The village playground is shaped as a square of side 40 metres. A path of width s metres is created around the playground for people to walk. Find an expression for the area of the path in terms of s.
Step-by-step solution
To find: Area of the path, as an expression in s
Idea: The playground and the path together make a bigger square. Area of the path = area of the big square − area of the playground. Expand the big square with (a + b)2 = a2 + 2ab + b2.
- The path goes around the playground, so it adds s metres on both sides of each edge. Side of the outer square = 40 + s + s = (40 + 2s) m.½ mark
- Area of the path = area of outer square − area of playground
= (40 + 2s)2 − 402½ mark - Use (a + b)2 = a2 + 2ab + b2 with a = 40 and b = 2s:
(40 + 2s)2 = 1600 + 2(40)(2s) + (2s)2 = 1600 + 160s + 4s21 mark - Area of the path = 1600 + 160s + 4s2 − 1600 = 4s2 + 160s
Taking 4s common: 4s(s + 40) m21 mark - Note: if the path were inside the playground instead, the inner square would have side (40 − 2s) m, and the area of the path would be 402 − (40 − 2s)2 = 160s − 4s2 m2.
Check: Take s = 5: outer side 50 m, so 2500 − 1600 = 900 m2. Formula: 4(25) + 160(5) = 100 + 800 = 900 m2 ✓. Another way to see it: the path is four strips of 40 m × s m (160s) plus four corner squares of s × s (4s2), the same answer.
Answer to write in the exam
Side of playground = 40 m, width of path = s m
Side of outer square = (40 + 2s) m
Area of path = (40 + 2s)2 − 402
= 402 + 2(40)(2s) + (2s)2 − 402 [(a + b)2 = a2 + 2ab + b2]
= 1600 + 160s + 4s2 − 1600
= 4s2 + 160s
∴ Area of the path = 4s(s + 40) m2
Common mistakes that cost marks
- Taking the outer side as 40 + s. The path is on both sides, so the outer side is 40 + 2s.
- Writing (40 + 2s)2 = 1600 + 4s2, which leaves out the middle term 2ab = 160s.
- Writing the unit as m instead of m2. This is an area.
How this can come in the exam
A square playground of side 40 m has a path 5 m wide all around it (outside). The area of the path is
- 225 m2
- 800 m2
- 900 m2
- 1000 m2
Show answer
(C) 900 m2
Outer side = 40 + 2 × 5 = 50 m. 502 − 402 = 2500 − 1600 = 900 m2.
Assertion (A): The area of a path of width s m around a square of side 40 m is 4s(s + 40) m2.
Reason (R): (a + b)2 − a2 = 2ab + b2.
- Both A and R are true, and R is the correct explanation of A.
- Both A and R are true, but R is not the correct explanation of A.
- A is true but R is false.
- A is false but R is true.
Show answer
(A) Both A and R are true, and R is the correct explanation of A.
With a = 40 and b = 2s: (a + b)2 − a2 = 2(40)(2s) + (2s)2 = 160s + 4s2 = 4s(s + 40). R gives A directly.
Try one yourself
A square park of side 30 m has a path 2 m wide around it (outside). Find the area of the path.
Show answer
Outer side = 30 + 4 = 34 m. Area = 342 − 302 = 1156 − 900 = 256 m2. (Using 4s2 + 120s with s = 2: 16 + 240 = 256.)
More questions like this
- If a number plus its reciprocal equals 103, find the number.
- A rectangular pool has area 2x2 + 7x + 3 square hastas. If its width is 2x + 1 hastas, find its length. Hasta was a unit used to measure length.
- If both x − 2 and x − 12 are factors of px2 + 5x + r, show that p = r.
- If a + b + c = 5 and ab + bc + ca = 10, then prove that a3 + b3 + c3 − 3abc = −25.
- By factoring the expression, check that n3 − n is always divisible by 6 for all natural numbers n. Give reasons.