(c) 3.75 cm2
Let the areas of ΔAXY and ΔBXY be A1 and A2 respectively. The heights of the ΔAXY and ΔBXY are equal, so
\(\frac{A_1}{A_2}\) = \(\frac{\frac{1}{2}\times{AX}\times{height}}{\frac{1}{2}\times{BX\times{height}}}\)
⇒ \(\frac{A_1}{A_2}\) = \(\frac{AX}{BX} = \frac{3}{5}\) .......(i)
Also, ΔAXY ~ ΔABC
⇒ \(\frac{\text{Area of ΔAXY}}{\text{Area of ΔABC}}\) = \(\big(\frac{AX}{AB}\big)^2\)
⇒ \(\frac{A_1}{16}\) = \(\bigg(\frac{3x}{3x+5x}\bigg)^2\) ⇒ \(\frac{A_1}{16}\) = \(\frac{9}{64}\)
⇒ A1 = \(\frac{9}{64}\) x 16 = \(\frac{9}{4}\)
∴ From (i) and (ii) A2 = \(\frac{5}{3}\) x A1 = \(\frac{5}{3}\) x \(\frac{9}{4}\) = \(\frac{15}{4}\) = 3.75 cm2.