Perimeter and Area of Composite Figures

Find missing lengths, perimeters and areas of figures made of rectangles, triangles, parallelograms and trapeziums, by adding parts or by subtracting them from a larger shape.

Before you start

Area of a rectangle $=$ length $\times$ breadth; area of a triangle $= \frac{1}{2} \times$ base $\times$ height; area of a parallelogram $=$ base $\times$ height, $A = bh$ ; area of a trapezium $= \frac{1}{2} \times$ (sum of parallel sides) $\times$ height, $A = \frac{1}{2}(a + b)h$ . In each, the height is the perpendicular height, not a slant side.
The perimeter of a figure is the total length of its boundary.
Change lengths to one unit before you multiply. Since 1 m = 100 cm, a rectangle 1.2 m by 45 cm has area $120 \times 45 = 5400\text{ cm}^{2}$ .

What a Composite Figure Is

A composite figure is a figure made up of two or more simple shapes: rectangles, squares, triangles, parallelograms and trapeziums. The floor plan of a flat, the front wall of a hut and a lawn with a path round it are all composite figures.

To find its area, cut it into simple shapes, or see it as a larger simple shape with pieces taken away. In the figure below, the dashed line cuts the figure into a rectangle and a triangle. The dashed line helps with the area, but it is inside the figure: it is not part of the boundary, so it is not part of the perimeter.

Composite figures

A composite figure is a figure made up of two or more simple shapes.

Its area is found by adding the areas of its parts, or by subtracting the areas of the parts removed from a larger shape.

Its perimeter is the total length of its boundary. A line where two parts meet is inside the figure and is not part of the perimeter.

Finding Missing Lengths

Often not every edge of a composite figure is labelled. When all its corners are right angles, each missing length can be found from the given ones by adding or subtracting.

Look at the figure from above: the edges you see across the top cover its whole width. Look from below: the edges you see along the bottom cover the same width. In the same way, the upright edges seen from the left and those seen from the right each cover the whole height.

Missing lengths

In a figure whose corners are all right angles, the horizontal edges seen from above and the horizontal edges seen from below each add up to the full width.

The vertical edges seen from the left and the vertical edges seen from the right each add up to the full height.

Example 1: Missing lengths in an L-shape

The L-shaped figure $ABCDEF$ is made of two rectangles, and all its corners are right angles. Find the lengths of $AF$ and $CD$ .

Solution

Seen from the left, the full height is $AF$ . Seen from the right, the full height is made of $BC$ and $DE$ :

$AF = BC + DE = 8 + 4 = 12\text{ cm}$

Seen from above, the full width is $FE$ . Seen from below, the full width is made of $AB$ and $CD$ , so $CD$ is what is left when $AB$ is taken from $FE$ :

$CD = FE - AB = 12 - 4 = 8\text{ cm}$

Check: $AB + CD = 4 + 8 = 12\text{ cm}$ , which is $FE$ .

$AF = 12\text{ cm}$ and $CD = 8\text{ cm}$ .

Perimeter: the Boundary Only

To find a perimeter, start at a corner and go round the boundary, edge by edge, until you are back where you started. Write down every edge as you pass it, find any missing length first, then add. A line inside the figure is never on the boundary.

Perimeter of a composite figure

The perimeter is the total length of the boundary: go round it edge by edge and add the lengths.

When two shapes are joined edge to edge, the length along which they touch is inside the new figure. Perimeter of the figure $=$ sum of the two perimeters $-$ $2 \times$ (length along which they touch).

Example 2: The perimeter of the L-shape

Find the perimeter of the L-shaped figure $ABCDEF$ of Example 1.

Solution

From Example 1, $CD = 8$ cm and $AF = 12$ cm. Go round from $A$ : $A \to B \to C \to D \to E \to F \to A$ .

$P = AB + BC + CD + DE + EF + FA$

$P = 4 + 8 + 8 + 4 + 12 + 12$

$P = 48\text{ cm}$

Check a second way: $AB$ and $CD$ together are as long as $FE$ , and $BC$ and $DE$ together are as long as $AF$ . So the perimeter is the same as that of a square 12 cm by 12 cm: $2 \times (12 + 12) = 48$ cm.

The perimeter of the figure is $48\text{ cm}$ .

Example 3: Joining two rectangles

The T-shaped figure is made of a bar, a rectangle 20 cm by 5 cm, with a stem, a rectangle 8 cm by 10 cm, joined to the middle of its lower side. Ali says its perimeter is the perimeter of the bar added to the perimeter of the stem, $50 + 36 = 86$ cm. Find the perimeter, and explain where Ali went wrong.

Solution

The stem is in the middle, so the two short edges beside it are equal (the ticks show this). Each is $(20 - 8) \div 2 = 6$ cm.

Go round clockwise from the top left corner: the top 20, the right side of the bar 5, the edge beside the stem 6, the right side of the stem 10, the bottom 8, the left side of the stem 10, the edge beside the stem 6, the left side of the bar 5.

$P = 20 + 5 + 6 + 10 + 8 + 10 + 6 + 5 = 70\text{ cm}$

Where Ali went wrong: the bar and the stem touch along 8 cm (the pink dashed line). That line is in the bar's perimeter, 50 cm, and again in the stem's, 36 cm, but it is inside the T-shape, so it is not on the boundary at all. It must come off twice:

$50 + 36 - 8 - 8 = 70\text{ cm}$ , which agrees with going round.

The perimeter is $70\text{ cm}$ . Ali counted the 8 cm line where the two rectangles touch, twice, although it is not on the boundary at all.

Example 2 showed a figure whose perimeter is the same as that of the rectangle round it. This is true of every staircase figure: a rectangle with steps cut from one or more of its corners.

Example 4: A staircase figure

All the corners of this figure are right angles. Only two lengths are given. (a) Find its perimeter. (b) Can its area be found from these two lengths?

Solution

(a) Seen from above, the tops of the three steps cover the whole width, so together they are as long as the bottom edge, 15 m. Seen from the right, the fronts of the three steps cover the whole height, so together they are as long as the left side, 10 m.

Go round from the bottom left corner: the bottom 15, then the fronts and tops of the steps, then the left side 10.

$P = 15 + (\text{fronts of the steps}) + (\text{tops of the steps}) + 10$

$P = 15 + 10 + 15 + 10 = 50\text{ m}$

This is the perimeter of the 15 m by 10 m rectangle round the figure: $2 \times (15 + 10) = 50$ m.

(b) No. The area depends on how wide and how tall each step is, and these are not given. Two staircases 15 m wide and 10 m tall can have different areas, but they always have the same perimeter, 50 m.

(a) The perimeter is $50\text{ m}$ . (b) No: the area needs the sizes of the steps.

Staircase figures

A figure that is a rectangle with steps cut from its corners has the same perimeter as the rectangle.

This is not true when a piece is cut from the middle of a side (a notch): the two sides of the notch are added to the boundary, as in Example 6.

Area: Adding or Subtracting

There are two ways to find the area of a composite figure. Adding: cut it into simple shapes with dashed lines and add their areas. Subtracting: see it as a larger simple shape with pieces removed, and subtract the pieces. Both ways give the same answer, so one way checks the other. Use whichever needs fewer steps.

Example 5: The L-shape, both ways

Find the area of the L-shaped figure $ABCDEF$ of Example 1, (a) by adding, (b) by subtracting.

Solution

(a) By adding. A dashed line from $C$ to $G$ on $AF$ cuts the figure into rectangle X, 12 cm by 4 cm, and rectangle Y, 4 cm by 8 cm.

Area of X $= 12 \times 4 = 48\text{ cm}^{2}$

Area of Y $= 4 \times 8 = 32\text{ cm}^{2}$

Area $= 48 + 32 = 80\text{ cm}^{2}$

(b) By subtracting. Draw the square $AHEF$ round the figure, 12 cm by 12 cm. The figure is this square with the square $BHDC$ , 8 cm by 8 cm, removed.

Area $= 12 \times 12 - 8 \times 8 = 144 - 64 = 80\text{ cm}^{2}$

The area is $80\text{ cm}^{2}$ , the same both ways.

Example 6: A U-shaped flower bed

In a community garden, a flower bed is a rectangle 16 m by 10 m with a rectangular piece removed from the middle of one long side. The piece is 8 m along the side and 6 m deep. Find (a) the perimeter and (b) the area of the flower bed.

Solution

(a) The piece is in the middle, so the two parts of the long side left beside it are equal: each is $(16 - 8) \div 2 = 4$ m.

Go round anticlockwise from the bottom left corner: the bottom 16, the right side 10, the top right part 4, down the notch 6, across the bottom of the notch 8, up the notch 6, the top left part 4, the left side 10.

$P = 16 + 10 + 4 + 6 + 8 + 6 + 4 + 10 = 64\text{ m}$

Check: the whole rectangle has perimeter $2 \times (16 + 10) = 52$ m. The notch takes 8 m of the long side off the boundary and puts back the bottom of the notch, also 8 m, and adds its two sides, $6 + 6 = 12$ m. So $P = 52 + 12 = 64$ m. The perimeter is more than the rectangle's, because the piece was cut from the middle of a side.

(b) By subtracting: Area $= 16 \times 10 - 8 \times 6 = 160 - 48 = 112\text{ m}^{2}$

Check by adding: the strip along the bottom is 16 m by $10 - 6 = 4$ m, and each arm is 4 m by 6 m: $16 \times 4 + 2 \times (4 \times 6) = 64 + 48 = 112\text{ m}^{2}$

(a) The perimeter is $64\text{ m}$ . (b) The area is $112\text{ m}^{2}$ .

Figures with Triangles, Parallelograms and Trapeziums

A composite figure may have slanted edges. Cut it so that every part is a shape whose area you know, and use each part's own perpendicular height. A missing height or parallel side is often found by subtracting, as before.

The perimeter needs the length of every slant side on the boundary: these are given in the question. Go round the boundary as before; the dashed lines you drew for the area are not on it.

Example 7: The front of a hut

The front wall of a hut is a rectangle 8 m wide and 6 m tall with a triangle on top. The triangle's perpendicular height is 300 cm and its two slant sides are 5 m each. Find (a) the area and (b) the perimeter of the front wall.

Solution

The height is in centimetres and every other length is in metres, so change it first: 300 cm = 3 m.

(a) The dashed line cuts the wall into a rectangle 8 m by 6 m and a triangle with base 8 m and perpendicular height 3 m.

Area of rectangle $= 8 \times 6 = 48\text{ m}^{2}$

Area of triangle $= \frac{1}{2} \times 8 \times 3 = 12\text{ m}^{2}$

Area of wall $= 48 + 12 = 60\text{ m}^{2}$

(b) The dashed lines are inside the figure, so they are not on the boundary, and the height of 3 m is not part of the perimeter either. Go round from the bottom left corner: the bottom 8, the right side 6, the two slant sides 5 and 5, the left side 6.

$P = 8 + 6 + 5 + 5 + 6 = 30\text{ m}$

(a) The area of the front wall is $60\text{ m}^{2}$ . (b) Its perimeter is $30\text{ m}$ .

Example 8: A garden plot, both ways, with costs

Mr Tan's plot in a community garden has the shape shown; the three corners marked are right angles, and the arrows show that the top edge is parallel to the bottom edge.

(a) Find the area of the plot in two ways.

(b) He covers the plot with turf at $4.50 per $\text{m}^{2}$ and puts a fence round it at $12 per metre. Find the total cost.

Solution

(a) By adding. A dashed line down from the top of the slant side cuts the plot into rectangle X and trapezium Y.

Rectangle X is 8 m by 6 m: area $= 8 \times 6 = 48\text{ m}^{2}$ .

Trapezium Y has parallel sides 6 m and 3 m, both upright (marked with double arrows, since they are a different pair from the top and bottom). Its height is the distance between them, $12 - 8 = 4$ m: area $= \frac{1}{2} \times (6 + 3) \times 4 = 18\text{ m}^{2}$ .

Area $= 48 + 18 = 66\text{ m}^{2}$

By subtracting. The plot is a rectangle 12 m by 6 m with a right-angled triangle cut from its top right corner. The triangle's two shorter sides are $12 - 8 = 4$ m and $6 - 3 = 3$ m.

Area $= 12 \times 6 - \frac{1}{2} \times 4 \times 3 = 72 - 6 = 66\text{ m}^{2}$

(b) Turf: $66 \times 4.50 = 297$ , so the turf costs $297.

Fence: go round from the bottom left corner: $P = 12 + 3 + 5 + 8 + 6 = 34\text{ m}$ . Then $34 \times 12 = 408$ , so the fence costs $408.

Total: $297 + 408 = 705$ , so the total cost is $705.

(a) The area of the plot is $66\text{ m}^{2}$ , both ways. (b) The total cost is $705.

Example 9: A rectangle and a parallelogram

A sticker is made of a rectangle and a parallelogram, as shown. The arrows mark the parallelogram's two pairs of parallel sides, and the dashed line is its perpendicular height. Find (a) the area and (b) the perimeter of the sticker.

Solution

(a) The dashed line across the figure cuts it into a rectangle 10 cm by 3 cm and a parallelogram. The parallelogram's base is the top of the rectangle, 10 cm, and its perpendicular height is 4 cm; the slant side, 5 cm, is not its height.

Area of rectangle $= 10 \times 3 = 30\text{ cm}^{2}$

Area of parallelogram $= 10 \times 4 = 40\text{ cm}^{2}$

Area of sticker $= 30 + 40 = 70\text{ cm}^{2}$

(b) Opposite sides of a parallelogram are equal, so its top edge is 10 cm and its other slant side is 5 cm. The left side of the rectangle is 3 cm, like the right side. The line where the two shapes meet and the dashed height are inside the figure, so they are not on the boundary.

Go round from the bottom left corner: the bottom 10, the right side 3, the slant side 5, the top 10, the other slant side 5, the left side 3.

$P = 10 + 3 + 5 + 10 + 5 + 3 = 36\text{ cm}$

(a) The area of the sticker is $70\text{ cm}^{2}$ . (b) Its perimeter is $36\text{ cm}$ .

Shaded Regions

When a question asks for the area of a shaded region, see which shapes are taken away from which. The shaded area is the area of the whole shape minus the areas of the unshaded parts. Where the holes are does not matter, only their sizes.

Example 10: A stencil

Ravi makes a stencil by cutting a right-angled triangle and a square out of a rectangular card 16 cm by 10 cm. The triangle's two shorter sides are 6 cm and 5 cm, and the square's sides are 3 cm. Find the area of the card that is left (shaded).

Solution

Area of card $= 16 \times 10 = 160\text{ cm}^{2}$

Area of triangle $= \frac{1}{2} \times 6 \times 5 = 15\text{ cm}^{2}$

Area of square $= 3 \times 3 = 9\text{ cm}^{2}$

Shaded area $= 160 - 15 - 9 = 136\text{ cm}^{2}$

The area of the card left is $136\text{ cm}^{2}$ .

A path round a lawn, a frame round a picture and a border of tiles round a pool are shaded regions of a special kind: the region between two rectangles, one inside the other. They are the subject of the lesson on the area of paths and borders.

Common mistakes

Counting a line inside the figure. The line where two parts meet is not on the boundary. Adding the perimeters of the parts counts it twice (Example 3).
Forgetting what a notch does. When a piece is cut from the middle of a side, the part of the side it took is no longer on the boundary, and the three edges of the notch (its two sides and its bottom) are. The flower bed of Example 6 has perimeter 64 m: not the rectangle's 52 m, and not $52 + 6 + 8 + 6 = 72$ m, which keeps the 8 m of the side that was removed.
Using a slant side as a height. The area of a triangle, parallelogram or trapezium uses its perpendicular height; a slant side is used only in the perimeter.
Multiplying lengths in different units. Change 300 cm to 3 m (or 3 m to 300 cm) before you multiply (Example 7).
Thinking the same perimeter means the same area. Staircase figures 15 m by 10 m all have perimeter 50 m, but their areas differ (Example 4).

Practice

All the corners of these figures are right angles unless the figure shows otherwise. Work each question before you open its solution.

Practice 1: The floor plan of a flat

The figure is the floor plan of the L-shaped living and dining area of an HDB flat.

(a) Find $CD$ and $DE$ . (b) Find the perimeter of the floor. (c) Find the area of the floor. (d) Vinyl flooring costs $38 per $\text{m}^{2}$ . Find the cost of covering the floor.

ShowHide solution

(a) Seen from above, the width is $FE + CD$ ; seen from below it is $AB$ : $CD = AB - FE = 9 - 5 = 4$ m.

Seen from the left, the height is $AF$ ; seen from the right it is $BC + DE$ : $DE = AF - BC = 7 - 4 = 3$ m.

(b) $P = AB + BC + CD + DE + EF + FA = 9 + 4 + 4 + 3 + 5 + 7 = 32\text{ m}$

Check: it is a staircase figure, so $P = 2 \times (9 + 7) = 32$ m.

(c) By adding: a line from $D$ across to $AF$ cuts it into a rectangle 9 m by 4 m and a rectangle 5 m by 3 m: $9 \times 4 + 5 \times 3 = 36 + 15 = 51\text{ m}^{2}$

Check by subtracting: $9 \times 7 - 4 \times 3 = 63 - 12 = 51\text{ m}^{2}$

(d) $51 \times 38 = 1938$ , so the flooring costs $1938.

(a) $CD = 4\text{ m}$ , $DE = 3\text{ m}$ (b) $32\text{ m}$ (c) $51\text{ m}^{2}$ (d) $1938

Practice 2: Two ways to join

Siti has two rectangles of card, each 9 cm by 3 cm. She joins them edge to edge to make a T-shape, with the stem joined to the middle of the bar, then moves the stem to one end to make an L-shape, as shown.

(a) Find the perimeter of each figure. (b) Find the area of each figure. (c) Siti says the perimeter of each figure is $24 + 24 = 48$ cm. Explain her mistake.

ShowHide solution

(a) T-shape: the stem is in the middle (the ticks), so each edge beside it is $(9 - 3) \div 2 = 3$ cm. Clockwise from the top left corner:

$P = 9 + 3 + 3 + 9 + 3 + 9 + 3 + 3 = 42\text{ cm}$

L-shape: the left side is $3 + 9 = 12$ cm and the edge beside the stem is $9 - 3 = 6$ cm. Clockwise from the top left corner:

$P = 9 + 3 + 6 + 9 + 3 + 12 = 42\text{ cm}$

(b) Each figure is the two rectangles: $2 \times (9 \times 3) = 54\text{ cm}^{2}$ .

(c) Each rectangle has perimeter $2 \times (9 + 3) = 24$ cm. In both figures the rectangles touch along 3 cm (the dashed line), which is in both perimeters but inside the figure. It must come off twice: $48 - 3 - 3 = 42$ cm.

(a) Both perimeters are $42\text{ cm}$ . (b) Both areas are $54\text{ cm}^{2}$ . (c) She counted the 3 cm line where the rectangles touch, twice, although it is not on the boundary.

Practice 3: A cross

A badge is a cross made of two rectangles, each 15 cm by 5 cm, crossing at right angles at their middles.

(a) Find how far each arm reaches beyond the square in the middle. (b) Find the perimeter of the badge. (c) Find its area in two ways.

ShowHide solution

(a) Each rectangle is 15 cm long; the square in the middle takes 5 cm of it, and the two arms left are equal: $(15 - 5) \div 2 = 5$ cm.

(b) Going round, each of the 4 arms gives three edges of 5 cm: a side, the end and the other side. $P = 4 \times (5 + 5 + 5) = 60\text{ cm}$

(c) By adding: the cross is 5 squares of side 5 cm (the middle and the 4 arms): $5 \times (5 \times 5) = 125\text{ cm}^{2}$

From the two rectangles: $15 \times 5 + 15 \times 5 = 150\text{ cm}^{2}$ counts the middle square twice, so the area is $150 - 5 \times 5 = 125\text{ cm}^{2}$ .

(a) 5 cm (b) $60\text{ cm}$ (c) $125\text{ cm}^{2}$

Practice 4: Steps up to a stage

The side wall of a flight of four concrete steps up to the stage of a school hall has the shape shown. Each step is 20 cm high and 30 cm deep. Find (a) the perimeter and (b) the area of the side wall.

ShowHide solution

(a) The bottom is $4 \times 30 = 120$ cm and the back is $4 \times 20 = 80$ cm.

Go round from the bottom left corner: the bottom 120, then four fronts of 20 and four tops of 30, then down the back 80.

$P = 120 + 4 \times 20 + 4 \times 30 + 80 = 120 + 80 + 120 + 80 = 400\text{ cm}$

Check: it is a staircase figure, so $P = 2 \times (120 + 80) = 400$ cm.

(b) By adding: cut it into four upright strips 30 cm wide, of heights 80 cm, 60 cm, 40 cm and 20 cm:

Area $= 30 \times 80 + 30 \times 60 + 30 \times 40 + 30 \times 20 = 2400 + 1800 + 1200 + 600 = 6000\text{ cm}^{2}$

Check by subtracting: the rectangle round it is $120 \times 80 = 9600\text{ cm}^{2}$ , and the empty space above the steps is three strips 30 cm wide and 20 cm, 40 cm and 60 cm tall, $30 \times (20 + 40 + 60) = 3600\text{ cm}^{2}$ : $9600 - 3600 = 6000\text{ cm}^{2}$ .

(a) $400\text{ cm}$ (b) $6000\text{ cm}^{2}$

Practice 5: A T-shaped void deck

The floor of a void deck is T-shaped: a rectangle 18 m by 6 m with a rectangle 8 m wide and 10 m long joined to the middle of one long side, as shown. (a) Find the perimeter of the floor. (b) Find the area of the floor in two ways. (c) The floor is tiled at $42 per $\text{m}^{2}$ . Find the cost.

ShowHide solution

(a) The lower part is in the middle, so each edge beside it is $(18 - 8) \div 2 = 5$ m. Clockwise from the top left corner:

$P = 18 + 6 + 5 + 10 + 8 + 10 + 5 + 6 = 68\text{ m}$

Check: it is a staircase figure, a rectangle 18 m by $6 + 10 = 16$ m with a step cut from each lower corner, so $P = 2 \times (18 + 16) = 68$ m.

(b) By adding: $18 \times 6 + 8 \times 10 = 108 + 80 = 188\text{ m}^{2}$

By subtracting: from the rectangle 18 m by 16 m round it, take the two pieces beside the lower part, each 5 m by 10 m: $18 \times 16 - 2 \times (5 \times 10) = 288 - 100 = 188\text{ m}^{2}$

(a) $68\text{ m}$ (b) $188\text{ m}^{2}$ (c) $7896

Practice 6: A garden with a notch

A school garden is a rectangle 20 m by 12 m with a rectangular piece, 6 m along the side and 5 m deep, removed from the middle of one long side for a shed. (a) Find the perimeter of the garden. (b) Find its area in two ways. (c) An edging strip is laid all round the garden at $9 per metre. Find the cost.

ShowHide solution

(a) The piece is in the middle (the ticks), so each part of the long side left beside it is $(20 - 6) \div 2 = 7$ m.

Anticlockwise from the bottom left corner: the bottom 20, the right side 12, the top right part 7, down the notch 5, across its bottom 6, up the notch 5, the top left part 7, the left side 12.

$P = 20 + 12 + 7 + 5 + 6 + 5 + 7 + 12 = 74\text{ m}$

Check: the whole rectangle has perimeter $2 \times (20 + 12) = 64$ m. The bottom of the notch, 6 m, replaces the 6 m of side it took, and its two sides are added: $64 + 5 + 5 = 74$ m.

(b) By subtracting: $20 \times 12 - 6 \times 5 = 240 - 30 = 210\text{ m}^{2}$

By adding: the strip along the bottom is 20 m by $12 - 5 = 7$ m, and each arm is 7 m by 5 m: $20 \times 7 + 2 \times (7 \times 5) = 140 + 70 = 210\text{ m}^{2}$

(a) $74\text{ m}$ (b) $210\text{ m}^{2}$ (c) $666

Practice 7: A corner cut off

A door sign is made from a rectangle of wood 20 cm by 12 cm by cutting off a right-angled triangle at one corner. The triangle's two shorter sides are 8 cm and 6 cm, and the cut edge is 10 cm long. Find (a) the area and (b) the perimeter of the sign.

ShowHide solution

(a) Area $= 20 \times 12 - \frac{1}{2} \times 8 \times 6 = 240 - 24 = 216\text{ cm}^{2}$

(b) The top edge left is $20 - 8 = 12$ cm and the right edge left is $12 - 6 = 6$ cm. Anticlockwise from the bottom left corner: the bottom 20, the right edge 6, the cut 10, the top 12, the left side 12.

$P = 20 + 6 + 10 + 12 + 12 = 60\text{ cm}$

Check: the rectangle's perimeter is $2 \times (20 + 12) = 64$ cm. The cut takes 8 cm and 6 cm off the boundary and adds the 10 cm edge: $64 - 8 - 6 + 10 = 60$ cm.

(a) $216\text{ cm}^{2}$ (b) $60\text{ cm}$

Practice 8: A plate with two holes

In a design and technology lesson, Ali cuts a square hole of side 4 cm and a right-angled triangular hole, with shorter sides 6 cm and 5 cm, out of a rectangular metal plate 20 cm by 15 cm. Find the area of metal left (shaded).

ShowHide solution

Area of plate $= 20 \times 15 = 300\text{ cm}^{2}$

Area of square hole $= 4 \times 4 = 16\text{ cm}^{2}$

Area of triangular hole $= \frac{1}{2} \times 6 \times 5 = 15\text{ cm}^{2}$

Area of metal left $= 300 - 16 - 15 = 269\text{ cm}^{2}$

Where the holes are does not matter: only their sizes.

The area of metal left is $269\text{ cm}^{2}$ .

Practice 9: A parallelogram and a triangle

The figure is made of parallelogram $ABCD$ and triangle $DEC$ . The dashed lines are perpendicular heights. Find (a) the area and (b) the perimeter of the figure.

ShowHide solution

(a) Parallelogram $ABCD$ : base $AB = 10$ cm and perpendicular height 6 cm. The slant side $BC = 7.5$ cm is not the height.

Area of parallelogram $= 10 \times 6 = 60\text{ cm}^{2}$

Triangle $DEC$ : its base $DC = AB = 10$ cm (opposite sides of a parallelogram are equal) and its perpendicular height is 4.8 cm.

Area of triangle $= \frac{1}{2} \times 10 \times 4.8 = 24\text{ cm}^{2}$

Area of figure $= 60 + 24 = 84\text{ cm}^{2}$

(b) $AD = BC = 7.5$ cm (opposite sides of a parallelogram are equal). $DC$ is inside the figure, so it is not on the boundary. Go round from $A$ : $AB$ , $BC$ , $CE$ , $ED$ , $DA$ .

$P = 10 + 7.5 + 8 + 6 + 7.5 = 39\text{ cm}$

(a) The area of the figure is $84\text{ cm}^{2}$ . (b) Its perimeter is $39\text{ cm}$ .

Practice 10: The front of a pavilion

The front of a pavilion in a park is a rectangle 14 m wide and 4 m tall with a trapezium on top. The trapezium's parallel sides are 14 m and 6 m, its perpendicular height is 3 m and its slant sides are 5 m each. (a) Find the area of the front. (b) A string of lights is fixed all round the edge of the front at $8.50 per metre. Find the cost.

ShowHide solution

(a) The dashed line cuts the front into a rectangle 14 m by 4 m and a trapezium with parallel sides 14 m and 6 m and perpendicular height 3 m.

Area of rectangle $= 14 \times 4 = 56\text{ m}^{2}$

Area of trapezium $= \frac{1}{2} \times (14 + 6) \times 3 = 30\text{ m}^{2}$

Area of front $= 56 + 30 = 86\text{ m}^{2}$

(b) The 14 m line where the shapes meet and the dashed height are not on the boundary. Go round from the bottom left corner: the bottom 14, the right side 4, a slant side 5, the top 6, the other slant side 5, the left side 4.

$P = 14 + 4 + 5 + 6 + 5 + 4 = 38\text{ m}$

$38 \times 8.50 = 323$ , so the lights cost $323.

(a) $86\text{ m}^{2}$ (b) $323

Practice 11: A corner for a bin centre

The grass patch beside a block of HDB flats is a rectangle 12 m by 8 m. A right-angled triangle is cut from one corner for a bin centre; its two shorter sides are 3 m and 400 cm, and its third side is 5 m. (a) Find the area of grass left. (b) The grass left is covered with new turf at $6.80 per $\text{m}^{2}$ , and a low fence is put all round its edge at $15 per metre. Find the total cost.

ShowHide solution

Change to metres first: 400 cm = 4 m.

(a) Area of grass $= 12 \times 8 - \frac{1}{2} \times 3 \times 4 = 96 - 6 = 90\text{ m}^{2}$

(b) The bottom edge left is $12 - 3 = 9$ m and the right edge left is $8 - 4 = 4$ m. Going round anticlockwise from the bottom left corner: the bottom 9, the cut 5, the right edge 4, the top 12, the left side 8.

$P = 9 + 5 + 4 + 12 + 8 = 38\text{ m}$

Check: the rectangle's perimeter is $2 \times (12 + 8) = 40$ m; the cut takes 3 m and 4 m off the boundary and adds the 5 m edge: $40 - 3 - 4 + 5 = 38$ m.

Turf: $90 \times 6.80 = 612$ , so the turf costs $612. Fence: $38 \times 15 = 570$ , so the fence costs $570.

Total: $612 + 570 = 1182$ , so the total cost is $1182.

(a) $90\text{ m}^{2}$ (b) $1182

Practice 12: A shaded triangle

In rectangle $ABCD$ , $AB = 12$ cm and $AD = 8$ cm. $P$ is on $BC$ with $BP = 3$ cm, and $Q$ is on $DC$ with $CQ = 4$ cm. Find the area of the shaded triangle $APQ$ .

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No side of $\triangle APQ$ lies along a side of the rectangle, so subtract the three unshaded right-angled triangles from the rectangle.

$PC = 8 - 3 = 5$ cm and $DQ = 12 - 4 = 8$ cm.

Area of $\triangle ABP = \frac{1}{2} \times 12 \times 3 = 18\text{ cm}^{2}$

Area of $\triangle PCQ = \frac{1}{2} \times 5 \times 4 = 10\text{ cm}^{2}$

Area of $\triangle QDA = \frac{1}{2} \times 8 \times 8 = 32\text{ cm}^{2}$

Shaded area $= 12 \times 8 - 18 - 10 - 32 = 96 - 60 = 36\text{ cm}^{2}$

The area of $\triangle APQ$ is $36\text{ cm}^{2}$ .

Key points

A composite figure is made up of two or more simple shapes.
Area: add the areas of the parts, or subtract the parts removed from a larger shape; one way checks the other.
Perimeter: go round the boundary edge by edge; a line where two parts meet is not on it.
Missing lengths: when every corner is a right angle, the edges seen from opposite sides add up to the same width or height.
A staircase figure has the same perimeter as the rectangle round it; a notch in the middle of a side adds its two sides.
A triangle, parallelogram or trapezium uses its perpendicular height for area; slant sides are used only in the perimeter.
Change lengths to one unit before multiplying. Cost $=$ area $\times$ price per $\text{m}^{2}$ , or perimeter $\times$ price per metre.