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ML Aggarwal Solutions Class 9 Mathematics Solutions for Problems on Simultaneous Linear Equations Exercise 6 in Chapter 6 - Problems on Simultaneous Linear Equations

Question 14 Problems on Simultaneous Linear Equations Exercise 6

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.

Answer:

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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