McGraw Hill Glencoe Geometry, 2012
MH
McGraw Hill Glencoe Geometry, 2012 View details
1. Geometric Mean
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Exercise 57 Page 545

We can start with evaluating the length of AD by using the Pythagorean Theorem.

C

Practice makes perfect

Let's take a look at the given diagram. It is a given that AD is perpendicular to BC, and AB is perpendicular to AC. Let x be the length of AD and y be the length of DC.

Our first step will be to evaluate the value of x. Since we are dealing with a right triangle and we already know the measures of two side lengths, let's use the Pythagorean Theorem. According to this theorem, the sum of the squared legs of a right triangle is equal to its squared hypotenuse.
5^2+x^2= 10^2
â–Ľ
Solve for x
25+x^2=100
x^2=75
sqrt(x^2)=sqrt(75)
x=sqrt(75)
x=sqrt(25*3)
x=sqrt(25)*sqrt(3)
x=5*sqrt(3)
x=5sqrt(3)

The length of AD is 5sqrt(3). Let's add this information to our picture.

Now, let's recall the definition of the Geometric Mean Altitude Theorem.

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.

Using this theorem, we can solve for the value of y.
5sqrt(3)=sqrt(5*y)
â–Ľ
Solve for y
5sqrt(3)=sqrt(5y)
(5sqrt(3))^2=(sqrt(5y))^2
5^2*(sqrt(3))^2=(sqrt(5y))^2
5^2*3=5y
25*3=5y
75=5y
15=y
y=15
We found that the length of DC is 15.

Since we are asked to find the length of BC, we will add the lengths of BD and DC, according to the Segment Addition Postulate. BC=5+15=20 The length of BC is 20. This corresponds with answer C.