McGraw Hill Integrated II, 2012
MH
McGraw Hill Integrated II, 2012 View details
1. Geometric Mean
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Exercise 7 Page 623

≈18 ft 11 in.

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 represent the height of the statue that is above Corey'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. Let's express the given lengths only in inches using the fact that there are 12 inches in one foot. DB=sqrt(AD * DC) 8*12+ 8=sqrt(x( 5*12+ 8)) Next, we will solve the above equation for x using inverse operations.
8*12+8=sqrt(x(5*12+8))
Solve for x
96+8=sqrt(x(60+8))
104=sqrt(x(68))
104=sqrt(68x)
104^2=(sqrt(68x))^2
104^2=68x
10816=68x
68x=10816
x=159.05...
x≈159
The length of AD is approximately 159 inches. Now, we will rewrite it using feet and inches.
159in.
Rewrite inches as feet and inches
156in.+3in.
13*12in+3in.
13*1ft+3in.
13ft+3in.
13ft3in.
Finally, to evaluate the approximate height of the statue, we will add the lengths of AD and DC. 13ft3in.+ 5ft8in.=18ft11in. The height of the statue is approximately 18 feet 11 inches.