we worked it out the remaining area of the square to be negative 28
but this can not be right, please help!!!
Yes, negative 28 cannot be the area of the square.
Why?
Firstly, the area cannot be negative. There are no negative areas.
Secondly, the side of the square is 5+5 = 10cm. So, the area is 10*10 = 100 sq.cm.
The two identical circles touch each other on one point only--the midpoint of one of the side of the square.
--------------------
Edit.
Ooppss, sorry, I did not see the fine prints below the figure.
Now that I saw them, let me rectify my answer.
My poor eyesight cannot see which portion of the circle is shaded. (I think I have to buy a pair of all-seeing eyeglasses.) So let me assume that the portion in question is that portion not occupied by the circles.
The two circles occupied portions of the square that is a combined semicircle. (Two quadrants is one semicircle.)
Hence, area of shaded portion, A, is area of square of 10cm a side, minus (1/2)pi(5^2).
A = 10*10 -(1/2)pi(25)
A = 100 -39.27
A = 60.73 sq.cm. --------------answer.