The perimeter of a rectangle is 26 cm. The length and the breadth of the rectangle are integers. Find the number of possible pairs of length and breadth. Also find the maximum possible area of the rectangle.

The correct answer is: Total 6 pairs of length & breadth in cm are possible. Also, maximum possible area of the rectangle is 42 cm2.

Hint:-
Perimeter of a rectangle = 2 (length + breadth)
Area of a rectangle = length × breadth
Step-by-step solution:-
We know that Perimeter of a rectangle = 2 (length + breadth)
∴ Perimeter of the given rectangle = 2 (length + breadth)
∴ 26 = 2 (length + breadth)
∴ 26 / 2 = length + breadth
∴ 13 = length + breadth
i.e. length + breadth = 13 cm ..................................... (Equation i)
We need to find all the possible combination of lengths & breadths that would add up to 13 i.e. satisfy Equation i.
∴ The possible combinations are-
(Length, breadth) = (12,1); (11,2); (10,3); (9,4); (8,5); (7,6)
∴ Total 6 pairs of length & breadth in cm are possible.
Area of rectangle at (length, breadth) = (12,1) = length × breadth = 12 × 1 = 12 cm2
Area of rectangle at (length, breadth) = (11,2) = length × breadth = 11 × 2 = 22 cm2
Area of rectangle at (length, breadth) = (10,3) = length × breadth = 10 × 3 = 30 cm2
Area of rectangle at (length, breadth) = (9,4) = length × breadth = 9 × 4 = 36 cm2
Area of rectangle at (length, breadth) = (8,5) = length × breadth = 8 × 5 = 40 cm2
Area of rectangle at (length, breadth) = (7,6) = length × breadth = 7 × 6 = 42 cm2
∴ The maximum possible area of the given rectangle is 42 cm2.
Final Answer:-
∴ Total 6 pairs of length & breadth in cm are possible. Also, maximum possible area of the rectangle is 42 cm2.
Note:-
Theoretically, (1,12); (2,11); (3,10); (4,9); (5,8); (6,7) also add up to 13 but for a rectangle, length should always be greater than breadth.
∴ these combinations are not possible for the given question.