Rectangles are drawn on line segments of fixed lengths. When the breadths are 6 m and 5 m
respectively the sum of the areas of the rectangles is 83 m². But if the breadths are 5 m and 4 m
respectively the sum of the areas is 68 m². Find the sum of the areas of the squares drawn on the
line segments.
Let’s consider the length of the room as x meter
And the breadth of the room = y meter
So, the area of the room = length x breadth = xy sq. meter
Now, according to the first condition given in the problem, we have
xy = (x + 1) (y + 1) + 1 – 21
xy = xy + x + y + 1 – 21
x + y = 20 … (i)
And, according to the second condition, we have
xy = (x - 1) (y + 2) – 14
xy = xy – y + 2x – 2 – 14
2x – y = 16 … (ii)
Adding (i) and (ii), we get
x + y = 20
2x – y = 16
3x = 36
x = 36/3
x = 12
On substituting the value of x in equation (i), we get
12 + y = 20
y = 20 – 12
y = 8
Thus, the length of the room = 12 m and the breadth of the room = 8 m
Therefore, the perimeter of the room = 2 x (length + breadth)
= 2 x (12 + 8) = 2 x 20
= 40 m
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