An isosceles triangle with equal sides of 5 inches and a base of 6 inches is inscribed in a circle. What is the distance between the center of the circle and the base?

Guest May 29, 2021

#1**+1 **

Melody, CPhill, and Tigsy already did an amazing job answering this problem. :))

https://web2.0calc.com/questions/math_51432

=^._.^=

catmg May 29, 2021

#3**+1 **

I just thought of a nice different solution, so I'll post it here. :))

I'm going to refrence Melody's graph since I'm not sure how to make my own yet.

Note than if you split this triangle in half, you get a 3, 4, 5 traingle.

Let's call the radius r.

The point (3, 4) a.

The point (0, 0) b.

The point (6, 0) c.

The point (3, 0) d.

And the center C.

Since aC = r, Cd = 4 - r.

bC = r too.

ad = 3.

By pythagreon

(4-r)^2 + 3^2 = r^2

r^2 - 8r + 16 + 9 = r^2

8r = 25

r = 25/8

=^._.^=

catmg May 29, 2021

#4**+2 **

This isn't EXACTLY the same problem, but you can use the answer that catmg referenced to find the distance between the center of the circle and the base as

sqrt ( 3.125^2 - 3^2) =

0.875 units

CPhill May 29, 2021

#7**+3 **

An isosceles triangle with equal sides of 5 inches and a base of 6 inches is inscribed in a circle. What is the distance between the center of the circle and the base?

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

BC = 5 MC = 3 BN = 2.5

BM = sqrt(5^{2} - 3^{2})

BM / BC = BN / BO BO = 3 ^{1}/_{8}

**OM = BM - BO = ^{7}/_{8}**

jugoslav May 29, 2021