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

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

Let’s construct the radii OP and OK to the tangency point.

Then, by the property of tangents: MR = ML, HL = HK, AR = AK.

AM + MH + AH = AM + ML + HL + AH = AM + MR + HK + AH = (AM + MR) + (AH + HK) = AR + AK = AB / 2 + AD / 2 = AB.

The perimeter of the triangle AMH is equal to the length of the side of the square.

Then AM + AH = 12 – MH = 12 – 5 = 7 cm.

Let AM = X cm, then AH = (7 – X) cm.

By the Pythagorean theorem: MH ^ 2 = AM ^ 2 + AH ^ 2.

25 = X ^ 2 + (7 – X) ^ 2 = X ^ 2 + 49 –14 * X + X ^ 2.

2 * X ^ 2 – 14 * X + 24 = 0.

X ^ 2 – 7 * X + 12 = 0.

Let’s solve the quadratic equation.

X1 = 3 cm.

X2 = 4 cm.

Let AM = 3 cm, then AH = 7 – 3 = 4 cm.

Then Samn = AM * AH / 2 = 3 * 4/2 = 6 cm2.

Answer: The area of ​​the AMH triangle is 6 cm2.




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