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

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

2. 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.

3. 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

4. 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?

5. $\displaystyle Angle ACB=sin^-1\frac{5}{13} $$\displaystyle =22.6198 Let length of radius be \displaystyle x and point of centre of circle be \displaystyle O \displaystyle Angle OCA=\frac{22.6198}{2} \displaystyle =11.3099 (external point of tangent of circle) \displaystyle tan11.3099=\frac{x}{13} \displaystyle x=13tan11.3099 \displaystyle =2.6cm 6. Originally Posted by Punch \displaystyle Angle ACB=sin^-1\frac{5}{13}$$\displaystyle =22.6198$

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

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

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

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

solve for r

7. Originally Posted by Punch

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

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

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

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

Check this again, it's not over 13.

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

8. thanks i overlooked that

9. 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.

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

solve for r
i didnt quite understand the points u said to draw

the part where u said to each vertex