Question

Given :AD is an altitude of ABC BD =3x DC =5x-14 AC =8x-10 ADC=5x+20 DAC=2x+2

Answer 1

Given that line AD is an altitude of triangle ABC.

Equation to find the altitude AD is:

`(AD)^2=(BD)(DC)`

Where

BD = 3x

DC = 5x - 14

Let's solve for x:

m∠ADC = 5x + 20

m∠DAC = 2x + 2

We know that m∠ADC is 90 degrees

Thus,

5x + 20 = 90

5x = 90 - 20

5x = 70

`\begin{gathered} (5x)/(5)=(70)/(5) \\ \\ x=14 \end{gathered}`

Since x = 14, we have:

BD = 3x = 3(14) = 42

DC = 5x - 14 = 5(14) - 14 = 70 - 14 = 56

Therefore

`\begin{gathered} (AD)^2=^{}(BD)(CD) \\ \\ (AD)^2=(42)(56) \\ \\ (AD)^2=2352 \end{gathered}`

Take the square root of both sides:

`\begin{gathered} \sqrt[]{(AD)^2}=\sqrt[]{2352} \\ \\ AD\text{ = }48.5 \end{gathered}`

AD = 48.5

ANSWER:

Equation setup = (AD)² = (BD)(DC)