Hello kandyfloss Originally Posted by

**kandyfloss** Two discs,each of length 11.6 cm are laid on a table with their centers 14cm apart.

find the area common to the two discs?

i used the formula 1/2(r^2)theta to find the area of sector and then subtract the area of triangle from it 1/2(r^2)sin.theta

the values i get are 60.14 for area of sector

and 105.62 (i'm in radians)

whats going wrong here.are my value right?

Have a look at the diagram I've attached.

I assume that it's the radius of the discs that is 11.6 cm. So in the $\displaystyle \triangle ABO$: $\displaystyle \cos \theta = \frac{7}{11.6}$

$\displaystyle \Rightarrow \theta = 0.9230$ radians

and the area of the overlap is 4 times the area of the shape bounded by the straight lines AB and BP and the arc AP; which is:$\displaystyle 4\times\big(\tfrac12\cdot11.6^2\cdot0.9230 - \tfrac12\cdot11.6\cdot7\cdot\sin(0.9230)\big)$ (that's the area of sector OAP minus the area of triangle OAB)$\displaystyle =118.9 \text{ cm}^2$, if my arithmetic is correct

Grandad