Exercise 52 Page 626

Let's recall that in a right triangle the orthocenter is located in the vertex of the right angle. In our exercise, this means that the orthocenter is located at point B, and the length of BD is 6.4. Let x represent the length of DC.

According to the Geometric Mean Altitude Theorem, the length of BD is a geometric mean between the lengths of AD and DC. Let's substitute appropriate values to write an equation. BD=sqrt(AD* DC) ⇓ 6.4=sqrt(5.3*x)Now, we will solve this equation to find the value of x.

6.4=sqrt(5.3* x)

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Solve for x

RaiseEqn

LHS^2=RHS^2

6.4^2=(sqrt(5.3* x))^2

PowSqrt

( sqrt(a) )^2 = a

6.4^2=5.3* x

CalcPow

Calculate power

40.96=5.3* x

DivEqn

.LHS /5.3.=.RHS /5.3.

40.96/5.3=x

RearrangeEqn

Rearrange equation

x=40.96/5.3

CalcQuot

Calculate quotient

x=7.7283...

RoundDec

Round to 2 decimal place(s)

x≈ 7.73

We found that the length of DC is approximately 7.73. We will add this information to our picture. Let y represents the length of BC.

To find the value of y, let's recall that, according to the Geometric Mean Leg Theorem, the length of the leg of a right triangle is the geometric mean between the length of the hypotenuse and the segment of the hypotenuse adjacent to that leg. BC=sqrt(AC* DC) ⇓ y=sqrt((5.3+7.73)*7.73) Let's solve the above equation for y.

y=sqrt((5.3+7.73)*7.73)

AddTerms

Add terms

y=sqrt(13.03*7.73)

Multiply

Multiply

y=sqrt(100.7219)

CalcRoot

Calculate root

y=10.0360...

RoundDec

Round to 2 decimal place(s)

y≈10.04

The length of BC is approximately 10.04 units.