 36.9° (c) 150 m
**Surds in algebra.** A square has area $50$ cm². (a) Find the side length as a simplified surd. (b) Find the diagonal of the square. (c) Find the perimeter.
(a) $5\sqrt{2}$ cm (b) 10 cm (c) $20\sqrt{2}$ cm
**Application: angle of elevation.** A pole of height 12 m casts a shadow 5 m long. (a) Find the distance from the tip of the shadow to the top of the pole. (b) Find the angle of elevation of the sun. (c) Identify any Pythagorean triple.
(a) 13 m (b) ≈ 67.4° (c) 5-12-13
**Rationalising denominators.** Rationalise each: (a) $\dfrac{5}{\sqrt{2}}$ (b) $\dfrac{1}{\sqrt{3} - 1}$ (c) $\dfrac{3}{\sqrt{5} + \sqrt{2}}$
(a) $\tfrac{5\sqrt{2}}{2}$ (b) $\tfrac{\sqrt{3} + 1}{2}$ (c) $\sqrt{5} - \sqrt{2}$
**Pythagoras in space.** A box has dimensions $a, b, c$. The space diagonal $d$ satisfies $d^2 = a^2 + b^2 + c^2$. (a) Express $d$ in surd form for $a = 1, b = 2, c = 2$. (b) Find the smallest cube containing a stick of length 10 m. (c) Investigate: can the space diagonal of a cube equal twice the side length?
(a) $d = 3$ (b) Side $\approx 5.77$ m (c) No — would require side $= s\sqrt{3}/2 \approx 0.577 s$, contradicting $s = $ side