Perimeter and Area of a Rectangle

Worked example

1. 1

   $l=9,w=4$ — The rectangle is 9 cm long and 4 cm wide. Write both down first so you use the right numbers.

2. 2

   $9+4=13$ — Add one length and one width. That is half the way round: two of the four sides.

3. 3

   $P=2\times 13=26\text{ cm}$ — Double it, because the other length and width are the same. Perimeter is a length, so the unit is cm.

4. 4

   $A=9\times 4=36{\text{ cm}}^2$ — Length times width: 4 rows of 9 make 36 squares. Area counts squares, so the unit is cm².

Worked example

1. 1

   $9\text{ whole squares}$ — Count the squares fully inside the shape first. Each one is 1 cm².

2. 2

   $4\times \frac{1}{2}=2$ — The shape also cuts 4 squares in half along their diagonals. Two halves make one whole square.

3. 3

   $A=9+2=11{\text{ cm}}^2$ — Add the whole squares and the halves for the total area.

Worked example

1. 1

   $80+55=135$ — Add one length and one width. This is only half the way round.

2. 2

   $2\times 135=270\text{ m}$ — Double it to cover all four sides. That is one walk round the field.

3. 3

   $4\times 270=1080\text{ m}$ — She walks round 4 times, so multiply one lap by 4.

Worked example

1. 1

   $12+4=16$ — Add one length and one width of the rectangle.

2. 2

   $P=2\times 16=32\text{ cm}$ — Double it for all four sides. The square has this same perimeter.

3. 3

   $\text{side}=32÷4=8\text{ cm}$ — A square has 4 equal sides, so share the perimeter between 4.

4. 4

   $8\times 8=64{\text{ cm}}^2$ — Area of the square: side times side.

5. 5

   $12\times 4=48{\text{ cm}}^2$ — Area of the rectangle: length times width.

6. 6

   $64-48=16{\text{ cm}}^2$ — The question asks for the difference, so take the smaller area from the larger.

Worked example

1. 1

   $A=529{\text{ cm}}^2$ — A square has area 529 cm². You need the side before you can find the perimeter.

2. 2

   $s^2=529$ — The area of a square is side times side, so the side squared is 529.

3. 3

   $s=\sqrt{529}=23\text{ cm}$ — Square root undoes squaring. Check: 23 × 23 = 529. Use 23, not −23: a length cannot be negative.

4. 4

   $P=4\times 23=92\text{ cm}$ — Four equal sides, so multiply the side by 4.

Worked example

1. 1

   $A=45,l=9$ — The area is 45 cm² and the length is 9 cm. The width is missing.

2. 2

   $9\times w=45$ — Area is length times width, so write that with w for the missing side.

3. 3

   $w=45÷9=5\text{ cm}$ — Divide by 9 to undo the times. Check: 9 × 5 = 45.

4. 4

   $P=30,l=9$ — A new question: the perimeter is 30 cm and the length is 9 cm. Find the width.

5. 5

   $30-2\times 9=12$ — The perimeter holds two lengths. Take both off: 30 − 18 = 12 is left for the two widths.

6. 6

   $w=12÷2=6\text{ cm}$ — Two equal widths share the 12, so halve it. Check: 2(9 + 6) = 30.

Worked example

1. 1

   $l+w=26÷2=13$ — Half the perimeter is one length plus one width.

2. 2

   $l\times w=40$ — The area is length times width.

3. 3

   $5+8=13,5\times 8=40$ — Find two numbers that add to 13 and multiply to 40. Pairs that multiply to 40: 1 and 40, 2 and 20, 4 and 10, 5 and 8.

4. 4

   $8\text{ cm by }5\text{ cm}$ — Draw it 8 squares long and 5 squares wide on the grid.

Worked example

1. 1

   $3\times 16=48{\text{ cm}}^2$ — The area of A. B has this same area.

2. 2

   $\text{length of B}=48÷6=8\text{ cm}$ — Area ÷ width undoes the times and gives B's missing side.

3. 3

   $P_A=2(3+16)=38\text{ cm}$ — Perimeter of A: add a length and a width, then double.

4. 4

   $P_B=2(6+8)=28\text{ cm}$ — Perimeter of B, the same way.

5. 5

   $38-28=10\text{ cm}$ — A has the longer perimeter, by 10 cm. A long thin rectangle has more edge for the same area.

Worked example

1. 1

   $2((x+5)+x)=26$ — Perimeter is twice (length + width). Put both expressions in and set it equal to 26.

2. 2

   $2(2x+5)=26$ — Collect the x's inside the bracket: x + x = 2x.

3. 3

   $4x+10=26$ — Multiply out the bracket: 2 × 2x = 4x and 2 × 5 = 10.

4. 4

   $4x=16$ — Take 10 from both sides.

5. 5

   $x=4$ — Divide both sides by 4.

6. 6

   $\text{length}=4+5=9,\text{width}=4$ — Put x = 4 back into each side. Check: 2(9 + 4) = 26.

7. 7

   $A=9\times 4=36{\text{ cm}}^2$ — The question asks for the area, so finish with length times width.