# Surface area of a diagram

The picture shows a chest which has a uniform cross-section $ABCDE$, in which $ABCE$ is a rectangle . $M$ is the midpoint of $AB$. $CDE$ is an arc of a circle with $M$ as the centre . $AB = 32$cm , $BC = 12$cm, $BQ = 50$cm . Find the surface area of the lid , $CDETSR$, leaving answer correct to the nearest whole number .

Here's what I did before I got stuck .

I found $EM=MC = 20$ cm

I found that angle $EMC$ = $1.8545$ radian.

Then I found Arc length $EDC$ = $37.1$ cm .

Till here I got stuck... Can I get help? Thanks alot !

• We can assume that CDETSR is flat and rectangular in shape.That is, (37.1)(50) square units. – Mick Jul 31 '16 at 14:39
• I find 283 cm.I will show my result below. – onurcanbektas Jul 31 '16 at 15:03

and the angle EMC=0.283, since $(\pi / 2) -2 * \arctan(3/4) = 0.283 radian$