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McGraw Hill Integrated II, 2012

MH

McGraw Hill Integrated II, 2012 View details

1. Geometric Mean

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Exercise 37 Page 625

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

4

Practice makes perfect

Let's take a look at the given triangle.

Since we know the length of the altitude, and the expressions that represent the partial segments of the hypotenuse, we will use the Geometric Mean (Altitude) Theorem to find the value of m.

We want to compare the theorem to the expressions in our figure. In our case, 6 is the length of the altitude, and m and m+5 are the lengths of the partial segments of the hypotenuse. AD/CD = CD/DB ⇔ m/6 = 6/m+5 Now, we can find the value of m.

m/6 = 6/m+5

▼

Solve for m

LHS * 6=RHS* 6

m/6 * 6 = 6/m+5 * 6

FracMultDenomToNumber

a/6* 6 = a

m = 6/m+5 * 6

MoveRightFacToNum

a/c* b = a* b/c

m = 6 * 6/m+5

Multiply

m = 36/m+5

LHS * (m+5)=RHS* (m+5)

m(m+5) = 36/m+5 * (m+5)

FracMultDenomToNumber

a/(m+5)* (m+5) = a

m(m+5) = 36

Distribute m

m^2+5m= 36

LHS-36=RHS-36

m^2+5m-36=0

Write as a difference

m^2+9m-4m-36=0

Factor out m

m(m+9)-4m-36=0

Factor out - 4

m(m+9)-4(m+9)=0

Factor out (m+4)

(m-4)(m+9)=0

Use the Zero Product Property

lcm-4=0 & (I) m+9=0 & (II)

(I): LHS+4=RHS+4

lm=4 m+9=0

(II): LHS-9=RHS-9

lm_1=4 m_2=- 9

Since lengths cannot be negative, m is equal to 4.

Subchapter links

Exercises p.623-627