Problem on right angle triangle and sem-circle inside.

Hello maths bugs(🐞) and hivers(🐝)
I hope you are strong and stout and doing great in life.

Today I have come up with a problem based on a right angle triangle and a semi-circle inside as shown in the figure below. We have to find out radius of it.

We have given two side of the right angle triangle. We can easily find out the 3rd side or the hypotenuse of the triangle. Check details in the below:

Details and short cut of Pythagorean triplet.

METHOD 1:

We need to copy the right angle triangle.Consider BC as axis, we need to rotate the triangle anti-clock wise and thus we find the following figure. You can see the circle (⭕) inside the triangle and touching every side of the triangle. For in-circle we know the relation between radius(r) of the circle and area (∆) & semi peri-meter (s) of the triangle. It is given by ∆= r.s. We can also find out the area of the right angle triangle using 1/2× base× height formula.
Now equation both the formula, we can easily find the required radius(r) as you can see in the figure above. So the in-radius(r) is 3/2 cm.

For more information tap In-circle.

METHOD 2:

Let's do this using similarity.For that we have join DE which is in-radius (r). Now we get two similiar triangles. Namely ∆ABC and ∆EDC. Angle C is common for both of them and angle ABC & angle CDE equal to 90°. So ∆ABC ~ ∆EDC. From similarity we can have AC/AB equals to CD/DE or 5/3= (4-r)/r. That seams 5r =12 - 3r or 8r = 12 or r= 3/2.

For similarity and how to use it,tap similarity please.

Please ignore the mistakes if you can. Make a comment if you have suggestion or if you need further detailing about the topic or if you want more of this kind of tricky problem.I will be glad to reply you.

Thanks for visiting and supporting. It means a lot.

Have a nice day.

All is well.

Regards: @meta007