Exercise 38 Page 625

Let's begin with applying the given information to the diagram. We will express all the dimensions only in feet.

We are asked to evaluate the length of segment

\overline{A E} .

To do this, notice that, according to the Segment Addition Postulate, we can rewrite the length of

\overline{A E}

as a sum of the lengths of

\overline{A D}, \overline{D B}

and

\overline{B E} .

\overline{A E} = \overline{A D} + \overline{D B} + \overline{B E}

Let's start with evaluating the length of

\overline{A D}

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 segments.

Looking at

△ A B C,

we can see that, according to this theorem, the length of

\overline{D C}

is the geometric mean between the lengths of

\overline{A D}

and

\overline{D B} .

D C = A D \cdot D B ⇓ 6 \frac{4}{1 2} = A D \cdot 5

We will solve the above equations for

A D .

Our first step will be to rewrite the mixed number on the left-hand side as a fraction. Let

x

represent the length of

\overline{A D} .

6 \frac{4}{1 2} = x \cdot 5

▼

Solve for

x

$\frac{a}{b} = \frac{a / 4}{b / 4}$

6 \frac{1}{3} = x \cdot 5

$a \frac{b}{c} = \frac{a \cdot c + b}{c}$

\frac{1 9}{3} = x \cdot 5

\frac{1 9}{3} = 5 x

$LHS^{2} = RHS^{2}$

(\frac{1 9}{3})^{2} = (5 x)^{2}

$(a)^{2} = a$

(\frac{1 9}{3})^{2} = 5 x

$(\frac{a}{b})^{m} = \frac{a ^{m}}{b ^{m}}$

\frac{1 9 ^{2}}{3 ^{2}} = 5 x

Calculate power

\frac{3 6 1}{9} = 5 x

$LHS / 5 = RHS / 5$

\frac{3 6 1}{9} / 5 = x

$\frac{a}{b} / c = \frac{a}{b \cdot c}$

\frac{3 6 1}{9 \cdot 5} = x

\frac{3 6 1}{4 5} = x

Rearrange equation

x = \frac{3 6 1}{4 5}

Calculate quotient

x = 8.0222 \dots

Round to nearest integer

x \approx 8

The length of

\overline{A D}

is approximately

8

feet. Now, we will find the length of

\overline{B E} .

To do this, let's recall the Geometric Mean Leg Theorem.

Geometric Mean Leg Theorem
The altitude drawn to the hypotenuse of a right triangle separates the hypotenuse into two segments. The length of a leg of this triangle is the geometric mean between the length of the hypotenuse and the segment of the hypotenuse adjacent to that leg.

Using this theorem, we can write an equation.