Perimeter, area and volume – Year 7 maths worksheet answers to download as PDF

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1. Answer to Question 1

   For a rectangle, perimeter = 2 × (length + width) and area = length × width. Check that both lengths are in the same unit before you calculate.

   1. Perimeter \(2(9+5)=\)28 cm. Area \(9\times5=\)45 cm².
   2. Perimeter \(4\times7.5=\)30 cm. Area \(7.5^{2}=\)56.25 cm².
   3. 85 cm = 0.85 m. Perimeter \(2(2.4+0.85)=\)6.5 m. Area \(2.4\times0.85=\)2.04 m².

2. Answer to Question 2

   1. Area of a triangle = \(\dfrac{1}{2}\) × base × height: \(\dfrac{1}{2}\times12\times7=\)42 cm².
   2. \(\dfrac{1}{2}\times9.6\times5=4.8\times5=\)24 m².
   3. Area of a parallelogram = base × perpendicular height: \(15\times8=\)120 cm². Do not use the slanting side.
   4. Work backwards: \(12\times h=96\), so \(h=96\div12=\)8 cm.

3. Answer to Question 3

   1. \(\dfrac{1}{2}(6+10)\times4=\)32 cm².
   2. \(\dfrac{1}{2}(8.5+11.5)\times6=\)60 m².
   3. \(60-12=\)48 cm².
   4. Cutting a corner does not change the perimeter of the rectangle: \(2(10+6)=\)32 cm.

   In part (d) the two cut edges (4 cm and 3 cm) replace exactly the two pieces of the rectangle’s edge that were removed, which is why the perimeter stays 32 cm even though the area is smaller.

4. Answer to Question 4

   Volume of a cuboid = length × width × height. Surface area = 2 × (lw + lh + wh), the total area of the six faces.

   1. Volume \(8\times5\times3=\)120 cm³. Surface area \(2(40+24+15)=\)158 cm².
   2. Volume \(4^{3}=\)64 cm³. Surface area \(6\times16=\)96 cm².
   3. Volume \(2.5\times1.2\times0.8=\)2.4 m³.
   4. \(120\div24=\)5 cm.

5. Answer to Question 5

   1. 1 m² = 100 cm × 100 cm = 10 000 cm², so \(3.5\times10000=\)35 000 cm².
   2. Divide by 10 000: \(45000\div10000=\)4.5 m².
   3. 1 litre = 1 000 cm³, so \(2.4\times1000=\)2 400 cm³.
   4. Divide by 1 000: \(5000\div1000=\)5 litres.
   5. 1 m³ = 100 × 100 × 100 = 1 000 000 cm³, and \(1000000\div1000=\)1 000 litres.

6. Answer to Question 6

   In each part, write the formula, put in the values you know and use inverse operations to find the missing length.

   1. Width \(84\div12=\)7 cm. Perimeter \(2(12+7)=\)38 cm.
   2. Side 9 cm. Area 81 cm².
   3. \(\dfrac{1}{2}\times12\times h=30\), \(6h=30\), \(h=\)5 cm.
   4. Width \(17-10=\)7 cm. Area \(10\times7=\)70 cm².

7. Answer to Question 7

   1. Area of the whole wall = length × height: \(6.5\times2.8=\)18.2 m².
   2. Window \(1.5\times1.2=1.8\) m² and door \(2\times0.9=1.8\) m². Subtract both: \(18.2-1.8-1.8=\)14.6 m².
   3. \(14.6\div12\approx1.22\). One tin is not enough, and you cannot buy part of a tin, so round up to 2 tins.
   4. \(2\times16.50=\)£33.

8. Answer to Question 8

   1. Perimeter = 2 × (length + width) = \(2(90+60)=\)300 m.
   2. The outside line is 300 m and the centre line goes across the width (60 m): \(300+60=\)360 m.
   3. Area = length × width: \(90\times60=\)5 400 m².
   4. \(5400\div150=36\) exactly, so 36 bags of seed are needed.

9. Answer to Question 9

   1. Change to centimetres first: 2.4 m = 240 cm and 1.8 m = 180 cm. Area \(240\times180=\)43 200 cm².
   2. Each tile is a square: \(30\times30=\)900 cm².
   3. \(43200\div900=\)48 tiles. Check: 8 tiles fit along 240 cm and 6 along 180 cm, and \(8\times6=48\).
   4. 48 tiles need 5 boxes (4 boxes only hold 40). Cost \(5\times14=\)£70.

10. Answer to Question 10

    1. Volume = length × width × depth: \(12\times6\times1.5=\)108 m³.
    2. 1 m³ = 1 000 litres, so \(108\times1000=\)108 000 litres.
    3. \(108000\div4500=\)24 hours, a whole day.
    4. No top. Floor \(12\times6=72\) m², long walls \(2\times12\times1.5=36\) m², short walls \(2\times6\times1.5=18\) m². Total 126 m².

11. Answer to Question 11

    1. \(30\times20\times10=\)6 000 cm³.
    2. Three pairs of faces: \(30\times20=600\), \(30\times10=300\), \(20\times10=200\). Surface area \(2(600+300+200)=\)2 200 cm².
    3. Add 10% by multiplying by 1.1: \(2200\times1.1=\)2 420 cm².
    4. 2 420 cm² is 2.42 lots of 1 000 cm²: \(2.42\times0.60=1.452\), so about £1.45.

12. Answer to Question 12

    1. Area = \(14\times9=\)126 m².
    2. The patio is \(5\times4=20\) m², so the lawn is \(126-20=\)106 m².
    3. \(\dfrac{106}{126}\); divide top and bottom by 2: \(\dfrac{53}{63}\), about 84% of the garden.
    4. Perimeter \(2(14+9)=46\) m. Cost \(46\times2.30=\)£105.80.

13. Answer to Question 13

    1. \(60\times30\times40=\)72 000 cm³.
    2. 1 litre = 1 000 cm³, so \(72000\div1000=\)72 litres.
    3. Use the water depth as the height: \(60\times30\times32=57600\) cm³, which is 57.6 litres.
    4. The base area is \(60\times30=1800\) cm². The rock pushes up 1 800 cm³ of water, so the rise is \(1800\div1800=\)1 cm.

14. Answer to Question 14

    1. \(90\times60\times100=540000\) cm³. Divide by 1 000: 540 litres.
    2. \(540000\div600=\)900 bottles.
    3. 80% of 900: \(900\times0.8=\)720 bottles.
    4. Front and back \(2\times90\times100=18000\) cm², sides \(2\times60\times100=12000\) cm², top \(90\times60=5400\) cm². Total 35 400 cm².

15. Answer to Question 15

    1. Volume = \(3\times2\times0.6=\)3.6 m³ (3 600 litres).
    2. Path = outer rectangle − pond: \(4\times3-3\times2=12-6=\)6 m².
    3. One slab is \(0.5\times0.5=0.25\) m². \(6\div0.25=\)24 slabs.
    4. 24 slabs need 3 packs (2 packs only give 20). Cost \(3\times18=\)£54.

16. Answer to Question 16

    1. \(20\times12+8\times5=240+40=\)280 m².
    2. Perimeter of the big rectangle \(2(20+12)=64\). The small rectangle adds two sides of 5 m and one side of 8 m but removes 8 m of the big rectangle, so perimeter = \(64+5+5=\)74 m.
    3. \(280\times4.50=\)£1 260.
    4. \(\dfrac{40}{280}\times100=14.3\), so about 14%.

17. Answer to Question 17

    1. Perimeter \(2(5+4)=18\) m, area \(18\times2.5=\)45 m².
    2. \(45-8=\)37 m².
    3. \(37\div10\times2=\)7.4 litres.
    4. Buy one 5 litre tin (£34) and one 2.5 litre tin (£19): 7.5 litres for £53. (Three 2.5 litre tins cost £57 and give 7.5 litres; two 5 litre tins cost £68.)

18. Answer to Question 18

    1. Halve it: 30 cm².
    2. Multiply: 24 cm³.
    3. 1 m² = 10 000 cm², so 2 m² = 20 000 cm².
    4. Not necessarily. A 6 cm by 6 cm square has perimeter 24 cm, but a 9 cm by 4 cm rectangle also has area 36 cm² and perimeter 26 cm.

19. Answer to Question 19

    1. Use the water depth: \(50\times40\times20=40000\) cm³, which is 40 litres.
    2. \(40-6=\)34 litres.
    3. 34 litres = 34 000 cm³. Depth = volume ÷ base area: \(34000\div2000=\)17 cm.
    4. Full: \(50\times40\times25=50000\) cm³ = 50 litres. Space left: \(50-34=\)16 litres.

20. Answer to Question 20

    1. The fence is two widths and one length: \(2x+\text{length}=40\), so the length is \(40-2x\) m.
    2. Area = width × length = \(x(40-2x)\) m².
    3. \(x=5\): \(5\times30=150\). \(x=8\): \(8\times24=192\). \(x=10\): \(10\times20=200\). \(x=12\): \(12\times16=192\). \(x=15\): \(15\times10=150\).
    4. The areas rise then fall symmetrically. The greatest is 200 m² at \(x=10\) m (length 20 m).