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Year 7 Maths Area and Volume

Area of Rectangle

20 practice questions 1 video lesson Theory + worked examples

The area of a rectangle is the number of unit squares that cover it, found quickly as \(\text{length} \times \text{width}\). In NSW Year 7 maths (NESA outcome MA4‑ARE‑C‑01) this is the first area formula, and the formulas for triangles and parallelograms grow out of it.

You will count squares in rows, use \(A = l \times w\) and \(A = s \times s\) for a square, find a missing side by dividing the area by the known side, change both sides to the same unit before multiplying, cost jobs such as painting a stage at a price per square metre, and compare rectangles made from the same 36 unit squares. It is part of Year 7 Area and Volume.

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Theory

The area of a shape is the amount of flat surface it covers, measured in square units. For a rectangle, \(A = l \times w\) (length times width, also written \(A = L \times B\) with \(B\) for breadth). This Year 7 maths page shows where the formula comes from, how to find a missing side and how to work with mixed units.

A square centimetre (\(1~\text{cm}^2\)) is a square \(1\) cm by \(1\) cm. The area is the number of these squares that cover the shape.

Count in rows. A rectangle \(5\) cm long and \(3\) cm wide holds \(3\) rows of \(5\) squares: \(3 \times 5 = 15~\text{cm}^2\). The shortcut is the formula \(A = l \times w\). You may also see \(A = L \times B\) or \(A = LB\), where \(B\) is the breadth (another word for width).

Square: length and width are equal, so \(A = s \times s\).

Missing side: divide the area by the side you know. If \(A = 56~\text{cm}^2\) and \(l = 8\) cm, then \(w = 56 \div 8 = 7\) cm. For a square, find the number that times itself gives the area.

Units: the sides must be in the same unit before you multiply. Write the answer in square units.

Rectangle 5 cm by 3 cm on a centimetre grid A rectangle 5 squares long and 3 squares high, made of 15 one-centimetre squares.
\(3\) rows of \(5\) unit squares: \(A = 5 \times 3 = 15~\text{cm}^2\).
Length and breadth A rectangle with the length along the bottom and the breadth up the side. length (l or L) breadth A = l × w
\(A = l \times w\), also written \(A = L \times B\).

Formulas:

\[ A = l \times w \]
A = l × w
\[ A = s \times s \]
A = s × s ( square )
\[ w = A \div l \]
w = A ÷ l
RectangleAreaPerimeter
\(12\) by \(1\)\(12~\text{cm}^2\)\(26\) cm
\(6\) by \(2\)\(12~\text{cm}^2\)\(16\) cm
\(4\) by \(3\)\(12~\text{cm}^2\)\(14\) cm
Same area, different perimeter. Rectangles with the same area can have different perimeters. The one closest to a square has the smallest perimeter.
Doubling both sides does not double the area. A \(4\) cm by \(3\) cm rectangle (\(12~\text{cm}^2\)) becomes \(8\) cm by \(6\) cm (\(48~\text{cm}^2\)): \(2 \times 2 = 4\) times bigger.

How to find the area of a rectangle

  1. Change both sides to the same unit.
  2. Multiply length by width.
  3. Write the answer in square units.

How to find a missing side

  1. Substitute into \(A = l \times w\).
  2. Divide the area by the known side.
  3. Check by multiplying back.
Example 1 — Rectangle and square
Find the area of
(i) a rectangle \(14\) cm long and \(9\) cm wide
(ii) a square sandpit with sides of \(2.5\) m
Solution
\(14 \times 9\)\(=\)\(126\)
\(2.5 \times 2.5\)\(=\)\(6.25\)
14×9=126,2.5×2.5=6.25

(i) \(126~\text{cm}^2\)
(ii) \(6.25~\text{m}^2\)

Example 2 — Length and breadth in different units
Find the area of a rectangle with length \(5\) m and breadth \(240\) cm.
Solution
\(240 \div 100\)\(=\)\(2.4\)
\(5 \times 2.4\)\(=\)\(12\)
240 cm =2.4 m ,5×2.4=12

\(12~\text{m}^2\)

Example 3 — Missing side
(i) A garden bed has an area of \(91~\text{m}^2\) and a length of \(13\) m. Find its width.
(ii) A square chessboard has an area of \(1600~\text{cm}^2\). How long is each side?
Solution
\(91 \div 13\)\(=\)\(7\)
\(40 \times 40\)\(=\)\(1600\)
91÷13=7,40×40=1600

(i) \(7\) m (check: \(13 \times 7 = 91\))
(ii) \(40\) cm

Example 4 — A yoga mat
A yoga mat is \(1.8\) m long and \(60\) cm wide. Find its area in square metres.
Solution
\(60 \div 100\)\(=\)\(0.6\)
\(1.8 \times 0.6\)\(=\)\(1.08\)
60 cm =0.6 m ,1.8×0.6=1.08

\(1.08~\text{m}^2\). Check in cm: \(180 \times 60 = 10\,800~\text{cm}^2 = 1.08~\text{m}^2\).

Example 5 — Perimeter, then area and cost
A square backyard has a perimeter of \(48\) m. Turf costs \(\$15\) per square metre. How much does it cost to turf the backyard?
Solution
\(48 \div 4\)\(=\)\(12\)
\(12 \times 12\)\(=\)\(144\)
\(144 \times 15\)\(=\)\(2160\)
48÷4=12,12×12=144,144×15=2160

\(\$2160\)

Example 6 — A path around a pool
A rectangular pool is \(10\) m by \(4\) m. A path \(2\) m wide goes all the way around it. Find the area of the path.
Solution
\(10 + 2 + 2\)\(=\)\(14\)
\(4 + 2 + 2\)\(=\)\(8\)
\(14 \times 8\)\(=\)\(112\)
\(10 \times 4\)\(=\)\(40\)
\(112 - 40\)\(=\)\(72\)
outer 14×8=112, pool 10×4=40, path 112−40=72

The outer rectangle is \(14\) m by \(8\) m. Path \(= 112 - 40 = 72~\text{m}^2\).

Example 7 — Perimeter to area
A rectangle has a perimeter of \(26\) cm and a length of \(8\) cm. Find its area.
Solution
\(26 - 2 \times 8\)\(=\)\(10\)
\(10 \div 2\)\(=\)\(5\)
\(8 \times 5\)\(=\)\(40\)
26−16=10, width =10÷2=5, area =8×5=40

Width \(5\) cm, area \(40~\text{cm}^2\).

Common pitfalls

Area is not perimeter. For an \(8\) m by \(5\) m rectangle, \(A = 40~\text{m}^2\); \(8 + 5 + 8 + 5 = 26\) m is the perimeter.
Different units. For \(2\) m by \(30\) cm, use \(2 \times 0.3 = 0.6~\text{m}^2\), not \(2 \times 30 = 60\).
Linear units. Write \(40~\text{m}^2\), not \(40\) m.
Halving instead of dividing. A missing side is area \(\div\) known side, not area \(\div\,2\).

Frequently asked questions

What is the formula for the area of a rectangle?

\(A = l \times w\) (length times width, or breadth).

How do you find the area of a square?

Multiply the side by itself: \(A = s \times s\).

How do you find the width if you know the area?

Divide the area by the length. \(91 \div 13 = 7\).

Why must the units be the same?

Otherwise the number is wrong: \(2\) m by \(30\) cm is \(0.6~\text{m}^2\), not \(60\).

Which rectangle with a given area has the smallest perimeter?

The one closest to a square.