Concept:
(i) In general, the length of any rectangle is always greater than its width.
(ii) Area of rectangle (A) = L × B
(iii) Perimeter of rectangle (P) = 2 ( L + B)
Given: Area of rectangle 'A' = 400 square meters
Let 'L' and 'B' be the length and width of the given rectangle.
According to the problem,
L = 3 B - 23 ------------------(1)
Area of the rectangle = 400 m²
or, L× B = 400 m²
or, (3 B - 23 )×B = 400
or, 3 B² - 23 B - 400 = 0
Apply, discriminant method,
Here, we will calculate only the positive value of B because width or length will never be negative.
[tex] B = \frac{23 +\sqrt{23^{2}+(4)(3)(400)}}{2(3)}\\ B = 16 [/tex]
Now, using equation (1),
L = 3 B -23 = 3×16 - 23 = 25 m
Hence, length (L) = 25 m and Width (B) = 16 m
Now, we shall calculate the perimeter (P) of the rectangle
P = 2( L + B)
or, P = 2 (25 + 16 ) m
or, P = 82 m
Hence, the required perimeter of the given rectangle will be 82 m
