# Thread: sss or sas ?

1. ## sss or sas ?

In the triangle ABC AC = BC = 20 inches and AB = 10 inches. Circle is inscribed in the triangle, what is the radius of the circle...see attached...#71

2. Where are you stuck?

3. well i'm pretty sure the one radius pointing down is the midpoint of AB so those 2 segments are 5 inches, then i bisected the 3 angles in the circle to figure out that Ar and Br should be 5 inches each. i dont know what to do next

4. Originally Posted by igottaquestion
In the triangle ABC AC = BC = 20 inches and AB = 10 inches. Circle is inscribed in the triangle, what is the radius of the circle...see attached...#71
sketch lines from each vertex to the incenter.

note that the area of the three triangles formed sums to the area of the large triangle ...

$\displaystyle \frac{1}{2}(20)r + \frac{1}{2}(20)r + \frac{1}{2}(10)r = \frac{1}{2}(10)h$

$\displaystyle 25r = 5h$

$\displaystyle r = \frac{h}{5}$

find the height of the triangle from the vertex angle to the base and you can find r

5. ok so the height is 5 sq rt of 15, so r=sq rt of 15?

6. That's it!

7. Well...let equal sides = a and base = b

radius incircle = bSQRT(4a^2 - b^2) / [2(2a + b)]

OR (if you prefer the SQRT in denominator!):

radius incircle = b(2a - b) / [2SQRT(4a^2 - b^2)]

No need to calculate height.

8. Originally Posted by igottaquestion
In the triangle ABC AC = BC = 20 inches and AB = 10 inches. Circle is inscribed in the triangle, what is the radius of the circle...see attached...#71
1. You are dealing with 2 right triangles. (see attachment)

2. Grey triangle: $\displaystyle h^2 + 5^2 = 20^2$

3. The right triangle at the top of the grey triangle:

$\displaystyle r^2+15^2 = (h-r)^2$

4. Calculate h from 2. and plug in this term into the equation in 3. Solve for r.

Spoiler:
You should come out with $\displaystyle r = \sqrt{15}$

Second attempt: The 2 right triangles are similar. So use proportions:

$\displaystyle \dfrac r{15} = \dfrac5h$

First calculate h, then r using the proportion.