Scale Drawings and Maps – Area Scale Factors | Pearson Edexcel International GCSE Maths

Scale drawings are useful for representing large real-life objects and areas using smaller measurements. When calculating an area from a scale drawing, the scale factor must be squared because area is two-dimensional.

Question – Area of a Park from a Scale Drawing

A map has a scale of 1 : 40 000. A rectangular park is shown on the map as measuring 5.8 cm by 4.4 cm.

Calculate the area of the park in real life, giving the answer in km² to 3 significant figures.

Working

1 : 4 × 10⁴

5.8 : 5.8 × 4 × 10⁴
4.4 : 4.4 × 4 × 10⁴

Area of park in real life:

5.8 × 4.4 × 4² × (10⁴)² cm²

= 408.32 × 10⁸ cm²

1 m = 100 cm = 10² cm
1000 m = 1 km = 10⁵ cm
1 cm = 10⁻⁵ km
(1 cm)² = (10⁻⁵ km)²
1 cm² = 10⁻¹⁰ km²

Area = 408.32 × 10⁸ × 10⁻¹⁰ km²

= 408.32 × 10⁻² km²

= 4.08 km² to 3 s.f.
Answer: 4.08 km²

Why the Scale Factor Is Squared

The map scale is:

1 : 40 000

This scale applies to lengths. Since area has two dimensions, the scale factor must be applied twice:

(40 000)²

Equivalently:

(4 × 10⁴)² = 4² × (10⁴)²

Converting cm² to km²

1 km = 10⁵ cm

1 cm = 10⁻⁵ km

(1 cm)² = (10⁻⁵ km)²

1 cm² = 10⁻¹⁰ km²

Key Results

Map scale:
1 : 40 000

Area scale factor:
(40 000)²

Conversion:
1 cm² = 10⁻¹⁰ km²

Area of the park:
4.08 km²

Source: Question adapted from Harry Smith, Revise Pearson Edexcel International GCSE (9–1) Mathematics A – Higher Tier Revision Guide, Pearson.

Solutions and workings: Tiago Hands.

Popular posts from this blog

A Geometric Way to Visualise sin(x + y) and cos(x + y)

Image
The angle–addition identities for sine and cosine often appear as algebraic formulas, but they can also be understood by combining two right triangles in a simple geometric construction. The calculations for the side lengths follow directly from the definitions of sine and cosine. sin(x + y) = sin x cos y + cos x sin y cos(x + y) = cos x cos y − sin x sin y Start with a right triangle of angle y and hypotenuse 1. From basic trigonometry, its horizontal and vertical sides are: cos y and sin y. Next, attach a second right triangle with angle x . Its hypotenuse is the side of length cos y from the first triangle, so its adjacent and opposite sides become: adjacent = cos x · cos y opposite = sin x · cos y Likewise, if the first triangle's vertical side sin y is used as a hypotenuse in a similar way, it contributes: adjacent = cos x · sin y opposite = sin x · sin y When the horizontal components are combined, they give the expression for cos(x + y...
Read more

The Method of Differences — A Clean Proof of the Sum of Cubes

The Method of Differences — A Clean Proof of the Sum of Cubes The method of differences is a remarkably elegant tool for evaluating finite sums. When each term of a series can be written in the form f(r+1) − f(r) , the sum “collapses” — all interior terms cancel, leaving only a boundary expression. This behaviour is called a telescoping sum . 1) Telescoping Sums Assume the general term u r can be written as: u r = f(r+1) − f(r). Then the finite sum from r = 1 to r = n becomes: Σ u r = Σ ( f(r+1) − f(r) ). To see what happens, write out a few terms: u₁ = f(2) − f(1) u₂ = f(3) − f(2) u₃ = f(4) − f(3) ⋮ uₙ = f(n+1) − f(n) When these are added, everything cancels except the first and last pieces: Σ u r = f(n+1) − f(1). This is the essence of the method: interior structure disappears, leaving just the difference between the final and initial states. 2) A Classic Application — The Sum of Cubes We will use this technique to prove the well-known ...
Read more

The Shortest Distance Between Two Skew Lines in ℝ³

The Shortest Distance Between Two Skew Lines in ℝ³ This post derives, from first principles, a vector formula for the shortest distance between two skew lines in ℝ³. The argument uses only the definitions and basic properties of the dot product and cross product; no higher results are assumed. 1. Vector Equations of the Lines Let the two lines be given in vector form by r = a + λb r = c + μd where: a , c are position vectors of fixed points on each line, b , d are non-zero direction vectors, λ, μ ∈ ℝ are parameters. The lines are skew if they are neither parallel nor intersecting. Our goal is to find a closed-form expression for the minimum distance between them. 2. Direction of the Common Perpendicular The segment that realises the shortest distance lies along a line perpendicular to both direction vectors b and d . A vector perpendicular to both is given by their cross product: b × d. Assuming b and d are not parallel, b × d ≠ 0. A unit ...
Read more