Algebra 5
5 marks · HardA garden is to be designed with a rectangular lawn of width m and length m. A path of uniform width m is built all around the lawn. The total area (lawn plus path) is m.
Show that .
Given that the width of the path, , is 2 m, form and solve a quadratic equation to find the dimensions of the lawn.
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
Show answer
(a) (x + 2y)(x + 4 + 2y) = 168
(b) x = 8\text{ m} (or) x = -12,\text{ but width cannot be negative so } x=8\text{ m},\text{ length }=12\text{ m}
Common mistakes: Omitting to add 2y on each side; Arithmetic slips expanding and rearranging
Show mark scheme and worked solution
Marking points
M1 (a) · 1 mark · Correctly showing area of total rectangle with path as (x+2y)(x+4+2y) and equate to 168
M2 (b) · 1 mark · Lets y=2 and substitutes correctly to obtain (x+4)(x+8)=168 or equivalent
M3 (b) · 1 mark · Expands and rearranges to quadratic in standard form (follow-through)
A1 (b) · 1 mark · Solves the quadratic for x, obtaining correct possible values, x=8, x=-12 (follow-through)
B1 (b) · 1 mark · Selects and states the valid physical solution, x=8 m and finds length=12 m (follow-through)
Worked solution
a) The total outer rectangle has width (x + 2y) and length (x + 4 + 2y). So area is:
Setting this equal to the given area:
b) Putting y = 2:
Expanding:
Rearranging:
Factorise or use quadratic formula ( ):
Width must be positive: m. Length = 8 + 4 = 12 m.