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Perimeter and Area

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Perimeter and Area PERIMETER AND AREA In this unit, we will develop and apply the formulas for the perimeter and area of various two-dimensional figures. Perimeter Perimeter The perimeter of a polygon, denoted by P, is the distance around the polygon, i.e. the sum of the lengths of its sides. Examples Answer the following. Remember to include units. 1. Find the perimeter of an equilateral triangle with side length 7 m. 2. Find the perimeter of a square with side length 5 in. 3. Find the perimeter of a regular hexagon with side length 10 m. 4. Find the perimeter of a rectangle with length 8 cm and width 5 cm. 5. If the perimeter of a regular pentagon is 75 in, find the length of one side. 6 If a tennis court has a perimeter of 210 ft and a length of 78 ft, find its width. 7 Find the perimeter of the following swimming pool. (Assume that all intersecting sides in the diagram are perpendicular to each other.) 13 m 3 m 10 m Pool 9 m The solutions to the above examples can be found below. Solutions to Examples: 1. Find the perimeter of an equilateral triangle with side length 7 m. 7 m P = 7773(7)21++= = The perimeter of the equilateral triangle is 21 m. 2. Find the perimeter of a square with side length 5 in. P = 5555+++= 4(5)20 = 5 in The perimeter of the square is 20 in. 3. Find the perimeter of a regular hexagon with side length 10 m. 10 m P =10+++++= 10 10 10 10 10 6(10) = 60 The perimeter of the regular hexagon is 60 m. 4. Find the perimeter of a rectangle with length 8 cm and width 5 cm. 5 cm P = 8+++= 8 5 5 2(8) + 2(5) = 36 8 cm The perimeter of the rectangle is 36 cm. Note: The perimeter of a rectangle with length l and width w can be represented by the formula: Plw= 22+ . 5. If the perimeter of a regular pentagon is 75 in, find the length of one side. Let x = the length of a side of the regular pentagon Since the perimeter of the pentagon is 75 inches, we can say that xxxxx+ +++=75 , or equivalently 575x = . x in Next, we solve for x by dividing both sides by 5: x =15 The length of one side of the regular pentagon is 15 inches. 6 If a tennis court has a perimeter of 210 ft and a length of 78 ft, find its width. Let x = the width of the tennis court We know that P = 210 . x ft Using the formula 22lwP+=, 2(78)+ 2x = 210 . 78 ft Solving for x: 156+ 2x = 210 254x = x = 27 Therefore, the width of the tennis court is 27 ft. 7. Find the perimeter of the following swimming pool. (Assume that all intersecting sides in the diagram are perpendicular to each other.) 13 m 3 m 10 m Pool 9 m Let x and y represent the unlabeled sides of the pool, as shown in the diagram below. 13 m 3 m x m 10 m Pool y m 9 m The horizontal distance across the pool is 13 m (as seen along the top edge of the pool). Therefore, 913+ x = , so x = 4 . The vertical distance across the pool is 10 m (as seen along the left edge of the pool). Therefore, 3+ y = 10 , so y = 7 . Now, adding the side lengths to find the perimeter of the figure: P =+91013347 + +++= 46 The perimeter of the pool is 46 m. Circumference The circumference of a circle is the distance around the circle. If a circle is compared to a many-sided polygon, the circumference of a circle can be likened to the perimeter of a polygon. Let us first begin with an exploration regarding circumference. Consider the table below, which contains a listing of various circular objects (lid, soda can, drinking glass, jar, etc.) In each of these three-dimensional objects, a plane containing a circle has been isolated, and the diameter and circumference of that circle has been listed. Measurements have been rounded to the nearest tenth of a centimeter. OBJECT DIAMETER CIRCUMFERENCE CIRCUMFERENCE DIAMETER Lid (outer rim) 6.1 cm 19.3 cm Soda Can (top rim) 5.5 cm 17.2 cm Drinking Glass (top rim) 8.2 cm 25.7 cm Jar (top rim) 4.7 cm 14.9 cm Mixing Bowl (top rim) 21.6 cm 67.8 cm Bucket (top rim) 31.0 cm 97.5 cm Exercises Circumference 1. Use a calculator to compute the ratio Diameter for each object in the last column of the table above. Round this ratio to the nearest thousandth. 2. Find three other circular objects. Add them to the table above, measure the Circumference circumference and diameter for each, and then compute the ratio Diameter for each object. (Note: To measure the circumference of an object, you may want to wrap a piece of string around the object, and then measure the string.) 3. Analyze the last column of numbers in the table. What do you notice? A completed chart can be found below. OBJECT DIAMETER CIRCUMFERENCE CIRCUMFERENCE DIAMETER Lid (outer rim) 6.1 cm 19.3 cm 3.164 Soda Can (top rim) 5.5 cm 17.2 cm 3.127 Drinking Glass (top rim) 8.2 cm 25.7 cm 3.134 Jar (top rim) 4.7 cm 14.9 cm 3.170 Mixing Bowl (top rim) 21.6 cm 67.8 cm 3.139 Bucket (top rim) 31.0 cm 97.5 cm 3.145 Notice that the numbers in the last column are very close to each other in their numerical Circumference value. Ancient mathematicians noticed that the ratio Diameter always seemed to give the same number, and it became a challenge to try to determine the value of this ratio. Since we rounded to the nearest tenth (and measurement is not completely accurate due to human error and/or measuring devices), the values in our table are not exactly the same. If we were to be able to measure perfectly, we would find that the ratio Circumference Diameter is always equal to the number π (pronounced “pi”). π is an irrational number, but its value is shown below to 50 decimal places: π ≈ 3.14159265358979323846264338327950288419716939937510... Note: For more information on irrational numbers, see the “Irrational Numbers” tutorial in the Additional Materials section. Circumference Since the ratio Diameter is equal to π , we can write the following equation: C = π (Where C represents circumference and d represents diameter.) d Multiplying both sides of the equation by d, we obtain the following equation: Cd= π Since the diameter d of a circle is twice the radius r, i.e. dr= 2 , we can also write: Cd==π ππ(2)2 r = r Circumference The circumference of a circle, denoted by C, is the distance around the circle. The circumference is given by either of the following formulas: Cd= π , or equivalently, Cr= 2π , where d and r represent the diameter and radius of the circle, respectively. Examples Find the circumference of the following circles. First, write the circumference of each circle in terms of π . Then use the approximation π ≈ 3.14 and compute the circumference of each circle to the nearest hundredth. (Be sure to include units in each answer.) 1. 5 ft 2. 12 cm Solutions: 1. The radius of the circle is 5 ft (r = 5) so we use the equation Cr= 2π . Cr==22510π ππ( ) = Exact answer: C =10π ft Using 3.14 for π : C =≈10π 10() 3.14 Approximate answer: C ≈ 31.4 ft 2. The diameter of the circle is 12 cm (d =12) so we use the equation Cd= π . Cd==π ππ()12 = 12 Exact answer: C =12π cm Using 3.14 for π : C =≈12π 12() 3.14 Approximate answer: C ≈ 37.68 cm Area of a Circle The formula for the area of a circle can be found below. Area of a Circle The area of a circle with radius r is given by the formula Ar= π 2 Examples Find the area of the following circles. First, write the area of each circle in terms of π . Then use the approximation π ≈ 3.14 and compute the area of each circle to the nearest hundredth. (Be sure to include units in each answer.) 1. 2. 10 cm 7 ft Solutions: 1. Ar==⋅=ππ22749ft. π 2 Exact answer: A = 49π ft2 Using 3.14 for π : A =≈49π 49() 3.14 Approximate answer: A ≈153.86 ft 2 2. The diameter of the circle is 10 cm (d =10) so r = 5 cm. Ar==⋅=ππ22525cm. π 2 Exact answer: A = 25π cm2 Using 3.14 for π : A =≈25π 25() 3.14 Approximate answer: A ≈ 78.5 cm2 Special Right Triangles (Needed for upcoming sections involving area) There are two special right triangles which are useful to us as we study the area of polygons. These triangles are named by the measures of their angles, and are known as 45o-45o-90o triangles and 30o-60o-90o triangles. A diagram of each triangle is shown below: o 45 30o hypotenuse longer hypotenuse leg leg o 45 60o leg shorter leg Tutorial: For a more detailed exploration of this section along with additional examples and exercises, see the tutorial entitled “Special Right Triangles” in the Additional Material section. The theorems relating to special right triangles can be found below, along with examples of each.
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