In a right angle triangle, length of perpendicular is 16 cm and length of hypotenuse is 20 cm. A square is inscribed in a triangle. One of the vertices of square coincides with the vertex of triangle. What is the maximum possible area of the square (in cm²)?

\(\frac{2304}{49}\)

Using pythagoras theorem

BC = 12 cm

ΔCXY ∼ ΔCBA

\(\frac{CX}{CB}\) = \(\frac{XY}{BA}\) ⇒ \(\frac{12-a}{12}\) = \(\frac{a}{16}\)

                  ⇒ 48 - 4a = 3a

                 ⇒ 48 = 7a

                 ⇒  a = \(\frac{48}{7}\) cm

Now,

Side of sqaure = \(\frac{P × B}{P + B}\) = \(\frac{16 × 12}{28}\) = \(\frac{48}{7}\) cm

Area of square = \( {\left({\frac{48}{7}}\right)}^{2}\) = \(\frac{2304}{49}\) cm²