Hello, SpGcAr1117!
These have little to do with inscribed angles ... mostly Pythagorus.
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Code:
11.
* * *
* *
* *
* *
* O *
* * *
* | \ *
/ | \ r
* / 10 | 10 \ *
A * - - - * - - - * B
* D|8 *
* * *
C
We have a circle with center \(\displaystyle O.\)
Radii \(\displaystyle OA\,=\,OB\,=\,OC\,=\,r\)
Radius \(\displaystyle OC\) bisects chord \(\displaystyle AB\) at \(\displaystyle D\)
\(\displaystyle \;\;\)and \(\displaystyle OC\,\perp\,AB\)
We are told that: \(\displaystyle \,DC\,=\,8,\;AD\,=\,DB\,=\,10\)
We see that: \(\displaystyle \,OD\,=\,r\,-\,8\)
In right triangle \(\displaystyle ODB:\;(r\,-\,8)^2\,+\,10^2\:=\:r^2\)
We have: \(\displaystyle \,r^2\,-\,16r\,+\,64\,+\,100\:=\:r^2\;\;\Rightarrow\;\;16r\,=\,36\;\;\Rightarrow\;\;r\,=\,\frac{9}{4}\)
Code:
12.
C
* * *
* 6 | 6 *
A * + * B
* \ |D / *
\ | / r
* \ | / *
* * *
* |O *
|
* | *
* | *
* | *
* * *
E
We have a circle with center \(\displaystyle O.\)
Radii \(\displaystyle OA\,=\,OB\,=,OC\,=\,OE\,=\,r.\)
\(\displaystyle OC\) bisects chord \(\displaystyle AB\) at \(\displaystyle D\)
\(\displaystyle \;\;\)and \(\displaystyle OC\,\perp\,AB.\)
We are told that \(\displaystyle DE\,=\,10,\;AD\,=\,DB\,=\,6\)
We see that: \(\displaystyle \,DO\,=\,10\,-\,r\)
In right triangle \(\displaystyle ODB:\;(10\,-\,r)^2\,+\,6^2\;=\;r^2\)
We have: \(\displaystyle \,100\,-\,20r\,+\,r^2\,+\,36\;=\;r^2\;\;\Rightarrow\;\;20r\,=\,136\;\;\Rightarrow\;\; r\,=\,\frac{34}{5}\)
Edit: Too fast for me, Denis!