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Use Properties of Rectangles, Triangles, and Trapezoids

Use Properties of Rectangles, Triangles, and Trapezoids

By the end of this section, you will be able to: understand linear, square, and cubic measure, use properties of rectangles, use properties of triangles, and use properties of trapezoids.

In this section, we’ll continue working with geometry applications. We will add some more properties of triangles, and we’ll learn about the properties of rectangles and trapezoids.

Understand linear, square, and cubic measure

When you measure your height or the length of a garden hose, you use a ruler or tape measure. A tape measure might remind you of a line — you use it for linear measure, which measures length. Inch, foot, yard, mile, centimeter, and meter are units of linear measure.

When you want to know how much tile is needed to cover a floor, or the size of a wall to be painted, you need to know the area, a measure of the region needed to cover a surface. Area is measured in square units. We often use square inches, square feet, square centimeters, or square miles. A square centimeter is a square that is one centimeter on each side, and a square inch is a square that is one inch on each side.

Picture a rectangular rug that is 22 feet long by 33 feet wide, made of 11-foot squares. The rug is made of 66 squares, so its area is 66 square feet.

When you measure how much it takes to fill a container, such as the amount of gasoline that can fit in a tank, or the amount of medicine in a syringe, you are measuring volume. Volume is measured in cubic units such as cubic inches or cubic centimeters. When measuring the volume of a rectangular solid, you measure how many cubes fill the container. A cubic centimeter is a cube that measures one centimeter on each side, and a cubic inch is a cube that measures one inch on each side.

Suppose a cube measures 33 inches on each side and is cut into unit cubes. If we took the big cube apart, we would find 2727 little cubes, each measuring one inch on all sides. So each little cube has a volume of 11 cubic inch, and the volume of the big cube is 2727 cubic inches.

Example. For each item, state whether you would use linear, square, or cubic measure: (a) amount of carpeting needed in a room, (b) extension cord length, (c) amount of sand in a sandbox, (d) length of a curtain rod, (e) amount of flour in a canister, (f) size of the roof of a doghouse.

(a) You are measuring how much surface the carpet covers, which is the area — square measure. (b) You are measuring how long the extension cord is, which is the length — linear measure. (c) You are measuring the volume of the sand — cubic measure. (d) You are measuring the length of the curtain rod — linear measure. (e) You are measuring the volume of the flour — cubic measure. (f) You are measuring the area of the roof — square measure.

A can of paint covers a wall with area 200 square feet using 2 cans. How many square feet does 1 can cover? (This is testing the square-measure idea: paint coverage is an area rate.)

A rectangular bedroom floor measures 12 feet by 10 feet. How many square feet of floor space does it have?

Many geometry applications involve finding the perimeter or the area of a figure, and it’s important to understand what each means. Picture a room that needs new floor tiles. The tiles come in squares that are a foot on each side — one square foot. How many of those squares are needed to cover the floor? This is the area of the floor. Next, think about putting new baseboard around the room once the tiles are laid. To figure out how many strips are needed, you must know the distance around the room — this distance is the perimeter.

Perimeter and area. The perimeter is a measure of the distance around a figure. The area is a measure of the surface covered by a figure.

A square tile that is 11 inch on each side has a perimeter of 44 inches (if an ant walked around its edge, it would walk 44 inches) and an area of 11 square inch.

Example. Each of two square tiles is 11 square inch, shown together side by side as a 2×12 \times 1 rectangle of tiles. (a) What is the perimeter of the figure? (b) What is the area?

(a) The perimeter is the distance around the figure. Walking around the outside edge of the two joined tiles covers 66 inches, so the perimeter is 66 inches. (b) The area is the surface covered by the figure. There are 22 square-inch tiles, so the area is 22 square inches.

Each box in a figure is 1 square inch. The figure is a row of 3 such boxes side by side (a 3 by 1 rectangle of unit squares). Find the perimeter of the figure.

Each box in a figure is 1 square inch. The figure is a 2 by 2 block of such boxes (four unit squares arranged in a square). Find the area of the figure.

Use the properties of rectangles

A rectangle has four sides and four right angles. The opposite sides of a rectangle are the same length. We refer to one side of the rectangle as the length, LL, and the adjacent side as the width, WW.

The perimeter, PP, of the rectangle is the distance around the rectangle. If you started at one corner and walked around the rectangle, you would walk L+W+L+WL + W + L + W units, or two lengths and two widths. The perimeter then is P=L+W+L+WP = L + W + L + W, or P=2L+2WP = 2L + 2W.

What about the area of a rectangle? Remember the rectangular rug from the beginning of this section — it was 22 feet long by 33 feet wide, and its area was 66 square feet. Since 6=236 = 2 \cdot 3, the area, AA, is the length, LL, times the width, WW, so the area of a rectangle is A=LWA = L \cdot W.

Properties of rectangles.

  • Rectangles have four sides and four right (90°90°) angles.
  • The lengths of opposite sides are equal.
  • The perimeter, PP, of a rectangle is the sum of twice the length and twice the width: P=2L+2WP = 2L + 2W.
  • The area, AA, of a rectangle is the length times the width: A=LWA = L \cdot W.

For easy reference, here is the Problem Solving Strategy for Geometry Applications restated:

Use a problem-solving strategy for geometry applications.

  1. Read the problem and make sure you understand all the words and ideas. Draw the figure and label it with the given information.
  2. Identify what you are looking for.
  3. Name what you are looking for. Choose a variable to represent that quantity.
  4. Translate into an equation by writing the appropriate formula or model for the situation. Substitute in the given information.
  5. Solve the equation using good algebra techniques.
  6. Check the answer in the problem and make sure it makes sense.
  7. Answer the question with a complete sentence.

Example. The length of a rectangle is 3232 meters and the width is 2020 meters. Find (a) the perimeter, and (b) the area.

(a) Let P=P = the perimeter. Substituting into P=2L+2WP = 2L + 2W:

P=2(32)+2(20)P=64+40P=104 P = 2(32) + 2(20) \qquad\Rightarrow\qquad P = 64 + 40 \qquad\Rightarrow\qquad P = 104

Checking: 20+32+20+32=10420 + 32 + 20 + 32 = 104. The perimeter of the rectangle is 104104 meters.

(b) Let A=A = the area. Substituting into A=LWA = L \cdot W:

A=3220A=640A = 32 \cdot 20 \qquad\Rightarrow\qquad A = 640

Checking: 3220=64032 \cdot 20 = 640. The area of the rectangle is 640640 square meters.

The length of a rectangle is 120 yards and the width is 50 yards. Find the perimeter.

The length of a rectangle is 120 yards and the width is 50 yards. Find the area.

In the next example, the width is defined in terms of the length, so we wait to draw the figure until we have expressions for both.

Example. The width of a rectangle is two inches less than the length. The perimeter is 5252 inches. Find the length and width.

Let L=L = length, so the width is L2L - 2. Substituting into P=2L+2WP = 2L + 2W with P=52P = 52:

52=2L+2(L2)52=2L+2L452 = 2L + 2(L - 2) \qquad\Rightarrow\qquad 52 = 2L + 2L - 452=4L456=4L14=L 52 = 4L - 4 \qquad\Rightarrow\qquad 56 = 4L \qquad\Rightarrow\qquad 14 = L

The length is 1414 inches. The width is L2=12L - 2 = 12 inches. Checking: 14+12+14+12=5214 + 12 + 14 + 12 = 52. The length is 1414 inches and the width is 1212 inches.

The width of a rectangle is seven meters less than the length. The perimeter is 58 meters. Find the length.

The length of a rectangle is eight feet more than the width. The perimeter is 60 feet. Find the width.

Example. The length of a rectangle is four centimeters more than twice the width. The perimeter is 3232 centimeters. Find the length and width.

Let w=w = width, so 2w+4=2w + 4 = length. Substituting into P=2L+2WP = 2L + 2W with P=32P = 32:

32=2(2w+4)+2w32=4w+8+2w32 = 2(2w + 4) + 2w \qquad\Rightarrow\qquad 32 = 4w + 8 + 2w32=6w+824=6w4=w 32 = 6w + 8 \qquad\Rightarrow\qquad 24 = 6w \qquad\Rightarrow\qquad 4 = w

The width is 44 cm. The length is 2w+4=2(4)+4=122w + 4 = 2(4) + 4 = 12 cm. Checking: P=2(12)+2(4)=32P = 2(12) + 2(4) = 32. The length is 1212 cm and the width is 44 cm.

The length of a rectangle is eight more than twice the width. The perimeter is 64 feet. Find the width.

The width of a rectangle is six less than twice the length. The perimeter is 18 centimeters. Find the length.

Example. The area of a rectangular room is 168168 square feet. The length is 1414 feet. What is the width?

Let W=W = width. Substituting into A=LWA = LW: 168=14W168 = 14W. Dividing both sides by 1414 gives W=12W = 12. Checking: 1412=16814 \cdot 12 = 168. The width of the room is 1212 feet.

The area of a rectangle is 598 square feet. The length is 23 feet. What is the width?

The width of a rectangle is 21 meters. The area is 609 square meters. What is the length?

Example. The perimeter of a rectangular swimming pool is 150150 feet. The length is 1515 feet more than the width. Find the length and width.

Let w=w = width, so w+15=w + 15 = length. Substituting into P=2L+2WP = 2L + 2W with P=150P = 150:

150=2(w+15)+2w150=2w+30+2w150 = 2(w + 15) + 2w \qquad\Rightarrow\qquad 150 = 2w + 30 + 2w150=4w+30120=4w30=w 150 = 4w + 30 \qquad\Rightarrow\qquad 120 = 4w \qquad\Rightarrow\qquad 30 = w

The width of the pool is 3030 feet, and the length is w+15=45w + 15 = 45 feet. Checking: 150=2(45)+2(30)150 = 2(45) + 2(30). The length of the pool is 4545 feet and the width is 3030 feet.

The perimeter of a rectangular swimming pool is 200 feet. The length is 40 feet more than the width. Find the width.

The length of a rectangular garden is 30 yards more than the width. The perimeter is 300 yards. Find the width.

Use the properties of triangles

We now know how to find the area of a rectangle. We can use this fact to help us visualize the formula for the area of a triangle: label the rectangle’s length bb and width hh, so its area is bhbh. We can divide this rectangle into two congruent triangles (triangles with identical side lengths and angles, so their areas are equal). The area of each triangle is one-half the area of the rectangle, or 12bh\tfrac{1}{2}bh. This is why the formula for the area of a triangle is A=12bhA = \tfrac{1}{2}bh.

To find the area of a triangle, you need to know its base and height. The base is the length of one side of the triangle, usually the side at the bottom. The height is the length of the line that connects the base to the opposite vertex, and makes a 90°90° angle with the base.

Triangle properties. For any triangle ΔABC\Delta ABC, the sum of the measures of the angles is 180°180°:

mA+mB+mC=180°m\angle A + m\angle B + m\angle C = 180°

The perimeter of a triangle is the sum of the lengths of the sides:

P=a+b+cP = a + b + c

The area of a triangle is one-half the base, bb, times the height, hh:

A=12bhA = \tfrac{1}{2}bh

Example. Find the area of a triangle whose base is 1111 inches and whose height is 88 inches.

Let A=A = the area. Substituting into A=12bhA = \tfrac{1}{2}bh:

A=12118A=44 square inches A = \tfrac{1}{2} \cdot 11 \cdot 8 \qquad\Rightarrow\qquad A = 44 \text{ square inches}

Checking: 44=?12(11)(8)44 \overset{?}{=} \tfrac{1}{2}(11)(8), and 44=4444 = 44. The area is 4444 square inches.

Find the area of a triangle with base 13 inches and height 2 inches.

Find the area of a triangle with base 14 inches and height 7 inches.

Example. The perimeter of a triangular garden is 2424 feet. The lengths of two sides are 44 feet and 99 feet. How long is the third side?

Let c=c = the third side. Substituting into P=a+b+cP = a + b + c with P=24P = 24, a=4a = 4, b=9b = 9:

24=4+9+c24=13+c11=c 24 = 4 + 9 + c \qquad\Rightarrow\qquad 24 = 13 + c \qquad\Rightarrow\qquad 11 = c

Checking: 4+9+11=244 + 9 + 11 = 24. The third side is 1111 feet long.

The perimeter of a triangular garden is 48 feet. The lengths of two sides are 18 feet and 22 feet. How long is the third side?

The lengths of two sides of a triangular window are 7 feet and 5 feet. The perimeter is 18 feet. How long is the third side?

Example. The area of a triangular church window is 9090 square meters. The base of the window is 1515 meters. What is the window’s height?

Let h=h = the height. Substituting into A=12bhA = \tfrac{1}{2}bh with A=90A = 90, b=15b = 15:

90=1215h90=152h12=h 90 = \tfrac{1}{2} \cdot 15 \cdot h \qquad\Rightarrow\qquad 90 = \tfrac{15}{2}h \qquad\Rightarrow\qquad 12 = h

Checking: 90=?12151290 \overset{?}{=} \tfrac{1}{2} \cdot 15 \cdot 12, and 90=9090 = 90. The height of the triangle is 1212 meters.

The area of a triangular painting is 126 square inches. The base is 18 inches. What is the height?

A triangular tent door has an area of 15 square feet. The height is 5 feet. What is the base?

Isosceles and equilateral triangles

Besides the right triangle, some other triangles have special names. A triangle with two sides of equal length is called an isosceles triangle — the third side is the base. A triangle that has three sides of equal length is called an equilateral triangle.

Isosceles and equilateral triangles. An isosceles triangle has two sides the same length. An equilateral triangle has three sides of equal length.

Example. The perimeter of an equilateral triangle is 9393 inches. Find the length of each side.

Let s=s = length of each side. Substituting into P=a+b+cP = a + b + c, with all three sides equal to ss:

93=s+s+s93=3s31=s 93 = s + s + s \qquad\Rightarrow\qquad 93 = 3s \qquad\Rightarrow\qquad 31 = s

Checking: 31+31+31=9331 + 31 + 31 = 93. Each side is 3131 inches.

Find the length of each side of an equilateral triangle with perimeter 39 inches.

Find the length of each side of an equilateral triangle with perimeter 51 centimeters.

Example. Arianna has 156156 inches of beading to use as trim around a scarf. The scarf will be an isosceles triangle with a base of 6060 inches. How long can she make the two equal sides?

Let s=s = the length of each equal side. Substituting into P=a+b+cP = a + b + c with P=156P = 156 and base 6060:

156=s+60+s156=2s+60156 = s + 60 + s \qquad\Rightarrow\qquad 156 = 2s + 6096=2s48=s96 = 2s \qquad\Rightarrow\qquad 48 = s

Checking: 48+60+48=15648 + 60 + 48 = 156. Arianna can make each of the two equal sides 4848 inches long.

A backyard deck is in the shape of an isosceles triangle with a base of 20 feet. The perimeter of the deck is 48 feet. How long is each of the equal sides of the deck?

A boat's sail is an isosceles triangle with base of 8 meters. The perimeter is 22 meters. How long is each of the equal sides of the sail?

Use the properties of trapezoids

A trapezoid is a four-sided figure, a quadrilateral, with two sides that are parallel and two sides that are not. The parallel sides are called the bases. We call the length of the smaller base bb, and the length of the bigger base BB. The height, hh, of a trapezoid is the distance between the two bases.

bBh

The formula for the area of a trapezoid is:

Areatrapezoid=12h(b+B)\text{Area}_{\text{trapezoid}} = \tfrac{1}{2}h(b + B)

Splitting the trapezoid into two triangles (each with height hh, one with base bb and one with base BB) may help you understand the formula: the area of the trapezoid is the sum of the areas of the two triangles, 12bh+12Bh\tfrac{1}{2}bh + \tfrac{1}{2}Bh, which factors to 12h(b+B)\tfrac{1}{2}h(b + B).

Properties of trapezoids.

  • A trapezoid has four sides.
  • Two of its sides are parallel and two sides are not.
  • The area, AA, of a trapezoid is A=12h(b+B)A = \tfrac{1}{2}h(b + B).

Example. Find the area of a trapezoid whose height is 66 inches and whose bases are 1414 and 1111 inches.

Let A=A = the area. Substituting into A=12h(b+B)A = \tfrac{1}{2}h(b + B) with h=6h = 6, b=11b = 11, B=14B = 14:

A=126(11+14)A=126(25)A=3(25)A=75 square inches A = \tfrac{1}{2} \cdot 6 \cdot (11 + 14) \qquad\Rightarrow\qquad A = \tfrac{1}{2} \cdot 6(25) \qquad\Rightarrow\qquad A = 3(25) \qquad\Rightarrow\qquad A = 75 \text{ square inches}

Checking: this should be reasonable — a rectangle with the same big base B=14B = 14 and height 66 has area 8484 square inches, and a rectangle with the same little base b=11b = 11 and height 66 has area 6666 square inches, so the trapezoid’s area should be between 6666 and 8484. Indeed 66<75<8466 < 75 < 84. The area of the trapezoid is 7575 square inches.

The height of a trapezoid is 14 yards and the bases are 7 and 16 yards. What is the area?

The height of a trapezoid is 18 centimeters and the bases are 17 and 8 centimeters. What is the area?

Example. Find the area of a trapezoid whose height is 55 feet and whose bases are 10.310.3 and 13.713.7 feet.

A=125(10.3+13.7)A=125(24)A=125A=60 square feet A = \tfrac{1}{2} \cdot 5 \cdot (10.3 + 13.7) \qquad\Rightarrow\qquad A = \tfrac{1}{2} \cdot 5(24) \qquad\Rightarrow\qquad A = 12 \cdot 5 \qquad\Rightarrow\qquad A = 60 \text{ square feet}

Checking: this is reasonable since it’s less than a rectangle with base 13.713.7 and height 55 (68.568.5 sq ft), and more than a rectangle with base 10.310.3 and height 55 (51.551.5 sq ft). The area of the trapezoid is 6060 square feet.

The height of a trapezoid is 7 centimeters and the bases are 4.6 and 7.4 centimeters. What is the area?

The height of a trapezoid is 9 meters and the bases are 6.2 and 7.8 meters. What is the area?

Example. Vinny has a garden shaped like a trapezoid, with a height of 3.43.4 yards and bases of 8.28.2 and 5.65.6 yards. How many square yards will be available to plant?

A=123.4(5.6+8.2)A=12(3.4)(13.8)A=23.46 square yards A = \tfrac{1}{2} \cdot 3.4 \cdot (5.6 + 8.2) \qquad\Rightarrow\qquad A = \tfrac{1}{2}(3.4)(13.8) \qquad\Rightarrow\qquad A = 23.46 \text{ square yards}

Checking: this is reasonable — less than a rectangle with base 8.28.2 and height 3.43.4 (27.8827.88 sq yd), and more than a rectangle with base 5.65.6 and height 3.43.4 (19.0419.04 sq yd). Vinny has 23.4623.46 square yards in which he can plant.

Lin wants to sod his lawn, which is shaped like a trapezoid. The bases are 10.8 yards and 6.7 yards, and the height is 4.6 yards. How many square yards of sod does he need?

Kira wants to cover her patio with concrete pavers. The patio is shaped like a trapezoid whose bases are 18 feet and 14 feet and whose height is 15 feet. How many square feet of pavers will she need?

Key terms

linear measure — a measure of length, in units such as inches, feet, or centimeters. area (square measure) — a measure of the surface covered by a figure, in square units. volume (cubic measure) — a measure of the space filled by a solid, in cubic units. perimeter — the distance around a figure. rectangle — a four-sided figure with four right angles and equal opposite sides. congruent — having identical side lengths and angles. base and height (of a triangle) — the side a triangle sits on, and the perpendicular distance from that side to the opposite vertex. isosceles triangle — a triangle with two sides of equal length. equilateral triangle — a triangle with three sides of equal length. trapezoid — a four-sided figure with exactly two parallel sides (the bases).


This section is adapted from Prealgebra 2e, Section 9.4: Use Properties of Rectangles, Triangles, and Trapezoids by Lynn Marecek, MaryAnne Anthony-Smith, and Andrea Honeycutt Mathis, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: recreated the trapezoid figure as an accessible inline graphic and described other figures (tape measures, cubes, tile grids) in prose instead of hotlinking images; omitted the Be Prepared quiz, Manipulative Mathematics and Links to Literacy callouts, Media links, and end-of-section exercises; and converted the practice problems (“Try Its”) into interactive exercises with instant feedback.