8  Rectangles and squares

A rectangle is a flat shape with four sides and four right angles. Its two dimensions are the length \(L\) and the width \(W\).

A labeled rectangle showing length L along the bottom and width W along the right side, with double-headed arrows indicating each dimension

We have two ways to measure the size of a rectangle.

Perimeter is measured in units of length (feet, meters). Area is measured in square units (square feet, square meters).

8.1 Perimeter

A rectangle has two sides of length \(L\) and two sides of length \(W\), so the perimeter is

\[\text{perimeter} = \text{length} + \text{width} + \text{length} + \text{width} = 2 \cdot \text{length} + 2 \cdot \text{width}\]

\[P = 2L + 2W\]

Example 8.1 The Taos Inn has a conference room that is 14 feet long and 10 feet wide. How much decorative trim is needed to go around the entire perimeter of the room?

\[P = 2(14) + 2(10) = 28 + 20 = 48 \text{ feet.}\]

Squares

A square is a special rectangle where \(L = W\). If we call the common side length \(S\), the perimeter formula simplifies: \[P = 2S + 2S = 4S.\]

For example, a square patio with \(S = 11\) feet has perimeter \(P = 4(11) = 44\) feet.

Example 8.2 (Activity: Marking off park spaces) The town of Red River is laying out two new park spaces.

  • Space A is a rectangular picnic area, 20 feet long and 12 feet wide.
  • Space B is a square performance stage with side length 15 feet.
  1. How much rope is needed to mark the boundary of Space A?
  1. How much rope is needed to mark the boundary of Space B?

  1. \(P = 2(20) + 2(12) = 40 + 24 = 64\) feet of rope.

  2. \(P = 4(15) = 60\) feet of rope.

8.2 Area

The area of a rectangle is the length multiplied by the width:

\[\text{area} = \text{length} \times \text{width}\]

\[A = L \cdot W\]

Example 8.3 The conference room at the Taos Inn is 14 feet long and 10 feet wide. How much carpet is needed to cover the floor?

\[A = 14 \cdot 10 = 140 \text{ square feet.}\]

Squares

For a square with side length \(S\), the area formula simplifies: \[A = S \cdot S = S^2.\]

For example, a square tile with \(S = 4\) inches has area \(A = 4^2 = 16\) square inches.

Example 8.4 (Activity: Covering the park spaces) Recall the two park spaces from Example 8.2.

  1. How many square feet of sod are needed to cover Space A?
  1. How many square feet of paving stones are needed for Space B?
  1. What is the total area of both spaces combined?

  1. \(A = 20 \cdot 12 = 240\) square feet.

  2. \(A = 15^2 = 225\) square feet.

  3. \(240 + 225 = 465\) square feet total.

8.3 Practice: perimeter and area

Example 8.5 (Activity: Computing perimeter and area) Complete the following table. The first row is completed for you as a reference.

\(L\) \(W\) \(P = 2L + 2W\) \(A = L \cdot W\)
14 ft 10 ft 48 ft 140 sq ft
9 ft 5 ft
20 ft 8 ft
6 ft 6 ft
7 ft 3 ft
12 ft 12 ft
4.5 ft 8 ft

Note: rows 4 and 6 are squares.

\(L\) \(W\) \(P\) \(A\)
14 ft 10 ft 48 ft 140 sq ft
9 ft 5 ft 28 ft 45 sq ft
20 ft 8 ft 56 ft 160 sq ft
6 ft 6 ft 24 ft 36 sq ft
7 ft 3 ft 20 ft 21 sq ft
12 ft 12 ft 48 ft 144 sq ft
4.5 ft 8 ft 25 ft 36 sq ft

8.4 Finding a missing dimension

So far we have computed \(P\) and \(A\) when both \(L\) and \(W\) are given. We can also run this process in reverse: if we know the area or perimeter and one dimension, we can set up an equation and solve for the missing dimension. This is the same equation-solving process from the previous chapter, now applied to geometry.

Example 8.6 A storage shed near Taos has a rectangular floor that is 18 feet wide. The total floor area is 270 square feet. How long is the shed?

We use the area formula with \(A = 270\) and \(W = 18\): \[270 = L \cdot 18\] Dividing both sides by 18: \[L = 270 \div 18 = 15 \text{ feet.}\]

Example 8.7 A rancher near Mora wants to fence a rectangular alfalfa field. The field is 200 feet long, and there is 740 feet of fencing available. How wide can the field be?

We use the perimeter formula with \(P = 740\) and \(L = 200\): \[ 740 = 2(200) + 2W \Rightarrow 740 = 400 + 2W \Rightarrow 340 = 2W \Rightarrow W = 170 \text{ feet.} \]

Example 8.8 (Activity: Rectangle problems) For each problem, identify which formula to use, set up the equation, and solve.

  1. A classroom at UNM-Taos is 32 feet long. The room requires 864 square feet of new flooring. How wide is the classroom?
  1. Maria is planting a rectangular vegetable garden at the Taos Farmers Market. She has 52 feet of fencing and wants the garden to be 16 feet long. How wide can the garden be?
  1. A rancher near Cimarron wants to build a rectangular horse corral. The corral will be 45 feet wide, and 200 feet of fencing is available. How long can the corral be?
  1. A square tile used at a Taos Pueblo art market has an area of 81 square inches. What is the side length of the tile? (Hint: what number times itself equals 81?)

  1. \(A = L \cdot W\), so \(864 = 32 \cdot W\). Dividing both sides by 32: \(W = 27\) feet.

  2. \(P = 2L + 2W\), so \(52 = 2(16) + 2W \Rightarrow 52 = 32 + 2W \Rightarrow 20 = 2W \Rightarrow W = 10\) feet.

  3. \(P = 2L + 2W\), so \(200 = 2L + 2(45) \Rightarrow 200 = 2L + 90 \Rightarrow 110 = 2L \Rightarrow L = 55\) feet.

  4. \(A = S^2\), so \(S^2 = 81\). Since \(9 \cdot 9 = 81\), the side length is \(S = 9\) inches.

Example 8.9 (Activity: Missing dimensions) Complete the following table. Some entries are given; find the rest. For the last two rows the rectangle is a square (\(L = W = S\)): find \(S\) first, then complete the row.

\(L\) \(W\) \(P = 2L + 2W\) \(A = L \cdot W\)
8 ? 56
? 5 60
15 ? 46
? 6 40
\(S\) \(S\) 64
\(S\) \(S\) 36

Row 1: \(8 \cdot W = 56 \Rightarrow W = 7\). Then \(P = 2(8) + 2(7) = 30\).

Row 2: \(L \cdot 5 = 60 \Rightarrow L = 12\). Then \(P = 2(12) + 2(5) = 34\).

Row 3: \(2(15) + 2W = 46 \Rightarrow 30 + 2W = 46 \Rightarrow W = 8\). Then \(A = 15 \cdot 8 = 120\).

Row 4: \(2L + 2(6) = 40 \Rightarrow 2L + 12 = 40 \Rightarrow 2L = 28 \Rightarrow L = 14\). Then \(A = 14 \cdot 6 = 84\).

Row 5: \(S^2 = 64\). Since \(8 \cdot 8 = 64\), \(S = 8\). Then \(P = 4(8) = 32\).

Row 6: \(4S = 36 \Rightarrow S = 9\). Then \(A = 9^2 = 81\).

8.5 Square roots

In the missing dimension problems above, the last two rows involved a square where we knew the area \(A\) and needed to find the side length \(S\). This means solving \(S^2 = A\) — finding a number that, when multiplied by itself, gives \(A\).

This operation is called taking the square root of \(A\), written \(\sqrt{A}\): \[\sqrt{A} = S \quad \text{means} \quad S \cdot S = A.\]

Example 8.10 \(\sqrt{64} = 8\), because \(8 \cdot 8 = 64\).

\(\sqrt{36} = 6\), because \(6 \cdot 6 = 36\).

\(\sqrt{100} = 10\), because \(10 \cdot 10 = 100\).

The table below lists the square roots of the first several perfect squares — numbers whose square root is a whole number.

\(A\) \(\sqrt{A}\) \(A\) \(\sqrt{A}\)
1 1 49 7
4 2 64 8
9 3 81 9
16 4 100 10
25 5 121 11
36 6 144 12

For areas that are not perfect squares, \(\sqrt{A}\) is not a whole number. In those cases, use the \(\sqrt{\phantom{x}}\) button on your calculator.

For example, a square with area 50 square feet has side length \(S = \sqrt{50} \approx 7.07\) feet.

Example 8.11 (Activity: Finding side lengths from area) For each square, find the side length \(S\).

  1. A square has area 49 square feet. What is \(S\)?
  1. A square has area 144 square meters. What is \(S\)?
  1. A square courtyard has area 225 square feet. What is \(S\)? (Hint: try \(S = 15\).)
  1. A square patio has area 75 square feet. What is \(S\)? Use a calculator and round to two decimal places.

  1. \(\sqrt{49} = 7\) feet, since \(7 \cdot 7 = 49\).

  2. \(\sqrt{144} = 12\) meters, since \(12 \cdot 12 = 144\).

  3. \(\sqrt{225} = 15\) feet, since \(15 \cdot 15 = 225\).

  4. \(\sqrt{75} \approx 8.66\) feet.

8.6 Homework exercises

Exercise 8.1 For each problem, identify which formula to use, set up the equation, and solve.

  1. A rectangular storage unit in Taos is 9 feet wide and has a floor area of 108 square feet. How long is the storage unit?
  1. The Taos County Fair is setting up a rectangular exhibition space. The space is 60 feet long, and 220 feet of rope is available to mark the boundary. How wide can the exhibition space be?
  1. A square room at the UNM-Taos student center has a perimeter of 56 feet. What is the area of the room?
  1. A rancher near Abiquiú wants to build a rectangular sheep pen. The pen is 25 feet long, and 110 feet of fencing is available. How wide is the pen? What is the area of the pen?

  1. \(A = L \cdot W\), so \(108 = L \cdot 9\). Dividing both sides by 9: \(L = 12\) feet.

  2. \(P = 2L + 2W\), so \(220 = 2(60) + 2W \Rightarrow 220 = 120 + 2W \Rightarrow 100 = 2W \Rightarrow W = 50\) feet.

  3. \(P = 4S = 56 \Rightarrow S = 14\) feet. Then \(A = 14^2 = 196\) square feet.

  4. \(P = 2L + 2W\), so \(110 = 2(25) + 2W \Rightarrow 110 = 50 + 2W \Rightarrow 60 = 2W \Rightarrow W = 30\) feet. The area is \(A = 25 \cdot 30 = 750\) square feet.

Exercise 8.2 Complete the following table. Some entries are given; find the rest. For the last two rows the rectangle is a square (\(L = W = S\)): find \(S\) first, then complete the row.

\(L\) \(W\) \(P = 2L + 2W\) \(A = L \cdot W\)
11 4
6 ? 48
? 7 84
20 ? 62
? 9 50
\(S\) \(S\) 100
\(S\) \(S\) 44

Row 1: \(P = 2(11) + 2(4) = 30\). \(A = 11 \cdot 4 = 44\).

Row 2: \(6 \cdot W = 48 \Rightarrow W = 8\). Then \(P = 2(6) + 2(8) = 28\).

Row 3: \(L \cdot 7 = 84 \Rightarrow L = 12\). Then \(P = 2(12) + 2(7) = 38\).

Row 4: \(2(20) + 2W = 62 \Rightarrow 40 + 2W = 62 \Rightarrow W = 11\). Then \(A = 20 \cdot 11 = 220\).

Row 5: \(2L + 2(9) = 50 \Rightarrow 2L + 18 = 50 \Rightarrow 2L = 32 \Rightarrow L = 16\). Then \(A = 16 \cdot 9 = 144\).

Row 6: \(S^2 = 100\). Since \(10 \cdot 10 = 100\), \(S = 10\). Then \(P = 4(10) = 40\).

Row 7: \(4S = 44 \Rightarrow S = 11\). Then \(A = 11^2 = 121\).