Question

A wire would enclose an area of 1936 $m^2$, if it is bent into a square. The wire is cut into two pieces. The longer piece is thrice as long as the shorter piece. The long and the short pieces are bent into a square and a circle, respectively. Which of the following choices is closest to the sum of the areas enclosed by the two pieces in square meters?

• A.
  1096
• B.
  1111
• C.
  1243
• D.
  2486

Answer 1

The correct option is C

1243
$Area=1936m^2$
$a^2=1936m^2$
$a=44m$

$Lengthofwire=4a$
$=4\times 44=176m$

$Now,Part-1length$
$=3\times 44=132m$

$Part-2length$
$=1\times 44=44m$

$Longwireisbentinsquare$

$So,4a=132$
$a=33m$

$Areaofsquare={33}^2=1089m^2$

$Now,smallwireisbentintoacircle$

$So,$
$\pi D=44$
$\frac{22}{7}\times D=44$
$D=14m$

$Areaofcircle$
$=\frac{\pi }{4}\times D^2=\frac{\pi }{4}\times {14}^2$

$=153.94m^2$

$Totalareaenclosed$

$=AreaofSquare+AreaofCircle$

$=1089+153.94$

$=1242.97=1243m^2$