Exercise 25 Page 624

Let's begin with recalling the 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 two segments .

Now, let's take a look at the given picture. We will label the vertices with consecutive letters. Let

x

represents the height of the waterfall above Makayla's eye level.

As we can see,

\overline{B D}

A B C .

Therefore, we can write that the length of

\overline{B D}

is the geometric mean between the lengths of

\overline{A D}

and

\overline{D C} .

D B = A D \cdot D C 28 = x \cdot 5

Next, we will solve the above equation for

x

28 = x \cdot 5

▼

Solve for

x

28 = 5 x

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

2 8^{2} = (5 x)^{2}

$(a)^{2} = a$

2 8^{2} = 5 x

Calculate power

784 = 5 x

Rearrange equation

5 x = 784

$LHS / 5 = RHS / 5$

x = 156.8

The length of

\overline{A D}

is

156.8

feet. With this information, we can evaluate the height of the watefall. To do this, we will add the lengths of

\overline{A D}

and

\overline{D C} .

156.8 ft + 5 ft = 161.8 ft

The height of the waterfall is

161.8

feet.