Exercise 21 Page 624

We want to find the values of x, y, and z.

Notice that x is the altitude of the given right triangle, y is a partial segment of the hypotenuse, and z is a leg. We will find their values one at a time.

Finding y and z

Since we know the length of one leg and the length of the hypotenuse, we will use the Geometric Mean (Leg) Theorem to find the values of y and z.

We will start by finding the value of y, which corresponds to DB on this figure. AB/CB = CB/DB ⇔ 14/5 = 5/y Now, we can solve the equation for y.

14/5 = 5/y

▼

Solve for y

MultEqn

LHS * 5=RHS* 5

14/5 * 5 = 5/y * 5

FracMultDenomToNumber

a/5* 5 = a

14 = 5/y * 5

MoveRightFacToNum

a/c* b = a* b/c

14 = 5 * 5/y

MultEqn

LHS * y=RHS* y

14y = 5 * 5/y * y

FracMultDenomToNumber

a/y* y = a

14y = 5 * 5

DivEqn

.LHS /14.=.RHS /14.

y = 5 * 5/14

Multiply

Multiply

y = 25/14

CalcQuot

Calculate quotient

y ≈ 1.8

Having found y, we can now calculate the value of z, which corresponds to AC. AC = sqrt(AD * AB) ⇔ z=sqrt(( 14 - 1.8)14) We will evaluate the right-hand side to find z.

z = sqrt((14 - 1.8)14)

▼

Evaluate right-hand side

SubTerm

Subtract term

z = sqrt(12.2 * 14)

Multiply

Multiply

z=sqrt(170.8)

UseCalc

Use a calculator

z ≈ 13.1

Finding x

Let's go back to the given figure.

Since we know the lengths of both partial segments of the hypotenuse, we will use the Geometric Mean (Altitude) Theorem to find the value of x.

We want to compare the theorem to the expressions in our figure. In our case, x is the length of the altitude, and ( 14-1.8) and 1.8 are the lengths of the partial segments of the hypotenuse. CD = sqrt(AD * DB) ⇔ x = sqrt(( 14-1.8) 1.8) Finally, we can evaluate the right-hand side to find the value of x.

x=sqrt((14-1.8)1.8)

▼

Evaluate right-hand side

SubTerm

Subtract term

x=sqrt((12.2)1.8)

Multiply

Multiply

x=sqrt(21.96)

UseCalc

Use a calculator

x ≈ 4.7