McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
1. Geometric Mean
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Exercise 25 Page 624

161.8 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 waterfall above Makayla's eye level.
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) 28=sqrt(x* 5) Next, we will solve the above equation for x using inverse operations.
28=sqrt(x*5)
â–Ľ
Solve for x
28=sqrt(5x)
28^2=(sqrt(5x))^2
28^2=5x
784=5x
5x=784
x=156.8
The length of AD is 156.8 feet. With this information, we can evaluate the height of the watefall. To do this, we will add the lengths of AD and DC. 156.8ft+ 5ft=161.8ft The height of the waterfall is 161.8 feet.