McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
1. Geometric Mean
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Exercise 24 Page 624

18 ft

Practice makes perfect

Let's begin with recalling the Geometric Mean Altitude Theorem. The altitude drawn to the hypotenuse of a right triangle separates the hypotenuse into two segments. The length of this altitude is the geometric mean between the lengths of these two segments. Now, let's take a look at the given picture. We will label the vertices with consecutive letters. Let x represents the height of the gym ceiling above Evelina's line of vision.

As we can see, BD is an altitude of a right triangle ABC. Therefore, we can write that the length of BD is the geometric mean between the lengths of AD and DC. DB=sqrt(AD * DC) 10=sqrt(x* 5) Next, we will solve the above equation for x using inverse operations.
10=sqrt(x*5)
â–Ľ
Solve for x
10=sqrt(5x)
10^2=(sqrt(5x))^2
10^2=5x
100=5x
5x=100
x=20
The length of AD is 20 feet. With this information, we can evaluate the height of the gym ceiling. To do this, we will add the lengths of AD and DC. 20ft+ 5ft=25ft The height of the gym ceiling is 25 feet. Since Evelina wants the ends of the strings where the stars will be attached to be 7 feet from the floor, we need to subtract 7 from the height of the gym ceiling to evaluate the strings length. 25- 7=18 She should make the strings of the length of 18 feet.