Worked Solutions
Measurement & Geometry — Worked Solutions (Year 7 Maths)
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Worked examples for Year 7 Maths Measurement & Geometry. Each shows where the marks are awarded, the key idea, and the full solution explained by your choice of tutor — Stella, Ella or Cassie.
How to use these
Try each question first, then check your working. Use the tutor tabs to read the full solution in the style that suits you: Stella is direct and challenging, Ella is warm and explains the why, and Cassie is concise and analytical.
Example 1 — Area and perimeter of an L-shape
Question
A rectangular garden bed is 8 m long and 5 m wide. A small square section measuring 2 m by 2 m is removed from one corner. Find the area of the remaining garden bed.
Solution
Find the full rectangle, then take away the square.
Rectangle area: $8 \times 5 = 40 \text{ m}^2$.
Square removed: $2 \times 2 = 4 \text{ m}^2$.
Remaining area: $40 - 4 = 36 \text{ m}^2$.
The answer is $36 \text{ m}^2$. Always keep the units squared for area — drop them and you lose an easy mark.
The trick with an awkward shape is to start with a shape you already know, then adjust. Here we have a full rectangle with a corner taken out.
First the whole rectangle. Area of a rectangle is length times width, so $8 \times 5 = 40 \text{ m}^2$.
The piece removed is a square 2 m on each side, so its area is $2 \times 2 = 4 \text{ m}^2$.
Because that square is gone, we subtract it from the rectangle: $40 - 4 = 36 \text{ m}^2$.
So the remaining garden bed is $36 \text{ m}^2$. We write $\text{m}^2$ because area measures a flat surface in two directions, length and width together.
Whole minus removed piece.
- Rectangle: $8 \times 5 = 40 \text{ m}^2$
- Square removed: $2 \times 2 = 4 \text{ m}^2$
- Remaining: $40 - 4 = 36 \text{ m}^2$
Answer: $36 \text{ m}^2$.
Where the marks go
- 1 mark: Correct rectangle area: $8 \times 5 = 40 \text{ m}^2$
- 1 mark: Correct area of the removed square: $2 \times 2 = 4 \text{ m}^2$
- 1 mark: Correct remaining area with units: $36 \text{ m}^2$
Key idea
For a composite shape, find the area of the whole rectangle and subtract any piece removed; keep area in square units.
Example 2 — Angles on a straight line
Question
Three angles meet at a point on a straight line. They measure $x$, $48^\circ$ and $77^\circ$. Find the value of $x$.
Solution
Angles on a straight line add to $180^\circ$, so write the equation and solve.
$x + 48 + 77 = 180$.
Add the known angles: $48 + 77 = 125$, so $x + 125 = 180$.
Subtract: $x = 180 - 125 = 55^\circ$.
So $x = 55^\circ$. State the reason "angles on a straight line sum to $180^\circ$" — the reason is part of the mark.
When angles sit side by side along a straight line, they fill a half-turn, which is $180^\circ$. So all three angles together must add to $180^\circ$.
That gives us $x + 48 + 77 = 180$.
Let's tidy the numbers we know: $48 + 77 = 125$. Now the equation is $x + 125 = 180$.
To find $x$ we take away the $125$: $x = 180 - 125 = 55^\circ$.
So $x = 55^\circ$. The whole idea rests on a straight line being a half-turn, which is why the three angles must total $180^\circ$.
Angles on a straight line sum to $180^\circ$.
- $x + 48 + 77 = 180$
- $48 + 77 = 125$
- $x = 180 - 125 = 55^\circ$
Answer: $x = 55^\circ$.
Where the marks go
- 1 mark: Sets up the equation $x + 48 + 77 = 180$ (angles on a straight line)
- 1 mark: Solves correctly to get $x = 55^\circ$
Key idea
Angles on a straight line add to $180^\circ$, so subtract the known angles from $180^\circ$ to find the unknown.