so hmm notice the picture below
for the first figure, what you really have is 3 3x3 squares, and then 2 half-squares... now.. if you take the 2 halves and put them together, they can make up a square, so their area is the same as one square of 3x3
and surely, you know how to get the area of a square
or you can just use the half-squares as triangles, with a base of 3 and height of 3, and get their area that way as well
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now.. for 8
notice the picture below, is really just 4 circles of radius 4, and one square overlapping them at the center
the area of that, will be, the area of the circles PLUS the yellow part right in the middle of the square
because the square is overlapping all 4 circles, the overlapped area is not included since the circle's area would already include that, so the area is just, the circles plus the yellow part, the part the in the middle of the 4 circles
now, to get the area of a circle, well, that's peanuts, is just πr² r= radius=4
now, to get the yellow area part, get the area of the square, then, subtract the "circle's sector" overlapping it, that is, subtract the corner of each circle from the square's area
the square is 2radius long on one side and the other, so is a 8x8 square
[tex]\bf \textit{area of a sector of a circle}\\\\
A=\cfrac{\theta\pi r^2}{360}\qquad
\begin{cases}
r=radius\\
\theta=\textit{angle in degrees}\\
----------\\
\theta=90^o\\
r=4
\end{cases}\implies A=\cfrac{90\pi \cdot 4^2}{360}
\\\\\\
A=4\pi \impliedby \textit{4 sectors will be}\implies 4\cdot 4\pi \implies 16\pi \\\\
-----------------------------\\\\
\textit{the yellow part area is then }(8\cdot 8) - (16\pi )[/tex]
now.... the area of all 4 circles plus that yellow part will then be
[tex]\bf \textit{area of a circle}\\\\
A=\pi r^2\\\\
-----------------------------\\\\
A=\pi 4^2\implies A=16\pi \qquad \textit{4 circles}\implies 4\cdot 16\pi =\boxed{64\pi}
\\\\\\
\textit{plus the yellow area }64\pi +(64+16\pi)\implies 64+80\pi [/tex]
on another note, if you notice the part that's overlapping the square, is one-quarter of each circle, 4 circles, 4 quarters, so, all 4 quarters do have the area of one whole circle, and you can simply subtract one-whole-circle with radius of 4 area, from the square as well
and you could do it that way as well
