Question

Two lawns have the same area, $121$ sq. m. One is a square and the other is a rectangle. Four times the perimeter of the rectangle is equal to five times that of the square. The length and breadth of the rectangle are

• A. $11$ m, $11$ m
• B. $22$ m, $5\frac{1}{2}$ m
• C. $55$ m, $2\frac{1}{2}$ m
• D. $33$ m, $3\frac{2}{3}$ m

Answer 1

Answer: (B)

$22$ m, $5\frac{1}{2}$ m

Let the side of the square be $x$.
Hence, $x^2=121$
$\therefore x=\pm 11$
Hence, side length of the square is $11$ m.
Now let $l$ and $b$ be the length and breadth of the rectangle.
Hence, $4\left(2\right(l+b\left)\right)=5(4\times 11)$

$2(l+b)=55$

$l=\frac{55}{2}-b$

And Area is $=121=l.b$

$=b(\frac{55}{2}-b)$

$\therefore 55b-2b^2=242$

$2b^2-55b+242=0$

$b=22m$ and $b=\frac{11}{2}m$

If $b=22m$ then $l=\frac{55-44}{2}$= $\frac{11}{2}m$