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Composite and Shaded Figures: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 The clipped corner

An $18$ cm by $8$ cm rectangle has its upper right corner removed along a diagonal. The removed right triangle has perpendicular legs $5$ cm and $6$ cm along the rectangle sides. Find the retained area.

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

Text description of this figure

Schematic rectangle 18 cm wide and 8 cm high with its upper right corner removed along a diagonal. The removed right triangle has horizontal leg 5 cm and vertical leg 6 cm. The shaded polygon is retained. No scale is supplied.
1. Hint 1
  Identify the familiar whole and the part that is absent.
2. Hint 2
  Use the perpendicular legs to find the missing triangle area.
Answer

$129\text{ cm}^2$ .

Full solution

The outer rectangle has area $18(8)=144$ square centimeters.

The removed right triangle has area $\tfrac12(5)(6)=15$ square centimeters.

Subtract the excluded region once.
$A=144-15$ $A=129\text{ cm}^2.$

Answer

$129\text{ cm}^2$ .

Key idea

A convenient containing figure can make subtraction shorter than splitting the retained region.
Problem 2 The joined rectangles

A $7$ cm by $9$ cm rectangle and a $12$ cm by $4$ cm rectangle have disjoint interiors. Their lower sides are aligned, and they meet along the entire $4$ cm side of the second rectangle, with no other contact. Find the area and perimeter of their union.

Only solid boundary edges contribute to the union perimeter.

Text description of this figure

Two rectangles with disjoint interiors, 7 by 9 cm and 12 by 4 cm, joined along the entire side of length 4 cm of the second rectangle. Their lower edges are aligned. The shared edge is dashed. No scale is supplied.
1. Hint 1
  The same join has different roles in area and perimeter.
2. Hint 2
  Add rectangular areas, but remove both copies of the shared side from the sum of their perimeters.
Answer

$\begin{aligned}A&=111\text{ cm}^2\\P&=56\text{ cm}\end{aligned}$

Full solution

The shared segment has no area.
The union area is
$A=7(9)+12(4)$ $A=111\text{ cm}^2.$

The rectangle perimeters are $32$ cm and $32$ cm.
Their sum counts the $4$ cm shared segment twice, although it is internal.

Remove both copies.
$P=32+32-2(4)$ $P=56\text{ cm}.$

Answer

$\begin{aligned}A&=111\text{ cm}^2\\P&=56\text{ cm}\end{aligned}$

Key idea

A join contributes zero area and removes two copies of its length from the perimeter sum.
Problem 3 The staircase outline

An orthogonal polygon has the boundary steps shown. In order from its upper left vertex, travel right $6$ cm, down $4$ cm, right $9$ cm, down $7$ cm, right $5$ cm, down $5$ cm, left $x$ cm, and up $y$ cm to the start. Find $x$ , $y$ , its perimeter, and its area.

Schematic; use the labeled steps rather than measuring.

Text description of this figure

A schematic orthogonal staircase boundary. Starting at the upper left and following the boundary: right 6 cm, down 4 cm, right 9 cm, down 7 cm, right 5 cm, down 5 cm, left x cm, and up y cm back to the start. No scale is supplied.
1. Hint 1
  A closed path has equal total distances in opposite horizontal directions and in opposite vertical directions.
2. Hint 2
  Use vertical cuts at the two inward corners to find three rectangles.
3. Hint 3
  The rectangle widths are the three rightward steps; determine each full height above the common bottom.
Answer

$\begin{aligned}x&=20\text{ cm}\\y&=16\text{ cm}\\P&=72\text{ cm}\\A&=229\text{ cm}^2\end{aligned}$

Full solution

Closure horizontally gives
$x=6+9+5$ $x=20.$
Closure vertically gives
$y=4+7+5$ $y=16.$

The total rightward travel is $20$ cm and total downward travel is $16$ cm.
Returning to the start adds the same distances.
$P=2(20+16)$ $P=72\text{ cm}.$

Vertical cuts from the inward corners to the bottom give rectangles of widths $6$ , $9$ , and $5$ cm.
Their heights are $16$ , $12$ , and $5$ cm respectively, since the second top is $4$ cm below the first and the third top is another $7$ cm down.

Their interiors do not overlap and cover the polygon.
$A=6(16)+9(12)+5(5)$ $A=229\text{ cm}^2.$

Each strip reaches the same bottom line.

Answer

$\begin{aligned}x&=20\text{ cm}\\y&=16\text{ cm}\\P&=72\text{ cm}\\A&=229\text{ cm}^2\end{aligned}$

Key idea

Aligned steps recover hidden full lengths and the heights needed for an additive split.
Problem 4 The recessed entrance

A room follows a $9.6$ m by $7.4$ m rectangle except for a rectangular entrance recess $3.2$ m wide and $2.5$ m deep, opening into the middle of one $9.6$ m side. All edges are horizontal or vertical. Find the floor area and the baseboard length if a $1.1$ m doorway occupies part of the recess bottom. All other boundary edges receive baseboard.

The notch opens to the outside.

Text description of this figure

Schematic rectangular outline 9.6 by 7.4 m with a rectangular notch 3.2 m wide and 2.5 m deep opening into the middle of its top side. All sides are horizontal or vertical. The notch is not a closed hole. No scale is supplied.
1. Hint 1
  The recess is open to the outside rather than a closed interior hole.
2. Hint 2
  Its bottom replaces an equal length on the original side, and its two depth edges add boundary.
3. Hint 3
  Subtract the doorway only after finding the complete new boundary.
Answer

$\begin{aligned}A&=63.04\text{ m}^2\\\text{baseboard}&=37.9\text{ m}\end{aligned}$

Full solution

Subtract the rectangular recess from the full floor rectangle.
$A=9.6(7.4)-3.2(2.5)$ $A=63.04\text{ m}^2.$

The original rectangle perimeter is $34$ m.
The missing $3.2$ m top segment is replaced by the $3.2$ m recess bottom.
The two $2.5$ m depth edges are additional.
$P=34+2(2.5)$ $P=39\text{ m}.$

The doorway takes $1.1$ m of that boundary but receives no baseboard.
$39-1.1=37.9\text{ m}.$

Answer

$\begin{aligned}A&=63.04\text{ m}^2\\\text{baseboard}&=37.9\text{ m}\end{aligned}$

Key idea

An open side recess adds two depth edges even when the removed and replacement widths cancel.
Problem 5 The two unit systems

A $35$ cm by $22$ cm rectangle loses an upper right rectangular corner measuring $120$ mm by $80$ mm. Its horizontal notch width is $120$ mm. Find the remaining area in square meters and its perimeter in meters. Use $10$ mm equal to $1$ cm and $100$ cm equal to $1$ m. Explain the conversion factors.

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

Text description of this figure

Schematic L-shaped polygon with horizontal and vertical sides. Total width 35 cm, total height 22 cm. The removed upper right rectangular corner is 120 mm wide and 80 mm deep. Dashed lines show the original top and right edges of that corner. No scale is supplied.
1. Hint 1
  Convert the notch lengths into the same unit as the outside lengths before finding any area.
2. Hint 2
  A corner notch replaces equal horizontal and vertical boundary lengths.
3. Hint 3
  A one-meter square has one hundred centimeters in each of two directions.
Answer

$\begin{aligned}A&=0.0674\text{ m}^2\\P&=1.14\text{ m}\end{aligned}$

Full solution

The notch is $12$ cm by $8$ cm.
Both follow from dividing the millimeter lengths by $10$ .

Subtract areas in square centimeters.
$A=35(22)-12(8)$ $A=674\text{ cm}^2.$

A one-meter square has $100$ cm along each side, so it contains $100(100)$ square centimeters.
$674\div10000=0.0674\text{ m}^2.$

The corner removal replaces a $12$ cm horizontal part and an $8$ cm vertical part by equal new lengths.
The perimeter is
$P=2(35+22)$ $P=114\text{ cm}.$

Perimeter is one length, so divide by $100$ only once.
$114\text{ cm}=1.14\text{ m}.$

Answer

$\begin{aligned}A&=0.0674\text{ m}^2\\P&=1.14\text{ m}\end{aligned}$

Key idea

Square-unit conversion uses both directions, while converting a boundary length uses one.
Problem 6 The shared boundary rule

Two simple polygons have disjoint interiors and meet along exactly one boundary segment of positive length $s$ . They have no other contact. Their perimeters are $P_1$ and $P_2$ . Prove that the perimeter of their union is $P_1+P_2-2s$ .

The perimeter symbols refer to the original polygons, not to individual sides.

Text description of this figure

Two schematic simple polygons joined along a single vertical segment s. Their interiors lie on opposite sides of that segment. Their perimeters before joining are P one and P two. The common segment is dashed and there is no other contact. No scale is supplied.
1. Hint 1
  Follow how many times the common segment occurs before and after joining.
2. Hint 2
  First test the counting with perimeters $34$ cm and $46$ cm and a shared $9$ cm side.
3. Hint 3
  Every nonshared boundary segment remains exposed; only the two counted copies of the join disappear.
Answer

$P=P_1+P_2-2s$ .

Full solution

For the numerical instance, the sum $34+46$ is $80$ cm and includes the $9$ cm join twice.
Both copies become internal, leaving
$80-2(9)=62\text{ cm}.$

For the general polygons, each nonshared boundary part belongs to exactly one original perimeter and remains on the union boundary, because their interiors are disjoint and they have no other contact.
The shared segment belongs to each perimeter, so the sum counts its length twice.

No part of that shared segment is exposed after joining.
Removing its two counted copies gives
$P=P_1+P_2-2s.$
Nothing in this boundary accounting depended on the sample lengths.

Answer

$P=P_1+P_2-2s$ .

Key idea

Perimeter addition must account for how often each exposed or internal segment was counted.
Problem 7 The notched trapezoid

A trapezoid has parallel bases $21$ cm and $13$ cm and perpendicular height $10$ cm. A rectangular notch $5$ cm wide opens into the middle of the longer base and extends perpendicularly into the trapezoid. The notch lies entirely within it, and the retained area is $135$ square centimeters. Find the notch depth.

The two bases are parallel; the notch sides are perpendicular to them.

Text description of this figure

Schematic trapezoid with horizontal parallel bases 13 cm at the top and 21 cm at the bottom, and perpendicular height 10 cm. A rectangular notch of width 5 cm and unknown depth d opens into the middle of the bottom base. The notch lies wholly within the trapezoid. No scale is supplied.
1. Hint 1
  Compare the retained area with the area of the complete trapezoid.
2. Hint 2
  The missing area is five times the notch depth.
Answer

$7$ cm.

Full solution

The complete trapezoid area is
$A=\frac{21+13}{2}(10)$ $A=170\text{ cm}^2.$

The removed rectangle therefore has area
$170-135=35\text{ cm}^2.$
If its depth is $d$ cm,
$5d=35$ $d=7.$

The depth is positive and below the full $10$ cm height.
With the stated containment of the notch, it gives the described retained region.

Answer

$7$ cm.

Key idea

An area deficit gives a missing product, which can reveal a notch dimension.
Problem 8 The two openings

A $24$ cm by $17$ cm rectangle contains two holes strictly inside it. They have no contact. One hole is a right triangle with perpendicular legs $6$ cm and $8$ cm; the other is a square of side $5$ cm. Find the retained area and the total boundary length including both holes.

Both holes lie strictly inside and have no contact with each other.

Text description of this figure

Schematic rectangular region, outer dimensions 24 cm by 17 cm. Two separated interior holes are a right triangle with perpendicular legs 6 cm and 8 cm and a square of side 5 cm. The triangle has a right-angle square. Dashed outlines indicate removed regions. No scale is supplied.
1. Hint 1
  Area loses each hole, while total boundary includes each hole outline.
2. Hint 2
  Find the third side of the right triangle before computing its perimeter.
Answer

$\begin{aligned}A&=359\text{ cm}^2\\P_{\text{total}}&=126\text{ cm}\end{aligned}$

Full solution

The triangular hole has area $24$ square centimeters and the square hole has area $25$ square centimeters.
They are separate, so
$A=24(17)-24-25$ $A=359\text{ cm}^2.$

Pythagoras gives the triangle hypotenuse.
$\sqrt{6^2+8^2}=10\text{ cm}.$
Its boundary is $24$ cm.

The outer rectangle perimeter is $82$ cm, and the square hole perimeter is $20$ cm.
Include both inner boundaries.
$P_{\text{total}}=82+24+20$ $P_{\text{total}}=126\text{ cm}.$

Answer

$\begin{aligned}A&=359\text{ cm}^2\\P_{\text{total}}&=126\text{ cm}\end{aligned}$

Key idea

Separate holes are subtracted for area and included once each in total boundary length.
Problem 9 The outer width

A rectangle is $19$ cm high and has unknown width $w$ cm. It contains two holes strictly inside, with disjoint interiors: a $7$ cm by $6$ cm rectangle and a convex kite with perpendicular diagonals $12$ cm and $15$ cm. The retained area is $533$ square centimeters. Find $w$ and the outside boundary length only.
1. Hint 1
  The retained area and both missing areas together fill the outer rectangle.
2. Hint 2
  Use the perpendicular diagonals for the kite area, then recover the outer width.
Answer

$\begin{aligned}w&=35\text{ cm}\\P_{\text{outside}}&=108\text{ cm}\end{aligned}$

Full solution

The rectangular hole area is $42$ square centimeters.
The kite area uses its perpendicular diagonals.
$A_{\text{kite}}=\tfrac12(12)(15)$ $A_{\text{kite}}=90\text{ cm}^2.$

The complete outer area includes both holes.
$19w=533+42+90$ $19w=665$ $w=35.$

Only the outside boundary is requested, so omit both hole outlines.
$P_{\text{outside}}=2(35+19)$ $P_{\text{outside}}=108\text{ cm}.$

Answer

$\begin{aligned}w&=35\text{ cm}\\P_{\text{outside}}&=108\text{ cm}\end{aligned}$

Key idea

Recover the whole area before its dimension, and keep the requested boundary convention fixed.
Problem 10 The plate specification

A rectangular plate is $0.45$ m wide and $0.28$ m high. A rectangular opening lies strictly inside it; its height is $12$ cm and its width is $x$ cm. The retained plate area is $1080$ square centimeters. Find $x$ , the total boundary length including the opening in meters, and the retained area in square meters. Use $1$ m equal to $100$ cm.

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

Text description of this figure

Schematic rectangular region with an interior rectangular hole. Outer dimensions 45 by 28 cm; hole dimensions x by 12 cm. The hole lies strictly inside. Dashed edges mark the removed region, which has no fill. No scale is supplied.
1. Hint 1
  Start by expressing all dimensions in a common length unit.
2. Hint 2
  Recover the opening area from the outside area and the retained area.
3. Hint 3
  A length converts once, while a square unit contains two length directions.
Answer

$\begin{aligned}x&=15\text{ cm}\\P_{\text{total}}&=2\text{ m}\\A&=0.108\text{ m}^2\end{aligned}$

Full solution

The outside dimensions are $45$ cm by $28$ cm.
Their area is $1260$ square centimeters.
The opening therefore has area
$1260-1080=180\text{ cm}^2.$

Its height is $12$ cm, so
$12x=180$ $x=15.$
This width is positive and below $45$ cm, and the height is below $28$ cm, so an interior placement is possible.

The outer rectangle perimeter is
$P_{\text{outside}}=2(45+28)$ $P_{\text{outside}}=146\text{ cm}.$
The opening perimeter is
$P_{\text{opening}}=2(15+12)$ $P_{\text{opening}}=54\text{ cm}.$

The total boundary includes both rectangles.
$P_{\text{total}}=146+54$ $P_{\text{total}}=200\text{ cm}.$

Perimeter is a length.
$200\text{ cm}=2\text{ m}.$

A one-meter square is $100$ cm by $100$ cm, so divide its square-centimeter measure by $10000$ .
$1080\text{ cm}^2=0.108\text{ m}^2.$

Answer

$\begin{aligned}x&=15\text{ cm}\\P_{\text{total}}&=2\text{ m}\\A&=0.108\text{ m}^2\end{aligned}$

Key idea

An area equation, a boundary trace, and a unit conversion must each refer to the same retained region.

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