Given a right-angled triangle with legs 2 and 3 cm. Find the area of the square inscribed in this triangle.

From the vertex of the right angle we draw the bisector CM, and from the point M straight lines parallel to BC and AC.

CКMН is a square, since the opposite sides are parallel, and the diagonal forms an angle of 45 with the sides.

Let us prove that triangles ABC and ВKM are similar.

Both triangles are rectangular, and they have a common angle B, which means they are similar in acute angle.

Let CM = KM = MH = CX = X cm, then BK = (3 – X) cm.

BC / AC = ВK / KM.

3/2 = (3 – X) / X.

3 * X = 6 – 2 * X.

5 * X = 6.

X = 6/5.

Let’s define the area of the square.

S = CK ^ 2 = (6/5) ^ 2 = 36/25 = 1 (11/25) cm2.

Answer: The area of the square is 1 (11/25) cm2.