McGraw Hill Glencoe Geometry, 2012
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McGraw Hill Glencoe Geometry, 2012 View details
1. Geometric Mean
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Exercise 29 Page 542

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

x= 3sqrt(3)2≈ 2.6
y= 32
z=3

Practice makes perfect

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

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

Finding x and z

Since we know the lengths of one leg and of the hypotenuse, we will use the Geometric Mean (Leg) Theorem to find the values of x and z.

We will start by finding the value of x, which corresponds to DB on this figure. Notice that the difference between AB and DB is AD. AB/AC = AC/AD ⇔ 2sqrt(3)/sqrt(3) = sqrt(3)/2sqrt(3)-x Now, we can find the value of x.
2sqrt(3)/sqrt(3) = sqrt(3)/2sqrt(3)-x
Solve for x
2sqrt(3)/sqrt(3) * sqrt(3) = sqrt(3)/2sqrt(3)-x * sqrt(3)
2sqrt(3) = sqrt(3)/2sqrt(3)-x * sqrt(3)
2sqrt(3) = sqrt(3) * sqrt(3)/2sqrt(3)-x
2sqrt(3) = 3/2sqrt(3)-x
2sqrt(3) * (2sqrt(3)-x) = 3/2sqrt(3)-x * (2sqrt(3)-x)
2sqrt(3)(2sqrt(3)-x) = 3
12-2sqrt(3)x = 3
- 2sqrt(3)x = - 9
2sqrt(3)x = 9
x = 9/2sqrt(3)
x = 9 * sqrt(3)/2 * sqrt(3) * sqrt(3)
x = 9 * sqrt(3)/2*3
x = 3sqrt(3)/2
Using a calculator, we can express x, which corresponds to DB, as about 2.6. Having found the value of x, we can find z, which corresponds to CB. CB = sqrt(DB * AB) ⇕ z=sqrt(3sqrt(3)/2 * 2sqrt(3)) We will evaluate the right-hand side to find the value of z.
z = sqrt(3sqrt(3)/2* 2sqrt(3))
Evaluate right-hand side
z = sqrt(3sqrt(3)* 2sqrt(3)/2)
z = sqrt(6sqrt(3)*sqrt(3)/2)
z = sqrt(6 * 3/2)
z = sqrt(18/2)
z = sqrt(9)
z = 3

Finding y

Let's go back to the given figure.

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

We want to compare the theorem to the expressions in our figure. In our case, y is the length of the altitude, and ( 2sqrt(3)- 3sqrt(3)2) and 3sqrt(3)2 are the lengths of the partial segments of the hypotenuse. CD = sqrt(AD * DB) ⇕ y = sqrt((2sqrt(3)- 3sqrt(3)2)3sqrt(3)2) Finally, we can evaluate the right-hand side to find y.
y=sqrt((2sqrt(3)-3sqrt(3)/2)3sqrt(3)/2)
Evaluate right-hand side
y=sqrt((4sqrt(3)/2-3sqrt(3)/2)3sqrt(3)/2)
y=sqrt((4sqrt(3)-3sqrt(3)/2)3sqrt(3)/2)
y=sqrt((sqrt(3)/2)3sqrt(3)/2)
y=sqrt(9/4)
y=sqrt(9)/sqrt(4)
y=3/2