Improve Tuition · Maths · Perimeter & Area · Assessment
Challenging assessment paper
2D Shapes: Perimeter and Area
Test mastery — mixed problems and reasoning.
18 questions · 60 marks · 45–60 minutes · feedback shown after you submit
How this assessment works
This paper tests whether the skills are secure. Work through all 18 questions, then press Submit & Mark — unlike the practice paper, the answers and feedback stay hidden until you submit. Numbers can be entered with or without units.
Q1.A patio is made from regular hexagonal tiles. Each outside edge is 18 cm long. The patio has 24 outside edges. Work out the perimeter.
Your answer:(2 marks)
24 × 18 = 432 cm.
Q2.The area of a rectangle is 96 cm². Its length is 12 cm. Find the width, then the perimeter.
Your answer:(3 marks)
Width = 96 ÷ 12 = 8 cm.
Perimeter = 2 × (12 + 8) = 40 cm.
Q3.The length of a rectangle is 8 cm longer than its width. The area is 128 cm². Both sides are whole numbers. Find the perimeter.
Your answer:(4 marks)
8 × 16 = 128, difference 8.
Perimeter = 2 × (8 + 16) = 48 cm.
Compound area
11 marks in this section
Q4.An L-shaped garden: left rectangle width 7 m, height 12 m; right rectangle width 5 m, height 6 m. Work out the total area.
Split the shape into two rectangles, find each area, then add them together.
Your answer:(3 marks)
Left = 7 × 12 = 84 m².
Right = 5 × 6 = 30 m².
Total = 114 m².
Q5.An L-shaped playground has a total bottom length of 18 m. Left rectangle: width 8 m, height 14 m. Right rectangle height 5 m. Work out the total area.
Your answer:(4 marks)
Right width = 18 − 8 = 10 m.
Left = 8 × 14 = 112 m². Right = 10 × 5 = 50 m².
Total = 162 m².
Q6.A rectangular field is 70 m by 45 m. Inside it: a building 24 m by 15 m and a pond 12 m by 8 m. The rest is grass. Work out the grass area.
Find the whole area, then subtract the shapes inside it.
Your answer:(4 marks)
Field = 70 × 45 = 3150 m².
Building = 360 m², Pond = 96 m².
Grass = 3150 − 360 − 96 = 2694 m².
Triangle area
9 marks in this section
Q7.A triangular section of land has base 16 m and height 9 m. Work out its area.
A triangle is exactly half of the rectangle around it.
Your answer:(2 marks)
½ × 16 × 9 = 72 m².
Q8.A triangle has an area of 84 cm². Its base is 14 cm. Find the height.
Your answer:(3 marks)
Area = ½ × base × height, so 84 = ½ × 14 × h = 7h.
h = 84 ÷ 7 = 12 cm.
Q9.A garden is a rectangle (length 18 m, width 7 m) with a triangle attached (base 18 m, height 6 m). Work out the total area.
Your answer:(4 marks)
Rectangle = 18 × 7 = 126 m².
Triangle = ½ × 18 × 6 = 54 m².
Total = 180 m².
Surface-area style problems
9 marks in this section
Q10.A cardboard tunnel has no base: length 28 cm, width 9 cm, height 11 cm. The outside is the top and two sides. Work out the total outside area.
Count only the faces that are there: top + two sides (no base).
Your answer:(4 marks)
Top = 28 × 9 = 252 cm².
Two sides = 28 × 11 × 2 = 616 cm².
Total = 868 cm².
Q11.A tunnel (no base) is length 35 cm, width 14 cm, height 10 cm. The outside is painted. One tin covers 500 cm². How many tins are needed?
Your answer:(5 marks)
Top = 35 × 14 = 490 cm². Two sides = 35 × 10 × 2 = 700 cm².
Total = 1190 cm². 1190 ÷ 500 = 2.38.
Round UP: 3 tins.
Compound perimeter reasoning
9 marks in this section
Q12.An L-shape has: bottom length 16 cm, left height 12 cm, top short length 7 cm, right short height 5 cm. Work out the perimeter. (Find the two missing sides first.)
Your answer:(5 marks)
Missing vertical = 12 − 5 = 7 cm. Missing horizontal = 16 − 7 = 9 cm.
Perimeter = 16 + 5 + 9 + 7 + 7 + 12 = 56 cm.
Q13.Two rectangles are joined along one full side of 6 cm. Rectangle A is 10 cm by 6 cm; rectangle B is 4 cm by 6 cm. Work out the perimeter of the new shape.
Your answer:(4 marks)
Perimeters: A = 32 cm, B = 20 cm, total 52 cm.
The joined side (6 cm) is hidden on both shapes: subtract 12 cm.
52 − 12 = 40 cm.
Coverage & real-life application
10 marks in this section
Q14.A room has 4 walls: two are 9 m wide and 3 m tall, two are 6 m wide and 3 m tall. One tin covers 20 m². How many tins are needed?
Add up the area of every wall, divide by what one tin covers, then round UP — a part-tin still means buying a whole tin.
Your answer:(5 marks)
Two walls = 9 × 3 × 2 = 54 m². Two walls = 6 × 3 × 2 = 36 m².
Total = 90 m². 90 ÷ 20 = 4.5.
Round UP: 5 tins.
Q15.A school hall is 20 m by 12 m. A stage takes up 6 m by 4 m. The rest is covered with tiles. Each tile covers 0.5 m². How many tiles are needed?
Your answer:(5 marks)
Hall = 20 × 12 = 240 m². Stage = 24 m².
Floor to tile = 240 − 24 = 216 m².
216 ÷ 0.5 = 432 tiles.
Reasoning
3 marks in this section
Q16.A pupil says: “The area of a triangle is base × height.” Explain why this is not correct.
Your explanation:
(1 mark)
Compare with the model answer, then mark yourself:
Model answer
Base × height gives the area of the whole rectangle around the triangle. A triangle is only half of that rectangle, so you must halve it: area = ½ × base × height.
Q17.A pupil finds the area of a car park by adding the supermarket area to the whole site area. Explain the mistake.
Your explanation:
(1 mark)
Compare with the model answer, then mark yourself:
Model answer
The supermarket sits inside the site, so it should be subtracted, not added. The car park is what is left: whole site − supermarket (and any other buildings).
Q18.A pupil says 4 tins are enough because the wall area is 90 m² and each tin covers 20 m². Explain why 5 tins are needed.
Your explanation:
(1 mark)
Compare with the model answer, then mark yourself:
Model answer
4 tins cover only 4 × 20 = 80 m², which is less than 90 m². The leftover 10 m² still needs paint, so you must round up and buy a 5th tin even though it is only partly used.