Perimeter

Add the lengths of all three sides.

Opposite sides of a rectangle have equal lengths.

A square has four equal sides.

A pentagon has five sides. A regular polygon has equal side lengths.

Subtract the total of the three known sides from the perimeter.

Use the equal-side marks to find the unlabelled side.

1. The two marked sides are each 10 cm.
2. Perimeter =10+10+8=28 cm.

First find the total length, then convert millimetres to centimetres.

1. 3\times65=195 mm.
2. 195\div10=19.5 cm.

The perimeter contains two widths and two lengths.

1. The two widths total 6+6=12 cm.
2. The two lengths total 38-12=26 cm, so x=26\div2=13 cm.

Find the two unlabelled lengths before adding the outside edges.

1. The inner horizontal length is 10-6=4 cm.
2. The inner vertical length is 8-3=5 cm.
3. Perimeter =10+3+4+5+6+8=36 cm.

Count only edges on the outside, not edges shared by two squares.

1. There are 10 outside edges, each 2 cm long.
2. Perimeter =10\times2=20 cm.

Find the whole perimeter, then account for the gate.

1. Perimeter =2(28+19)=94 m.
2. Fencing =94-3=91 m.

A shared edge is inside the combined shape, so it is not part of the outside boundary.

1. The 8 cm by 5 cm rectangle has perimeter 2(8+5)=26 cm.
2. The 3 cm by 4 cm rectangle has perimeter 2(3+4)=14 cm.
3. The shared 3 cm edge was counted twice. Perimeter =26+14-2\times3=34 cm.

Use the perimeter to find how many metres of edging are required.

1. Perimeter =2(12.4+8.6)=42 m.
2. Cost =42\times2.75=\text{£}115.50.

The horizontal boundary lengths add to twice the total width. Consider the vertical lengths in the same way.

1. Total horizontal boundary length =16+16=32 cm.
2. The vertical boundary lengths total 54-32=22 cm.
3. They add to twice h, so h=22\div2=11 cm.

Find how much wire the triangle uses, then how much is left for the rectangle.

1. Triangle: 9+12+15=36 cm of wire.
2. Rectangle: 108-36=72 cm of wire.
3. 2x+2\times11=72, so 2x=50 and x=25 cm.

Find one tile side, then trace the entire boundary, including the indentation.

1. Each tile side is 18\div4=4.5 cm.
2. The outside boundary contains 16 tile sides; the edges shared by tiles are not counted.
3. Perimeter =16\times4.5=72 cm.

Write a perimeter expression that includes both equal sloping sides.

1. The perimeter is (x+2)+3x+(2x-1)+(2x-1)=8x.
2. 8x=64, so x=8.
3. Check: the side lengths are 10 cm, 24 cm, 15 cm and 15 cm; they total 64 cm.

Count how many widths of fencing are used in total, including the internal fence.

1. Let the width be w metres. The length is 2w metres.
2. The fencing total is 2(2w)+2w+w=7w.
3. 7w=84 gives w=12. The length is 2\times12=24 m.

Find the fencing length, then the number of whole panels that must be bought.

1. Boundary length =2(24.8+14)=77.6 m.
2. Fencing length =77.6-2.6=75 m.
3. 75\div2.4=31.25, so 32 whole panels are needed.
4. Cost =32\times18.50=\text{£}592.

Use the equal areas to find the missing width of Bed B, then consider the separate boundaries.

1. Bed A has area 4\times3=12 square metres. Bed B has the same area, so its width is 12\div6=2 m.
2. Outside boundary =2(18+12)=60 m.
3. First bed boundary =2(4+3)=14 m.
4. Second bed boundary =2(6+2)=16 m.
5. Total edging =60+14+16=90 m.