Answer:
[tex]\mathrm{130cm^2}\text{ is actually not the correct answer. }[/tex]
Here's why:
[tex]\text{Solution:}\\\text{Diagonal of small square }(d)=4\text{cm}\\\text{Lenght of small square }(l)=\ ?\\\text{Diagonal of big square }(D)=9\text{cm}\\\text{Length of big square}(L) =\ ?[/tex]
[tex]\text{Using pythagoras theorem, }\\d^2=l^2+l^2\\\text{or, }d^2=2l^2\\\text{or, }\dfrac{4^2}{2}=l^2\\\\\text{or, }l^2=8\\\therefore\ \text{Area of small square}(a)=8\text{cm}^2[/tex]
[tex]D^2=L^2+L^2\\\text{or, }9^2=2L^2\\\\\text{or, }L^2=\dfrac{81}{2}\\\\\text{or, Area of big square}(A)=40.5\text{cm}^2[/tex]
[tex]\therefore\ \text{Area of shaded region = Area of big square - area of small square}\\\text{}\hspace{3.83cm}=40.5-8=32.5\text{cm}^2[/tex]
As you can see, the area of the big square is 40.5sq. cm, so there's no way that a part of the big square (area of the shaded region) will have area equal to 130sq. cm.
