Big Ideas Math Geometry, 2014
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Big Ideas Math Geometry, 2014 View details
1. Similar Polygons
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Exercise 17 Page 424

Consider Theorem 8.1, Perimeters of Similar Polygons.

Perimeter of NCAA Court: 288feet
Perimeter of School Court: 259.2feet

Practice makes perfect

We are given two similar basketball courts, which can be modeled in the following diagram.

We want to find the perimeter of both courts. Let's start with the perimeter of the NCAA basketball court.

Perimeter of NCAA Basketball Court

Looking at the diagram, we can see that both courts have a rectangular shape. We know that the perimeter of a rectangle is twice the length plus twice the width of the rectangle. Let P_N be the perimeter of the NCAA basketball court. P_N&=2l_N+2w_N Now, we can substitute l_N=94 and w_N=50 in our equation to find the perimeter of the NCAA basketball court.
P_N=2l_N+2w_N
P_N=2( 94)+2( 50)
P_N=188+100
P_N=288
Therefore, the perimeter of the NCAA basketball court is 288 feet.

Perimeter of School Court

To find the perimeter of the school court we will recall Theorem 8.1.

Theorem 8.1: Perimeters of Similar Polygons

If two polygons are similar, then the ratio of their perimeters is equal to the ratios of their corresponding side lengths.

Looking at the rectangles, we can see that w_S corresponds to w_N. Let P_S be the perimeter of the school court. From here, using the theorem we can write the following proportion. P_S/P_N=w_S/w_N Next, we will substitute w_S=45, w_N=50, and P_N=288 into our proportion to find the perimeter of school court.
P_S/P_N=w_s/w_N
P_S/288=45/50
P_S=12 960/50
P_S=259.2
Therefore, the perimeter of the school court is 259.2 feet.