The perimeter of a rectangle is 82 m and its area is 400 m2.

Solution:

(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,

$\therefore b=16 \mathrm{~m}$