Area, perimeter & composite shapes

What you will learn

break a composite shape into rectangles and triangles (decompose),
build a complex outline by starting from a bigger rectangle and subtracting (compose),
approximate the area of an irregular shape using a grid,
handle trapeziums and L-shapes confidently.

1. Core area formulas (recap)

Standard area formulas

Rectangle

A = L \times W.

Triangle

A = \tfrac{1}{2} \times b \times h,

where $b$ is the base and $h$ the perpendicular height.

Parallelogram

A = b \times h.

Trapezium

A = \tfrac{1}{2}(a + b) \times h,

where $a, b$ are the parallel sides and $h$ the distance between them.

2. Composite shapes: decompose and add

Worked example 1 L-shape area

An L-shape is made by joining a $6$ m by $4$ m rectangle and a $3$ m by $2$ m rectangle.

L-shape: two rectangles joined. Area = sum of the two rectangles.

Area $= 6 \times 4 + 3 \times 2 = 24 + 6 = 30$ m $^2$ .

3. Composite shapes: start big and subtract

Worked example 2 Cut corners

A rectangular piece of cardboard $40$ cm by $30$ cm has a $5$ cm square cut from each of its four corners. Find the remaining area.

Start with a big rectangle; subtract the four removed squares.

Original rectangle area: $40 \times 30 = 1200$ cm $^2$ .
Four squares removed: $4 \times (5 \times 5) = 100$ cm $^2$ .
Remaining area: $1200 - 100 = 1100$ cm $^2$ .

4. Approximating an irregular shape on a grid

Worked example 3 Grid estimate

An irregular lake is drawn on a $1$ km grid. Count the fully-covered squares (say $28$ ) and the partial squares (say $14$ , counted as half).

Area estimate:

28 + \tfrac{1}{2} \times 14 = 35 \text{ km}^2.

Finer grids (say $100$ m instead of $1$ km) give a more accurate estimate.

5. Perimeter of composite shapes

Trace the outline. Do not miss any hidden sides when two rectangles meet.

Worked example 4 Staircase perimeter

A shape is formed by attaching a $3$ m by $2$ m rectangle to the corner of a $6$ m by $4$ m rectangle, forming a staircase. The visible outline has sides $6, 4, 3, 2, 3, 2$ (walking around the L).

Perimeter $= 6 + 4 + 3 + 2 + 3 + 2 = 20$ m.

Practice: Year 8 core

Fluency

Simple composite areas

A rectangle $8$ m by $5$ m has a $2$ m by $3$ m rectangle cut from one corner. Find the remaining area.
Two rectangles, $4 \times 2$ and $3 \times 5$ , are joined to form a single L. Find the total area.
A trapezium has parallel sides $8$ cm and $14$ cm and height $5$ cm. Find the area.
A floor plan is a $10$ m by $6$ m rectangle with a triangular bay window (base $6$ m, height $2$ m) added to one long side. Find the total floor area.
A shape is made of a rectangle $12 \times 5$ with a semicircle of diameter $5$ on one short side (use $\pi \approx 3.14$ ). Find the area.

Fluency

Perimeter

A rectangle $12$ by $7$ has a $3$ by $2$ notch cut into one long side (flush with the top-right). Find the new perimeter.
An L-shape has outer dimensions $10$ m by $8$ m, with a $4$ m by $3$ m notch removed from the top-right corner. Find the perimeter.
A rectangle $20$ by $15$ has a square of side $5$ cut from each corner. Find the perimeter of the remaining cross shape.

Fluency

Approximate by grid

On a $1$ cm grid, a shape covers $18$ full squares and $10$ partial squares. Estimate its area (count partial squares as half).
Using a finer $0.5$ cm grid, a shape covers $70$ full small squares and $20$ partial ones. Estimate its area.
Explain why approximating with a finer grid gives a better area estimate.

Reasoning

Explain and spot the mistake

Sam says the area of an L-shape is always the sum of the two rectangles that make it. Is that right? Give an example where it fails.
A student computes the perimeter of a shape by adding the two outer dimensions and multiplying by $2$ , ignoring the notch. Explain why this can be wrong.
Without calculating, decide which has the larger area: a $10 \times 10$ square with a $3 \times 3$ square cut from one corner, or an $11 \times 9$ rectangle. Justify.
A trapezium has parallel sides $3$ m and $7$ m and height $4$ m. Another has parallel sides $5$ m and $5$ m (so a rectangle) and height $4$ m. Which has the larger area? What does this tell you about the average of the parallel sides?

Problem solving

Real contexts

A backyard is L-shaped: a $15$ m by $12$ m rectangle with a $6$ m by $4$ m rectangle cut out of one corner (a shed). Grass seed covers $1$ m $^2$ per gram. How much seed is needed (in kg)?
A concrete slab is a rectangle $8$ m by $6$ m with a semicircular pool alcove (radius $1.5$ m) cut out of one long side. Find the slab area ( $\pi \approx 3.14$ ).
A picture frame has outer $25$ cm by $20$ cm and an inner window $19$ cm by $14$ cm. What is the area of the frame material?
A paddock on a map has shape of a trapezium (parallel sides $400$ m and $600$ m, height $250$ m). Find its area in hectares ( $1$ ha $= 10\,000$ m $^2$ ).

Challenge

Reasoning

Harder composites

A shape consists of a semicircle ( $r = 4$ cm) sitting atop a rectangle $8$ cm by $5$ cm (semicircle on an $8$ cm side). Find the area and the perimeter.
A square of side $20$ cm has a circle inscribed in it (touching all four sides). Find the area between the circle and the square (use $\pi \approx 3.14$ ).
Find the area of a hexagonal stop-sign-like shape with side $6$ cm, treated as a rectangle $12 \times 6\sqrt{3}/2$ with two triangles bolted to the short sides. (Extension: this requires $\sqrt{3}$ ; accept $\sqrt{3} \approx 1.73$ .)

Answers

Answer key

Attempt the practice first. When you're ready to check, expand the answers below.

Show the full answer key

Year 8 core - answers

Fluency

Simple composite areas

$34$ m $^2$ . Method: $40 - 6$ .
$23$ . Method: $8 + 15$ .
$55$ cm $^2$ . Method: $\tfrac{1}{2}(8 + 14) \times 5$ .
$66$ m $^2$ . Method: rectangle $60$ m $^2$ ; triangle $\tfrac{1}{2} \times 6 \times 2 = 6$ ; total $60 + 6$ .
About $69.8$ cm $^2$ . Method: rectangle $60$ cm $^2$ ; semicircle $\tfrac{1}{2} \pi (2.5)^2 \approx 9.8$ cm $^2$ .

Fluency

Perimeter

$42$ . Method: the notch adds two $2$ -cm inward cuts and a new $3$ cm edge parallel to the top; original perimeter $38$ plus $2 \times 2 = 4$ from the notch = $42$ . (Alternative walking-around count gives the same total.)
$36$ m. Method: walk the L: $10 + 8 + 4 + 3 + (10 - 4) + (8 - 3) = 10 + 8 + 4 + 3 + 6 + 5 = 36$ .
$80$ . Method: each corner removes a square but adds two new edges of length $5$ instead of one $5$ -edge; net change $= 0$ per corner. Original perimeter $= 2(20 + 15) = 70$ ; actually each cut increases the perimeter by $5 + 5 - 5 - 5 = 0$ . Wait - each corner cut replaces a corner with two $5$ -cm edges, so perimeter stays $70$ . Correct answer: $70$ .

Fluency

Approximate by grid

$23$ cm $^2$ . Method: $18 + 10/2$ .
$20$ cm $^2$ . Method: each small square is $0.25$ cm $^2$ ; $(70 + 10) \times 0.25$ .
Fewer partial squares, each contributing less error. As the grid gets finer the estimate approaches the true area.

Reasoning

Explain and spot the mistake

Correct for the decomposition approach, but fails if the two rectangles overlap. Example: two $4 \times 4$ rectangles that overlap by $2 \times 2$ ; naive sum $32$ , actual area $28$ .
The formula $P = 2(L + W)$ is only for a plain rectangle. A notch changes the outline; the perimeter can be the same, larger, or (rarely) smaller depending on the notch shape.
Both have area $91$ - equal. $10 \times 10 - 3 \times 3 = 91$ ; $11 \times 9 = 99$ . Wait: second is $99$ . So the plain rectangle is larger.
Both equal $\tfrac{1}{2}(3 + 7) \times 4 = 20$ m $^2$ and $5 \times 4 = 20$ m $^2$ . Equal - the average of the parallel sides ( $5$ ) equals the rectangle’s constant width.

Problem solving

Real contexts

Grass area: $15 \times 12 - 6 \times 4 = 180 - 24 = 156$ m $^2$ . Seed $= 156$ g $= 0.156$ kg.
$44.47$ m $^2$ approx. Method: slab $= 48$ m $^2$ ; semicircle $= \tfrac{1}{2} \pi (1.5)^2 \approx 3.53$ ; $48 - 3.53$ .
$234$ cm $^2$ . Method: $500 - 266$ .
$12.5$ ha. Method: area $= \tfrac{1}{2}(400 + 600) \times 250 = 125\,000$ m $^2$ ; divide by $10\,000$ .

Challenge - answers

Reasoning

Harder composites

Area $\approx 65.12$ cm $^2$ (rectangle $40$ + semicircle $\tfrac{1}{2} \pi \times 16 \approx 25.12$ ). Perimeter $\approx 5 + 8 + 5 + \pi \times 4 \approx 30.57$ cm.
$86$ cm $^2$ approx. Method: square $400$ ; circle $\pi \times 10^2 = 314$ ; $400 - 314$ .
About $93.5$ cm $^2$ . Method: $12 \times 6 \times \dfrac{\sqrt{3}}{2} \approx 12 \times 5.196 = 62.35$ … this is an approximation problem; any reasonable decomposition and arithmetic is fine.

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