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A. $30\,{{m}^{2}}$

B. $40\,{{m}^{2}}$

C. $50\,{{m}^{2}}$

D. $60\,{{m}^{2}}$

Answer

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Step 1: Rate of A=$52\,m/\min $ =$\dfrac{52}{60}\,m/\sec $=$\dfrac{13}{15}\,m/\sec $

Rate of B=$68\,m/\min $ =$\dfrac{68}{60}\,m/\sec $ =$\dfrac{17}{15}\,\,m/\sec $

Step 2: Distance travelled by A= $Rate\,of\,A\,\times \,\,time\,taken$=$\dfrac{13}{15}\,\times 15$ =$13\,m$

Distance travelled by B= $Rate\,of\,B\,\times \,time\,taken$=$\dfrac{17}{15}\,\times 15$ =$17\,m$

Step 3: Since B crosses the field along its sides therefore, it is basically sum of two sides of rectangle i.e

$x+y=17$

Squaring both sides we get,

\[\begin{align}

& \\

& {{(x+y)}^{2}}={{(17)}^{2}} \\

& {{x}^{2}}+{{y}^{2}}+2xy={{(17)}^{2}} \\

& {{(D)}^{2}}+2xy={{(17)}^{2}}\,\,\,[\because {{x}^{2}}+{{y}^{2}}={{D}^{2}}]\, \\

&2xy={{(17)}^{2}}\,-{{(D)}^{2}} \\

&2xy={{(17)}^{2}}\,-{{(13)}^{2}}\,\,\,\,\,\,[\because D\,is\,dis\tan ce\,traveled\,by\,A]\,\,\, \\

&2xy=289-169 \\

&2xy=120 \\

&xy=60{{m}^{2}} \\

& \\

& \, \\

\end{align}\]

∴$xy$ is the area of the rectangular field. It comes out to be $60\,{{m}^{2}}$