Question

if Bc=15 and DC=28 what is the diameter of the circle O to the nearest tenth A.13.1 B.37.3 C.52.3 D.67.3

Answer 1

@jim_thompson5910

Answer 2

have a look at this page

http://www.regentsprep.org/Regents/math/geometry/GP14/CircleSegments.htm

and use theorem 3

Answer 3

i still dont understand what to do though ?

Answer 4

let AB = x

Answer 5

we want to find the length of AB, ie solve for x

Answer 6

using theorem 3, we would get

BC*AC = (DC)^2

15*(15+x) = 28^2

x = ???

Answer 7

TY

Answer 8

what did you get?

Answer 9

52.3

Answer 10

Try it again. That is incorrect.

Answer 11

how is that incorrect , i work the problem out and got 52.2 rounded to the nearest tenth it would be 52.3

Answer 12

I'll show you how to solve for x

15*(15+x) = 28^2

15*(15+x) = 784

15*15+15*x = 784

225 + 15x = 784

15x = 784 - 225

15x = 559

x = 559/15

x = 37.2666666666667

which rounds to 37.3

Answer 13

question why would you put 15 twice if BC=15 then formula should be 15(x)=28^2 which would give you 15x=784. Not really understanding how you got another 15.

Answer 14

if we use theorem 3, we will get this

BC*AC = (DC)^2

Answer 15

AC is composed of the segments AB and BC

so

AC = AB + BC

AC = x + 15

AC = 15 + x

Answer 16

that's how I'm getting 15 + x instead of just "x"

Answer 17

notice how they state

"(whole secant)×(external part) =
(tangent)^2"

Answer 18

The AC is the whole secant