High School Maths · Year 2 · Unit 15 · Lesson 5 of 5

Scale Drawings and Real-Life Loci: From Maps to Masts

About 15 minutes · scales written as 1 : n and "1 cm to …" · map distances both ways · areas from plans · bearings on scale drawings · a dog on a lead, a goat by a wall, a mast's signal range · answers to 3 significant figures · calculator allowed · exit quiz

Hook

On a map with a scale of 1 : 25,000, two schools are 6 cm apart.

How far apart are they in real life?

Each 1 cm on the map stands for 25,000 cm. So 6 cm stands for 6 × 25,000 = 150,000 cm = 1,500 m = 1.5 km. In Year 1 you used bearings; now you will put bearings, scales and loci together, as surveyors, sailors and phone engineers do.

Learn

Master the essentials

Left: half a disc, ½πr². Right: three-quarters of a disc, ¾πr².

Scale 1 : n

Every 1 cm on the drawing stands for n cm in real life.

1 : 500 means 1 cm stands for 500 cm = 5 m

Drawing to real life

Multiply by n, then change the units.

7 cm × 20,000 = 140,000 cm = 1.4 km

Real life to drawing

Change the real length to centimetres first, then divide by n.

1 km = 1,000 m = 100,000 cm

"1 cm to …"

Write the real length in centimetres to get the 1 : n form.

1 cm to 2 km is 1 : 200,000

Areas

Lengths are multiplied by n, so areas are multiplied by n². It is often easier to find the real lengths first.

1 : 100: 1 cm² stands for 1 m²

Bearings on a scale drawing

Draw a north line. Measure the bearing clockwise with a protractor, then mark the length using the scale.

Accurate to 2° and 2 mm

A dog on a lead

Tied to a post: a disc. Tied to a ring that slides along a straight rail: a rectangle with a semicircle at each end.

The "running track" of Lesson 3, filled in

A goat by a wall

Tied to a point on a long wall: half a disc. Tied at the outside corner of a building: three-quarters of a disc.

½πr² and ¾πr²

Worked examples

See it done

1. A map has a scale of 1 : 50,000. Two villages are 7.2 km apart. How far apart are they on the map?

  1. Change to centimetres: 7.2 km = 7,200 m = 720,000 cm.
  2. Divide by the scale: 720,000 ÷ 50,000 = 14.4.
  3. They are 14.4 cm apart on the map.

14.4 cm

2. A ship sails 12 km due north from P to Q, then 9 km due east from Q to R. Make a scale drawing with 1 cm to 3 km. Find the distance PR and the bearing of R from P.

  1. PQ: 12 ÷ 3 = 4 cm, drawn due north. QR: 9 ÷ 3 = 3 cm, drawn due east.
  2. Measure PR on the drawing: 5 cm. (Check: √(4² + 3²) = 5.)
  3. Real distance: 5 × 3 = 15 km.
  4. Angle QPR = tan⁻¹(9 ÷ 12) = 36.87…°, so the bearing of R from P is 037° (to the nearest degree).

PR = 15 km, on a bearing of 037°

3. A goat is tied by a rope 7 m long to a post on a long straight wall. Find the area the goat can reach, correct to 3 significant figures.

  1. The wall stops the goat, so it reaches half a disc of radius 7 m.
  2. Area = ½ × π × 7² = 24.5π = 76.969… m².
  3. To 3 significant figures: 77.0 m². (Keep the 0: it is the third figure.)

77.0 m²

Common mistake: Mixing units. Change to the same unit before you divide by the scale, and remember 1 km = 100,000 cm. On a 1 : 25,000 map, 4 cm stands for 100,000 cm = 1 km, not 100 km. And round only at the very end: keep the full calculator value until the last step.

PractiseCalculator allowed

10 questions with instant feedback

Choose the correct answer. A calculator is allowed. Check the units at every step, and round only at the end. Questions 1–3 are warm-ups, 4–7 are at exam standard and 8–10 are challenge questions. After you answer, the feedback shows the full solution.

Score: 0 / 10Answered: 0 / 10

Warm-up (The basic skill: one or two steps.)

1.A house plan has a scale of 1 : 200. A wall is 3.5 cm long on the plan. How long is the real wall?

Not quite. This changes cm to m by dividing by 10. There are 100 cm in a metre.

Solution

  1. Real length = 3.5 × 200 = 700 cm.
  2. 700 cm ÷ 100 = 7 m.

Answer: C. 7 m.

Not quite. This divides by 1,000. There are 100 cm in a metre.

Solution

  1. Real length = 3.5 × 200 = 700 cm.
  2. 700 cm ÷ 100 = 7 m.

Answer: C. 7 m.

Correct.

Solution

  1. Real length = 3.5 × 200 = 700 cm.
  2. 700 cm ÷ 100 = 7 m.

Not quite. This is the length in centimetres. Change it to metres.

Solution

  1. Real length = 3.5 × 200 = 700 cm.
  2. 700 cm ÷ 100 = 7 m.

Answer: C. 7 m.

2.A map has a scale of 1 : 40,000. Two towns are 9 km apart. How far apart are they on the map?

Correct.

Solution

  1. 9 km = 9 × 100,000 = 900,000 cm.
  2. Map distance = 900,000 ÷ 40,000 = 22.5 cm.

Not quite. This uses 1 km = 10,000 cm. In fact 1 km = 100,000 cm.

Solution

  1. 9 km = 9 × 100,000 = 900,000 cm.
  2. Map distance = 900,000 ÷ 40,000 = 22.5 cm.

Answer: A. 22.5 cm.

Not quite. This uses 1 km = 1,000,000 cm. In fact 1 km = 100,000 cm.

Solution

  1. 9 km = 9 × 100,000 = 900,000 cm.
  2. Map distance = 900,000 ÷ 40,000 = 22.5 cm.

Answer: A. 22.5 cm.

Not quite. This multiplies by the scale. Going from real life to the map, divide.

Solution

  1. 9 km = 9 × 100,000 = 900,000 cm.
  2. Map distance = 900,000 ÷ 40,000 = 22.5 cm.

Answer: A. 22.5 cm.

3.On a scale drawing with a scale of 1 cm to 5 km, PQ measures 6.4 cm and the bearing of Q from P is 125°. Find the real distance PQ and the bearing of P from Q.

Not quite. This is the bearing of Q from P. From Q back to P, add 180°.

Solution

  1. Real distance: 6.4 × 5 = 32 km.
  2. Back bearing: 125° + 180° = 305°.

Answer: D. 32 km, on a bearing of 305°.

Not quite. This divides 6.4 by 5. Each 1 cm stands for 5 km, so multiply.

Solution

  1. Real distance: 6.4 × 5 = 32 km.
  2. Back bearing: 125° + 180° = 305°.

Answer: D. 32 km, on a bearing of 305°.

Not quite. This subtracts 125° from 360°. A back bearing differs by 180°.

Solution

  1. Real distance: 6.4 × 5 = 32 km.
  2. Back bearing: 125° + 180° = 305°.

Answer: D. 32 km, on a bearing of 305°.

Correct.

Solution

  1. Real distance: 6.4 × 5 = 32 km.
  2. Back bearing: 125° + 180° = 305°.

Exam standard (Several steps, like real exam questions.)

4.A dog's lead is 3.5 m long. It is clipped to a ring that slides along a straight rail 5 m long, in the middle of a lawn. Find the area of lawn the dog can reach. Give your answer correct to 3 significant figures.

Not quite. This counts the rectangle on one side of the rail only. The dog reaches 3.5 m on EACH side.

Solution

  1. The region is a rectangle 5 m by 7 m, with a semicircle of radius 3.5 m at each end.
  2. Rectangle: 5 × 7 = 35 m². Two semicircles make one circle: π × 3.5² = 12.25π m².
  3. 35 + 12.25π = 73.484… = 73.5 m² (3 s.f.).

Answer: B. 73.5 m².

Correct.

Solution

  1. The region is a rectangle 5 m by 7 m, with a semicircle of radius 3.5 m at each end.
  2. Rectangle: 5 × 7 = 35 m². Two semicircles make one circle: π × 3.5² = 12.25π m².
  3. 35 + 12.25π = 73.484… = 73.5 m² (3 s.f.).

Not quite. This makes the ends square: a 12 m by 7 m rectangle. Round each end the dog sweeps a semicircle.

Solution

  1. The region is a rectangle 5 m by 7 m, with a semicircle of radius 3.5 m at each end.
  2. Rectangle: 5 × 7 = 35 m². Two semicircles make one circle: π × 3.5² = 12.25π m².
  3. 35 + 12.25π = 73.484… = 73.5 m² (3 s.f.).

Answer: B. 73.5 m².

Not quite. This uses 2πr, which is a length. Area needs πr².

Solution

  1. The region is a rectangle 5 m by 7 m, with a semicircle of radius 3.5 m at each end.
  2. Rectangle: 5 × 7 = 35 m². Two semicircles make one circle: π × 3.5² = 12.25π m².
  3. 35 + 12.25π = 73.484… = 73.5 m² (3 s.f.).

Answer: B. 73.5 m².

5.Two coastguard stations are 30 km apart, with B due east of A. A ship is on a bearing of 045° from A and on a bearing of 315° from B. How far is the ship from A? Give your answer correct to 3 significant figures.

Not quite. This is half of AB. The ship is not on AB.

Solution

  1. From A, the bearing 045° makes an angle of 90° − 45° = 45° with AB.
  2. From B, 315° is 45° west of north, so it also makes 45° with BA.
  3. The angle at the ship is 90°, and the two shorter sides are equal: 2 × AS² = 30² = 900.
  4. AS = √450 = 21.21… = 21.2 km (3 s.f.).

Answer: C. 21.2 km.

Not quite. This multiplies by √2. AB is the hypotenuse, so the shorter sides are smaller.

Solution

  1. From A, the bearing 045° makes an angle of 90° − 45° = 45° with AB.
  2. From B, 315° is 45° west of north, so it also makes 45° with BA.
  3. The angle at the ship is 90°, and the two shorter sides are equal: 2 × AS² = 30² = 900.
  4. AS = √450 = 21.21… = 21.2 km (3 s.f.).

Answer: C. 21.2 km.

Correct.

Solution

  1. From A, the bearing 045° makes an angle of 90° − 45° = 45° with AB.
  2. From B, 315° is 45° west of north, so it also makes 45° with BA.
  3. The angle at the ship is 90°, and the two shorter sides are equal: 2 × AS² = 30² = 900.
  4. AS = √450 = 21.21… = 21.2 km (3 s.f.).

Not quite. This assumes an equilateral triangle. The angles at A and B are 45°, not 60°.

Solution

  1. From A, the bearing 045° makes an angle of 90° − 45° = 45° with AB.
  2. From B, 315° is 45° west of north, so it also makes 45° with BA.
  3. The angle at the ship is 90°, and the two shorter sides are equal: 2 × AS² = 30² = 900.
  4. AS = √450 = 21.21… = 21.2 km (3 s.f.).

Answer: C. 21.2 km.

6.The table shows the signal range of three masts and their distances from a farm. The farm receives a mast's signal if it is within that mast's range. Which masts can the farm receive?

MastSignal range (km)Distance from the farm (km)
A129.5
B88.4
C1514.9

Not quite. Mast B reaches 8 km, but the farm is 8.4 km away.

Solution

  1. Compare each distance with the range.
  2. A: 9.5 ≤ 12 ✓. B: 8.4 > 8 ✗. C: 14.9 ≤ 15 ✓.
  3. The farm receives A and C only.

Answer: B. A and C only.

Correct.

Solution

  1. Compare each distance with the range.
  2. A: 9.5 ≤ 12 ✓. B: 8.4 > 8 ✗. C: 14.9 ≤ 15 ✓.
  3. The farm receives A and C only.

Not quite. Mast C also reaches it: 14.9 km is less than 15 km.

Solution

  1. Compare each distance with the range.
  2. A: 9.5 ≤ 12 ✓. B: 8.4 > 8 ✗. C: 14.9 ≤ 15 ✓.
  3. The farm receives A and C only.

Answer: B. A and C only.

Not quite. Mast A also reaches it: 9.5 km is less than 12 km.

Solution

  1. Compare each distance with the range.
  2. A: 9.5 ≤ 12 ✓. B: 8.4 > 8 ✗. C: 14.9 ≤ 15 ✓.
  3. The farm receives A and C only.

Answer: B. A and C only.

7.A rectangular field is 4.5 cm by 3.2 cm on a plan with a scale of 1 : 2,000. Which working gives its real area?

Not quite. This multiplies the AREA by 2,000. Areas are multiplied by 2,000², so find the real lengths first.

Solution

  1. Real lengths: 4.5 × 2,000 = 9,000 cm = 90 m, and 3.2 × 2,000 = 6,400 cm = 64 m.
  2. Area = 90 × 64 = 5,760 m².

Answer: D. 90 m × 64 m = 5,760 m².

Not quite. This answer is in cm², not m². Divide by 10,000.

Solution

  1. Real lengths: 4.5 × 2,000 = 9,000 cm = 90 m, and 3.2 × 2,000 = 6,400 cm = 64 m.
  2. Area = 90 × 64 = 5,760 m².

Answer: D. 90 m × 64 m = 5,760 m².

Not quite. This is the perimeter of the field, in metres.

Solution

  1. Real lengths: 4.5 × 2,000 = 9,000 cm = 90 m, and 3.2 × 2,000 = 6,400 cm = 64 m.
  2. Area = 90 × 64 = 5,760 m².

Answer: D. 90 m × 64 m = 5,760 m².

Correct.

Solution

  1. Real lengths: 4.5 × 2,000 = 9,000 cm = 90 m, and 3.2 × 2,000 = 6,400 cm = 64 m.
  2. Area = 90 × 64 = 5,760 m².

Challenge (Harder problems that mix ideas or need reasoning.)

8.A goat is tied by a rope 5 m long to the outside corner of a large barn. The walls meet at 90° and each is longer than 5 m. Find the area the goat can reach. Give your answer correct to 3 significant figures.

Not quite. This is a quarter of a disc. The barn blocks only one quarter, so the goat reaches three-quarters.

Solution

  1. The barn fills one quarter of the circle round the corner.
  2. The goat reaches three-quarters of a disc: ¾ × π × 5² = 18.75π.
  3. 18.75π = 58.904… = 58.9 m² (3 s.f.).

Answer: C. 58.9 m².

Not quite. This is the whole disc. The barn blocks one quarter of it.

Solution

  1. The barn fills one quarter of the circle round the corner.
  2. The goat reaches three-quarters of a disc: ¾ × π × 5² = 18.75π.
  3. 18.75π = 58.904… = 58.9 m² (3 s.f.).

Answer: C. 58.9 m².

Correct.

Solution

  1. The barn fills one quarter of the circle round the corner.
  2. The goat reaches three-quarters of a disc: ¾ × π × 5² = 18.75π.
  3. 18.75π = 58.904… = 58.9 m² (3 s.f.).

Not quite. This is half a disc, as for a goat tied to a straight wall.

Solution

  1. The barn fills one quarter of the circle round the corner.
  2. The goat reaches three-quarters of a disc: ¾ × π × 5² = 18.75π.
  3. 18.75π = 58.904… = 58.9 m² (3 s.f.).

Answer: C. 58.9 m².

9.On a map, a road 2.4 km long is 16 cm long. What is the scale of the map, in the form 1 : n?

Not quite. This uses 1 km = 1,000 cm. In fact 1 km = 100,000 cm.

Solution

  1. 2.4 km = 240,000 cm.
  2. 16 cm on the map stands for 240,000 cm, so 1 cm stands for 240,000 ÷ 16 = 15,000 cm.
  3. The scale is 1 : 15,000.

Answer: D. 1 : 15,000.

Not quite. This uses 1 km = 10,000 cm. In fact 1 km = 100,000 cm.

Solution

  1. 2.4 km = 240,000 cm.
  2. 16 cm on the map stands for 240,000 cm, so 1 cm stands for 240,000 ÷ 16 = 15,000 cm.
  3. The scale is 1 : 15,000.

Answer: D. 1 : 15,000.

Not quite. This is the real length in centimetres. Divide it by the 16 cm on the map.

Solution

  1. 2.4 km = 240,000 cm.
  2. 16 cm on the map stands for 240,000 cm, so 1 cm stands for 240,000 ÷ 16 = 15,000 cm.
  3. The scale is 1 : 15,000.

Answer: D. 1 : 15,000.

Correct.

Solution

  1. 2.4 km = 240,000 cm.
  2. 16 cm on the map stands for 240,000 cm, so 1 cm stands for 240,000 ÷ 16 = 15,000 cm.
  3. The scale is 1 : 15,000.

10.A mast stands 12 km from a straight coastline. Its signal reaches 20 km. How long is the stretch of coastline that receives the signal?

Correct.

Solution

  1. The perpendicular from the mast to the coast bisects the stretch that is in range.
  2. Half-length² = 20² − 12² = 400 − 144 = 256, so half the stretch is 16 km.
  3. The stretch is 2 × 16 = 32 km.

Not quite. This is only half of the stretch, on one side of the mast.

Solution

  1. The perpendicular from the mast to the coast bisects the stretch that is in range.
  2. Half-length² = 20² − 12² = 400 − 144 = 256, so half the stretch is 16 km.
  3. The stretch is 2 × 16 = 32 km.

Answer: A. 32 km.

Not quite. This is the diameter of the signal circle. The coast does not pass through the mast.

Solution

  1. The perpendicular from the mast to the coast bisects the stretch that is in range.
  2. Half-length² = 20² − 12² = 400 − 144 = 256, so half the stretch is 16 km.
  3. The stretch is 2 × 16 = 32 km.

Answer: A. 32 km.

Not quite. This adds the squares. The 20 km line is the hypotenuse, so subtract.

Solution

  1. The perpendicular from the mast to the coast bisects the stretch that is in range.
  2. Half-length² = 20² − 12² = 400 − 144 = 256, so half the stretch is 16 km.
  3. The stretch is 2 × 16 = 32 km.

Answer: A. 32 km.

Apply

Plan a rescue

Two coastguard stations: B is 40 km due east of A. A ship in trouble is within 25 km of A and nearer to B than to A. Use a ruler, compasses and a calculator, then compare with the model.

1. Using a scale of 1 cm to 5 km, how long is AB on your drawing?

2. Describe the two loci you draw, and shade the region where the ship can be.

3. The ship is on the line AB. Between which distances from A can it be?

  1. 40 ÷ 5 = 8 cm.
  1. Within 25 km of A: a circle, centre A, radius 25 ÷ 5 = 5 cm (solid), and its inside.
  2. Nearer to B: the side of the perpendicular bisector of AB that contains B (bisector dashed).
  3. Shade the part of the circle on B's side of the bisector.
  1. The bisector crosses AB 20 km from A, and the circle crosses AB 25 km from A.
  2. The ship is more than 20 km and at most 25 km from A.

Next: the Lesson 5 exit quiz, then the unit test. The unit test has 18 multiple-choice questions (no calculator, answers with π left in terms of π) and five structured problems (a calculator is allowed in two of them), 45 marks in all. Bring a ruler, compasses and a protractor. Allow 75 minutes.

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