Question

Question 58

(a) Find all the possible dimensions (in natural numbers) of a rectangle with a perimeter 36 cm and find their areas.

(b) Find all the possible dimensions (in natural numbers) of a rectangle with an area of 36 sq cm and find their perimeters.

Answer 1

(a) Let x and y be the length and breadth of the given rectangle whose perimeter is 36 cm

$\therefore$ Perimeter of rectangle = 2(Length + Breadth) = 2(x + y)

$\Rightarrow$ 2(x + y) = 36

$\Rightarrow$ $\frac{2x+y}{2}=\frac{36}{2}$ [dividing both sides by 2]

$\Rightarrow$ x + y = 18

Now, if the value of both length and breadth are natural numbers, then possible dimensions and thier areas are as follows

$\begin{array}{cc}\text{Dimensions(in cm)}& \text{Area(in cm)}\\ 17\times 1& 17\\ 16\times 2& 32\\ 15\times 3& 45\\ 14\times 4& 56\\ 13\times 5& 65\\ 12\times 6& 72\\ 11\times 7& 77\\ 10\times 8& 80\\ 9\times 9& 81\end{array}$

(b) Let x and y be the length and breadth of the given rectangle whose area is 36 sq cm.

$\therefore$ Area of rectangle = Length$\times$Breadth = x $\times$ y

$\Rightarrow$ x $\times$ y = 36 sq cm

If the value of both length and breadth are natural numbers, then possible dimensions and their perimeters are as follows

$\begin{array}{cc}\text{Dimensions(in cm)}& \text{Perimeter(in cm)}\\ 36\times 1& 74\\ 18\times 2& 40\\ 9\times 4& 26\\ 12\times 3& 30\\ 6\times 6& 24\end{array}$