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)?
Answer & explanation
Correct answer: option 2
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