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 !
You will find CM=EM=20cm
and the angle EMC=0.283, since $(\pi / 2) -2 * \arctan(3/4) = 0.283 radian$
So, from these the arc length EDC=0.283 * 20 =5,66 cm,
50 * 5.66 = 283 cm.