Effects of Changing Dimensions on Perimeter and Area

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Presentation transcript:

Effects of Changing Dimensions on Perimeter and Area Lesson 96 Effects of Changing Dimensions on Perimeter and Area

Theorem 87-1 states that two similar figures with a similarity ratio of 𝑎:𝑏 have perimeters in the ratio 𝑎:𝑏 and areas in the ratio 𝑎 2 : 𝑏 2 . This theorem can be used when an entire figure is dilated. Sometimes, however, we may want to find the area or perimeter of a figure when only one dimension is altered, or when the dimensions are changed by different scale factors. If one dimension of a polygon is changed, the ratio of its original perimeter to its new perimeter can be found by applying the formula for perimeter of a polygon. Geometry Lesson 96

Example 1 Changing Perimeter of a Polygon A rectangle is half as tall as it is long. If its height is reduced by half its original height, what is the ratio of the new rectangle’s perimeter to the original rectangle’s perimeter? SOLUTION Let the length of the rectangle be x. Since the rectangle’s height is half its length, its height is 0.5x. Determine its perimeter by adding the sides together. 𝑃=𝑥+𝑥+0.5𝑥+0.5𝑥 𝑃=3𝑥 When the height is reduced to one half its original height, it will be half of 0.5x, or 0.25x. The length of the rectangle does not change. Find the perimeter of the new rectangle by adding its sides together. 𝑃=𝑥+𝑥+0.25𝑥+0.25𝑥 𝑃=2.5𝑥 Therefore, the ratio of the new rectangle’s perimeter to the original rectangle’s perimeter is 2.5:3, or 5:6. Geometry Lesson 96

The same method can be applied to find the ratio of the area of two polygons when one dimension is altered. Geometry Lesson 96

Example 2 Changing Area of a Polygon Find the area of each polygon. Describe how each change affects the area. a. Triangle ABC has a base that is congruent to its height. If the base is dilated by a factor of 2, what is the ratio of the new triangle’s area to the original triangle’s area? SOLUTION The diagram illustrates this problem. Use the formula for area of a triangle to find the area of the original triangle. 𝐴= 1 2 𝑏ℎ 𝐴= 1 2 𝑦∙𝑦 𝐴= 1 2 𝑦 2 Now find the area of the altered triangle. 𝐴= 1 2 ∙2𝑦∙𝑦 𝐴= 𝑦 2 Compare the two expressions for area. The ratio of the triangles’ areas is 2:1. Geometry Lesson 96

Example 2 Changing Area of a Polygon Find the area of each polygon. Describe how each change affects the area. b. A parallelogram’s base is twice as long as its height. If the length of the base is doubled, and the height is halved, what is the ratio of the new parallelogram’s area to the original parallelogram’s area? SOLUTION The diagram illustrates this problem. Use the formula for area of a parallelogram to find the original parallelogram’s area. 𝐴=2𝑥∙𝑥 𝐴=2 𝑥 2 Now find the area of the altered parallelogram. 𝐴=4𝑥∙ 1 2 𝑥 The ratio of the areas is 1:1. Geometry Lesson 96

Example 4 Application: Home Improvements Bev is having a pool installed in her backyard. Her backyard is a rectangle with a length that is twice as long as its width. Bev decides that the pool will also be a rectangle, but it will run only three-fourths the length of the backyard and be half as wide. What is the ratio of the pool’s area to the backyard’s area? SOLUTION Draw a diagram to illustrate this situation. Notice that the length of the pool is three-fourths of 2x, or 1.5x. Find the area of the pool and Bev’s backyard. 𝐴=𝑏ℎ 𝐴=𝑏ℎ 𝐴=2𝑥∙𝑥 𝐴=1.5𝑥∙0.5𝑥 𝐴=2 𝑥 2 𝐴=0.75 𝑥 2 So the ratio is 0.75:2. To simplify this, multiply by 4 to eliminate the decimal, which results in the ratio 3:8. Geometry Lesson 96

Example 3 Altering the Dimensions of a Circle A circle’s radius is increased by a factor of 3. Find the ratio of the circle’s new area and circumference to its original circumference and area. SOLUTION Call the length of the initial radius x. The new radius will have a length of 3x. Find the area of each circle. 𝐴=𝜋 𝑟 2 𝐴=𝜋 𝑟 2 𝐴=𝜋 𝑥 2 𝐴=𝜋 3𝑥 2 𝐴=𝜋 𝑥 2 𝐴=9𝜋 𝑥 2 The ratio of the areas is 9:1. Now, find the circumference of each circle. 𝐶=2𝜋𝑟 𝐶=2𝜋𝑟 𝐶=2𝜋𝑥 𝐶=2𝜋 3𝑥 𝐶=2𝜋𝑥 𝐶=6𝜋𝑥 The ratio of the circumferences is 3:1. As you can see, circles conform to the ratios given in Theorem 87-1. Geometry Lesson 96

You Try!!!! a. One pair of opposite sides of a square are dilated by a factor of 4 while the other sides remain the same. What is the ratio of the new figure’s perimeter to that of the original? 5:2 b. What is the ratio of the first trapezoid’s area to the second trapezoid’s area? 8:15 Geometry Lesson 96

You Try!!!! c. The radius of a circle is x. If the radius is changed by a factor of 1 2 , what is the ratio between the original area and the new area? What is the ratio between the original circumference and the new circumference? 𝑅𝑎𝑡𝑖𝑜 𝑜𝑓 𝐴𝑟𝑒𝑎 4:1 𝑅𝑎𝑡𝑖𝑜 𝑜𝑓 𝐶𝑖𝑟𝑐𝑢𝑚𝑓𝑒𝑟𝑒𝑛𝑐𝑒 2:1 Geometry Lesson 96

You Try!!!! d. A block of apartments must be constructed in the shape of a square. In the middle of construction, it is discovered that the apartment lot must be shortened to make way for a road expansion. Due to this fact, the length of the lot is nine-tenths of what it was before. What is the ratio of the apartment block’s new area to its original planned area? 81:100 Geometry Lesson 96

Assignment Page 627 Lesson Practice (Ask Mr. Heintz) Practice 1-30 (Do the starred ones first) Geometry Lesson 96