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?
+2407 Melody, CPhill, and Tigsy already did an amazing job answering this problem. :))
https://web2.0calc.com/questions/math_51432
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+2407 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
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+129899 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
+1641 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?
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BC = 5 MC = 3 BN = 2.5
BM = sqrt(52 - 32)
BM / BC = BN / BO BO = 3 1/8
OM = BM - BO = 7/8
