Absolutely correct. My bad!
I have made the necessary changes

I got it. Thank you very much KarunMendiratta sir from my deeper heart

Another easy way for circle Radius is by the formulae

For right angled triangle inradius=(Hypotenuse-(sum of other two sides))/2

Not trigonometry is required here!

You can reduce the size of the triangle ABC to an equilateral triangle ABG of sides 6. The circle in this triangle must be smaller than the one in the bigger triangle ABC.

However, this new equilateral triangle ABG of sides 6 still definitely contains a bigger circle than the straight up smaller triangle DEF. Therefore, if the circle in ABC > the circle in ABG, and the circle in ABG > the circle in DEF; the circle in ABC must also > the circle in DEF

Step 1 – Quantity A
For a right triangle with sides a, b, and hypotenuse c, the radius r of the inscribed circle is given by:
r = (a + b - c) / 2

For sides a = 6, b = 8, and c = 10:
r = (6 + 8 - 10) / 2 = 4 / 2 = 2

Area of the inscribed circle:
A = πr2 = π(2)2 = 4π


Step 2 – Quantity B
The second triangle is equilateral with side length a = 4.
Height h of an equilateral triangle:
h = (√3 / 2)a = (√3 / 2) × 4 = 2√3
If you forget this formula, you can find h using the Pythagorean theorem- you can either solve for the height or realize that the triangle formed by the height is a 30-60-90 triangle.
The perpendicular from a vertex divides the triangle into two 30°–60°–90° right triangles. Each has half the base = 2 and hypotenuse = 4, so the height is 2√3.

The radius r of the inscribed circle in an equilateral triangle is:
r = h / 3 = (2√3) / 3
Area of this circle:
A = πr2 = π((2√3)/3)2 = π × (12/9) = 4π / 3



--> You can also easily get the radius(r) of the circle inscribed inside any equilateral triangle with side 'a', by using this formula--> r = a/(2√3)

Step 3 – Comparison
Quantity A = 4π, Quantity B = (4π) / 3
Therefore, Quantity A > Quantity B