Answer:
10-5[tex]\sqrt{2}[/tex]
Step-by-step explanation:
As per the attached figure, right angled [tex]\triangle MDL[/tex] has an inscribed circle whose center is [tex]I[/tex].
We have joined the incenter [tex]I[/tex] to the vertices of the [tex]\triangle MDL[/tex].
Sides MD and DL are equal because we are given that [tex]\angle M = \angle L = 45 ^\circ.[/tex]
Formula for area of a [tex]\triangle = \dfrac{1}{2} \times base \times height[/tex]
As per the figure attached, we are given that side a = 10.
Using pythagoras theorem, we can easily calculate that side ML = 10[tex]\sqrt{2}[/tex]
Points P,Q and R are at [tex]90 ^\circ[/tex] on the sides ML, MD and DL respectively so IQ, IR and IP are heights of [tex]\triangle[/tex]MIL, [tex]\triangle[/tex]MID and [tex]\triangle[/tex]DIL.
Also,
[tex]\text {Area of } \triangle MDL = \text {Area of } \triangle MIL +\text {Area of } \triangle MID+ \text {Area of } \triangle DIL[/tex]
[tex]\dfrac{1}{2} \times 10 \times 10 = \dfrac{1}{2} \times r \times 10 + \dfrac{1}{2} \times r \times 10 + \dfrac{1}{2} \times r \times 10\sqrt2\\\Rightarrow r = \dfrac {10}{2+\sqrt2} \\\Rightarrow r = \dfrac{5\sqrt2}{\sqrt2+1}\\\text{Multiplying and divinding by }(\sqrt2 +1)\\\Rightarrow r = 10-5\sqrt2[/tex]
So, radius of circle = [tex]10-5\sqrt2[/tex]
