Another Interesting Conclusion from a Non-hypotenuse Side and In-radius of a Right Triangle

Hello math bugs(🐞) and hivers(🐝)
I hope everything with okay with you.
Well come to another geometric problem and it's solution.

We have a right angle triangle. If in-radius(r) of circle inside the triangle and a side (a=non hypotenuse) are given, we can find the area using both of them. Area in terms of the a and r can be given by : ∆= ar(a-r)/(a-2r)

To prove the above consideration, what we need are follows:

☑️ Lentgh of a common tangents from a fixed point outside a circle are equal in lenght:

☑️☑️ The second thing we need here is the legendary Pythagoras theorem:

Hypotenuse² = Perpendicular² + Base². Details

LET'S PROVE ∆ = ar( a - r)/( a - 2r) :

As we know common tangents from outer point of a circle are equal in length, we can have three pairs of common tangents from three outer points (A, B & C) of O-centric in-circle with radius r. Check the following figure: 👇

Now, from right ∆ABC above, we can say

AB² + BC² = AC²

Or, (x + r)² + a² = {x + (a - r)}²

Or, x² + r² + 2rx + a² = x² + (a - r)² + 2x(a -r)

Or, r² + 2rx + a² = (a - r)² + 2ax - 2rx

[ x² is removed from both side ]

Or, r² + a² + 4rx = a² + r² -2ar +2ax

Or, 4rx = 2ax - 2ar

[(a² + r²) is removed from both side]

Or, 2rx = ax - ar [ Dividing both side by 2]

Or, 2rx - ax = - ar

Or, ax - 2rx = ar

Or, x(a - 2r) = ar

Or, x = ar/(a - 2r)

     LET'S GO TO THE FIGURE AGAIN

Height AB = (x + r) unit = r + ar/(a - 2r) unit

Base BC = a

Ar. ∆ABC = 1/2 × BC × AB unit²

Or, ∆= 1/2 × a × { r + ar/(a - 2r)} unit ²

Check the following figure now :

🎤🎤All the figures and drawing are made by me using android only. There may be some silly mistakes; excuse me for that and the figure may not be accurate ; please try considering the data only.

Thank you so much guys for stoping by

I hope you have liked it

Have a great day

All is well

Regards: @meta007