# 6.2 Use Properties of Rectangles, Triangles, and Trapezoids

**Learning Objectives**

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
- Use properties of trapezoids

**Try It**

Before you get started, take this readiness quiz:

1) The length of a rectangle is [latex]3[/latex] less than the width. Let [latex]w[/latex] represent the width. Write an expression for the length of the rectangle.

2) Simplify: [latex]\frac{1}{2}(6h)[/latex]

3) Simplify: [latex]\frac{5}{2}(10.3-7.9)[/latex]

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 Figure 6.2.1. A tape measure might remind you of a line—you use it for linear measure, which measures length. Inch, foot, yard, mile, centimetre 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 is square units. We often use square inches, square feet, square centimetres, or square miles to measure area. A square centimetre is a square that is one centimetre (cm) on each side. A square inch is a square that is one inch on each side Figure 6.2.2.

Square measures have sides that are each 1 unit in length.

Figure 6.2.3 shows a rectangular rug that is [latex]2[/latex] feet long by [latex]3[/latex] feet wide. Each square is [latex]1[/latex] foot wide by [latex]1[/latex] foot long, or [latex]1[/latex] square foot. The rug is made of [latex]6[/latex] squares. The area of the rug is [latex]6[/latex] square feet.

The rug contains six squares of 1 square foot each, so the total area of the rug is 6 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 centimetres. When measuring the volume of a rectangular solid, you measure how many cubes fill the container. We often use cubic centimetres, cubic inches, and cubic feet. A cubic centimetre is a cube that measures one centimetre on each side, while a cubic inch is a cube that measures one inch on each side Figure 6.2.4.

Suppose the cube in Figure 6.2.5 measures [latex]3[/latex] inches on each side and is cut on the lines shown. How many little cubes does it contain? If we were to take the big cube apart, we would find [latex]27[/latex] little cubes, with each one measuring one inch on all sides. So each little cube has a volume of [latex]1[/latex] cubic inch, and the volume of the big cube is [latex]27[/latex] cubic inches.

A cube that measures [latex]3[/latex] inches on each side is made up of [latex]27[/latex] one-inch cubes, or [latex]27[/latex] cubic inches.

**Example 6.2.1**

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.

**Solution**

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 |

**Try It**

4) Determine whether you would use linear, square, or cubic measure for each item.

a. amount of paint in a can

b. height of a tree

c. floor of your bedroom

d. diameter of bike wheel

e. size of a piece of sod

f. amount of water in a swimming pool

**Solution**

a. cubic

b. linear

c. square

d. linear

e. square

f. cubic

5) Determine whether you would use linear, square, or cubic measure for each item.

a. volume of a packing box

b. size of patio

c. amount of medicine in a syringe

d. length of a piece of yarn

e. size of housing lot

f. height of a flagpole

**Solution**

a. cubic

b. square

c. cubic

d. linear

e. square

f. linear

Many geometry applications will involve finding the perimeter or the area of a figure. There are also many applications of perimeter and area in everyday life, so it is important to make sure you understand what they each mean.

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 have been laid. To figure out how many strips are needed, you must know the distance around the room. You would use a tape measure to measure the number of feet 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

Figure 6.2.6 shows a square tile that is [latex]1[/latex] inch on each side. If an ant walked around the edge of the tile, it would walk [latex]4[/latex] inches. This distance is the perimeter of the tile.

Since the tile is a square that is [latex]1[/latex] inch on each side, its area is one square inch. The area of a shape is measured by determining how many square units cover the shape.

**Example 6.2.2**

Each of two square tiles is [latex]1[/latex] square inch. Two tiles are shown together.

a. What is the perimeter of the figure?

b. What is the area?

**Solution**

a. The perimeter is the distance around the figure. The perimeter is [latex]6[/latex] inches.

b. The area is the surface covered by the figure. There are [latex]2[/latex] square inch tiles so the area is [latex]2[/latex] square inches.

**Try It**

6) Find the: a. perimeter and b. area of the figure:

**Solution**

a. 8 inches

b. 3 sq. inches

7) Find the: a. perimeter b. area of the figure

**Solution**

a. 8 centimetres

b. 4 sq. centimetres

**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, [latex]L[/latex], and the adjacent side as the width, [latex]W[/latex]. See Figure 6.2.11.

A rectangle has four sides, and four right angles. The sides are labelled [latex]L[/latex] for length and [latex]W[/latex] for width.

The perimeter, [latex]P[/latex], of the rectangle is the distance around the rectangle. If you started at one corner and walked around the rectangle, you would walk [latex]L+W+L+W[/latex] units, or two lengths and two widths. The perimeter then is

What about the area of a rectangle? Remember the rectangular rug from the beginning of this section. It was [latex]2[/latex] feet long by [latex]3[/latex] feet wide, and its area was [latex]6[/latex] square feet. See Figure 6.2.13. Since [latex]A=2\times\ 3[/latex] we see that the area, [latex]A[/latex], is the length, [latex]L[/latex], times the width, [latex]W[/latex], so the area of a rectangle is [latex]A=L\times \ W[/latex]

The area of this rectangular rug is [latex]6[/latex] square feet, its length times its width.

**Properties of Rectangles**

- Rectangles have four sides and four right [latex]90^{\circ}[/latex] angles.
- The lengths of opposite sides are equal.
- The perimeter, [latex]P[/latex], of a rectangle is the sum of twice the length and twice the width. See Figure 6.2.11.

[latex]P=2L+2W[/latex]. - The area, [latex]A[/latex], of a rectangle is the length times the width.

[latex]A=L\times W[/latex]

For easy reference as we work the examples in this section, we will restate the Problem Solving Strategy for Geometry Applications here.

**HOW TO**

#### Use a Problem Solving Strategy for Geometry Applications

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

**Example 6.2.3**

The length of a rectangle is [latex]32[/latex] meters and the width is [latex]20[/latex] meters. Find a. the perimeter b. the area

**Solution**

*a.** Step 1: *Read the problem. Draw the figure and label it with the given information.

*Step 2:* Identify what you are looking for.

The perimeter of a rectangle.

*Step 3:* Name. Choose a variable to represent it.

Let [latex]P[/latex]= the perimeter.

*Step 4:* Translate.

Write the appropriate formula. Substitute.

[latex]\begin{eqnarray*}P\;&=&\;2L\;+\;2W\\P\;&=&\;2(32)\;+\;2(20)\end{eqnarray*}[/latex]

*Step 5:* Solve the equation.

[latex]\begin{eqnarray*}P\;&=&\;64\;+\;40\\P\;&=&\;104\end{eqnarray*}[/latex]

*Step 6:* Check.

[latex]\begin{eqnarray*}P\;&\overset?=&\;104\\20\;+\;32\;+\;20\;+\;32\;&\overset?=&\;104\\104\;&\overset?=&\;104\\\end{eqnarray*}[/latex]

*Step 7:* Answer the question.

The perimeter of the rectangle is [latex]104[/latex] meters.

*b.** Step 1: *Read the problem. Draw the figure and label it with the given information.

*Step 2:* Identify what you are looking for.

The area of a rectangle

*Step 3:* Name. Choose a variable to represent it.

Let [latex]A[/latex]= the area.

*Step 4:* Translate.

Write the appropriate formula. Substitute.

[latex]\begin{eqnarray*}A\;&=&\;L\;\times\;W\\A\;&=&\;32m\;\times\;20m\end{eqnarray*}[/latex]

*Step 5:* Solve the equation.

[latex]\begin{eqnarray*}A\;&=&\;640\end{eqnarray*}[/latex]

*Step 6:* Check.

[latex]\begin{eqnarray*}A\;&\overset?=&\;640\\32\;\times\;20\;&\overset?=&\;640\\640\;&\overset?=&\;640\end{eqnarray*}[/latex]

*Step 7:* Answer the question.

The area of the rectangle is [latex]60[/latex] square meters.

**Try It**

8) The length of a rectangle is [latex]120[/latex] yards and the width is [latex]50[/latex] yards. Find a. the perimeter and b. the area.

**Solution**

a. 340 yd

b. 6000 sq. yd

9) The length of a rectangle is [latex]62[/latex] feet and the width is [latex]48[/latex] feet. Find a. the perimeter and b. the area.

**Solution**

a. 220 ft

b. 2976 sq. ft

**Example 6.2.4**

Find the length of a rectangle with perimeter [latex]50[/latex] inches and width [latex]10[/latex] inches.

**Solution**

*Step 1: *Read the problem. Draw the figure and label it with the given information.

*Step 2:* Identify what you are looking for.

The length of the rectangle.

*Step 3:* Name. Choose a variable to represent it.

Let [latex]L[/latex]= the length.

*Step 4:* Translate.

Write the appropriate formula. Substitute.

[latex]\begin{align*}P&=2L+2W\\50&=2L+2(10)\end{align*}[/latex]

*Step 5:* Solve the equation.

[latex]\begin{eqnarray*}50\;{\color{red}{-\;20}}\;&=&\;2L\;+\;20\;{\color{red}{-20}}\\30\;&=&2L\\\frac{30}{\color{red}{2}}\;&=&\;\frac{2L}{\color{red}{2}}\\15\;&=&\;L\\\end{eqnarray*}[/latex]

*Step 6:* Check.

[latex]\begin{align*}P&=50\\15+10+15+10&\overset?=50\\50&=50\checkmark\end{align*}[/latex]

*Step 7:* Answer the question.

The length is [latex]15[/latex] inches.

**Try It**

10) Find the length of a rectangle with a perimeter of 80 inches and width of 25 inches.

**Solution**

15 in.

11) Find the length of a rectangle with a perimeter of 30 yards and width of 6 yards.

**Solution**

9 yd

In the next example, the width is defined in terms of the length. We’ll wait to draw the figure until we write an expression for the width so that we can label one side with that expression.

**Example 6.2.5**

The width of a rectangle is two inches less than the length. The perimeter is [latex]52[/latex] inches. Find the length and width.

**Solution**

*Step 1: *Read the problem.

*Step 2:* Identify what you are looking for.

The length of the rectangle.

*Step 3:* Name. Choose a variable to represent it.

Since the width is defined in terms of the length, we let [latex]L=[/latex] length. The width is two feet less that the length, so we let width [latex](W)=L-2[/latex].

Now we can create a visual to represent it.

*Step 4:* Translate.

Write the appropriate formula. The formula for the perimeter of a rectangle relates all the information.

Substitute in the given information.

[latex]\begin{eqnarray*}P\;&=&\;2L\;+\;2W\\52\;&=&\;2L\;+\;2(L-2)\end{eqnarray*}[/latex]

*Step 5:* Solve the equation.

Combine like terms.

[latex]\begin{align*}52&=2L+2(L-2)\\ 52&={\color{red}{2L+2L}}-4\\ 52&={\color{red}{4L}}-4 \end{align*}[/latex]

Add [latex]4[/latex] to each side.

[latex]\begin{align*}{\color{blue}{4+}}52&=4L-4{\color{blue}{+4}}\\ 56&=4L\cancel{-4+4} \end{align*}[/latex]

Divide by [latex]4[/latex] and solve.

[latex]\begin{align*} \frac{56}{{\color{blue}{4}}}&=\frac{4L}{{\color{blue}{4}}}\\ \frac{56}{{\color{blue}{4}}}&=\frac{\cancel4L}{{\color{blue}{\cancel4}}}\\ \frac{56}{4}&=L\\ 14&=L \end{align*}[/latex]

**Step 6: Find the width**.

[latex]\begin{align*}W&=L-2\\W&={\color{red}{14}}-2\\W&=12\end{align*}[/latex]

*Step 7:* Check.

[latex]\begin{align*}P&=2{\color{red}{L}}+2{\color{blue}{W}}\\52&=2({\color{red}{14}})+2({\color{blue}{12}})\\52&={\color{red}{28}}+{\color{blue}{24}}\\52&=52\checkmark\end{align*}[/latex]

*Step 8:* Answer the question.

The length is [latex]14[/latex] feet and the width is [latex]12[/latex] feet.

**Try It**

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

**Solution**

18 m, 11 m

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

**Solution**

11 ft , 19 ft

**Example 6.2.6**

The length of a rectangle is four centimetres more than twice the width. The perimeter is [latex]32[/latex] centimetres. Find the length and width.

**Solution**

*Step 1: *Read the problem.

*Step 2:* Identify what you are looking for.

The length and width.

*Step 3:* Name. Choose a variable to represent it. Then draw a diagram to show what we are solving for.

Let [latex]W=[/latex] width. The length is four more than twice the width. [latex]2w+4=[/latex] length.

*Step 4:* Translate.

Write the appropriate formula and substitute in the given information.

[latex]\begin{eqnarray*}P\;&=&\;{\color{blue}{2L}}\;+\;{\color{red}{2W}}\\32\;&=&\;{\color{blue}{2(2W\;+\;4)}}\;+\;{\color{red}{2W}}\end{eqnarray*}[/latex]

*Step 5:* Solve the equation.

[latex]\begin{align*} 32&={\color{blue}{2(2W+4)}}+{\color{red}{2W}}\\ 32&={\color{blue}{4W+8}}+{\color{red}{2W}}\\ 32&=6W+8\\ 24&=6W\\ {\color{red}{4}}&=W\\ 2W+4&=length\\ 2({\color{red}{4}})+4&=\;length\\12&=\;length\end{align*}[/latex]

*Step 6:* Check.

[latex]\begin{align*}P&=2L+2W\\32&\overset?=2\times12+2\times4\\32&=32\;\checkmark\end{align*}[/latex]

*Step 7:* Answer the question.

The length is [latex]12[/latex] cm and the width is [latex]4[/latex] cm.

**Try It**

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

**Solution**

8 ft, 24 ft

15) The width of a rectangle is six less than twice the length. The perimeter is 18 centimetres. Find the length and width.

**Solution**

5 cm, 4 cm

**Example 6.2.7**

The area of a rectangular room is [latex]168[/latex] square feet. The length is [latex]14[/latex] feet. What is the width?

**Solution**

*Step 1: *Read the problem.

*Step 2:* Identify what you are looking for.

The width of a rectangular room

*Step 3:* Name. Choose a variable to represent it.

Let [latex]W=[/latex] width.

*Step 4:* Translate.

Write the appropriate formula and substitute in the given information.

[latex]\begin{eqnarray*}A\;&=&\;LW\\168\;&=&\;14W\end{eqnarray*}[/latex]

*Step 5:* Solve the equation.

[latex]\begin{eqnarray*}\frac{168}{14}\;&=&\;\frac{14W}{14}\\12\;&=&\;W\end{eqnarray*}[/latex]

*Step 6:* Check.

[latex]\begin{align*}A&=LW\\168&\overset?=14\times12\\168&=168\;\checkmark\end{align*}[/latex]

*Step 7:* Answer the question.

The width of the room is [latex]12[/latex] feet.

**Try It**

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

**Solution**

26 ft

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

**Solution**

29 m

**Example 6.2.8**

Find the length of a rectangle with perimeter [latex]50[/latex] inches and width [latex]10[/latex] inches.

**Solution**

*Step 1: *Read the problem. Draw the figure and label it with the given information.

[latex]P=50[/latex] inches.

*Step 2:* Identify what you are looking for.

The length of the rectangle.

*Step 3:* Name. Choose a variable to represent it.

Let [latex]L=[/latex] the length.

*Step 4:* Translate.

Write the appropriate formula and substitute in the given information.

[latex]\begin{eqnarray*}\mathrm P\;&=&\;2\mathrm L\;+\;2\mathrm W\\50\;&=&\;2\mathrm L\;+\;2(10)\\\end{eqnarray*}[/latex]

*Step 5:* Solve the equation.

[latex]\begin{align*}50{\color{red}{-20}}&=2\mathrm{L}+20{\color{red}{-20}}\\ 30&=2\mathrm{L}\\\frac{30}2&=\frac{2\mathrm{L}}2\\ 15&=\mathrm{L}\end{align*}[/latex]

*Step 6:* Check.

[latex]\begin{eqnarray*}15+10+15+10&\overset?=&50\\50&=&50\checkmark\end{eqnarray*}[/latex]

*Step 7:* Answer the question.

The length is [latex]15[/latex] inches.

**Try It**

18) Find the length of a rectangle with: perimeter 80 and width 25.

**Solution**

15

19) Find the length of a rectangle with: perimeter 30 and width 6.

**Solution**

9

**Example 6.2.9**

The perimeter of a rectangular swimming pool is [latex]150[/latex] feet. The length is [latex]15[/latex] feet more than the width. Find the length and width.

**Solution**

*Step 1: *Read the problem. Draw the figure and label it with the given information.

*Step 2:* Identify what you are looking for.

The length and width of the pool.

*Step 3:* Name. Choose a variable to represent it.

Let [latex]W=[/latex] width.

Let [latex]W+15=[/latex]length.

*Step 4:* Translate.

Write the appropriate formula and substitute in the given information.

[latex]\begin{eqnarray*}\mathrm P\;&=&\;2\mathrm L\;+\;2\mathrm W\\150\;&=&\;2(\mathrm W\;+15)\;+\;2\mathrm W\\\end{eqnarray*}[/latex]

*Step 5:* Solve the equation.

[latex]\begin{eqnarray*}150\;&=&\;2\mathrm{W}\;+\;30\;+\;2\mathrm{W}\\150\;&=&\;4\mathrm{W}\;+\;30\\120\;&=&\;4\mathrm{W}\\{\color{red}{30}}\;&=&\;\mathrm{W}\;(\text{the width of the pool})\\\\\mathrm{W}\;+\;15\;&=&\;\text{length of pool}\\{\color{red}{30}}\;+\;15\;&=&\;\mathrm{L}\\45\;&=&\;\mathrm{L}\end{eqnarray*}[/latex]

*Step 6:* Check.

[latex]\begin{eqnarray*}\mathrm P\;&=&\;2\mathrm L\;+\;2\mathrm W\\150\;&\overset?=&\;2(45)\;+\;2(30)\\150\;&=&\;150✓\end{eqnarray*}[/latex]

*Step 7:* Answer the question.

The length of the pool is [latex]45[/latex] feet and the width is [latex]30[/latex] feet.

**Try It**

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

**Solution**

30 ft, 70 ft

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

**Solution**

60 yd, 90 yd

**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. In the rectangle in Figure 6.2.21, we’ve labelled the length [latex]b[/latex] and the width [latex]h[/latex], so it’s area is [latex]bh[/latex]*.*

We can divide this rectangle into two congruent triangles Figure 6.2.22. Triangles that are congruent have identical side lengths and angles, and so their areas are equal. The area of each triangle is one-half the area of the rectangle, or [latex]\frac{1}{2}bh[/latex]. This example helps us see why the formula for the area of a triangle is [latex]A=\frac{1}{2}bh[/latex].

A rectangle can be divided into two triangles of equal area. The area of each triangle is one-half the area of the rectangle.

The formula for the area of a triangle is [latex]A=\frac{1}{2}bh[/latex], where [latex]b[/latex] is the base and [latex]h[/latex] is the height.

To find the area of the 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 [latex]90^{\circ}[/latex] angle with the base. Figure 6.2.23 shows three triangles with the base and height of each marked.

The height [latex]h[/latex] of a triangle is the length of a line segment that connects the the base to the opposite vertex and makes a [latex]90^{\circ}[/latex] angle with the base.

**Triangle Properties**

For any triangle [latex]\bigtriangleup ABC[/latex], the sum of the measures of the angles is [latex]180^{\circ}[/latex].

[latex]m\angle A+m\angle B+m\angle C=180^{\circ}[/latex]

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

[latex]P=a+b+c[/latex]

**Example 6.2.10**

Find the area of a triangle whose base is [latex]11[/latex] inches and whose height is [latex]8[/latex] inches.

**Solution**

*Step 1: *Read the problem. Draw the figure and label it with the given information.

*Step 2:* Identify what you are looking for.

The area of the triangle.

*Step 3:* Name. Choose a variable to represent it.

Let [latex]A=[/latex] area of the triangle.

*Step 4:* Translate.

Write the appropriate formula and substitute in the given information.

[latex]\begin{eqnarray*}A\;&=&\;\frac12\times\;b\;\times\;h\\A\;&=&\;\frac12\times\;11\;\times\;8\end{eqnarray*}[/latex]

*Step 5:* Solve the equation.

[latex]\begin{eqnarray*}A\;&=&\;44\;\text{square inches}\end{eqnarray*}[/latex]

*Step 6:* Check.

[latex]\begin{eqnarray*}A\;&=&\frac12bh\\44\;&\overset?=&\;\frac12\;(11)8\\44\;&=&\;44\;\checkmark\\\end{eqnarray*}[/latex]

*Step 7:* Answer the question.

The area is [latex]44[/latex] square inches.

**Try It**

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

**Solution**

13 sq. in.

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

**Solution**

49 sq. in.

**Example 6.2.11**

The perimeter of a triangular garden is [latex]24[/latex] feet. The lengths of two sides are [latex]4[/latex] feet and [latex]9[/latex] feet. How long is the third side?

**Solution**

*Step 1: *Read the problem. Draw the figure and label it with the given information.

*Step 2:* Identify what you are looking for.

The length of the third side of a triangle.

*Step 3:* Name. Choose a variable to represent it.

Let [latex]c=[/latex] the third side.

*Step 4:* Translate.

Write the appropriate formula and substitute in the given information.

[latex]\begin{eqnarray*}P\;&=&\;a\;+\;b\;+\;c\\P\;&=&\;4\;+\;9\;+\;c\end{eqnarray*}[/latex]

*Step 5:* Solve the equation.

[latex]\begin{eqnarray*}24\;&=&\;13\;+\;c\\11\;&=&\;c\end{eqnarray*}[/latex]

*Step 6:* Check.

[latex]\begin{eqnarray*}P\;&=&\;a\;+\;b\;+\;c\\24\;&\overset?=&\;4\;+\;9\;+\;11\\24\;&=&\;24\;\checkmark\\\end{eqnarray*}[/latex]

*Step 7:* Answer the question.

The third side is [latex]11[/latex] feet long.

**Try It**

24) The perimeter of a triangular garden is 24 feet. The lengths of two sides are 18 feet and 22 feet. How long is the third side?

**Solution**

8 ft

25) 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?

**Solution**

6 ft

**Example 6.2.12**

The area of a triangular church window is [latex]90[/latex] square meters. The base of the window is [latex]15[/latex] meters. What is the window’s height?

**Solution**

*Step 1: *Read the problem. Draw the figure and label it with the given information.

*Step 2:* Identify what you are looking for.

The height of a triangle.

*Step 3:* Name. Choose a variable to represent it.

Let [latex]h=[/latex] the height.

*Step 4:* Translate.

Write the appropriate formula and substitute in the given information.

[latex]\begin{align*}A&=\frac12\times b\times h\\90&=\frac12\times15\times h\end{align*}[/latex]

*Step 5:* Solve the equation.

[latex]\begin{eqnarray*}90\;&=&\;\frac{15}2h\\12\;&=&\;h\\\\\end{eqnarray*}[/latex]

*Step 6:* Check.

[latex]\begin{eqnarray*}A\;&=&\;\frac12bh\\90\;&\overset?=&\;\frac12\;\times\;15\;\times\;12\\90\;&=&\;90\;\checkmark\\\\\end{eqnarray*}[/latex]

*Step 7:* Answer the question.

The height of the triangle is [latex]12[/latex] meters.

**Try It**

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

**Solution**

14 in.

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

**Solution**

6 ft

**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. A triangle that has three sides of equal length is called an equilateral triangle. Figure 6.2.28 shows both types of triangles.

In an isosceles triangle, two sides have the same length, and the third side is the base. In an equilateral triangle, all three sides have the same length.

**Isosceles and Equilateral Triangles**

An isosceles triangle has two sides the same length.

An equilateral triangle has three sides of equal length.

**Example 6.2.13**

The perimeter of an equilateral triangle is [latex]93[/latex] inches. Find the length of each side.

**Solution**

*Step 1: *Read the problem. Draw the figure and label it with the given information.

Perimeter = 93 in.

*Step 2:* Identify what you are looking for.

The length of the sides of an equilateral triangle.

*Step 3:* Name. Choose a variable to represent it.

Let [latex]s=[/latex] length of each side.

*Step 4:* Translate.

Write the appropriate formula and substitute in the given information.

[latex]\begin{eqnarray*}P\;&=&\;a\;+\;b\;+\;c\\93\;&=&\;s\;+\;s\;+\;s\\\\\end{eqnarray*}[/latex]

*Step 5:* Solve the equation.

[latex]\begin{eqnarray*}93\;&=&\;3s\\31\;&=&\;s\\\\\end{eqnarray*}[/latex]

*Step 6:* Check.

[latex]\begin{eqnarray*}93\;&\overset?=&\;31\;+\;31\;+\;31\\93\;&=&\;93\;\checkmark\end{eqnarray*}[/latex]

*Step 7:* Answer the question.

Each side is [latex]31[/latex] inches.

**Try It**

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

**Solution**

13 in.

29) Find the length of each side of an equilateral triangle with perimeter 51 centimetres.

**Solution**

17 cm

**Example 6.2.14**

Arianna has [latex]156[/latex] inches of beading to use as trim around a scarf. The scarf will be an isosceles triangle with a base of [latex]60[/latex] inches. How long can she make the two equal sides?

**Solution**

*Step 1: *Read the problem. Draw the figure and label it with the given information.

P = [latex]156[/latex] in

*Step 2:* Identify what you are looking for.

The lengths of the two equal sides.

*Step 3:* Name. Choose a variable to represent it.

Let [latex]s=[/latex] the length of each side.

*Step 4:* Translate.

Write the appropriate formula and substitute in the given information.

[latex]\begin{eqnarray*}P\;&=&\;a\;+\;b\;+\;c\\156\;&=&\;s\;+\;60\;+\;s\\\\\end{eqnarray*}[/latex]

*Step 5:* Solve the equation.

[latex]\begin{eqnarray*}156\;&=&\;2s\;+\;60\\90\;&=&\;2s\\48\;&=&\;s\\\\\end{eqnarray*}[/latex]

*Step 6:* Check.

[latex]\begin{eqnarray*}P\;&=&\;a\;+\;b\;+\;c\\156\;&\overset?=&\;48\;+\;60\;+\;48\\156\;&=&\;156\;\checkmark\\\\\end{eqnarray*}[/latex]

*Step 7:* Answer the question.

Arianna can make each of the two equal sides [latex]48[/latex] inches long.

**Try It**

30) 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?

**Solution**

14 ft

31) 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?

**Solution**

7 m

**Use the Properties of Trapezoids**

A trapezoid is 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 [latex]b[/latex], and the length of the bigger base [latex]B[/latex]. The height, [latex]h[/latex], of a trapezoid is he distance between the two bases as shown in Figure 6.2.31

A trapezoid has a larger base, [latex]B[/latex], and a smaller base, [latex]b[/latex]. The height [latex]h[/latex] is the distance between the bases.

**The formula for the area of a trapezoid is:**

Splitting the trapezoid into two triangles may help us understand the formula. The area of the trapezoid is the sum of the areas of the two triangles. See Figure 6.2.32.

Splitting a trapezoid into two triangles may help you understand the formula for its area.

The height of the trapezoid is also the height of each of the two triangles. See Figure 6.2.33.

**The formula for the area of a trapezoid is**

If we distribute, we get,

**Properties of Trapezoids**

- A trapezoid has four sides. See Figure 6.2.31
- Two of its sides are parallel and two sides are not.
- The area, [latex]A[/latex], of a trapezoid is [latex]{A}=\frac{1}{2}h(b+B)[/latex].

**Example 6.2.15**

Find the area of a trapezoid whose height is [latex]6[/latex] inches and whose bases are [latex]14[/latex] and [latex]11[/latex] inches.

**Solution**

*Step 1**: *Read the problem. Draw the figure and label it with the given information.

*Step 2:* Identify what you are looking for.

The area of the trapezoid.

*Step 3:* Name. Choose a variable to represent it.

Let [latex]A=[/latex]the area.

*Step 4:* Translate.

Write the appropriate formula and substitute in the given information.

[latex]\begin{eqnarray*}A\;&=&\;\frac12\times\;b\;\times\;h\\A\;&=&\;\frac12\times\;6\times\;(11+14)\\\\\end{eqnarray*}[/latex]

*Step 5:* Solve the equation.

[latex]\begin{eqnarray*}1A\;&=&\;\frac12\times\;6(25)\\A\;&=&\;3(25)\\A\;&=&\;75\;sq.\;inches\\\\\end{eqnarray*}[/latex]

*Step 6:* Check.

Is this answer reasonable?

The area of the larger rectangle is [latex]84[/latex] square inches and the area of the smaller rectangle is [latex]66[/latex] square inches. So it makes sense that the area of the trapezoid is between [latex]84[/latex] and [latex]66[/latex] square inches

**Step 7:****Answer** **the question.**

The area of the trapezoid is [latex]75[/latex] square inches.

**Try It**

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

**Solution**

161 sq. yd

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

**Solution**

225 sq. cm

**Example 6.2.16**

Find the area of a trapezoid whose height is [latex]5[/latex] feet and whose bases are [latex]10.3[/latex] and [latex]13.7[/latex] feet.

**Solution**

*Step 1: *Read the problem. Draw the figure and label it with the given information.

*Step 2:* Identify what you are looking for.

The area of the trapezoid.

*Step 3:* Name. Choose a variable to represent it.

Let [latex]A=[/latex] the area.

*Step 4:* Translate.

Write the appropriate formula and substitute in the given information.

[latex]\begin{eqnarray*}A\;&=&\;\frac12\;\times\;h\;\times\;(b+B)\\A\;&=&\;\frac12\;\times\;5\;\times\;(10.3\;+\;13.7)\end{eqnarray*}[/latex]

*Step 5:* Solve the equation.

[latex]\begin{eqnarray*}1A\;&=&\;\frac12\;\times\;5(24)\\A\;&=&\;12\;\times\;5\\A\;&=&\;60\;sq.\;ft\\\end{eqnarray*}[/latex]

*Step 6:* Check: Is this answer reasonable?

The area of the trapezoid should be less than the area of a rectangle with base [latex]13.7[/latex] and height [latex]5[/latex], but more than the area of a rectangle with base [latex]10[/latex].3 and height [latex]5[/latex].

*Step 7:* Answer the question.

The area of the trapezoid is [latex]6[/latex]0 square feet.

**Try It**

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

**Solution**

42 sq. cm

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

**Solution**

63 sq. m

**Example 6.2.17**

Vinny has a garden that is shaped like a trapezoid. The trapezoid has a height of [latex]3.4[/latex] yards and the bases are [latex]8.2[/latex] and [latex]5.6[/latex] yards. How many square yards will be available to plant?

**Solution**

*Step 1: *Read the problem. Draw the figure and label it with the given information.

*Step 2:* Identify what you are looking for.

The area of a trapezoid.

*Step 3:* Name. Choose a variable to represent it.

Let [latex]A=[/latex] the area.

*Step 4:* Translate.

Write the appropriate formula and substitute in the given information.

[latex]\begin{eqnarray*}A\;&=&\;\frac12\times\;b\;\times\;(b+B)\\A\;&=&\;\frac12\;\times\;3.4\;\times\;(5.6\;+\;8.2)\\\\\end{eqnarray*}[/latex]

*Step 5:* Solve the equation.

[latex]\begin{eqnarray*}A\;&=&\;\frac12(3.4)(13.8)\\A\;&=&\;23.46\;sq.\;yards\\\\\end{eqnarray*}[/latex]

[latex]\begin{align*} A_{\color{red}{\text{rectangle}}}&=Bh\\ &=(8.2)(3.4)\\ &=27.88\;yd^2 \end{align*}[/latex] | [latex]\begin{align*} A_\text{trapezpoid}&=\frac12(3.4\;yd)(5.6=8.2)\\ &=23.46\;yd^2 \end{align*}[/latex] | [latex]\begin{align*} A_{\color{blue}{\text{rectangle}}}&=bh\\ &=(5.6)(3.4)\\ &=19.04\;yd^2 \end{align*}[/latex] |

[latex]\begin{align*}A&_{\color{red}{\text{rectangle}}}&>\;\;\;A&_\text{trapezoid}&>\;\;\;A&_{\color{blue}{\text{rectangle}}}\\&{\color{red}{27.88}}&\;&23.46&\;&{\color{blue}{19.04}}\end{align*}[/latex] |

*Step 6:* Check: Is this answer reasonable?

Yes. The area of the trapezoid is less than the area of a rectangle with a base of [latex]8.2[/latex] yd and height [latex]3.4[/latex] yd, but more than the area of a rectangle with base [latex]5.6[/latex] yd and height [latex]3.4[/latex] yd.

*Step 7:* Answer the question.

Vinny has [latex]23.46[/latex] square yards in which he can plant.

**Try It**

36) 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?

**Solution**

40.25 sq. yd

37) Kira wants cover his patio with concrete pavers. If 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 he need?

**Solution**

240 sq. ft

*Spaghetti and Meatballs for All*will provide you with another view of the topics covered in this section.

**Key Concepts**

**Properties of Rectangles**- Rectangles have four sides and four right ([latex]90°[/latex]) angles.
- The lengths of opposite sides are equal.
- The perimeter, [latex]P[/latex], of a rectangle is the sum of twice the length and twice the width.
- [latex]P=2L+2W[/latex]

- The area, [latex]A[/latex], of a rectangle is the length times the width.
- [latex]A=L\times W[/latex]

**Triangle Properties**- For any triangle [latex]\bigtriangleup ABC[/latex], the sum of the measures of the angles is [latex]180°[/latex].
- [latex]m\angle A+m\angle B+m\angle C=180°[/latex]

- The perimeter of a triangle is the sum of the lengths of the sides.
- [latex]P=a+b+c[/latex]

- The area of a triangle is one-half the base, [latex]b[/latex], times the height, [latex]h[/latex].
- [latex]A=\frac{1}{2}bh[/latex]

- For any triangle [latex]\bigtriangleup ABC[/latex], the sum of the measures of the angles is [latex]180°[/latex].

**Self Check**

After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

On a scale of 1–10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?

**Glossary**

- area
- The area is a measure of the surface covered by a figure.

- equilateral triangle
- A triangle with all three sides of equal length is called an equilateral triangle.

- isosceles triangle
- A triangle with two sides of equal length is called an isosceles triangle.

- perimeter
- The perimeter is a measure of the distance around a figure.

- rectangle
- A rectangle is a geometric figure that has four sides and four right angles.

- trapezoid
- A trapezoid is four-sided figure, a quadrilateral, with two sides that are parallel and two sides that are not.

The area is a measure of the surface covered by a figure.

The perimeter is a measure of the distance around a figure.

A rectangle is a geometric figure that has four sides and four right angles.

A triangle with two sides of equal length is called an isosceles triangle

A triangle with all three sides of equal length is called an equilateral triangle.

A trapezoid is four-sided figure, a quadrilateral, with two sides that are parallel and two sides that are not.