1. P ETER ’ S G ARDENING SERVICE By Rebecca Buckley

2. I NTRODUCTION Peter is a professional gardener who tends the gardens and lawns of his many clients. One of his jobs is to fertilise the lawns regularly. He is trying a new fertiliser which, according to the directions, states that when the contents of the container are mixed with water the solution covers 64m 2 of lawn. To help him decide how many containers of fertiliser he may require, Peter needs to have some idea of possible sizes of lawn one container will cover. First Peter considers a square lawn. He wants to find the length of a square with area 64m 2.

3. Q UESTIONS 1) If the length of the square is represented by x, write an expression for the area of the square. 2) Since the area needs to be 64m 2, write an equation which shows that the area of the square is equal to 64m 2 using x. Part 1 will help. 3) Can you solve this equation? Remember to consider the different methods you have learnt. (You may wish to discuss this with a partner or with others in a small group.) What is the size of a square lawn which can be covered with one container of fertiliser?

4. S QUARE L AWN Q1 area = X² Q2 X = √64 Q3X = 8 X

5. R ECTANGULAR L AWN Q1area = L x W Q2 a) 64 = 2 x W 64 ÷ 2 = W b) 64 = 4 x W64 ÷ 4 = W c) 64 = 1 x W64 ÷ 1 = W d) 64 = 3 x W64 ÷ 3 = W e) 64 = 6 x W64 ÷ 6 = W Q3 a) 32 = W b) 16 = W c) 64 = W d) 21.3 = W e) 10.66˙ = W L W

6. T RIANGULAR L AWN Q1 area = 0.5 x B x H Q2 a) 64 = 0.5 x 6 x H64 = 3 x H 64 ÷ 3 = H b) 64 = 0.5 x 11 x H64 = 5.5 x H 64 ÷ 5.5 = H c) 64 = 0.5 x 9 x H64 = 4.5 x H 64 ÷ 4.5 = H d) 64 = 0.5 x 7 x H64 = 3.5 x H64 ÷ 3.5 = H e) 64 = 0.5 x 13 x H64 = 6.5 x H 64÷ 6.5 = H Q3 a) H = 21.33˙ b) H = 11.63 c) H = 7.11˙ d) H =14.22˙ e) H = 9.85 H B

7. C IRCULAR L AWN Q1area = Π R² Q2 64 = 3.142 x R²64 ÷ 3.142 = 20.37√ 20.37 = R Q3 R = 4.51 I’m pretty sure you can only get 1 circle with the area of 64m². R

8. L - S HAPED L AWN 3 little squares Q1 area = L² x 3 Q2 (64 ÷ 3)√ = L√21.4 = L Q3 L = 4.6 L L L L

9. M Y O WN S HAPE

10. H OUSE Q1 X² and 0.5 x B x H Q2 64 =.5 x 6 x 1564 = X² + 4564 - 45 = 19 √19= X Q3 B = 6 H = 15 X = 5.36 I solved it by doing the triangle first than subtracted it from 64 got 19, square rooted that to get the area of the square. X B H