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Math 7 · 3.MD.C.7D · 6.G.A.1

Area of composite figures and shaded regions

A floor, a flag or a lawn is rarely one plain rectangle or one circle. How do you find the area of a shape that has no formula of its own?

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Area of a circleMath 7Area of trianglesMath 6
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Surface areaGrades 6 to 8
Where you are 15 in Area and perimeter
Measurement12
Area and perimeter15
▴ 10 aboveArea of trianglesArea of any polygon by cutting it into triangles and rectangles
Area of composite figures and shaded regions
Why the circle area formula worksArea of a regular polygon

When you reach for it

There is no “area of an L” to look up for an L-shaped room, and none for a flag with a notch cut out or a pool with rounded ends. What such shapes do have is edges that belong to shapes you already know: rectangles, triangles, parallelograms and trapezoids, and the half and quarter circles of Area of a circle.

A composite figure is one that can be cut into pieces that each have a formula, or seen as a bigger known shape with a piece missing. A shaded region in a drawing is usually the second kind: what is left of one shape when another is taken out, such as the lawn round a pond.

The lesson takes two days. Part 1 (Examples 1 to 4) uses straight-sided pieces. Part 2 (Examples 5 to 7) brings in half and quarter circles and shaded regions.

The procedure

Cutting up and adding (Part 1)

Area counts the unit squares a shape covers (Area as covering). Cut a figure into pieces that do not overlap and leave no gaps, and every bit of it lies in exactly one piece, so the areas add:

\begin{aligned} \text{area of the figure} &= \text{area of piece 1} \\ &\quad + \text{area of piece 2} + \cdots \end{aligned}

Pieces may share an edge, since a line covers no surface, but they may not overlap, or the shared part is counted twice. Where to cut is your choice: the figure covers the same surface however it is sliced, so every cut gives the same total. Choose the simplest pieces, and cut a second way when you want a check. Splitting into rectangles alone is Splitting a figure into rectangles; here the pieces may also be triangles, trapezoids and parts of circles.

Completing and subtracting (Part 1)

If the figure and a missing piece together make a whole shape, the whole is the figure plus the missing piece. Taking the missing piece from both sides,

\begin{aligned} \text{area of the figure} &= \text{area of the whole} \\ &\quad - \text{area of the missing piece} \end{aligned}

This is often the shorter road, and it needs the missing piece to lie wholly inside the whole.

Finding the lengths a drawing leaves out (Part 1)

A cut makes new edges, and a drawing seldom labels them all. When every corner is a right angle, the edges run only across and up, and going round the figure, the edges running across on one side add up to those on the other side; the same holds for the edges running up. So a missing length is a difference: a bottom edge of 12 m below a top edge of 5 m and a step leaves 12 - 5 = 7 m for the step. Without right angles the lengths cannot be read off this way.

When the total area is known and a length is not, write the area as the sum of the pieces with the unknown in it, and solve that equation.

Half circles, quarter circles, exact answers (Part 2)

A diameter cuts a circle into two matching halves, so a half circle of radius r has area \displaystyle \frac{1}{2}\pi r^2; two radii at a right angle cut off a quarter, area \displaystyle \frac{1}{4}\pi r^2. The straight edge of a half circle is a diameter, so its radius is half of it; both straight edges of a quarter circle are radii. Two half circles of the same radius make one whole circle, and so do four quarter circles, so join them before computing.

Keep \pi as a symbol and the answer is a number plus a multiple of \pi, such as 50 + 9\pi: the exact area. The two terms do not combine, for the same reason that 3 + 2x is not 5x. For a decimal, use the calculator’s \pi key and round once, at the end, so rounding errors in the pieces cannot pile up. In this lesson rounded answers have one decimal place.

Worked examples

Example 1

An L-shaped room, three ways

Figure 1 is the floor plan of an L-shaped room. Every corner is a right angle. Find the floor area.

9\text{ m}
3\text{ m}
4\text{ m}
7\text{ m}
 A B
Figure 1The L-shaped room of Example 1. The dashed cut makes piece A and piece B.
Solution

Across the bottom the room is 9 m and across the top 4 m, so the unlabeled inner edge running across is 9 - 4 = 5 m. Up the left it is 7 m and up the right 3 m, so the inner edge running up is 7 - 3 = 4 m.

Cut across. The dashed line makes piece A, 9 m by 3 m, and piece B, 4 m by 4 m.

9 \times 3 + 4 \times 4 = 27 + 16 = 43

Cut up and down along the inner edge that runs up: a strip 4 m by 7 m and a block 5 m by 3 m.

4 \times 7 + 5 \times 3 = 28 + 15 = 43

Complete and subtract. The room is a 9 m by 7 m rectangle with a 5 m by 4 m corner missing.

9 \times 7 - 5 \times 4 = 63 - 20 = 43

The floor area is 43\text{ m}^2. The routes agree because each counts the same floor.

Example 2

The side of a skate ramp

The side of a skate ramp has a vertical back 1.2 m tall, a flat deck 1.5 m long on top, and a straight slope from the end of the deck down to the ground. Along the ground it is 5.5 m long. How much plywood covers one side?

Solution

A vertical line from the end of the deck to the ground cuts the side into a rectangle 1.5 m by 1.2 m and a right triangle of the same height whose base is the rest of the ground, 5.5 - 1.5 = 4 m. A triangle’s area is half its base times its height, so

1.5 \times 1.2 + \frac{1}{2} \times 4 \times 1.2 = 1.8 + 2.4 = 4.2

One side needs 4.2\text{ m}^2 of plywood. Check. The deck and the ground are both level, so they are parallel and the side is a trapezoid: \displaystyle \frac{1}{2}(1.5 + 5.5) \times 1.2 = 3.5 \times 1.2 = 4.2.

Example 3

A flag with a notch

A swallowtail flag (Figure 2) is a rectangle 90 cm long and 60 cm tall with a triangle cut out of its free end. The triangle’s base is the whole 60 cm end, and its tip is halfway up the flag, 30 cm in from the end. Find the area of cloth.

90\text{ cm}
60\text{ cm}
30\text{ cm}
Figure 2The flag of Example 3 is a rectangle with a triangle taken out.
Solution

Closing the notch along the free end makes a whole rectangle, 90 \times 60 = 5400. The missing triangle has base 60 cm, and its height, measured at a right angle to that base, is the dashed 30 cm, so its area is \displaystyle \frac{1}{2} \times 60 \times 30 = 900.

5400 - 900 = 4500

The flag has 4500\text{ cm}^2 of cloth. Adding also works: a line from the pole to the tip cuts the flag into two trapezoids, each \displaystyle \frac{1}{2}(90 + 60) \times 30 = 2250, but subtracting one triangle is shorter.

Example 4

A T-shaped room with a length missing

A T-shaped room has a floor area of 54\text{ m}^2. Its top bar is 10 m long and 3 m deep, and the stem, 4 m wide and x m long, hangs from the middle of the bar. Every corner is a right angle. Find x, and the depth of the room from the top of the bar to the end of the stem.

Solution

The bar has area 10 \times 3 = 30 and the stem 4 \times x = 4x, and together they make the floor:

30 + 4x = 54

Subtracting 30 from both sides gives 4x = 24, and dividing both sides by 4 gives x = 6. The stem is 6 m long, and the room is 3 + 6 = 9 m deep.

Check by subtracting. The room fills a 10 m by 9 m rectangle except for two blocks beside the stem. The stem is centered, so each block is (10 - 4) \div 2 = 3 m wide and 6 m tall.

10 \times 9 - 2 \times (3 \times 6) = 90 - 36 = 54
Example 5

A pool with rounded ends

A pool has two straight sides 18 m long joined by a half circle at each end, and it is 8 m across. Find the area of its surface, exactly and rounded.

Solution

Lines across where the curves begin cut off the ends, leaving a rectangle 18 \times 8 = 144. Each end is a half circle whose straight edge, the 8 m width, is a diameter, so the radius is 4 m, and the two ends together make one circle of area \pi \times 4^2 = 16\pi.

A = 144 + 16\pi \approx 194.3

The surface has area 144 + 16\pi\text{ m}^2, about 194.3\text{ m}^2. Check. The pool fits in a rectangle 18 + 8 = 26 m by 8 m, area 208\text{ m}^2, missing only four small corners, so a little under 208 is right.

Example 6

The part of a square outside a circle

Two square tiles each have side 16 cm (Figure 3).

(a) On the first, a circle is drawn touching all four sides, and the four corners outside the circle are painted. Find the painted area.

(b) On the second, four quarter circles of radius 8 cm are drawn, each centered at a corner, and the part of the tile outside all four is painted. Find the painted area.

16\text{ cm}
16\text{ cm}
Figure 3Example 6. Left, the corners outside a circle. Right, the middle outside four quarter circles.
Solution

(a) The circle touches all four sides, so its diameter is the 16 cm side and its radius 8 cm. The painted part is the square with the circle, which lies inside it, taken out:

16^2 - \pi \times 8^2 = 256 - 64\pi \approx 54.9

The painted area is 256 - 64\pi\text{ cm}^2, about 54.9\text{ cm}^2.

(b) A radius of 8 cm is half the side, so each quarter circle reaches exactly to the middles of the two sides at its corner, where it meets its neighbors without overlapping. The four together make one whole circle, \displaystyle 4 \times \frac{1}{4}\pi \times 8^2 = 64\pi, so the painted area is again

256 - 64\pi \approx 54.9

The tiles look nothing alike, yet each is the same square with one circle of radius 8 cm removed.

Example 7

A gravel path round a lawn

A lawn has the shape of the pool in Example 5: two straight edges 24 m long and a half circle at each end, and it is 10 m across. A gravel path 2 m wide runs all the way round it (Figure 4). Find the area of the path, exactly and rounded.

24\text{ m}
10\text{ m}
2\text{ m}
Figure 4The gravel path of Example 7 runs all the way round the lawn at the same width.
Solution

The path keeps the same width all round, so its outer edge is straight along the sides and, round each end, a half circle with the same center as the lawn’s end and a radius 2 m bigger. The lawn’s ends have radius 10 \div 2 = 5 m, so the outer ends have radius 7 m, and the outer shape is 10 + 2 + 2 = 14 m across.

Subtract. The path is the outer shape with the lawn taken out, each found as in Example 5:

\begin{aligned} \text{outer} &= 24 \times 14 + \pi \times 7^2 = 336 + 49\pi \\ \text{lawn} &= 24 \times 10 + \pi \times 5^2 = 240 + 25\pi \\ \text{path} &= 96 + 24\pi \end{aligned}

Add. The path is also two straight strips, each 24 m by 2 m, area 2 \times 48 = 96, and two curved ends that together make a ring between radii 7 and 5, area 49\pi - 25\pi = 24\pi. The total is 96 + 24\pi again.

96 + 24\pi \approx 171.4

The path covers 96 + 24\pi\text{ m}^2, about 171.4\text{ m}^2.

Boundaries

  • No overlaps, no gaps. The pieces added must cover the figure exactly once. Two bars of a plus sign added whole count the square where they cross twice.
  • The missing piece must lie inside the whole. Subtracting a pond’s area from a garden’s is right only when the whole pond is in the garden; if a shape sticks out past the edge, only the part inside is taken away.
  • Lengths come from facts, not from the look of the drawing. Reading a missing length as a difference needs right angles, or parallel edges, that are marked or stated. A drawing may not be to scale.
  • One unit throughout. Centimeters and meters must be brought to one unit before any area is worked out.
  • Every piece needs a formula. The method works for figures made of straight edges and arcs of circles. A pond with a wavy shore cannot be cut into finitely many known pieces; its area can be estimated by counting grid squares, and for many curves found exactly later by integration. A slice of a circle other than a half or a quarter is a sector, with its own formula.
  • Perimeters do not add. A cut line has no area but does have length, and it lies inside the figure. Adding the perimeters of the pieces counts each cut line twice, though it is not on the edge at all.
  • A rounded answer is not exact. 144 + 16\pi is the area; 194.3 is a close value of it.

Where this goes next

The same two moves, add the pieces or take a piece away, run through the rest of measurement. Surface area adds the areas of the flat faces of a solid. Composite solids such as a silo or a house with a pitched roof are split into boxes, cylinders and cones, and their volumes added or subtracted in exactly this way. On the coordinate plane the grid supplies the lengths, and a polygon with slanted sides is often easiest as a surrounding rectangle with right triangles taken off its corners. A shaded region divided by the whole is a probability: the chance that a dart landing anywhere on the tile in Example 6 lands on paint.

Exercises

Exercises 1 to 12 go with Part 1 and 13 to 20 with Part 2. Any corner described as square, or any turn described as a right angle, is 90^\circ. Give any area involving \pi exactly and rounded.

Part 1: add the pieces

  1. Ciaran walks round the edge of an L-shaped deck, turning through a right angle at each corner. From one corner he walks 8 m, turns left and walks 2 m, turns left and walks 5 m, turns right and walks 4 m, turns left and walks 3 m, then turns left and walks 6 m back to where he started. Find the area of the deck.
  2. A logo is a plus sign made of two bars, each 12 cm long and 4 cm wide, one across and one up and down, crossing at their centers. Find its area.
  3. The front of a birdhouse is a rectangle 18 cm wide and 20 cm tall with a triangle on top, whose base is the top edge of the rectangle and whose height is 9 cm. Find the area of the front.
  4. Cut straight across, a grassy bank beside a playing field has a slope up, a flat top and a slope down. The slope up is a right triangle 3 m along the ground and 1.4 m high, the top is a rectangle 2 m long and 1.4 m high, and the slope down is a right triangle 2.5 m along the ground and 1.4 m high. Find the area of the cross-section, then check it with the trapezoid formula.

Part 1: take away the missing part

  1. A rectangular garden 15 m by 9 m has a shed in it standing on a rectangular base 4 m by 3 m. How much of the garden is not under the shed?
  2. A notice board is a rectangle 40 cm by 30 cm: a cork panel 32 cm by 22 cm in the middle, and a wooden border of the same width all round it. Find the area of the border and its width.
  3. A sign is cut from a rectangle 1.5 m by 0.8 m by snipping off its two top corners. Each piece snipped off is a right triangle whose two shorter sides are both 0.2 m. Find the area of the sign.
  4. A logo is a parallelogram cut from a rectangle 12 cm by 5 cm by removing a right triangle from each of two opposite corners. Each triangle has one leg 3 cm along a long side of the rectangle and the other leg the whole 5 cm short side. Find the area of the logo by subtracting, then again from the parallelogram’s base and height.

Part 1: missing lengths

  1. The plan of an L-shaped studio has six walls meeting at right angles. Drawn with its longest wall along the bottom, its bottom edge is 11 m, its left edge 7 m, its top edge 6 m and its right edge 4 m. Find the lengths of the two edges that form the inner corner, and the floor area.
  2. The top of a U-shaped desk fits in a rectangle 2.4 m wide and 1.6 m deep. A rectangular gap 1.2 m wide and 1 m deep is cut from the middle of the front edge, leaving two arms of equal width. Find the width of each arm and the area of the desk top.
  3. A patio is an L made of two rectangles: one 8 m by 3 m, and a second, 4 m wide, standing on one end of the first rectangle’s 8 m side. The patio has area 40\text{ m}^2. How far does the second rectangle reach out from the first, and how deep is the patio at its deepest?
  4. A garden bed is a rectangle 6 m by 4 m with a triangle on one of its 6 m sides, that side being the triangle’s base. The bed has area 33\text{ m}^2. Find the height of the triangle.

Part 2: half and quarter circles

  1. Emeric’s arched window is a rectangle 80 cm wide and 150 cm tall with a half circle on top, whose straight edge is the top of the rectangle. Find the area of glass.
  2. A table top has two straight sides 2.4 m long and a half circle at each end, and it is 1 m across. Find its area.
  3. A garden bed is a square of side 4 m with a quarter circle of radius 4 m attached along one side, one straight edge of the quarter circle lying along that side. Find the area of the bed.
  4. A keyhole-shaped badge is a half circle of radius 4 mm sitting on a trapezoid. The trapezoid’s top edge is the half circle’s straight edge, its bottom edge is 4 mm, parallel to the top, and its height is 10 mm. Find the area of the badge.

Part 2: shaded regions

  1. Philippa’s rectangular garden is 12 m by 8 m, and a round pond of radius 2 m lies wholly inside it. The rest is lawn. Find the area of the lawn.
  2. A square is drawn inside a circle of radius 6 cm with all four of its corners on the circle. Find the area inside the circle but outside the square.
  3. A carpenter cuts a quarter circle of radius 30 cm, centered at one corner, out of a square board 30 cm by 30 cm. Find the area of the board left over.
  4. A rectangular lawn 10 m by 6 m has a path 1 m wide all round it. The path’s outside corners are rounded: each is a quarter circle of radius 1 m centered at a corner of the lawn. Find the area of the path.

Reasoning

  1. Benedetta adds the areas of the two bars of the plus sign in Exercise 2 and gets 96\text{ cm}^2. Explain why this is not the area of the logo, and state the condition that pieces must meet for their areas to add up to the area of the figure.
  2. Dagny-Rose says that a shape with two straight sides of length L and a half circle at each end, w across, has area \displaystyle Lw + \frac{1}{4}\pi w^2. Show that she is right, and check the formula against Example 5.
  3. A circle is drawn in a square, touching all four sides. Show that the part of the square outside the circle is always the same fraction of the square, whatever the size, and find that fraction exactly and to three decimal places.
  4. Alastair finds the perimeter of the room in Example 1 by adding the perimeter of piece A to the perimeter of piece B. Find both his total and the room’s real perimeter, and explain why areas of pieces add up but perimeters do not.
Answers
  1. 28\text{ m}^2. The first two walks mark a strip 8 m by 2 m, area 16, and above it stands a block 3 m wide and 4 m tall, area 12: 16 + 12 = 28. Check: an 8 m by 6 m rectangle with a 5 m by 4 m corner missing, 48 - 20 = 28.
  2. 80\text{ cm}^2. The bar across has area 12 \times 4 = 48. The other bar sticks out above and below it in two pieces, each 4 cm by (12 - 4) \div 2 = 4 cm, area 2 \times 16 = 32. 48 + 32 = 80.
  3. 441\text{ cm}^2. Rectangle 18 \times 20 = 360, triangle \displaystyle \frac{1}{2} \times 18 \times 9 = 81.
  4. 6.65\text{ m}^2. Slope up \displaystyle \frac{1}{2} \times 3 \times 1.4 = 2.1, top 2 \times 1.4 = 2.8, slope down \displaystyle \frac{1}{2} \times 2.5 \times 1.4 = 1.75. As a trapezoid with bottom 3 + 2 + 2.5 = 7.5 and top 2: \displaystyle \frac{1}{2}(2 + 7.5) \times 1.4 = 6.65.
  5. 123\text{ m}^2. 15 \times 9 - 4 \times 3 = 135 - 12 = 123.
  6. Area 496\text{ cm}^2, width 4 cm. 40 \times 30 - 32 \times 22 = 1200 - 704 = 496. The panel is 40 - 32 = 8 cm narrower, shared between two sides, so the width is 4 cm, and (30 - 22) \div 2 = 4 agrees.
  7. 1.16\text{ m}^2. 1.5 \times 0.8 = 1.2, each corner \displaystyle \frac{1}{2} \times 0.2 \times 0.2 = 0.02, and 1.2 - 2 \times 0.02 = 1.16.
  8. 45\text{ cm}^2. Subtracting: \displaystyle 12 \times 5 - 2 \times \frac{1}{2} \times 3 \times 5 = 60 - 15 = 45. The parallelogram’s base is 12 - 3 = 9 cm and its height 5 cm: 9 \times 5 = 45.
  9. The inner edges are 5 m and 3 m, and the area is 62\text{ m}^2. Across: 11 - 6 = 5. Up: 7 - 4 = 3. A bottom strip 11 by 4 (44) and the block above it, 6 by 3 (18), give 62. Check: 11 \times 7 - 5 \times 3 = 77 - 15 = 62.
  10. Each arm is 0.6 m wide, and the area is 2.64\text{ m}^2. (2.4 - 1.2) \div 2 = 0.6, and 2.4 \times 1.6 - 1.2 \times 1 = 3.84 - 1.2 = 2.64. Check: a back strip 2.4 by 0.6 (1.44) and two arms 0.6 by 1 (1.2).
  11. It reaches 4 m, and the patio is 7 m deep at its deepest. 8 \times 3 + 4x = 40, so 4x = 16 and x = 4; the depth is 3 + 4 = 7 m.
  12. 3 m. \displaystyle 6 \times 4 + \frac{1}{2} \times 6 \times h = 33, so 24 + 3h = 33, 3h = 9 and h = 3.
  13. 12000 + 800\pi\text{ cm}^2 \approx 14513.3\text{ cm}^2. The rectangle is 80 \times 150 = 12000. The half circle has radius 40 and area \displaystyle \frac{1}{2}\pi \times 40^2 = 800\pi.
  14. \displaystyle 2.4 + \frac{1}{4}\pi\text{ m}^2 \approx 3.2\text{ m}^2. Rectangle 2.4 \times 1 = 2.4. The ends have radius 0.5 and together make one circle, \displaystyle \pi \times 0.5^2 = \frac{1}{4}\pi.
  15. 16 + 4\pi\text{ m}^2 \approx 28.6\text{ m}^2. Square 4^2 = 16, quarter circle \displaystyle \frac{1}{4}\pi \times 4^2 = 4\pi.
  16. 60 + 8\pi\text{ mm}^2 \approx 85.1\text{ mm}^2. The half circle has area \displaystyle \frac{1}{2}\pi \times 4^2 = 8\pi. Its straight edge, the trapezoid’s top, is 8 mm, so the trapezoid has area \displaystyle \frac{1}{2}(8 + 4) \times 10 = 60.
  17. 96 - 4\pi\text{ m}^2 \approx 83.4\text{ m}^2. Garden 12 \times 8 = 96, pond \pi \times 2^2 = 4\pi.
  18. 36\pi - 72\text{ cm}^2 \approx 41.1\text{ cm}^2. The circle has area \pi \times 6^2 = 36\pi. The square’s diagonals are diameters, crossing at the center at right angles, so they cut the square into four right triangles with legs 6 and 6, area 4 \times 18 = 72.
  19. 900 - 225\pi\text{ cm}^2 \approx 193.1\text{ cm}^2. Board 30^2 = 900. The quarter circle has area \displaystyle \frac{1}{4}\pi \times 30^2 = 225\pi, and since its radius is the side, it lies inside the board.
  20. 32 + \pi\text{ m}^2 \approx 35.1\text{ m}^2. Two strips 10 by 1 and two strips 6 by 1 run along the sides, area 20 + 12 = 32. The four rounded corners are quarter circles of radius 1, together one circle of area \pi \times 1^2 = \pi.
  21. The two bars overlap in a 4 cm by 4 cm square at the center, and adding the bars whole counts that square twice: 96 - 16 = 80, the true area. Areas add only when the pieces do not overlap and together cover the whole figure, so that every bit of it is counted exactly once.
  22. The middle is a rectangle L by w, area Lw. Each end is a half circle with diameter w, radius \displaystyle \frac{w}{2}, and the two ends make one circle of area \displaystyle \pi\left(\frac{w}{2}\right)^2 = \frac{1}{4}\pi w^2. So the area is \displaystyle Lw + \frac{1}{4}\pi w^2. With L = 18 and w = 8: \displaystyle 144 + \frac{1}{4}\pi \times 64 = 144 + 16\pi, as in Example 5.
  23. The fraction is \displaystyle 1 - \frac{\pi}{4} \approx 0.215. If the circle has radius r, the square has side 2r and area 4r^2, and the part outside the circle has area 4r^2 - \pi r^2. Dividing, \displaystyle \frac{4r^2 - \pi r^2}{4r^2} = 1 - \frac{\pi}{4}, and r has canceled, so the fraction does not depend on the size.
  24. Alastair gets 40 m; the room’s perimeter is 32 m. Piece A has perimeter 2(9 + 3) = 24 and piece B has 2(4 + 4) = 16, total 40. Going round the room: 9 + 3 + 5 + 4 + 4 + 7 = 32. The 4 m cut line is counted twice in his total, once in each piece, and it is not on the edge of the room at all: 40 - 2 \times 4 = 32. Every bit of floor lies in exactly one piece, so areas add; the cut line holds no floor, but it does have length.
kg1128 · updated 2026-10-07 · CC-BY-SA-4.0