A circle is inscribed in a square with a side of 12. The segment MN with ends on sides AB and AD

A circle is inscribed in a square with a side of 12. The segment MN with ends on sides AB and AD touches this circle. It is known that MN = 5 Find the area of the triangle AMN.

Let point K lie in the middle of side AB, point O – the center of the circle, OK – the radius of the circle equal to 6 cm.
Let’s find AM, for this we calculate what the segment KM is equal to. The tangent line is perpendicular to the radius, and OS bisects, S is a point by MN. From the triangle MOS, we find the MO by the Pythagorean theorem:
MO = (6 ^ 2 – (5/2) ^ 2) ^ (1/2) = 6.5 cm.
Now we find KM from the triangle KOM, also by the Pythagorean theorem, since OK is perpendicular to AB, hence KM:
KM = (6.5 ^ 2 – 6 ^ 2) ^ (1/2) = 2.5.
Then AM = 12 – (6 + 2.5) = 3.5 cm.
From a right-angled triangle AMN according to the Pythagorean theorem:
AN = (5 ^ 2 – 3.5 ^ 2) ^ (1/2) ≈ 3.5 cm.
The AMN area is:
S = ½ * AM * AN = ½ * 3.5 * 3.5 ≈ 6.25 cm ^ 2.