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What is the geometric mean of the measures of the segments AD and DC? Show your work.

What is the geometric mean of the measures of the segments AD and DC? Show your work-example-1
User Aldorath
by
4.6k points

1 Answer

1 vote

Answer:

Mean = 7.525 In

Explanation:

Let BD =X

let dab = a

Let dcb = b

X/sin a= 12

X/sinb= 5

a+b = 90

b= 90-a

X/sinb = x/sin(90-a)= 5

Equating the value of x

12sina = 5sin(90-a)

12sina= 5sin90 - 5sina

Recall that Sin 90= 1

12sina= 5 -5sina

17sina= 5

Sina= 5/17

a= sin^-15/17

a= sin^-1 *0.2941

a= 17.1°

a+b=90

b = 90-a

b = 90-17.1

b =72.9°

X/Sina= 12

X/sin17.1= 12

X = 12*0.2940

X= 3.528

X= 3.5 In

12²-3.5²= AD²

5²-3.5²= DC²

144-12.25= AD²

25-12.25= DC²

131.75= AD²

12.75= DC²

AD= 11.48 In

DC= 3.57 In

Mean =( 11.48+3.57)/2

Mean = 15.05/2

Mean = 7.525 In

User Misha Slyusarev
by
3.8k points