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Total length of the parking area to be fenced up BC = DE + FG – (AM + LK)
= (AB + BC + CD + DE + EF + FG + GK + KL) – HJ = (35 + 30) – (8 + 32)
= (45 + 25 + 35 + 35 + 12 + 30 + 100 + 32) – 10 = 25 m
= 314 – 10 GK = LM + AB + CD + EF
= 304 m = 8 + 45 + 35 + 12
Total cost to fence up the parking area = 100 m
= 304 × RM80 Total length of fence × RM80
= RM24 320

Perimeter of the enclosed parking area C D
= 2 × [(LM + AB + CD + EF) + (DE + FG)]
= 2 × [(8 + 45 + 35 + 12) + (35 + 30)] A B
= 2 × (100 + 65) L M E F
= 2 × 165
= 330 m K G
Total length of the parking area to be fenced up J H
= 330 – HJ – LM – MA KG = LM + AB + CD + EF
CB + AM + LK = DE + FG
= 330 – 10 – 8 – 8
= 304 m

Self Practice 10.1c

1. The diagram shows the plan of a swimming pool. 20 m
What is the perimeter of the swimming pool?
5 m 13 m CHAPTER

10

2. The diagram shows the plan of Encik Yahya’s
house. Encik Yahya wants to install colorful LED 8 m
lights around his house as decorations to celebrate 10 m
Hari Raya. Calculate the installation cost if the
cost to install LED lights is RM20 per metre. 25 m

3. In the diagram, ABC is an equilateral triangle, A
BCFG is a square and CDEF is a rectangle. The
perimeter of the whole diagram is 65 cm, find the C D
length of GE. B
7 cm

G F E

229
Perimeter and Area

10 TB Math F1.indd 229 11/10/16 12:19 PM