3.10 Area of Rectangles and Triangles

Area of Rectangles and Triangles

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A diagram of a baseball field with its measurements marked. — **Figure 3.12** A historical baseball-field diagram with labeled lengths.A historical baseball-field diagram with labeled lengths. (credit: "Diagram of Regulation Diamond" by Erica Fischer from "Baseball" The World Book, 1920/Flickr, Public Domain, CC BY 2.0)

Area measures the size of a two-dimensional region, while perimeter measures the length of its boundary. We use lengths and area formulas to compare regions and to plan quantities such as flooring. In this lesson, the examples use triangles and rectangles with stated dimensions.

Area refers to a region measured in square units, like a square mile or a square foot. For example, to purchase tile for a kitchen floor, you would need to know how many square feet are needed because tile is sold by the square foot. Carpeting is sold by the square yard. Perimeter is measured in linear units. For example, fencing is sold in linear units, a linear foot or yard. Linear dimensions refer to an outline or a boundary. Square units refer to the area within that boundary. Different items may have different units, but either way, you must know the linear dimensions to calculate the area.

Many geometric shapes have formulas for calculating areas, such as triangles, regular polygons, and circles. To calculate areas for many irregular curves or shapes, we need calculus. However, in this section, we will only look at geometric shapes that have known area formulas. The notation for area, as mentioned, is in square units and we write sq in or sq cm, or use an exponent, such as ${in}^{2}$ or ${cm}^{2} .$ Note that linear measurements have no exponent above the units or we can say that the exponent is 1.

Area of Triangles

The formula for the area of a triangle is given as follows.

For example, consider the triangle in Figure 3.13.

A right triangle with its legs marked 5 centimeters and 4 centimeters. — **Figure 3.13** Triangle 1

The base measures 4 cm and the height measures 5 cm. Using the formula, we can calculate the area:

$A = \frac{1}{2} (4) (5) = \frac{1}{2} (20) = 10 {cm}^{2}$

In Figure 3.14, the triangle has a base equal to 7 cm and a height equal to 3.5 cm. Notice that we can only find the height by dropping a perpendicular to the base. The area is then

A 7 by 7 geoboard. A rubber band wraps three pegs at column 0 row 0, column 4 row 0 and column 1 row 3, making a triangle with a base of 4 units and a perpendicular height of 3 units. The board reports an area of 6 square units. The corners are draggable. Viewing window: x from -5.44 to 11.44, y from -2.22 to 8.22. — This triangle has a base of 4 units and perpendicular height 3 units, not the length of a slanted side. Its area is one-half × 4 × 3 = 6 square units. Keep the two base corners fixed and drag the top corner along the same row: the height stays 3 and the area stays 6. Move the top corner to a different row and the height, and therefore the area, changes.

$A = \frac{1}{2} (7) (3.5) = 12.25 {cm}^{2} .$

A triangle with its base marked b equals 7 centimeters and height marked h equals 3.5 centimeters. — **Figure 3.14** Triangle 2

Your Turn

Find the area of the triangle with a base equal to 4 cm and the height equal to 4 cm.

Show solution

$8\,\text{cm}^2$

The area of a triangle is $A = \frac{1}{2}bh$ , where b represents the base and h represents the perpendicular height to that base.

$A = \frac{1}{2}bh = \frac{1}{2}\left( {{\rm{4\ cm}}} \right)\left( {{\rm{4 \ cm}}} \right) = {\rm{ 8\ c}}{{\rm{m}}^2}$

The area is 8 square centimeters.

Rectangle

Similarly, the area for a rectangle is found by multiplying length times width. The rectangle in Figure 3.16 has width equal to 5 in and length equal to 12 in. The area is $A = 5 (12) = 60 {in}^{2} .$

A rectangle with its length and width marked 12 inches and 5 inches. It has four right angles. — **Figure 3.16** Area of a Rectangle

Many everyday applications require the use of the perimeter and area formulas. Suppose you are remodeling your home and you want to replace all the flooring. You need to know how to calculate the area of the floor to purchase the correct amount of tile, or hardwood, or carpet. If you want to paint the rooms, you need to calculate the area of the walls and the ceiling to know how much paint to buy, and the list goes on. Can you think of other situations where you might need to calculate area?

Your Turn

Find the area of a rectangular region with a length of 18 feet and a width 1/3 of the length.

Show solution

The area of a rectangle is $A = LW$ , where L represents the length and W represents the width.

The length is 18 feet.

Break the width expression into its components.

The width	is	one third of the length
width	=	$\frac{1}{3}\left( {18{\text{ ft}}} \right)$

width = $\frac{1}{3}\left( {18{\rm{\ ft}}} \right) = {\rm{ 6\ ft}}$

$A = LW = \left( {18{\rm{\ ft}}} \right)\left( {{\rm{6\ ft}}} \right) = 108{\rm{\ f}}{{\rm{t}}^2}$

The area is 108 square feet.

108 ft²

Your Turn

You remodel your kitchen and decide to change out the tile floor. The floor measures 30 ft by 15 ft. One box of tile costs $45 and covers 10 ft². How many boxes of tile should you buy, and what will it cost?

Show solution

45 boxes at a cost of $2,025.00

The area of a rectangle is $A = LW$ , where L represents the length and W represents the width.

$A = LW = \left( {30{\rm{\ ft}}} \right)\left( {{\rm{15\ ft}}} \right) = 450{\rm{\ f}}{{\rm{t}}^2}$

Make conversion factors out of the equivalences of one box costing $45 and covering 10 square feet to answer the questions.

$450\ {\rm{ }}\cancel{{{\rm{f}}{{\rm{t}}^2}}}\left( {\frac{{1\ {\rm{ box}}}}{{10\ {\rm{ }}\cancel{{{\rm{f}}{{\rm{t}}^2}}}}}} \right) = 45\ {\rm{ boxes}}$

$45\,\mathrm{boxes}\times\frac{\$45}{1\,\mathrm{box}}=\$2025$

You need 45 boxes of tiles. It will cost you $2,025.

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