Area Calculation MCQ 10 (Carpet in rectangle form)

But according to formula of area of rectangle ;

\[ Area Of Rectangle= Length(Longer side)\times Breadth (Shorter side) \]

i-e

\[ 60 = L\times S \]

or

\[ S=\frac{60}{L} \]

or

\[ L=\frac{60}{S} —-(2 \]

Now consider the right angled triangle ABC. Here Pythagoras theorem says;

\[ (AC)^2=(AB)^2+(BC)^2 \]

but

AC=D, AB=L and BC= S    so we can write;

\[ D^2=L^2+S^2 \]

\[ D=\sqrt(L^2+S^2) \]

Put this value of D in eq 1, we get;

\[ L+\sqrt(L^2+S^2=5S \]

\[ ==> \sqrt(L^2+S^2)=5S-L \]

Taking square on both sides, we get;

\[ L^2+S^2=(5S-L)^2 \]

\[ L^2+S^2=L^2+25S^2-10SL \]

\[ ==>24S^2-10SL=0 \]

Taking 2 S as common;

\[ ==> 2S(12S-5L)=0 \]

\[ L=\frac{12S}{5}  —(3 \]

Comparing eq 2 and 3

\[ \frac{12S}{5}=\frac{60}{S} \]

or

\[ s^2=25 \]

or

\[ S=5 \]

Here we ignore -5 because length can not be negative.