Exercise 35 Page 625

Let's take a look at the given triangle.

Since we know the length of one partial segment of the hypotenuse and the length of the altitude, we will use the Geometric Mean (Altitude) Theorem to find the value of t.

We want to compare the theorem to the expressions in our figure. In our case, 12 is the length of the altitude, and 24 and ( t+1) are the lengths of the partial segments of the hypotenuse. AD/CD = CD/DB ⇔ t+1/12 = 12/24 Now, we can find the value of t.

t+1/12 = 12/24

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Solve for t

ReduceFrac

a/b=.a /12./.b /12.

t+1/12 = 1/2

MultEqn

LHS * 12=RHS* 12

t+1/12 * 12 = 1/2 * 12

FracMultDenomToNumber

a/12* 12 = a

t+1= 1/2 * 12

MoveRightFacToNumOne

1/b* a = a/b

t+1= 12/2

CalcQuot

Calculate quotient

t + 1 = 6

SubEqn

LHS-1=RHS-1

t=5