Composite and Shaded Figures

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Learning goals

Decompose polygonal regions into nonoverlapping pieces or subtract contained regions
Recover hidden lengths from aligned sides and auxiliary segments
Trace exterior and hole boundaries without counting shared internal edges
Solve missing dimensions from a composite region's area
Convert perimeter and area using linear and square units correctly

Account for every piece once

A composite figure is a region assembled from simpler figures. To find its area, account for every included part exactly once. Two rectangles covering $6$ and $15$ square units, with no overlapping interiors, cover $21$ square units together. If a $4$ -square-unit part is then removed, $17$ square units remain. A shared edge has no area, so pieces may meet along their edges.

The area addition principle from Area from Rectangles to Triangles justifies this accounting. Use the triangle and quadrilateral formulas already established there and in Areas of Quadrilaterals.

An auxiliary segment is an added line that helps reveal familiar pieces. It does not add material or create a new outside edge. Choose cuts that leave pieces whose bases and perpendicular heights you can determine.

Worked example 1 Two ways to measure a floor plan

Problem. A floor plan has horizontal and vertical sides. Its total width is $17$ m, total height $11$ m, upper horizontal side $7$ m, and lower right side $4$ m. The upper right rectangular corner is absent. Find the floor area in two ways and its outside perimeter.

Solution. The missing width and height come from aligned sides. Across the full width, the upper side and missing corner span the same distance as the bottom. Vertically, the missing corner and lower right side span the full height.

\begin{aligned} \text{missing width}&=17-7\\ &=10\text{ m}\\ \text{missing height}&=11-4\\ &=7\text{ m}. \end{aligned}

Solid lines trace the actual boundary; dashed lines are auxiliary.

The vertical auxiliary segment divides the floor into a $7$ m by $11$ m rectangle and a $10$ m by $4$ m rectangle. Their interiors do not overlap, and together they fill the floor.

\begin{aligned} A&=7(11)+10(4)\\ A&=117\text{ m}^2. \end{aligned}

Solid lines trace the actual boundary; dashed lines are auxiliary.

Alternatively, complete the $17$ m by $11$ m rectangle and subtract the absent $10$ m by $7$ m corner.

\begin{aligned} A&=17(11)-10(7)\\ A&=117\text{ m}^2. \end{aligned}

The six solid edges form the outside boundary. The auxiliary segment inside the floor does not contribute.

\begin{aligned} P&=7+7+10+4+17+11\\ P&=56\text{ m}. \end{aligned}

The inward corner makes this a concave polygon, a simple polygon with an interior angle greater than $180^\circ$ . Concavity does not change the addition principle. The cuts must still cover the region without gaps or overlapping interiors. Do not assume that drawing from a concave vertex to any other vertex stays inside the polygon.

Check your understanding

Two rectangles with disjoint interiors cover a region. Their dimensions are $9$ cm by $3$ cm and $4$ cm by $5$ cm. What is the total area?

47\text{ cm}^2

21\text{ cm}^2

67\text{ cm}^2

94\text{ cm}^2

Subtract only what is absent

Subtraction is useful when completing a familiar outside shape is easier than splitting the retained region. The removed parts must lie inside that completed shape. If you subtract several parts separately, check that their interiors do not overlap, or the overlap would be removed twice.

Worked example 2 A shaded polygon with a slanted boundary

Problem. Start with a $19$ cm by $12$ cm rectangle. Remove its upper right corner along a diagonal. The removed corner is a right triangle with perpendicular legs $7$ cm and $6$ cm. Find the retained shaded area by subtraction and by addition.

Solution. The missing triangle has half the product of its perpendicular legs as its area.

The dashed corner is removed; its diagonal is part of the retained boundary.

\begin{aligned} A&=19(12)-\tfrac12(7)(6)\\ A&=207\text{ cm}^2. \end{aligned}

For an additive split, draw a perpendicular segment from the left end of the diagonal to the bottom. The left rectangle is $12$ cm wide because the full width is $19$ cm and the removed horizontal leg is $7$ cm. The right trapezoid has parallel vertical sides $12$ cm and $6$ cm, separated by $7$ cm.

The dashed corner is removed; its diagonal is part of the retained boundary.

\begin{aligned} A&=12(12)+\frac{12+6}{2}(7)\\ A&=207\text{ cm}^2. \end{aligned}

The diagonal is an actual boundary edge. The vertical segment in the additive split is internal. Neither the removed top edge nor the removed upper right edge belongs to the shaded polygon’s perimeter.

The same accounting handles overlapping figures. If two figures have areas $30$ and $22$ square units and overlap in $5$ square units, adding gives $52$ but counts those $5$ twice. Their union, meaning all points in either figure, has area $47$ square units. In general, add the two areas and subtract their common area once. This uses no new formula for the overlap: find that area with a familiar decomposition.

Check your understanding

An $18$ cm wide and $13$ cm high rectangle has its upper right corner removed along a diagonal. The retained top edge is $10$ cm long, and the retained right edge is $8$ cm long. What area remains?

194\text{ cm}^2

254\text{ cm}^2

214\text{ cm}^2

224\text{ cm}^2

Trace the boundary, including the requested holes

Perimeter measures boundary length, not the lengths of all lines in a drawing. Shared internal edges disappear when pieces are joined. For example, polygons with perimeters $26$ cm and $42$ cm joined along exactly one $8$ cm segment have union perimeter $52$ cm: the sum counts the shared segment twice, and neither copy remains exposed.

When polygonal pieces have disjoint interiors and meet only along shared boundary segments, subtract twice the total shared length from the sum of their perimeters. Do not use this shortcut for overlapping interiors without first identifying the actual union boundary.

A hole is an excluded region completely surrounded by the retained region. The remaining region has an outer boundary and an inner boundary; it is not one simple polygon. For a length around only the outside, omit the hole edges. For the total boundary length of the retained region, include them. A task about fencing or trim must identify which boundary is being covered.

Worked example 3 A courtyard gives two different boundary questions

Problem. A building occupies a $23$ m by $15$ m outer rectangle except for a $9$ m by $5$ m rectangular courtyard strictly inside it. Find the occupied area, the outside boundary length, and the total boundary length including the courtyard.

The inner rectangle is a hole. Include its four edges only when the task asks for them.

Solution. The courtyard contributes no occupied area.

\begin{aligned} A&=23(15)-9(5)\\ A&=300\text{ m}^2. \end{aligned}

The outside boundary is the outer rectangle.

\begin{aligned} P_{\text{outside}}&=2(23+15)\\ P_{\text{outside}}&=76\text{ m}. \end{aligned}

Include all four courtyard edges for the total boundary length.

\begin{aligned} P_{\text{total}}&=76+2(9+5)\\ P_{\text{total}}&=104\text{ m}. \end{aligned}

Check your understanding

Two polygons with disjoint interiors have perimeters $38$ cm and $30$ cm. They meet along exactly one complete $7$ cm segment. What is their union perimeter?

68

cm B

61

cm C

54

cm D

82

Square units contain two lengths

Changing the unit does not enlarge a region. It changes how many units describe the same region. A meter is $100$ centimeters long, but a square meter is a square with that length in each direction.

Why a square unit uses the length conversion twice#

A square $2$ m on each side has area $4$ square meters. In centimeters, each side is $200$ cm, so its area is $40{,}000$ square centimeters. Each of its four one-meter squares therefore contains $10{,}000$ one-centimeter squares.

Convert each side before counting the small squares.

More directly, each one-meter side divides into $100$ one-centimeter lengths. The one-meter square contains $100$ rows with $100$ small squares in each row.

1\text{ m}^2=100(100)\text{ cm}^2.

Nothing depends on the particular unit names. If one larger length unit equals $c$ smaller length units, where $c$ is positive, a square of one larger unit has $c$ smaller units along each side. By the rectangle area formula, one larger square unit equals $c^2$ smaller square units. This applies to the square units used to measure any polygonal area, whether or not that polygon tiles into whole unit squares.

Worked example 4 Convert an area and a perimeter

Problem. A polygonal region has area $0.37$ square meters and perimeter $3.9$ meters. Express its area in square centimeters and its perimeter in centimeters. Use $1$ m equal to $100$ cm.

Solution. The area conversion uses two factors of $100$ because a square unit contains two length directions.

0.37\text{ m}^2=3700\text{ cm}^2.

The perimeter is a length, so it uses one factor of $100$ .

3.9\text{ m}=390\text{ cm}.

To reverse these conversions, divide by $10{,}000$ for square centimeters to square meters, and by $100$ for centimeters to meters. Put all lengths in a common unit before using an area formula; do not multiply a length in meters directly by one in centimeters and label the result square centimeters.

Check your understanding

Use $1$ m equal to $100$ cm. How many square centimeters is $0.056$ square meters?

5.6

56

560

5600

Work backward from an area

A given area can determine an unknown length. Define the length, express the included and excluded areas in the same square unit, and solve the resulting equation. Then check that the length fits the geometry, not just the equation.

Worked example 5 Recover the width of a notch

Problem. A $24$ cm by $13$ cm rectangle has a rectangular notch removed from its upper right corner. The notch is $5$ cm deep and $x$ cm wide. The retained area is $267$ square centimeters. Find $x$ and the perimeter of the retained polygon.

Schematic; the unknown width comes from the area, not the drawing.

Solution. The whole rectangle has area $312$ square centimeters. The notch area is $5x$ square centimeters.

\begin{aligned} 312-5x&=267\\ 5x&=45\\ x&=9. \end{aligned}

The notch width is $9$ cm. It is positive and less than $24$ cm; its depth $5$ cm is less than $13$ cm, so the described corner notch is possible. Substituting gives the required retained area.

The remaining top side is $15$ cm, and the remaining right side is $8$ cm. Trace those, the two notch edges, the bottom, and the left side.

\begin{aligned} P&=15+5+9+8+24+13\\ P&=74\text{ cm}. \end{aligned}

Here the notch replaces a $9$ cm top segment and a $5$ cm right segment with equal horizontal and vertical lengths, so this corner removal leaves the original rectangle’s perimeter unchanged. A notch opening into the middle of a side has two added depth edges and behaves differently.

Check your understanding

A $21$ cm by $16$ cm rectangle contains an interior rectangular hole. The hole is $5$ cm wide, its height is $x$ cm, and the retained area is $296$ square centimeters. What is $x$ ?

6

8

10

12

Composite and Shaded Figures

Learning goals

Account for every piece once

Subtract only what is absent

Trace the boundary, including the requested holes

Square units contain two lengths

Work backward from an area

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice

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