Intersection Area

For radius(r) and distance between circles centers(d), use the following expressions for the common chord (c) and intersection area (A)

c^2=3r^2 - (d^2/2) -d{√(3r^2-(3/4)d^2)}

A=(√3/4)c^2+3[r^2(arcsin (c/2r))-(c/4){√(4r^2-c^2)}]

and using r=5 and d=6, Answer is A=9.865

Under laying idea is the same as that of Mr Marta Reece .

The centers of the circles, labeled $A,B,C$ form an equilateral triangle with side $6$ .

$\triangle ADC$ has sides $5, 5$ , and $6$ , so $\angle CAD=arccos(\frac{5^2+6^2-5^2}{2\times 5\times 6})=53.13^\circ$

$\angle DAB=60^\circ-\angle CAD=6.87^\circ$

$\angle EAD=60^\circ-2\times\angle DAB=46.26^\circ$

From $\triangle AED$ the distance $ED=\sqrt{5^2+5^2-2\times5\times5\times cos(46.26^\circ)}=3.93$

$\triangle DEF$ is equilateral with area $A_t=\frac{\sqrt{3}}{4}\times3.93^2=6.68$

To this we need to add areas of the three identical circular segments. Area of one of them is

$A_c=\frac{R^2}{2}(\frac{\alpha\pi}{180}-sin\alpha)=\frac{5^2}{2}(\frac{46.23^\circ\pi}{180}-sin(46.23^\circ))=1.06$

So the total area is $A_t+3\times A_c=6.68+3\times1.06=9.87$