Question

the sum of length and breadth of a rectangle is 15 metres. if the length is decrease by 2 metres and breadth increase by 1metre it, become a square find length and breadth of rectangle with the help of graph​

Answer 1

length and breadth of the rectangle = 7meter and 8 meters respectively

Answer 2

$\large\underline{\sf{Solution-}}$

$\begin{gathered}\begin{gathered}\bf\: Let-\begin{cases} &\sf{Length_{(rectangle)} = \: x \: m} \\ &\sf{Breadth_{(rectangle)} = \: y \: m} \end{cases}\end{gathered}\end{gathered}$

Case :- 1

According to statement,

The sum of length and breadth of a rectangle is 15 metres.

$\rm :\longmapsto\:x + y = 15 - - (1)$

Case :- 2

The length is decrease by 2 metres and breadth increase by 1metre.

We get,

$\begin{gathered}\begin{gathered}\bf\: Now-\begin{cases} &\sf{Length_{(rectangle)} = \: (x - 2) \: m} \\ &\sf{Breadth_{(rectangle)} = \: (y + 1) \: m} \end{cases}\end{gathered}\end{gathered}$

According to statement,

Rectangle becomes a square.

$\rm :\longmapsto\:Length_{(rectangle)} = Breadth_{(rectangle)}$

$\rm :\longmapsto\:x - 2 = y + 1$

$\rm :\implies\:x - y = 3 - - - (2)$

Now,

we have two linear equations,

$\sf \: \: \: \: \: \: \: \bull \: \: \: x + y = 15 - - - (1)$

and

$\sf \: \: \: \: \: \: \: \bull \: \: \: x - y = 3 - - - (2)$

Consider the equation of line x + y = 15

1. Substituting 'x = 0' in the given equation, we get

$\rm :\longmapsto\:0 + y = 15$

$\bf\implies \:y = 15$

2. Substituting 'y = 0' in the given equation, we get

$\rm :\longmapsto\:x + 0 = 15$

$\bf\implies \:x = 15$

Hᴇɴᴄᴇ,

➢ Pair of points of the given equation are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 15 \\ \\ \sf 15 & \sf 0 \end{array}} \\ \end{gathered}$

➢ Now draw a graph using the points (0 , 15) & (15 , 0)

➢ See the attachment graph. (Red line)

Consider the other line x - y = 3

1. Substituting 'x = 0' in the given equation, we get

$\rm :\longmapsto\:0 - y = 3$

$\bf\implies \:y = - 3$

2. Substituting 'y = 0' in the given equation, we get

$\rm :\longmapsto\:x - 0 = 3$

$\bf\implies \:x = 3$

Hᴇɴᴄᴇ,

➢ Pair of points of the given equation are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf - 3 \\ \\ \sf 3 & \sf 0 \end{array}} \\ \end{gathered}$

➢ Now draw a graph using the points (0 , - 3) & (3 , 0)

➢ See the attachment graph. (Purple line)

From graph, we conclude that the two lines intersect each other at (9, 6).

$\begin{gathered}\begin{gathered}\bf\: Hence-\begin{cases} &\sf{Length_{(rectangle)} = \: 9\: m} \\ &\sf{Breadth_{(rectangle)} = \: 6 \: m} \end{cases}\end{gathered}\end{gathered}$