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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 !

  • 1
    $\begingroup$ We can assume that CDETSR is flat and rectangular in shape.That is, (37.1)(50) square units. $\endgroup$ – Mick Jul 31 '16 at 14:39
  • $\begingroup$ I find 283 cm.I will show my result below. $\endgroup$ – onurcanbektas Jul 31 '16 at 15:03

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.


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