Area
Geometry and measures
- Foundation
- Non-calculator
- 8 questions
- 29 marks
These are original practice questions, not a past paper. This site is independent and not affiliated with AQA or Pearson.
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A rectangle is $7$ cm by $4$ cm. Find its area.
2 marks
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A triangle has base $10$ cm and height $6$ cm. Find its area.
2 marks
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A parallelogram has base $8$ cm and perpendicular height $5$ cm. Find its area.
3 marks
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A trapezium has parallel sides $6$ cm and $10$ cm, and height $4$ cm. Find its area.
3 marks
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The diagram shows an L-shaped floor. The full rectangle would be $10$ m by $8$ m, and a $4$ m by $3$ m rectangle has been removed from one corner. Find the area of the floor.
L-shaped floor, not to scale 4 marks
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A rectangular room is $5$ m by $4$ m. A square rug of side $1.5$ m lies on the floor. Find the area of the floor, the area of the rug, and the area not covered by the rug.
4 marks
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An isosceles triangle has two equal sides of $13$ cm and a base of $10$ cm. Find its perpendicular height and its area.
5 marks
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The L-shaped lawn in the diagram would be a $12$ m by $9$ m rectangle with a $5$ m by $4$ m rectangle removed. Turf costs £$6.50$ per m$^{2}$. Seed bags each cover $10$ m$^{2}$. Find the area, the cost of turf, and the number of seed bags needed to cover the lawn.
L-shaped lawn, not to scale 6 marks
Mark scheme
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Answer. $28$ cm$^{2}$
- $7 \times 4$
- $28$ cm$^{2}$
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Answer. $30$ cm$^{2}$
- $\dfrac{1}{2} \times 10 \times 6$
- $30$ cm$^{2}$
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Answer. $40$ cm$^{2}$
- Area $= \text{base} \times \text{height}$
- $40$ cm$^{2}$
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Answer. $32$ cm$^{2}$
- $\dfrac{6 + 10}{2} \times 4$
- $32$ cm$^{2}$
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Answer. $68$ m$^{2}$
- Full rectangle $= 10 \times 8 = 80$.
- Removed area $= 4 \times 3 = 12$.
- Floor $= 68$ m$^{2}$.
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Answer. $20$ m$^{2}$, $2.25$ m$^{2}$, and $17.75$ m$^{2}$
- Floor $= 5 \times 4 = 20$ m$^{2}$.
- Rug $= 1.5^{2} = 2.25$ m$^{2}$.
- Uncovered area $= 17.75$ m$^{2}$.
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Answer. Height $12$ cm, area $60$ cm$^{2}$
- Half of the base is $5$ cm.
- Height $= \sqrt{13^{2} - 5^{2}} = \sqrt{144} = 12$ cm.
- Area $= \dfrac{1}{2} \times 10 \times 12 = 60$ cm$^{2}$.
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Answer. Area $88$ m$^{2}$, turf £$572$, and $9$ bags
- Area $= 12 \times 9 - 5 \times 4 = 88$ m$^{2}$.
- Turf $= 88 \times 6.50 = £572$.
- $8$ bags cover $80$ m$^{2}$, so $9$ bags are needed.