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Here's me trying to do the problem. http://s10.postimg.org/bwwliium1/image.jpg

So this problem was from a textbook and it was in the chapter of the theorem: If $\theta$ is the angle subtended by a chord $PB$ at a point on a circle of a radius $r$, then $\sin \theta = PB/2r$.

I think the answer is 2, but I got 1.

Also, I know I can do area of triangle divided by half of the perimeter which is 2 in this case, but I want to see how I can apply the theorem to solving this.

Thanks.

!!EDIT!! I figured out what was wrong. The question says "...circle may be drawn through the three vertices of any triangle..." I think it meant circumscribed. That makes sense. Sorry about the confusion. Thanks for all your answers.

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Another approach using , as you did, the bisectrix of the right angle: if we call $\;M\;$ the vertex where the right angle is, and the upper one in $\;N\;$ and the rightmost one is $\;Q\;$ (next time do call names to all the vertices and intersection points! Whithout that geometry exercises become cumbersome), we have that

$$MP=MB=r\;\;\text{(why?)}$$

Since both tangents to a circle from the same exterior point have the same length, we have that:

$$BN=NA=6-r\implies AQ=4+r=QP$$

But we also know that $\;QP=8-r\;$ , so we we find that

$$8-r=4+r\implies r=2$$

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  • $\begingroup$ I don't understand how $QB=AQ$? $\endgroup$ – Kat Oct 20 '13 at 4:30
  • $\begingroup$ These two are tangents to the circle from the same exterior point ($\;Q\;$) to it... $\endgroup$ – DonAntonio Oct 20 '13 at 4:31
  • $\begingroup$ s10.postimg.org/bwwliium1/image.jpg $\endgroup$ – Kat Oct 20 '13 at 4:34
  • $\begingroup$ Yes @Kat , I already saw the diagram...so? $\endgroup$ – DonAntonio Oct 20 '13 at 4:35
  • $\begingroup$ s23.postimg.org/up6y1pimj/image.jpg Sorry about that. It's highlighted in purple. I don't get how BQ is tangent to the circle? $\endgroup$ – Kat Oct 20 '13 at 4:38
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The area of a triangle is half base times height - here $\frac 12 \times 6 \times 8 = 24$

Now join each of the vertices of the triangle to the in centre to get three triangles of height $r$ and base each of the sides - so the area, as you say, is $\frac 12 \times r \times (6+8+10)=12r$ so that $r=2$.

I can't see the logic of what you are doing in your diagram - you don't seem to be using the sides or angles of the triangle at all. It isn't obvious how you would use the theorem for this - but if you were looking for the circumradius instead of the inradius the sides of the triangle would be chords you could use in the application of the theorem.

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  • $\begingroup$ I used the 90degree angle, and made $\theta$ 45 degrees. I figure $PB = r\sqrt2$ via Pythagorean theorem. $\endgroup$ – Kat Oct 20 '13 at 4:00
  • $\begingroup$ So in general, I can't use the theorem to find the radius of the circle? $\endgroup$ – Kat Oct 20 '13 at 4:03
  • $\begingroup$ @Kat Well you have a square in the bottom left-hand corner, but that would be true whatever right-angled triangle you were working with. It is unclear whether the point you have labelled $A$ is supposed to be on the circle or on the hypotenuse. If you drew an accurate diagram you would see in can't be on both. Maybe that is what is confusing you? $\endgroup$ – Mark Bennet Oct 20 '13 at 5:50

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