1. Trig functions question

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
whats going wrong here.are my value right?

2. 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
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 $\triangle ABO$:
$\cos \theta = \frac{7}{11.6}$

$\Rightarrow \theta = 0.9230$ 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:
$4\times\big(\tfrac12\cdot11.6^2\cdot0.9230 - \tfrac12\cdot11.6\cdot7\cdot\sin(0.9230)\big)$ (that's the area of sector OAP minus the area of triangle OAB)
$=118.9 \text{ cm}^2$, if my arithmetic is correct

3. Yes 11.6 was the radius and the answer is right 118.9.
i tried doing the same thing but by using the angle AOC (pls see the picture)
as 90 deg..because it forms a square with equal radii's as its sides.
why can i not use 1/2r^2theta now to find the area of sector and area 1/2r^2sintheta of one circle ,subtract them and multiply by two to find the total common area?

4. Hello kandyfloss
Originally Posted by kandyfloss
If $\angle AOC = 90^o$, then $\theta$ would have to be $45^o$. Which it isn't. It's $\arccos (7/11.6)=52.88^o$