Find circle radius by given triangle inside

Question

So the triangle inside the circle:

$AB = 9\;cm$

$CB = 6\;cm$

$CH = 5\;cm$

I think solving this problem involves similar triangles.

Thanks in advance, I'd like to have a solution suitable for 9th grade.

EDIT: Thank you huys for the upvotes, didn't think this problem would prove to be so popular.

EDIT 2: Wait my book says that the answer should be $5,4$ degrees... Any thoughts?

Answer 1

you can find $$BH = \sqrt{BC^2 - CH^2} = \sqrt{11}, AH = AB - BH = 9 -\sqrt{11}$$ by the phythagoras theorem $AC = \sqrt{AH^2 + HC^2 } = \sqrt{25+81+11 - 18\sqrt {11}} = \sqrt{117-18\sqrt{11}} = 7.569$

the radius of the circumcircle is $$ \frac{AC}{2\sin\angle ABC}=\frac{7.569}{2\times 5/6} = 9.08367/2=4.5418$$

there should be an easier way to do this. i just can't see it now.

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$\bf p.s.$ here is a method that avoid the rule of $\sin$ altogether. let the center of the circumcircle be $O$ and $M$ the foot of the perp from $O$ to $AB.$ let $OM = x.$

now, we will find the square of the circumradius $R$ in two ways: from the right triangle $AMO,$ we have $$R^2 = (4.5)^2 + x^2 \tag 1$$ from the right triangle $OC?$, we have $$R^2 = (5-x)^2 +(4.5 - \sqrt{11})^2 \tag 2 $$

from $(1),(2)$ first find $x,$ then find $R.$

Answer 2

$ HB = \sqrt{ 36-25 }= \sqrt{11} $

$ AH = 9 - \sqrt{11} $

$ AC^2 = AH^2 + HC^2 ; AC = \sqrt{117 -18 \sqrt{11} } $

The area of the triangle is $12×AB×CH$ so you can calculate this easily as suggested by Mark Bennet.

$ R$ = product of sides/ (4 Area)

Answer 3

R=[AB][BC][CA]/4(Area of Triangle)

Area of triangle can be calculated by Heron's formula

So,by putting the values we get radius as 27/8 multiplied by root of 2.

The radius is the circumradius of the triangle as the circle is a circumcircle as it passes through the vertices of the triangle.