A string of length 24 cm is bent first into a square and then into a right-angled triangle by keeping one side of the square fixed as its base. Then the area of triangle equals to:

String of length 24 cm is bent into square, => Perimeter of square = 24 cm

Let side of square = $$a$$ cm

=> $$4a=24$$

=> $$a=\frac{24}{4}=6$$ cm

Let the other side of triangle be $$b$$ and hypotenuse be $$c$$ cm

=> Perimeter of triangle = $$a+b+c=24$$

=> $$b+c=24-6=18$$

=> $$c=18-b$$ ------------(i)

Also, using Pythagoras Theorem 

=> $$6^2+b^2=c^2$$

=> $$c^2-b^2=36$$ -----------(ii)

Solving equations (i) and (ii), we get : $$b=8$$ cm and $$c=10$$ cm

$$\therefore$$ Area of triangle = $$\frac{1}{2} ab$$

= $$\frac{1}{2}\times6\times8=24$$ $$cm^2$$

=> Ans - (A)