This set mixes the five skills from area, perimeter and surface area. Questions run from easy to harder.
Write full working on paper, then open the answer and compare. All questions are original.
Use π from your calculator and round final answers to 3 significant figures unless told otherwise.
Questions and worked answers
Q1. A rectangle is 9.5 cm by 6 cm. Find its area and perimeter.
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Area = 9.5 × 6 = 57 cm². Perimeter = 2 × (9.5 + 6) = 2 × 15.5 = 31 cm.
Q2. Find the area of (a) a triangle with base 14 cm and perpendicular height 9 cm, and (b) a trapezium with parallel sides 7 cm and 12 cm and height 6 cm.
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(a) ½ × 14 × 9 = 63 cm². (b) ½ × (7 + 12) × 6 = ½ × 19 × 6 = 57, so 57 cm².
Q3. A circle has radius 5.5 cm. Find its circumference and its area.
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Circumference = 2 × π × 5.5 = 11π = 34.56, so 34.6 cm. Area = π × 5.5² = 30.25π = 95.03, so 95.0 cm².
Q4. A garden is an L-shape. It fits inside a 15 m by 9 m rectangle, with a 6 m by 4 m rectangle cut from one corner. Find its area and perimeter.
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Area = 15 × 9 − 6 × 4 = 135 − 24 = 111 m². Cutting a corner rectangle leaves the perimeter the same as the outer rectangle: 2 × (15 + 9) = 48 m.
Q5. A classroom floor is 7.5 m long and 600 cm wide. (a) Find its area in m². (b) How many square carpet tiles of side 50 cm are needed to cover it?
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(a) 600 cm = 6 m, so area = 7.5 × 6 = 45 m². (b) A tile is 0.5 × 0.5 = 0.25 m². 45 ÷ 0.25 = 180 tiles. Check in cm: 750 × 600 = 450 000, and 450 000 ÷ 2500 = 180.
Q6. Convert (a) 3.6 cm² to mm², and (b) 0.05 m² to cm².
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(a) 1 cm² = 100 mm², so 3.6 × 100 = 360 mm². (b) 1 m² = 10 000 cm², so 0.05 × 10 000 = 500 cm².
Q7. A sector has radius 12 cm and angle 75°. Find the arc length, the area and the perimeter.
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Fraction = 75/360 = 5/24. Arc = 5/24 × 24π = 5π = 15.71, so 15.7 cm. Area = 5/24 × 144π = 30π = 94.25, so 94.2 cm². Perimeter = 15.708 + 24 = 39.71, so 39.7 cm.
Q8. A circular disc of radius 10 cm has a circular hole of radius 6 cm cut from the centre. Find the area of the remaining ring.
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Area = π × 10² − π × 6² = 100π − 36π = 64π = 201.06, so 201 cm².
Q9. A triangular prism has a right-angled triangle cross-section with sides 5 cm, 12 cm and 13 cm. The prism is 20 cm long. Find the total surface area.
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Two ends: 2 × ½ × 5 × 12 = 60. Perimeter of triangle = 5 + 12 + 13 = 30, so the side faces total 30 × 20 = 600. Total = 60 + 600 = 660 cm².
Q10. An open cylindrical tank has radius 4 cm and height 9 cm and no lid. Find the area of metal needed.
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Base: π × 4² = 16π. Curved surface: 2π × 4 × 9 = 72π. Total = 88π = 276.46, so 276 cm².
Q11. A student says the area of a circle with diameter 7 cm is π × 7² = 153.9 cm². Find the error and give the correct area. Use an estimate to show the first answer is too big.
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The radius is 7 ÷ 2 = 3.5 cm, not 7. Area = π × 3.5² = 12.25π = 38.48, so 38.5 cm². Estimate: a circle of diameter 7 fits inside a 7 by 7 square of area 49, so its area must be less than 49. The value 153.9 is far too big.
Q12. A sector has radius 8 cm and arc length 10 cm. Find the angle of the sector and its area.
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Circumference of the full circle = 2π × 8 = 16π = 50.27. Angle = 10 ÷ 50.27 × 360 = 71.62, so 71.6°. Sector area = angle/360 × π × 8² = 71.62/360 × 201.06 = 40.0 cm². Check: ½ × arc length × radius = ½ × 10 × 8 = 40.
If you got these wrong
| What went wrong | Questions | Go to |
|---|---|---|
| Wrong pieces, a missing length or an inside edge in the perimeter | Q4, Q8 | Decompose a compound shape |
| Area converted with a length factor, or mixed units | Q5, Q6 | Calculate area after converting units |
| Arc and sector formulas swapped, or radii missing from a perimeter | Q3, Q7, Q12 | Distinguish arc length from sector area |
| Missing a face, or a wrong height used | Q9, Q10 | Unfold a prism to find surface area |
| Answer has the wrong kind of unit, or diameter used as radius | Q11 | Check an area answer using dimensions |
Basic formulas (Q1, Q2) are revisited within the lessons above. Keep a note of each mistake in the mistake log and retest queue, and use the non-calculator working trainer for any arithmetic that slowed you down.
Next step
If the same error type appears after you have reread the lesson, it often helps to have someone watch you solve a fresh question. Our teachers do this in online one-to-one Mathematics tuition, which begins with a paid one-hour trial.