Find radius of circle

**Punch** · August 28th 2010, 06:15 AM

The only information provided is the lengths of the sides of the triangle

Question: Find the radius of the circle. Ans:2cm

**mathaddict** · August 28th 2010, 06:29 AM

Originally Posted by Punch

The only information provided is the lengths of the sides of the triangle

Question: Find the radius of the circle. Ans:2cm

Draw lines connecting the point where the circle touches the triangle and the center. Also, draw a line from C to the center of circle.

Use the fact that both tangents from a point to the same circle are equal.

**Punch** · August 28th 2010, 06:31 AM

hi, i tried that and the answer is 2.6... which is wrong!!!!

we probably couldnt assume that the lengths of the triangle are tangents as it is not stated in the question

**Isomorphism** · August 28th 2010, 06:39 AM

Originally Posted by Punch

hi, i tried that and the answer is 2.6... which is wrong!!!!

we probably couldnt assume that the lengths of the triangle are tangents as it is not stated in the question

Can you show us your computations?

**Punch** · August 28th 2010, 06:54 AM

$Angle ACB=sin^-1\frac{5}{13} $ $=22.6198$

Let length of radius be $x$ and point of centre of circle be $O$

$Angle OCA=\frac{22.6198}{2}$
$=11.3099$ (external point of tangent of circle)

$tan11.3099=\frac{x}{13}$

$x=13tan11.3099$
$=2.6cm$

**skeeter** · August 28th 2010, 07:14 AM

Originally Posted by Punch

$Angle ACB=sin^-1\frac{5}{13} $ $=22.6198$

Let length of radius be $x$ and point of centre of circle be $O$

$Angle OCA=\frac{22.6198}{2}$
$=11.3099$ (external point of tangent of circle)

$tan11.3099=\frac{x}{13}$ ratio is incorrect (not 13)

note that the area of triangle ABC = 30

sketch segments from the circle's center to each vertex. three triangles are formed.

$\frac{1}{2}(5 \cdot r) + \frac{1}{2}(12 \cdot r) + \frac{1}{2}(13 \cdot r) = 30$

solve for r

**mathaddict** · August 28th 2010, 07:20 AM

Originally Posted by Punch

$Angle ACB=sin^-1\frac{5}{13} $ $=22.6198$

Let length of radius be $x$ and point of centre of circle be $O$

$Angle OCA=\frac{22.6198}{2}$
$=11.3099$ (external point of tangent of circle)

$tan11.3099=\frac{x}{13}$

Check this again, it's not over 13.

$x=13tan11.3099$
$=2.6cm$

The circle does touch the triangle. Else, it can't be solved.

Lets call the point of intersection (for the one above ) be P and the one below, Q, and the one by the side, R.

AP=AR=x and RB=BQ=y

CP=13-x and CQ=12-y

x+y=5
13-x=12-y

Solving the simultaneous equation yields x=3, y=2.

angle ACB=22.62

angle COP=11.3

tan 11.3 = radius / 10

radius =2

**Punch** · August 28th 2010, 07:24 AM

thanks i overlooked that

**Punch** · August 28th 2010, 06:04 PM

Originally Posted by skeeter

note that the area of triangle ABC = 30

sketch segments from the circle's center to each vertex. three triangles are formed.

$\frac{1}{2}(5 \cdot r) + \frac{1}{2}(12 \cdot r) + \frac{1}{2}(13 \cdot r) = 30$

solve for r

i didnt quite understand the points u said to draw

the part where u said to each vertex

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