Exercise 15 Page 619

a 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 .

\frac{x}{h} = \frac{h}{y} or h = x \cdot y

Now, let's take a look at the given picture.

Using the theorem we recalled at the beginning, we can write an equation.

x = 8 \cdot 12.5

Now let's solve the equation.

x = 8 \cdot 12.5

Multiply

x = 100

CalcRoot

Calculate root

x = 10

The value of

x

10 .

b Let's begin with recalling the Pythagorean Theorem.

In a right triangle, the sum of the squares of the lengths of the legs is equal to the square of the length of the hypotenuse .

a^{2} + b^{2} = c^{2}

Now, let's look at the given picture.

Since an altitude in a triangle divides this triangle into two right triangles, we can use the Pythagorean Theorem to evaluate the value of

y .

10^{2} + 8^{2} = y^{2}

CalcPow

Calculate power

100 + 64 = y^{2}

AddTerms

Add terms

164 = y^{2}

RearrangeEqn

Rearrange equation

y^{2} = 164

Next we will take a square root of both sides of the equation. Notice that since

y

represents the side length, we will consider only the positive case when taking a square root of

y^{2} .

y^{2} = 164

SqrtEqn

$LHS = RHS$

y^{2} = 164

SqrtPowToNumber

$a^{2} = a$

y = 164

CalcRoot

Calculate root

y = 12.8062 \dots

RoundDec

Round to $1$ decimal place(s)

y \approx 12.8

The value of

y

is approximately

12.8 .

c Again, let's begin with recalling the Pythagorean Theorem.

In a right triangle, the sum of the squares of the lengths of the legs is equal to the square of the length of the hypotenuse .

a^{2} + b^{2} = c^{2}

Next we will look at the given picture for the last time.

Having in mind that an altitude in a triangle divides this triangle into two right triangles, we can again use the Pythagorean Theorem to evaluate the value of

z .

10^{2} + 12.5^{2} = z^{2}

CalcPow

Calculate power

100 + 156.25 = z^{2}

AddTerms

Add terms

256.25 = z^{2}

RearrangeEqn

Rearrange equation

z^{2} = 256.25

Next we will take a square root of both sides of the equation. Notice that since

z

represents the side length, we will consider only positive case when taking a square root of

z^{2} .

z^{2} = 256.25

SqrtEqn

$LHS = RHS$

z^{2} = 256.25

SqrtPowToNumber

$a^{2} = a$

z = 256.25

CalcRoot

Calculate root

z = 16.0078 \dots

RoundInt

Round to nearest integer

z \approx 16

The value of

z

is approximately

16 .

Subchapter links

Exercises p.618-619