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Big Ideas Math Geometry, 2014

BI

Big Ideas Math Geometry, 2014 View details

Chapter Test

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Exercise 12 Page 457

Use that the park and the model of the park are similar.

Perimeter: 8160 ft=2720 yd
Area: 4 032 000 ft^2=448 000 yd^2

Practice makes perfect

We are going to use that the park and the model of the park are similar in order to find the width of the park. When we know the park's width and height we can then calculate its perimeter and area.

Width of the Park

We are told that the park is 800 yards long. Here, we want to know its length in feet. Recall that one yard equals three feet. 800 yards=(3* 800) feet=2400 feet Let's make a diagram illustrating the situation.

Since we are making a scale model of the park, we know that the rectangles are similar. In similar polygons corresponding sides are proportional. Let's use this for the park and the model of the park. w/1.4=2400/2 ⇔ w= 1680

Perimeter of the Park

The perimeter of a rectangle is the sum of two times its length and two times its height. We know that the park's length is 2400 feet and that its width is 1680 feet. Let's use this to find its perimeter. P=2l+2w ⇒ P=2( 2400)+2( 1680)=8160 The park's perimeter is 8160 feet, which is equivalent to 81603=2720 yards.

Area of the Park

To find the park's area we will use the formula for the area of a rectangle. A=l w Let's substitute the park's length and width into the formula to find its area.

A=l w

l= 2400, w= 1680

A=( 2400) ( 1680)

Multiply

A=4 032 000

We have found that the park's area is 4 032 000 ft^2. Recall that nine square feet equals one square yard. A=4 032 000 ft^2=4 032 000/9=448 000 yd^2

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Exercises p.457