Worked examples on the Perimeter of Compound Shapes.

1. PERIMETER
COMPOUND SHAPES

2. WHAT IS
PERIMETER?
P e r i m e t e r i s t h e
d i s t a n c e a r o u n d a n
o b j e c t .
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3. PERIMETER OF
COMMON SHAPES
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4. EXAMPLES
The table gives
measurements for
some rectangles.
Fill in the missing
values.
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Length Width Perimeter
1) 4 𝑐𝑚 12 𝑐𝑚
2) 5 𝑚𝑚 60 𝑚𝑚
3) 5 𝑐𝑚 14 𝑐𝑚
4) 7 𝑐𝑚 8 𝑐𝑚
5) 3 𝑚 6 𝑚

5. 1 ) 𝑃 = 2 𝑙 + 𝑏
1 2 = 2 4 + 𝑏
1 2 = 8 + 2 𝑏
2 𝑏 = 1 2 − 8 = 4
𝑏 =
4
2
= 𝟐
2 ) 𝑃 = 2 𝑙 + 𝑏
6 0 = 2 𝑙 + 5
6 0 = 2 𝑙 + 1 0
2 𝑙 = 6 0 − 1 0 = 5 0
𝑙 =
5 0
2
= 𝟐 𝟓
3 ) 𝑃 = 2 ( 𝑙 + 𝑏 )
1 4 = 2 ( 5 + 𝑏 )
1 4 = 1 0 + 2 𝑏
2 𝑏 = 1 4 − 1 0 = 4
𝑏 =
4
2
= 𝟐
4 ) 𝑃 = 2 ( 𝑙 + 𝑏 )
𝑃 = 2 ( 7 + 8 )
= 2 ( 1 5 )
𝑃 = 𝟑 𝟎
5 ) 𝑃 = 2 ( 𝑙 + 𝑏 )
𝑃 = 2 6 + 3
= 2 ( 9 )
𝑃 = 𝟏 𝟖
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Length Width Perimeter
1) 4 𝑐𝑚 𝟐 𝒄𝒎 12 𝑐𝑚
2) 𝟐𝟓 𝒎𝒎 5 𝑚𝑚 60 𝑚𝑚
3) 5 𝑐𝑚 𝟐 𝒄𝒎 14 𝑐𝑚
4) 7 𝑐𝑚 8 𝑐𝑚 𝟑𝟎 𝒄𝒎
5) 6 𝑚 3 𝑚 𝟏𝟖 𝒎
SOLUTIONS:

6. EXAMPLE
W h a t i s t h e p e r i m e t e r o f t h e
f i g u r e s h o w n o n t h e r i g h t ?
To find the perimeter, we must add up all the lengths
around the edges of the shape.
We must calculate the unknown lengths, 𝑥 𝑎𝑛𝑑 𝑦.
Now, the length of the entire side is 6 𝑐𝑚 and the length of
the shorter side is 4 𝑐𝑚. Therefore, to get the value of
𝑥, subtract.
𝑥 = 6 − 4 = 2 𝑐𝑚
Similarly, subtract to get the value of 𝑦.
𝑦 = 6 − 4 = 2 𝑐𝑚
Perimeter = 6 + 6 + 2 + 2 + 4 + 4 = 𝟐𝟒 𝒄𝒎
6
𝒙
𝒚

7. EXAMPLE
W h a t i s t h e p e r i m e t e r o f
t h e f i g u r e s h o w n o n t h e
r i g h t ?
To find the perimeter, we must add up all the lengths around the
edges of the shape.
We must calculate the unknown lengths, 𝑝, 𝑞, 𝑟 𝑎𝑛𝑑 𝑠.
𝑝 is the total height, 𝑝 = 7𝑚
𝑞 = 3𝑚
𝑟 = 2𝑚
𝑠 = 4 𝑚
Perimeter = 7 + 4 + 1 + 3 + 2 + 2 + 1 + 2 + 2 + 3 + 1 + 4
= 𝟑𝟐 𝒎
7
𝒑
𝒒
𝒓
𝒔

8. EXAMPLE
C a l c u l a t e t h e p e r i m e t e r
o f t h e f i g u r e s h o w n .
Remember that in calculating perimeter,
inside lengths are ignored.
Add all outside lengths.
𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟
= 18.1 + 11.1 + 4.3 + 12.1 + 12.1 + 4.3 + 11.1
= 𝟕𝟑. 𝟏 𝒌𝒎
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9. EXAMPLE
C a l c u l a t e t h e p e r i m e t e r
o f t h e f i g u r e s h o w n .
Remember that in calculating perimeter, inside lengths are
ignored.
Add all outside lengths.
Circumference of circle = 2𝜋𝑟 𝜋 =
22
7
= 2 ×
22
7
× 10.1 = 63.5 𝑚𝑚
Since we half a semi-circle, this value must be divided by two.
The length 𝐵𝐶 = 3.1 𝑚𝑚
The length 𝐴𝐵 = 20.2 𝑚𝑚
𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 =
63.5
2
+ 3.1 + 20.2 + 3.1 = 𝟓𝟖. 𝟐 𝒎𝒎 to 1 d.p.
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10. • Elizabeth is going to make a triangular path in her
garden. The sides will be 20 metres, 36 metres, and
18 metres. What is the total length of the path?
𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 = 𝑎 + 𝑏 + 𝑐
𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑡𝑟𝑖𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑝𝑎𝑡ℎ = 20 + 36 + 18
= 74 𝑚
Therefore the total length of the path = 𝟕𝟒 𝒎
WORD PROBLEMS
• Johanna is building 2 square planter boxes
in her garden. Each box is 2 metres wide.
What is the total perimeter of the boxes?
𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑠𝑞𝑢𝑎𝑟𝑒 = 𝑠𝑖𝑑𝑒 × 4
𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑝𝑙𝑎𝑛𝑡𝑒𝑟 = 2 × 4 = 8𝑚
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑞𝑢𝑎𝑟𝑒 𝑝𝑙𝑎𝑛𝑡𝑒𝑟𝑠 = 2
𝑇𝑜𝑡𝑎𝑙 𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑏𝑜𝑥𝑒𝑠 = 2 × 8𝑚 = 𝟏𝟔𝒎
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11. • Trina hung a rectangular photograph with
decorative tape along the whole perimeter. She
used 24 centimetres of tape. The length of the
photograph is 7 centimetres. What is the width of
the photograph?
𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑙𝑒 = 2(𝑙 + 𝑤)
24 = 2(7 + 𝑤)
24 = 14 + 2𝑤
2𝑤 = 24 − 14 = 10
𝑤 =
10
2
= 5 𝑐𝑚
The width of the photograph = 𝟓 𝒄𝒎
WORD PROBLEMS
• Jesse is making a circular picture frame. He
wants the diameter of the frame to be
28 𝑐𝑚. What length of material does he
need?
𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑐𝑖𝑟𝑐𝑙𝑒 = 𝑐𝑖𝑟𝑐𝑢𝑚𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = 2𝜋𝑟
𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑓𝑟𝑎𝑚𝑒 = 2 ×
22
7
× 14 = 88
𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 = 𝟖𝟖 𝒄𝒎
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12. THE END
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