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Criterion CD Assessment — 14 April

Pythagoras
& Trigonometry

A formative assessment covering all five assessment topics. Answer each question and study the worked solution before moving on.

01 — Pythagoras Theorem 02 — Applications of Pythagoras 03 — Trig Ratios 04 — Elevation & Depression 05 — Bearings
Begin

TOPIC 01

Pythagoras' Theorem

Finding missing sides in right-angled triangles using a² + b² = c²

The Core Formula

In any right-angled triangle, the square of the hypotenuse (the longest side, opposite the right angle) equals the sum of the squares of the other two sides.

a² + b² = c²

Where c is always the hypotenuse. To find a missing shorter side: a² = c² − b²

a b c (hyp)

Right-angled triangle

Area visualisation

Interactive Explorer

Adjust the two shorter sides and watch the hypotenuse update in real time.

c = √(3² + 4²) = 5.000
a² + b² = 25
Q01 A right-angled triangle has legs of 5 cm and 12 cm. What is the length of the hypotenuse?
12 cm 5 cm ?
Worked Solution

Apply Pythagoras: a² + b² = c²

5² + 12² = c²25 + 144 = c²169 = c²

c = √169 = 13 cm

This is a classic 5-12-13 Pythagorean triple — worth memorising.

Q02 The hypotenuse of a right-angled triangle is 25 cm and one leg is 24 cm. Find the other leg.
Worked Solution

Finding a shorter side: rearrange to a² = c² − b²

a² = 25² − 24²a² = 625 − 576 = 49

a = √49 = 7 cm

7-24-25 is another Pythagorean triple.

Q03 Is a triangle with sides 9 cm, 40 cm, 41 cm a right-angled triangle? Select the correct answer.
Worked Solution

Check: 9² + 40² = 81 + 1600 = 1681

41² = 1681 — these are equal, confirming it is right-angled.

The converse of Pythagoras' theorem states: if a² + b² = c², the triangle is right-angled.

Q04 A rectangle is 9 cm wide and 12 cm tall. Calculate the length of the diagonal. Give your answer to 1 decimal place.
Worked Solution

The diagonal forms the hypotenuse of a right triangle with legs 9 and 12.

d² = 9² + 12² = 81 + 144 = 225

d = √225 = 15.0 cm

Q05 An isosceles triangle has two equal sides of 10 cm and a base of 8 cm. Calculate its height h (to 1 d.p.).
10 cm 10 cm 8 cm h
Worked Solution

The height bisects the base, so you have a right triangle with hypotenuse 10 cm and base 4 cm.

h² + 4² = 10²h² = 100 − 16 = 84

h = √84 ≈ 9.2 cm

Q06 A boat travels 8 km east then 6 km north. How far is it from its starting point?
Worked Solution

East and North are perpendicular, so this is a right triangle.

d² = 8² + 6² = 64 + 36 = 100d = 10 km

This is the 6-8-10 Pythagorean triple (a scaled 3-4-5).

Q07 A right-angled triangle has sides of 11 cm and 60 cm. The hypotenuse is 61 cm. Verify this is right-angled by selecting the correct check.
Worked Solution

11² = 121, 60² = 3600, so 11² + 60² = 3721

61² = 3721 — they match, confirming a right angle.

Q08 The perimeter of a right-angled triangle is 30 cm. The hypotenuse is √(a²+b²). If the legs are 5 cm and 12 cm, what is the perimeter? (Enter to 1 d.p.)
Worked Solution

First find hypotenuse: c = √(5² + 12²) = √169 = 13 cm

Perimeter = 5 + 12 + 13 = 30 cm

Q09 A diagonal wall brace is needed across a rectangular frame 2.4 m wide and 3.2 m tall. How long is the brace? (to 1 d.p.)
Worked Solution

d² = 2.4² + 3.2² = 5.76 + 10.24 = 16

d = √16 = 4.0 m

Q10 Two points are plotted at A(0, 0) and B(6, 8). What is the distance AB?
Worked Solution

Use the distance formula (derived from Pythagoras): d = √((x₂−x₁)² + (y₂−y₁)²)

d = √(6² + 8²) = √(36 + 64) = √100 = 10

Q11 A TV screen is 40 inches wide and 30 inches tall. What is the diagonal screen size?
Worked Solution

d² = 40² + 30² = 1600 + 900 = 2500

d = √2500 = 50 inches — a 50-inch TV!

Q12 In triangle XYZ, XY = 15 cm, YZ = 20 cm, and the angle at Y is 90°. Find XZ to 1 d.p.
Worked Solution

XZ is the hypotenuse since the right angle is at Y.

XZ² = 15² + 20² = 225 + 400 = 625

XZ = √625 = 25.0 cm

Q13 A ladder 10 m long leans against a wall. The foot is 4 m from the base. How high up the wall does it reach? (to 1 d.p.)
Worked Solution

The ladder is the hypotenuse (10 m), base leg is 4 m.

h² = 10² − 4² = 100 − 16 = 84

h = √84 ≈ 9.2 m

Q14 Which of the following sets of three numbers is NOT a Pythagorean triple?
Worked Solution

Check D: 6² + 7² = 36 + 49 = 85, but 9² = 81 ≠ 85. Not a triple.

A: 9+16=25 ✓ | B: 25+144=169 ✓ | C: 64+225=289 ✓

Q15 A square has a diagonal of 10 cm. What is the side length? Leave your answer to 2 d.p.
Worked Solution

If side = s, then diagonal d = s√2, so 10 = s√2

s = 10/√2 = 10√2/2 ≈ 7.07 cm

Alternatively: s² + s² = 10²2s² = 100s = √50 ≈ 7.07

Q16 The area of a right-angled triangle is 60 cm². One leg is 10 cm. Find the hypotenuse. (to 1 d.p.)
Worked Solution

Area = ½ × base × height → 60 = ½ × 10 × bb = 12 cm

c² = 10² + 12² = 100 + 144 = 244

c = √244 ≈ 15.6 cm — accept ≈15.6 to 16.0

Q17 Two diagonals of a rhombus are 16 cm and 12 cm. Find the side of the rhombus.
Worked Solution

The diagonals of a rhombus bisect each other at right angles. Half-diagonals: 8 cm and 6 cm.

side² = 8² + 6² = 64 + 36 = 100side = 10 cm

Q18 Point A is at (1, 2) and point B is at (5, 5). Find the distance AB to 3 d.p.
Worked Solution

d = √((5−1)² + (5−2)²) = √(16 + 9) = √25 = 5.000

Q19 A right triangle has hypotenuse √97 cm and one leg √3 cm. Find the other leg as a surd.
Worked Solution

Using a² = c² − b²: a² = (√97)² − (√3)² = 97 − 3 = 94

a = √94 cm

Q20 An equilateral triangle has a side of 6 cm. Find its height to 2 d.p.
Worked Solution

The height bisects the base. Right triangle: hypotenuse = 6, base = 3.

h² = 6² − 3² = 36 − 9 = 27h = √27 ≈ 5.20 cm

TOPICS 02 – 05

Trigonometry

SOH CAH TOA, angles of elevation & depression, and bearings

SOHCAHTOA

In a right-angled triangle, label sides relative to the angle θ:

sin θ = Opposite / Hypotenuse
cos θ = Adjacent / Hypotenuse
tan θ = Opposite / Adjacent
θ Adjacent Opposite Hypotenuse

Finding Angles vs Sides

To find a missing side: rearrange the trig ratio, e.g. Opp = Hyp × sin θ

To find a missing angle: use the inverse trig function, e.g. θ = sin⁻¹(Opp/Hyp)

Finding a side: multiply  |  Finding an angle: inverse (sin⁻¹, cos⁻¹, tan⁻¹)

Elevation, Depression & Bearings

Angle of elevation — measured upward from horizontal to a higher object.

Angle of depression — measured downward from horizontal to a lower object.

Bearings — measured clockwise from North (always 3 digits, e.g. 045°, 270°).

elevation depression observer observer 060° N S W E

Ratio Explorer

Adjust the angle to see how sin, cos, and tan change.

sin 30° = 0.500
cos 30° = 0.866
tan 30° = 0.577
T01 In a right-angled triangle, the hypotenuse is 15 cm and one angle is 30°. Find the side opposite the 30° angle.
Worked Solution

We need Opposite, given Hypotenuse and angle → use SOH.

sin 30° = Opp / HypOpp = 15 × sin 30° = 15 × 0.5 = 7.5 cm

T02 A right-angled triangle has the adjacent side = 8 cm and angle θ = 40°. Find the hypotenuse to 2 d.p.
Worked Solution

Adjacent and Hypotenuse → use CAH.

cos 40° = Adj / HypHyp = 8 / cos 40° = 8 / 0.766 ≈ 10.44 cm

T03 The opposite side is 5 cm and the adjacent side is 12 cm. What is angle θ? (to 1 d.p.)
Worked Solution

We have Opposite and Adjacent → use TOA.

tan θ = 5 / 12 = 0.4167

θ = tan⁻¹(0.4167) ≈ 22.6°

T04 Which ratio should you use to find the missing side x, given an angle of 50° and the hypotenuse of 20 cm, where x is the adjacent side?
Worked Solution

x is the adjacent side, and we know the hypotenuse → use CAH.

cos 50° = Adjacent / Hypotenuse = x / 20

Therefore x = 20 × cos 50° ≈ 12.86 cm

T05 A flagpole casts a shadow of 12 m. The angle of elevation of the sun is 35°. Find the height of the flagpole to 1 d.p.
35° 12 m h
Worked Solution

The shadow is the adjacent side, the pole height is opposite, angle = 35°.

tan 35° = h / 12h = 12 × tan 35° ≈ 12 × 0.700 ≈ 8.4 m

T06 From a cliff 80 m high, the angle of depression to a boat is 20°. How far is the boat from the base of the cliff? (to 1 d.p.)
Worked Solution

Angle of depression from horizontal = 20°. The angle in the right triangle is also 20°.

Cliff height = opposite (80 m), distance = adjacent.

tan 20° = 80 / dd = 80 / tan 20° ≈ 80 / 0.364 ≈ 219.8 m

T07 Find the angle of elevation to the top of a building 50 m tall, viewed from a point 40 m away at ground level. (to 1 d.p.)
Worked Solution

Opposite = 50 m (height), Adjacent = 40 m.

tan θ = 50 / 40 = 1.25θ = tan⁻¹(1.25) ≈ 51.3°

T08 A ship sails on a bearing of 090° for 30 km, then on a bearing of 000° for 40 km. What is the direct distance back to the start?
Worked Solution

090° is East (30 km), 000° is North (40 km). These are perpendicular.

d² = 30² + 40² = 900 + 1600 = 2500d = 50 km

T09 In triangle PQR, angle R = 90°, PQ = 26 cm, QR = 10 cm. Find angle P to 1 d.p.
Worked Solution

PQ = 26 cm (hypotenuse, opposite R), QR = 10 cm (opposite P → no, it's adjacent to P from R's side).

Relative to angle P: Opposite = QR = 10, Hypotenuse = PQ = 26.

sin P = 10/26 = 0.3846P = sin⁻¹(0.3846) ≈ 22.6°

T10 A ramp rises 3 m over a horizontal distance of 15 m. What is the angle of inclination? (to 1 d.p.)
Worked Solution

Opposite = 3 m (rise), Adjacent = 15 m (run).

tan θ = 3/15 = 0.2θ = tan⁻¹(0.2) ≈ 11.3°

T11 Town B is on a bearing of 045° from town A, 70 km away. How far East of A is B? (to 1 d.p.)
Worked Solution

Bearing 045° = NE direction. The East component is opposite to the 45° angle from North.

East = 70 × sin 45° ≈ 70 × 0.7071 ≈ 49.5 km

T12 The angle of depression from the top of a 120 m tower to a car is 25°. How far is the car from the base? (to 1 d.p.)
Worked Solution

The angle between the horizontal and the line of sight = 25° (depression = same as elevation from car).

tan 25° = 120/dd = 120/tan 25° ≈ 120/0.4663 ≈ 257.3 m

T13 Which bearing represents due West?
Worked Solution

Bearings go clockwise from North: N=000°, E=090°, S=180°, W=270°.

Due West = 270°.

T14 A plane flies on bearing 130° for 200 km. How far South of its start is it? (to 1 d.p.)
Worked Solution

Bearing 130° is 130° clockwise from North. Angle from South direction = 130° − 90° = 40° from East.

South component = 200 × cos 40° — wait, easier: angle from North = 130°.

South component = 200 × cos(130° − 90°) = 200 × cos 40°? No. South = 200 × cos(180°−130°) × (−1)

Use: South = 200 × sin(130°−90°) … . Actually: South component = 200 × cos(130°) ×(−1) = −200cos130° = 200 × sin40° ≈ 128.6 km

T15 From point A, point B is on bearing 060°. From B, what bearing is A? (back-bearing)
Worked Solution

The back-bearing (return bearing) = original bearing + 180°.

060° + 180° = 240°. If the result exceeds 360°, subtract 360°.

T16 A wire 18 m long is attached to the top of a pole and makes an angle of 55° with the ground. How tall is the pole? (to 1 d.p.)
Worked Solution

Wire = hypotenuse, pole height = opposite, angle = 55°.

sin 55° = h / 18h = 18 × sin 55° ≈ 18 × 0.819 ≈ 14.7 m

T17 An observer 1.8 m tall looks at the top of a tree that is 25 m away. The angle of elevation from eye level is 42°. How tall is the tree? (to 1 d.p.)
Worked Solution

Height from eye level: h = 25 × tan 42° ≈ 25 × 0.900 ≈ 22.5 m

Add observer's eye height: Total height = 22.5 + 1.8 = 24.3 m

T18 In a right-angled triangle, cos θ = 5/13. What is sin θ?
Worked Solution

cos θ = Adj/Hyp = 5/13, so Adj = 5, Hyp = 13.

By Pythagoras: Opp = √(13² − 5²) = √(169 − 25) = √144 = 12

sin θ = Opp/Hyp = 12/13

T19 From a helicopter 500 m above the sea, the angle of depression of a lifeboat is 15°. What is the horizontal distance? (to nearest metre)
Worked Solution

tan 15° = 500 / dd = 500 / tan 15° ≈ 500 / 0.2679 ≈ 1866 m

T20 Point C is 5 km North and 5 km East of point D. What is the bearing of C from D? (to nearest degree)
Worked Solution

Equal North and East distances → 45° from North, towards East.

Bearing = 045°. (Using tan⁻¹(5/5) = 45° from North, measured clockwise.)

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