McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
1. Geometric Mean
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Exercise 22 Page 624

Analyze what lengths you are given and use either the Geometric Mean (Altitude) Theorem or the Geometric Mean (Leg) Theorem.

x=40
y=10sqrt(5)≈ 22.4
z=20sqrt(5) ≈ 44.7

Practice makes perfect

We want to find the values of x, y, and z.

Notice that x is a partial segment of the hypotenuse, and y and z are the legs of the given right triangle. We will find their values one at a time.

Finding x

Since we know the lengths of the altitude and a partial segment of the hypotenuse, we will use the Geometric Mean (Altitude) Theorem to find the value of x.

We want to compare the theorem to the expressions in our figure. In our case, 20 is the length of the altitude, and 10 and x are the lengths of the partial segments of the hypotenuse. AD/CD = CD/DB ⇔ 10/20 = 20/x Now, we can find the value of x.
10/20 = 20/x
Solve for x
10/20 * 20 = 20/x * 20
10 = 20/x * 20
10 = 20 * 20/x
10x = 20 * 20/x * x
10x = 20 * 20
x = 20 * 2
x = 40

Finding y and z

Let's go back to the given figure.

Since we now know the lengths of both partial segments of the hypotenuse divided by the altitude, we will use the Geometric Mean (Leg) Theorem to find the values of y and z.

We will start by finding the value of y, which corresponds to AC on this figure. AC = sqrt(AD * AB) ⇔ y=sqrt(10(10+40)) Now, we can evaluate the right-hand side to find y.
y = sqrt(10(10+40))
Evaluate right-hand side
y = sqrt(10(50))
y = sqrt(10 * 10 * 5)
y = sqrt(10 * 10) * sqrt(5)
y = 10 sqrt(5)
Using a calculator, we can express y as about 22.4. Following the same reasoning, we can find z, which corresponds to CB. CB = sqrt(DB * AB) ⇔ z=sqrt(40(10+40)) Finally, we can evaluate the right-hand side to find the value of z.
z = sqrt(40 (10+40))
Evaluate right-hand side
z = sqrt(40(50))
z = sqrt(4 * 10 * 10 * 5)
z = sqrt(4) * sqrt(10 * 10) * sqrt(5)
z = 2 * 10 * sqrt(5)
z = 20sqrt(5)
Using a calculator, we can rewrite z as about 44.7.