Simple shapes, added or taken away
A composite shape is made of simpler shapes put together. There is no formula for its area as a whole, but there are formulas for its pieces: rectangles, triangles and circles. So its area is found by adding up the pieces that make it, or by starting with a bigger shape and taking a piece away.
This L-shape fits inside a rectangle 8 squares long and 5 squares high. Start with that whole rectangle. Its area is 8 × 5 = 40 squares.
The whole rectangle, 8 by 5, holds 40 squares.
The piece to take away
The L-shape is this rectangle with one corner missing. The missing corner is 3 squares across and 2 squares down, so it holds 3 × 2 = 6 squares.
The corner to be taken out is 3 across and 2 down: 3 × 2 = 6 squares.
Take it away
Take the corner away from the rectangle, and what is left is the L-shape, with a step in its top edge. Its area is 40 − 6 = 34 squares.
Cutting the L-shape into two rectangles gives the same answer. Cut it upright where the step is: the left piece is 5 across and 5 high, 25 squares, and the right piece is 3 across and 5 − 2 = 3 high, 9 squares, so 25 + 9 = 34. Pieces that do not overlap and leave no gaps always add up to the whole, however the shape is cut. Choose the way that needs the fewest missing lengths, and use another way to check.
box it: 40 − 2 = 38, the whole rectangle minus the missing corner
Make the notch 3 by 2 and box it
Drag the notch until it is 3 across and 2 down. The box is 8 × 5 = 40 and the notch is 3 × 2 = 6, so the L-shape is 40 − 6 = 34.
A slanted edge needs a triangle
Pieces do not have to be rectangles. This outline is a rectangle 6 units wide and 4 high, with a triangle on top. The triangle’s base is the top of the rectangle, 6 units, and its height is 3 units, straight up from that base to the top corner.
The rectangle is 6 × 4 = 24 square units, and the triangle is ½ × 6 × 3 = 9 square units, so the whole shape is 24 + 9 = 33 square units.
The line across splits the outline into a triangle with base 6 and height 3, on top of a rectangle 6 by 4: 9 + 24 = 33.
A curved edge needs part of a circle
A curved edge is usually part of a circle. A semicircle is half a circle, so its area is half of . When a semicircle stands on the end of a rectangle, that end is the semicircle’s diameter, so its radius is half the end.
A rectangle 10 m long and 4 m wide has a semicircle on one of its 4 m ends. The semicircle’s radius is , so its area is . The whole shape is . That is the exact area. With it is about ; keep the exact form until the last step, and round only the final answer.
A hole in the middle
A piece taken away does not have to be at a corner. A hole anywhere inside a shape, like a pond in a lawn, comes off in the same way: find the area of the whole outline, then subtract the area of the hole.
Two slips
Answering 40 gives the rectangle before the corner was taken out. Answering 40 + 6 = 46 adds the corner on, but the corner is missing from the shape, so it comes off: 40 − 6 = 34.
Worked example: A Lawn with a Semicircular End and a Pond, and the Rolls of Turf to Cover It
Question A lawn is a rectangle 12 m long and 8 m wide, with a semicircle on one of its 8 m ends. A rectangular pond 3 m by 2 m lies inside the rectangle and is not turfed. (a) Find the area to be turfed, as an exact expression in π and correct to 1 decimal place. (b) Turf is sold in rolls that each cover 1.5 m², at $6.40 a roll. How many rolls are needed to cover the lawn, and what do they cost?
1.Split the lawn into shapes whose areas we know: the 12 m by 8 m rectangle and the semicircle on its 8 m end, with the pond taken away. The 8 m end is the diameter of the semicircle, so its radius is 4 m.
The lawn is the 12 m by 8 m rectangle and a semicircle of radius 4 m, with the pond taken away. 2.The rectangle has area 12 × 8 = 96 m². The semicircle is half a circle of radius 4 m, so its area is 12 × π × 42 = 8π m². The pond has area 3 × 2 = 6 m².
The rectangle is 96 m², the semicircle 8π m² and the pond 6 m². 3.(a) The area to be turfed is 96 + 8π − 6 = 90 + 8π m². Since 8π ≈ 25.13, this is 115.13 m², which is 115.1 m² correct to 1 decimal place.
(a) The turfed area is 90 + 8π ≈ 115.1 m². 4.Each roll covers 1.5 m², so the lawn needs 115.13 ÷ 1.5 ≈ 76.76 rolls. Rolls are bought whole, and 76 rolls cover only 76 × 1.5 = 114 m², which is less than the lawn. So the number is rounded up to 77.
115.13 ÷ 1.5 ≈ 76.76, and 76 rolls cover only 114 m², so 77 rolls are needed. 5.(b) The lawn needs 77 rolls, which cost 77 × $6.40 = $492.80. Check: 77 rolls cover 77 × 1.5 = 115.5 m², just more than the 115.13 m² of lawn.
(b) 77 rolls cost 77 × $6.40 = $492.80.
Answer: (a) 90 + 8π m², which is 115.1 m² to 1 decimal place; (b) 77 rolls, costing $492.80
Common mistakes
- Taking the radius of the semicircle as 8 m, because the end of the lawn is 8 m long. The 8 m end is the diameter, so the radius is 4 m, and the semicircle's area is 8π m², not 32π m².
- Rounding 76.76 rolls down to 76. Seventy-six rolls cover 114 m² and leave more than 1 m² of bare soil, and part of a roll cannot be bought, so the number of rolls is rounded up.
More volume and surface area problems, worked step by step →