Question
Half the perimeter of a rectangular garden, whose length is $4\ m$ more than its width, is $36\ m$. find the dimensions of the garden.
✓
Answer
Let the dimensions (i.e., length and width) of the garden be $x$ and $y\ m$ respectively.
Then, $x = y + 4$ and $\frac{1}{2}(2x + 2y) = 36$
$\Rightarrow x - y = 4 ...(1)$
$x + y = 36 ...(2)$
Let us draw the graphs of equations $(1)$ and $(2)$ by finding two solutions for each of the equations. These two solution of the equations $(1)$ and $(2)$ are given below in table $1$ and table $2$ respectively.
For equation $(1)$
$x - y = 4$
$\Rightarrow y = x - 4$
Table $1$ of solutions
For equation $(2) x + y = 36$
$\Rightarrow y = 36 - x$
Table $2$ of solutions
We plot the points $A(4, 0)$ and $B(2, -2)$ on a graph paper and join these points to form the line $AB$ representing. The equation $(1)$ as shown in the figure.
Also, we plot the points $C(20, 16)$ and $D(16, 20)$ on the same graph paper and join these points to form the line $CD$ representing the equation $(2)$ as shown in the same figure.
In the figure, we observe that the two lines intersect at the point $C(20, 16)$ So $x = 20, y = 16$ is the required solution of the pair of linear equations formed. i.e., the dimensions of the garden are $20\ m$ and $16\ m.$
Verification : substituting $x = 20$ and $y = 16$ in $(1)$ and $(2),$ we find that both the equations are satisfied as shown below:
$20 - 16 = 4$
$20 + 16 = 36$
This verifies the solution.
Then, $x = y + 4$ and $\frac{1}{2}(2x + 2y) = 36$
$\Rightarrow x - y = 4 ...(1)$
$x + y = 36 ...(2)$
Let us draw the graphs of equations $(1)$ and $(2)$ by finding two solutions for each of the equations. These two solution of the equations $(1)$ and $(2)$ are given below in table $1$ and table $2$ respectively.
For equation $(1)$
$x - y = 4$
$\Rightarrow y = x - 4$
Table $1$ of solutions
| $x$ | $4$ | $2$ |
| $y$ | $0$ | $-2$ |
$\Rightarrow y = 36 - x$
Table $2$ of solutions
| $x$ | $20$ | $16$ |
| $y$ | $16$ | $20$ |
Also, we plot the points $C(20, 16)$ and $D(16, 20)$ on the same graph paper and join these points to form the line $CD$ representing the equation $(2)$ as shown in the same figure.
In the figure, we observe that the two lines intersect at the point $C(20, 16)$ So $x = 20, y = 16$ is the required solution of the pair of linear equations formed. i.e., the dimensions of the garden are $20\ m$ and $16\ m.$
Verification : substituting $x = 20$ and $y = 16$ in $(1)$ and $(2),$ we find that both the equations are satisfied as shown below:
$20 - 16 = 4$
$20 + 16 = 36$
This verifies the solution.
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