Find the radius of the inscribed circle of a right triangle. Find the radius of the inscribed circle of a right triangle. The triangle's height is $\sqrt{6} + \sqrt{2}$ while the bisector of the right angle is 4.
This seemed like a generic similar triangles problem but it isn't that simple when I tried to solve it. Any help?
EDIT: Here is a picture of the problem
 A: Because a right triangle's "height" is a leg of the triangle, one of the legs has length $\sqrt6+\sqrt2$. I am not completely sure about this assumption, but I think you meant to say that the length of the hypotenuse is $4$. If so, this implies that the length of the other leg is $\sqrt6-\sqrt2$.
An easy way to find the inradius of a triangle is to use the fact that the inradius multiplied by the semiperimeter (half the perimeter) of the triangle gives the area of the triangle. Here, the area is the two legs multiplied together and then halved to get $\dfrac{(\sqrt6+\sqrt2)(\sqrt6-\sqrt2)}{2}=\dfrac{6-2}{2}=2$, and the semiperimeter is $\dfrac{4+(\sqrt6+\sqrt2)+(\sqrt6-\sqrt2)}{2}=\sqrt6+2$. Dividing the area by the semiperimeter, we get that the inradius has length $\dfrac{2}{\sqrt6+2}=\dfrac{2(\sqrt6-2)}{6-4}=\boxed{\sqrt6-2}$.
Note: An easy way to prove that the inradius of a triangle multiplied by the semiperimeter gives the area is to draw line segments from the incenter to the vertices of the triangle as shown in the diagram (you may want to open it in a new tab to see it clearly). The incircle is tangent to all 3 sides of the triangle, so it is the altitude of each of the 3 triangles formed by the line segments we drew earlier. Each of those $3$ triangles has a base as one of the sides of the triangle (say lengths of $a,b,c$ and the inradius has length $i$), then their individual areas are $\frac{ai}{2},\frac{bi}{2},\frac{ci}{2}$. Adding them together, we get the area of the whole triangle, which is $\frac{ai}{2}+\frac{bi}{2}+\frac{ci}{2}=\frac{(a+b+c)i}{2}=\frac{a+b+c}{2}\cdot i$, which is the semiperimeter times the inradius. This holds true for any triangle.

A: Let $s=|CA|+|BC|$ and $p=|CA|\cdot|BC|$ then the double area of the triangle can be written as 
$$p=|CD|\cdot|AB|=|CE|s\sin(\pi/4).$$
Now in order to find the inradius $r$ recall that
$$r=\frac{p}{s+|AB|}=\frac{p}{\frac{\sqrt{2}p}{|CE|}+\frac{p}{|CD|}}=\frac{1}{\frac{\sqrt{2}}{|CE|}+\frac{1}{|CD|}}.$$
Can you take it from here?
A: 
Find the radius $r$ of the inscribed circle of a right triangle
$ABC$ 
given its height $|CD|=h_c=\sqrt6+\sqrt2$
and bisector $|CE|=4$.
Let $I$ be the center of the inscribed circle
and $A_t,B_t,C_t$ its touching points
\begin{align} 
\triangle CED,\ \triangle IEC_t,\triangle CB_tI:\quad
\frac{h_c}{\beta_c}
&=
\frac{r}{|IE|}
=
\frac{r}{\beta_c-|CI|}
=
\frac{r}{\beta_c-r\,\sqrt2}
,\\
r&=\frac{\beta_c\,h_c}{\beta_c+h_c\,\sqrt2}
=\frac{2\,\sqrt6}3
.
\end{align}
Edit
Additionally, the side lengths of the triangle can be found as follows.
We know that
\begin{align}
2r&=a+b-c
,\\
r\,(a+b+c)&=c\,h_c
,
\end{align}
so the hypotenuse is found as
\begin{align} 
c&=
\frac{2\,h_c\,\beta_c^2}{2\,h_c^2-\beta_c^2}
=
\tfrac{4\sqrt6}3\,(1+\sqrt3)
\approx 8.92284
.
\end{align}
The sizes of the legs can be found 
from system
\begin{align}
ab&=c\,h_c
,\\
a+b&=2\,r+c
.
\end{align}
Since $a<b$, we have
\begin{align}
a&=
r+\tfrac12\,\Big(c-\sqrt{(2r+c)^2-4\,c\,h_c}\Big)
= 
\tfrac{2\sqrt6}3\,(1+\sqrt3)
,\\
b&=
r+\tfrac12\,\Big(c+\sqrt{(2r+c)^2-4\,c\,h_c}\Big)
=
2\,\sqrt2\,(1+\sqrt3)
.
\end{align}
It follows that 
$\angle CAB=30^\circ$,
$\angle ABC=60^\circ$.
A: The hypotenuse is 4 and one of the legs is √6+√2.
Applying Pythagoras theorem we get the other leg as √6-√2.
Then we get the area as 1/2×{(√6+√2)(√6-√2)} = 1/2 × 4 = 2.
Perimeter= 4+√6+√2+√6-√2 = 4+2√6
Semi perimeter= 2+√6
Then 
Inradius = (Area)/(Semi perimeter)= 2/(√6+2) = {2(√6-2)}/(6-4) = 
(2√6-4)/2 = √6-2.
[I used the formula for inradius, which is AREA/SEMIPERIMETER]
A: Hint:
Notice that for $\triangle CDE$, $$\measuredangle DCE = \arccos\left(\frac{\sqrt6 + \sqrt2}4\right)$$
Then, finding any of the lengths $AC$, $AB$, $BC$, $BD$ is trivial.
A: Here is the exact answer:
$$r=2\sqrt{\frac{2}{3}}=1.6329931618554520654648560498039$$
The method I used:
First I found that the angle between $CD$ and $CE$ is equal to $15^0$. This was made possible because we knew $CD$ and $CE$ of the right-angled triangle $CDE$. Therefore, $\measuredangle CBA=60^0$ and $\measuredangle CAB=30^0$. 
$a=\frac{CD}{\cos30^0},\space\space\space b=\frac{CD}{\cos60^0},\space\space\space c=\sqrt{a^2+b^2},\space\space\space  $ and $\space\space\space r=\frac{a+b-c}{2}$
A: This is the answer to the problem you specified after giving the diagram:
First we do a bit of angle chasing
$\angle CBA= \theta$
$\angle BCD= \angle BAC = 90° - \theta$
In $∆CEA$ , $\angle CEA = 45° + \theta$
$\angle CED = 180° - \angle CEA = 135° - \theta$
So in $∆CED$ , $\angle DCE = \theta - 45°$
Applying Pythagoras Theorem in right $∆CDE$ we get 
$DE = √6 - √2$
$sin(\theta - 45°) = \frac{√6 - √2}{4}$
Solving this equation we get $\theta = 60°$ ,
$\angle CBA = 60°$ and $\angle CAB = \angle BCD = 30°$ 
After that it is just a lot of calculations, you can find out all the three sides using trigonometry, since we got very nice standard angles here. ( I am not showing the calculations, too much to write and anyways they are very simple)  :)
Then finally we get the inradius as $\frac{2√2}{√3}$.
[You can use $r = \frac{Area}{Semiperimeter}$ or since $∆ABC$ is right angled, $r = \frac{a+b-c}{2}$ .]
Hope this helps you!
