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Composite Figures and Shaded Regions

Find the area and perimeter of shapes built from rectangles, triangles and circles by adding and subtracting simple pieces.

Practice this skill 35 min Composite 2-D figures Leer en español
Key idea

What you'll learn

A composite figure is a shape made by sticking simple shapes together (an L-shaped room, a house-shaped wall, a window with a half-circle on top) or by cutting one shape out of another (a lawn with a round flower bed, a picture inside a frame).

In this lesson you will learn to:

  • add the areas of pieces that are joined together,
  • subtract the area of a piece that is cut out (a shaded region problem),
  • find the perimeter of a composite figure — the outside edge only, and
  • work backwards to a missing piece.

Why it matters on the GED: Composite figures are a favorite way to make an area question "GED-hard," because they combine two or three formulas from the formula sheet in one problem. Expect at least one, often with a circle or semicircle.

In real life: rooms are rarely perfect rectangles, and almost every flooring, sod or paving job has a cut-out (a kitchen island, a tree, a pool).

Words to know

  • Composite figure — a shape made of two or more simple shapes (an L-shape = two rectangles).
  • Decompose — split a figure into simple shapes you know how to measure.
  • Shaded region — the part of a figure that is colored in; usually "big shape minus the unshaded shape."
  • Cut-out (hole) — a piece removed from a shape (a square hole in a board).
  • Semicircle — half a circle; its area is \tfrac{1}{2}\pi r^2 and its curved edge is \tfrac{1}{2}\pi d.
  • Quarter circle — one fourth of a circle (a slice with a square corner).
  • Inner edge — a line where two pieces meet inside the figure; it is not part of the perimeter.
  • Outer boundary — the edge you would walk along to go around the whole figure; this is the perimeter.

All Math words in the glossary

Strategy 1: split and add

  1. Draw dashed lines to split the figure into rectangles, triangles, circles or half circles.
  2. Find any missing side lengths. In an L-shape, the long side equals the two short sides it lines up with. If the bottom is 12 ft and the top is 7 ft, the step on the right is 12 - 7 = 5 ft wide.
  3. Find each piece's area with its own formula.
  4. Add the pieces.

Example — L-shaped room: outer edges 14 ft by 10 ft, with a 6 ft by 4 ft corner missing.

  • Split with a vertical line: left piece 8 \times 10 = 80; right piece 6 \times 6 = 36. (The right piece is 10 - 4 = 6 ft tall.)
  • Total: 80 + 36 = 116\text{ ft}^2.

Example — "house" shape: a 10 ft square with a triangle on top (base 10 ft, height 4 ft).

  • Square 10 \times 10 = 100; triangle \tfrac{1}{2}(10)(4) = 20; total 120\text{ ft}^2.
A house shape: a 10 ft square with a triangle on top whose base is 10 ft and whose dashed height is 4 ft. Square 100 plus triangle ½ × 10 × 4 = 20 gives 120 square feet. 4 ft 10 × 10 = 100 10 ft 10 ft ½(10)(4) = 20 Total = 100 + 20 = 120 ft²

There is usually more than one way to split a figure. Every correct split gives the same total — a great way to check your work.

L-shaped figure: left side 9 ft, bottom 12 ft, right side 5 ft, top 7 ft, with a cut-out corner whose edges are 4 ft and 5 ft. A dashed vertical line splits it into rectangle A (left, 7 ft wide and 9 ft tall) and rectangle B (right, 5 ft wide and 5 ft tall) A B 7 ft 4 ft 5 ft 5 ft 12 ft 9 ft

An L-shape split into two rectangles. Rectangle A uses the full 9 ft height; rectangle B is only as tall as the step on the right.

Strategy 2: big shape minus the hole (shaded regions)

When a piece is missing or unshaded, it is often easiest to find the area of the whole outer shape and subtract the hole:

A_{\text{shaded}} = A_{\text{outer}} - A_{\text{hole}}
  • L-shape the other way: the 14 by 10 room is a full rectangle (140) minus the 6 by 4 corner (24): 140 - 24 = 116\text{ ft}^2 — the same as splitting ✓.
  • Lawn with a flower bed: yard 40 ft by 25 ft, round bed with radius 5 ft. Grass = 1{,}000 - 3.14(5^2) = 1{,}000 - 78.5 = 921.5\text{ ft}^2.
A 40 ft by 25 ft yard shaded green with a round flower bed of radius 5 ft cut out. Grass area = 1,000 − 3.14 × 25 = 921.5 square feet. 5 flower bed r = 5 ft grass 40 ft 25 ft Grass = 1,000 − 3.14(5²) = 921.5 ft²
  • Circle in a square: a circle that just fits inside a 10 cm square has diameter 10, so r = 5. The four corners have area 100 - 78.5 = 21.5\text{ cm}^2.
A circle that just fits inside a 10 cm square, so its diameter is 10 and radius 5. The four blue shaded corners outside the circle have area 100 − 78.5 = 21.5 square centimetres. d = 10, r = 5 10 cm 10 cm Corners = 100 − 78.5 = 21.5 cm² (shaded blue)
  • Border (frame or walkway): a 3 ft walkway around a 20 ft by 10 ft pool. The outer rectangle is 20 + 3 + 3 = 26 by 10 + 3 + 3 = 16. Walkway = 26 \times 16 - 20 \times 10 = 416 - 200 = 216\text{ ft}^2. (Remember to add the width on both sides.)
A 20 ft by 10 ft pool surrounded by a 3 ft walkway on every side, shaded green. The outer rectangle is 20 + 3 + 3 = 26 ft by 10 + 3 + 3 = 16 ft. Walkway area = 416 − 200 = 216 square feet. pool 20 × 10 20 ft 10 ft 3 3 3 3 26 ft = 20 + 3 + 3 16 ft Walkway = 26 × 16 − 20 × 10 = 416 − 200 = 216 ft²

Several holes? Subtract each one.

Perimeter of a composite figure

Perimeter is the distance around the outside only.

  • Do not include dashed lines or edges where two pieces meet — those are inside the figure.
  • For an L-shape, find every outside side (some may be missing and must be found by subtracting) and add them all.
  • Curved parts: a semicircle's curved edge is half a circumference: \tfrac{1}{2}\pi d. Two semicircles on opposite ends of a rectangle make one full circle together: \pi d.

Example — running track: two straight sides of 100 m and a semicircle on each end with diameter 60 m.

  • Straight parts: 100 + 100 = 200 m.
  • Two semicircles = one circle: \pi d = 3.14 \times 60 = 188.4 m.
  • Total: 200 + 188.4 = 388.4 m. The 60 m diameters are inside the track and are not added.
A running track: two straight sides of 100 m with a semicircle on each end of diameter 60 m. The blue outside edge is the perimeter: 200 m of straights plus one full circle, 3.14 × 60 = 188.4 m, for 388.4 m. The dashed 60 m diameters are inside the shape. 100 m 100 m 60 m 60 m semicircle P = 100 + 100 + π(60) = 200 + 188.4 = 388.4 m (the dashed 60 m diameters are inside, not added)

Area and perimeter use different pieces. For the area of a window with a semicircle on top, you add the rectangle and the half circle. For its perimeter, you add three sides of the rectangle and the curved arc — the bottom edge of the semicircle is inside.

Formula

Formulas you combine (the basic ones are on the GED formula sheet)

A_{\text{rectangle}} = lw \quad A_{\text{triangle}} = \tfrac{1}{2}bh \quad A_{\text{circle}} = \pi r^2 \quad C = \pi d = 2\pi r A_{\text{parallelogram}} = bh \qquad A_{\text{trapezoid}} = \tfrac{1}{2}h(b_1 + b_2)

Pieces of circles (not printed — work them out from the circle formulas):

A_{\text{semicircle}} = \tfrac{1}{2}\pi r^2 \qquad \text{curved edge of a semicircle} = \tfrac{1}{2}\pi d
  • l, w = length and width; b, h = base and perpendicular height; r = radius; d = diameter.

On the GED formula sheet: every basic area and circumference formula is on the sheet. There is no formula for composite figures — the skill being tested is choosing the pieces and deciding whether to add or subtract.

Worked example

Example 1 (easy): an L-shaped room

An L-shaped room is 14 ft by 10 ft on its outer edges, with a 6 ft by 4 ft corner missing. Find its area.

  1. Whole rectangle: 14 \times 10 = 140.
  2. Missing corner: 6 \times 4 = 24.
  3. Subtract: 140 - 24 = 116.
Two copies of an L-shaped room whose outer edges are 14 ft by 10 ft with a 6 ft by 4 ft corner missing at the top right. Left: the whole 14 by 10 rectangle (140) minus the red dashed 6 by 4 corner (24) gives 116 square feet. Right: a dashed vertical line splits it into rectangle A, 8 by 10 = 80, and rectangle B, 6 by 6 = 36; 80 + 36 = 116. 6 × 4 14 ft 10 ft 6 ft 4 ft Subtract: 140 − 24 = 116 ft² A 8 × 10 = 80 B 6 × 6 = 36 8 ft 6 ft 8 ft 10 ft 6 ft Split: 80 + 36 = 116 ft²

Answer: 116 ft²

Check: split instead: 8 \times 10 = 80 and 6 \times 6 = 36; 80 + 36 = 116 ✓.

Worked example

Example 2 (word problem): an arched window

A window is a rectangle 8 ft wide and 5 ft tall with a semicircle on top whose diameter is the 8 ft width. How much glass is needed? Use \pi \approx 3.14.

  1. Rectangle: 8 \times 5 = 40 ft².
  2. Semicircle: diameter 8, so r = 4. Half circle: \tfrac{1}{2} \times 3.14 \times 4^2 = \tfrac{1}{2} \times 50.24 = 25.12 ft².
  3. Add: 40 + 25.12 = 65.12.
An arched window: an 8 ft wide, 5 ft tall rectangle with a semicircle on top whose diameter is the 8 ft width (radius 4 ft). Rectangle 40 plus semicircle 25.12 gives 65.12 square feet of glass. r = 4 8 × 5 = 40 25.12 8 ft 5 ft Glass = 40 + ½(3.14)(4²) = 40 + 25.12 = 65.12 ft²

Answer: 65.12 ft² of glass

Check: the semicircle (25.12) is less than an 8 by 4 rectangle (32) that would hold it ✓.

Worked example

Example 3: the same window's perimeter

How much trim goes around the outside of the window in Example 2?

  1. Straight outside edges: the bottom (8 ft) and the two sides (5 ft each): 8 + 5 + 5 = 18 ft. The top of the rectangle is inside the window, so skip it.
  2. Curved edge: half a circumference: \tfrac{1}{2} \times 3.14 \times 8 = 12.56 ft.
  3. Add: 18 + 12.56 = 30.56.
The same arched window with its outer edge drawn thick in blue: bottom 8 ft, two sides 5 ft each, and the curved arc of ½ × 3.14 × 8 = 12.56 ft. The red dashed top of the rectangle is inside the window and is not trim. Total trim 30.56 ft. inside: not trim 8 ft 5 ft 5 ft arc = ½(3.14)(8) = 12.56 ft Trim = 8 + 5 + 5 + 12.56 = 30.56 ft

Answer: 30.56 ft of trim

Check: the arc (12.56) is longer than the straight 8 ft it replaces, as a curve should be ✓.

Worked example

Example 4 (work backwards): the missing square

An L-shaped patio was made by removing a square corner from a 12 ft by 15 ft rectangle. The patio's area is 144 ft². How long is each side of the removed square?

A 15 ft by 12 ft rectangle with a red dashed square corner of side s removed, leaving an L-shaped patio of 144 square feet. The side s is unknown. s × s s = ? patio 144 ft² 15 ft 12 ft
  1. Whole rectangle: 12 \times 15 = 180 ft².
  2. Area removed: 180 - 144 = 36 ft².
  3. Square side: s^2 = 36, so s = \sqrt{36} = 6.

Answer: 6 ft

Check: 180 - 6 \times 6 = 180 - 36 = 144 ✓.

Worked example

Example 5 (GED-hard): sodding a yard

A rectangular backyard is 30 ft by 20 ft. It contains an 8 ft by 8 ft square shed and a round above-ground pool with a radius of 6 ft. Sod costs $0.45 per square foot. How much will it cost to sod the rest of the yard? Use \pi \approx 3.14.

A 30 ft by 20 ft backyard shaded green for sod, containing a red 8 ft by 8 ft square shed and a blue round pool of radius 6 ft. shed 8 × 8 r = 6 pool sod 30 ft 20 ft
  1. Yard: 30 \times 20 = 600 ft².
  2. Shed: 8 \times 8 = 64 ft².
  3. Pool: 3.14 \times 6^2 = 3.14 \times 36 = 113.04 ft².
  4. Sod area: 600 - 64 - 113.04 = 422.96 ft².
  5. Cost: 422.96 \times 0.45 = 190.332 \approx 190.33.

Answer: about $190.33

Check: estimate: 600 - 64 - 113 \approx 423; 423 \times 0.45 \approx 190 ✓.

Calculator tip

Calculator tip: TI-30XS MultiView

Composite problems have several steps, so the calculator saves time — but organize the work first.

  • One line at a time: type the whole expression on one line: 30 × 20 − 8 x² − 3.14 × 6 x² enter → 422.96. The calculator does the squaring and multiplying before subtracting.
  • Semicircle: 0.5 × 3.14 × 4 x² enter → 25.12.
  • Use the \pi key only if the question says so; if it says "use 3.14," type 3.14 so your answer matches the choices.
  • Scroll up on the MultiView screen to reuse a previous answer instead of retyping it.

Don't let the calculator decide add-or-subtract for you. Sketch the pieces and write "+" or "−" beside each one before you type.

Common mistake

Common mistakes

  • Adding the hole instead of subtracting it. Rectangle 9 by 7 with a 3 by 3 hole: 63 - 9 = 54, not 63 + 9 = 72.
  • Forgetting the cut-out entirely. The L-shaped room is not the full 14 \times 10 = 140; the missing 6 by 4 corner must come off.
  • Using the wrong height for a piece. In an L-shape, the short rectangle is only as tall as the step, not the full height (5 \times 5, not 5 \times 9).
  • Treating a semicircle as a full circle (or forgetting the second semicircle). Two semicircles make one circle — not two circles.
  • Counting inside edges in the perimeter. For the running track, add the two straight sides and the curved ends; the 60 m diameters are inside.
  • Adding the walkway width only once. A 3 ft border around a 20 by 10 pool makes the outside 26 \times 16, not 23 \times 13.
Test strategy

GED test strategy

  • Sketch and label. Draw your dashed lines on your erasable note board and write each piece's dimensions before computing.
  • Decide add or subtract first. Joined pieces → add. Holes, unshaded parts, or "the rest of" → subtract.
  • Find missing sides by subtraction from the opposite side, and double-check that every side you use is on the figure.
  • Check with a second method. Split a different way, or use "whole minus hole" — the answers must match.
  • Estimate the final size. The shaded part must be smaller than the outer shape; a composite area must be bigger than each piece.
  • Read the last line again: area or perimeter? Cost? Number of bags (round up)?

Check yourself

Composite 2-D figuresNo calculator

The floor plan of an L-shaped room is shown.

L-shaped room. Left side 8 ft, bottom 10 ft, right side 5 ft, top 6 ft; the cut-out corner at the top right has a 3 ft vertical edge and a 4 ft horizontal edge 6 ft 3 ft 4 ft 5 ft 10 ft 8 ft

What is the area of the floor?

Answer choices
Hint

Picture the full 10 ft by 8 ft rectangle. What part of it is missing?

Check yourself

Composite 2-D figuresNo calculatorFill in

The front wall of a shed is a square with a triangle on top, as shown.

Shed front wall: a square 10 ft by 10 ft with a triangle on top whose base is 10 ft and whose dashed height is 4 ft 10 ft 10 ft 4 ft

What is the total area of the front wall?

ft²

Type your answer in the box. Fractions like 3/4 are fine.

Hint

Find the square's area and the triangle's area separately. The triangle's base is the top of the square.

Summary

  • Split a composite figure into rectangles, triangles and circle pieces; add joined pieces.
  • For shaded regions and holes: outer area − hole area.
  • Find missing sides by subtracting from the opposite side; use the right height for each piece.
  • Perimeter = outside edges only; a semicircle's arc is \tfrac{1}{2}\pi d, and two semicircles make one circle.
  • Check with a second split, estimate, and re-read whether the question wants area, perimeter or cost.
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