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# Solving Perimeter Problems

We now look at solving perimeter problems at a slightly advanced level, that also includes some concepts from other mathematics topics.

1. Calculate the perimeter of a rectangular field with dimensions 75 m x 20 m.

Perimeter = 2(l + b)

=  2(75 + 20)

= 2 x 95

= 190 m

2. A wire fence needs to be erected around the sides of a farm with dimensions 20.3 m x 9.2 m.

a) Find the length of the fencing required

b) Calculate the cost of the fencing wire at \$10 per metre

Perimeter of the farm = 2(l + b)

=  2(20.3 + 9.2)

= 2 x 29.5

= 59 m

Cost of wire fence = 59 x 10  =  \$590

3. The perimeter of a rectangular backyard is 34 m. If the yard is 6 m wide, find its length.

Perimeter = 2(l + b)

34 = 2(l + 6)

l + 6 = 17

l = 17 – 6

= 11 m

So the length of the backyard is 11 metres.

4. A square has the same perimeter as an equilateral triangle. If the triangle has sides of length 8 cm, what is the side length of the square.

Given equilateral triangle with side = 8 cm,

its perimeter = 8 + 8 + 8

=  24 cms

Perimeter of a square = 4s

4s  =  24

s  =  24 ÷ 4

=  6 cm

So the side length of the square is 6 cm

5. The length of a rectangle is twice its width. Find the dimensions of the rectangle if the perimeter is 36 cm
When length l is twice its width, we can express the length l = 2b.
So perimeter = 2(l + b) = 36
2(2b + b) = 36
3b = 18
b = 6
So the breadth of the rectangle is 6 cm, and its length is 12 cm (being twice its width).
6. A swimming pool has a 2 metre wide path around its edge. The outer dimensions of the path are 35 x 20 m. Find

a) the dimensions of the pool

b) the perimeter of the pool

Since there is a 2 metre wide path around the swimming pool, the actual dimensions of the pool will be 2 m less than the outer length and breadth on both sides.

So length of the pool = 35 – 2 – 2  =  31 m

and breadth of the pool = 20 – 2 – 2  =  16 m

Therefore the dimensions of the pool are 31 x 16 m

Perimeter of the pool = 2(31 + 16)

=  2 x 47

=  94 m

7. A school ground is 110 m long and 30 m wide. How far, in kilometres, will an athlete run by completing 20 laps of the track?

Perimeter of the school ground = 2(110 + 30)

=  2 x 140

=  280 m

Doing 20 laps around the track  =  280 x 20

=  5600 m

=  5.6 km  (1 km  =  1000 m)