The perimeter of a rectangle is 82 m and its area is 400 m2. The breadth of the rectangle is
(a) 25 m
(b) 20 m
(c) 16 m
(d) 9 m
(c) 16 m
Let the length and breadth of the rectangle be $l$ and $b$.
Perimeter of the rectangle $=82 \mathrm{~m}$
$\Rightarrow 2 \times(l+b)=82$
$\Rightarrow l+b=41$
$\Rightarrow l=(41-b)$ $\ldots(\mathrm{i})$
Area of the rectangle $=400 \mathrm{~m}^{2}$
$\Rightarrow l \times b=400 \mathrm{~m}^{2}$
$\Rightarrow(41-b) b=400 \quad($ using $(\mathrm{i}))$
$\Rightarrow 41 b-b^{2}=400$
$\Rightarrow b^{2}-41 b+400=0$
$\Rightarrow b^{2}-25 b-16 b+400=0$
$\Rightarrow b(b-25)-16(b-25)=0$
$\Rightarrow(b-25)(b-16)=0$
$\Rightarrow b=25$ or $b=16$
If $b=25$, we have :
$l=41-25=16$
Since, $l$ cannot be less than b,
$\therefore b=16 \mathrm{~m}$