Circle Geometry

Pls is there anyone who can help me solve this question.
An isosceles triangle ABC has its vertices on a circle. If /AB/=13cm, /BC/=13cm and /AC/=10cm. Calculate
1. The height /BM/ of the triangle.
2. The radius of the circle

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SIMPLE SUBJECT

Yet no answer to this question

the height BM is 12cm. This is from Pythograth theorem

1. The height |BM| is calculated using
Pythagoras Theorem (from ∆BMC)
Thus: |BM|2 = |BC|2 − |MC|2
= 132 − 52
= 169 – 25
= 144
∴|BM| = 12cm

2. The radius of the circle r is calculated
using Pythagoras Theorem (from ∆OMC, where O is the centre of the circle)
Note: |BO| = |AO| = |OC| = r(radius of the circle)
Also: |OM| = |BM| − r = (12 – r)
Thus:|OM|2 = |OC|2 − |MC|2
⇒ (12 – r)2 = r2 – 52
(12 – r)(12 – r) = r2 – 25
144 – 12r – 12r + r2 = r2 – 25
144 -24r + r2 – r2 = −25
−24r = −25 −144
−24r = − 169
∴ r = -169/-24
r ≈ 7cm

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Ashawo wrote:

Oh my God!! I hate math so much and I'm beginning to hate someone

very sweet

Spacetrotter wrote:

1. The height |BM| is calculated using
Pythagoras Theorem (from ∆BMC)
Thus: |BM|2 = |BC|2 − |MC|2
= 132 − 52
= 169 – 25
= 144
∴|BM| = 12cm

2. The radius of the circle r is calculated
using Pythagoras Theorem (from ∆OMC, where O is the centre of the circle)
Note: |BO| = |AO| = |OC| = r(radius of the circle)
Also: |OM| = |BM| − r = (12 – r)
Thus:|OM|2 = |OC|2 − |MC|2
⇒ (12 – r)2 = r2 – 52
(12 – r)(12 – r) = r2 – 25
144 – 12r – 12r + r2 = r2 – 25
144 -24r + r2 – r2 = −25
−24r = −25 −144
−24r = − 169
∴ r = -169/-24
r ≈ 7cm

great work bro