Area of Rectangles, Triangles, Parallelograms and Trapezia

Area of a rectangle = length × width.

A square has the same length and width.

Find base times height, then halve it.

Choose the length marked perpendicular to the base.

Add the parallel sides before multiplying by the height and halving.

Use length times width, keeping the decimal in the calculation.

1. A=3.5\times2=7\text{ m}^2.
2. The answer measures the surface of the mat.

The perpendicular height meets the extended line of the base at a right angle. Use this height in the triangle area formula.

1. The base is 9 cm and the perpendicular height is 4 cm. The height can meet an extension of the base; it does not have to lie inside the triangle.
2. A=\frac12\times9\times4=18\text{ cm}^2.

What number multiplied by 7 makes 63?

1. \text{length}\times7=63.
2. \text{length}=63\div7=9 cm.

Undo multiplying by the base.

1. 12\times h=72.
2. h=72\div12=6 m.

A trapezium is half of a parallelogram made from two matching copies.

1. Arun has found twice the required area. He must halve 60.
2. The area is 60\div2=30\text{ cm}^2.

The two sides meeting at the right angle can be the base and height.

1. Use 8 cm as the base and 6 cm as its perpendicular height.
2. A=\frac12\times8\times6=24\text{ cm}^2.
3. The 10 cm side is not a perpendicular height.

Find the trapezium’s area first, then multiply by the cost per square metre.

1. A=\frac12(5+11)\times7=56\text{ m}^2.
2. The cost is 56\times2=112, so the mulch costs £112.

Convert 75 cm to metres before multiplying.

1. 75\div100=0.75 m.
2. A=2.4\times0.75=1.8\text{ m}^2.

Double the area first to undo the halving in the triangle formula.

1. 42\times2=84, so 12\times h=84.
2. h=84\div12=7 cm.
3. Check: \frac12\times12\times7=42\text{ cm}^2.

Find the floor area, then check how many whole packs cover all of it.

1. The area is 8\times3.5=28\text{ m}^2.
2. 28\div3=9\frac13 packs.
3. Nine packs cover only 27\text{ m}^2, so 10 whole packs are needed.

Find the rectangle’s area, then work backwards with the triangle formula.

1. The rectangle’s area is 9\times6=54\text{ cm}^2.
2. For the triangle, \frac12\times12\times h=54.
3. h=54\times2\div12=9 cm.

Doubling the area and dividing by the height gives the sum of the parallel sides.

1. 60\times2\div6=20 cm is the sum of the parallel sides.
2. The missing side is 20-7=13 cm.
3. Check: \frac12(7+13)\times6=60\text{ cm}^2.

Use rectangle A to find the common perimeter, then use it to find the unknown side of B before comparing areas.

Part (a)

1. Rectangle A has perimeter 2(10+4)=28 cm.
2. For rectangle B, 2(x+6)=28, so x+6=14 and x=8 cm.

Part (b)

1. A_A=10\times4=40\text{ cm}^2.
2. A_B=8\times6=48\text{ cm}^2.
3. The difference is 48-40=8\text{ cm}^2. Equal perimeters do not guarantee equal areas.

Convert the height to metres, find one face, then include the second face.

1. 90\text{ cm}=0.9 m.
2. One face has area \frac12(1.8+2.6)\times0.9=1.98\text{ m}^2.
3. Both faces have total area 2\times1.98=3.96\text{ m}^2.

Compare the two areas first. Only the extra area needs to be multiplied by the price.

1. Triangle area: \frac12\times18\times11=99\text{ m}^2.
2. Rectangle area: 8\times13=104\text{ m}^2.
3. The rectangle has 104-99=5\text{ m}^2 more area.
4. The extra cost is 5\times12=60, so it costs £60 more.