Given that GK is a perpendicular bisector to JH which two triangles are congruent by the HL theorem
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The inscribed circle has its center at the point of intersection of the angle bisectors, which also happen to be the medians. Hence the altitude of the triangle is 3 times the radius, or 12 inches.
The side length of this triangle is 2/√3 times the altitude, so the area is
... Area = (1/2)·b·h = (1/2)·(24/√3 in)·(12 in)
... Area = 48√3 in² ≈ 83.1384 in²
