Question

The perimeter of a rectangle is

36

metres and the length is

2

metres more than three times the width. Find the dimension of rectangle by using the method of graph.

Accepted Answer

Let us form equations for the given statement.

Let us consider $l$ and $b$ as the length and breadth of the rectangle respectively.

Now let us frame the equation for the first statement

Perimeter of rectangle $= 36$

$2 (l + b) = 36$

$l + b = \frac{36}{2}$

$l = 18 - b . . . . . (1)$

$b$ $2$ $4$ $5$ $8$

$18$ $18$ $18$ $18$ $18$

$- b$ $- 2$ $- 4$ $- 5$ $- 8$

$l = 18 - b$ $16$ $14$ $13$ $10$

Points: $(2, 16), (4, 14), (5, 13), (8, 10)$

The second statement states that the length is $2$ metres more than three times the width which is a straight line written as $l = 3 b + 2 . . . . . . (2)$

Now we shall form table for the above equation $(2)$ .

$b$ $2$ $4$ $5$ $8$

$3 b$ $6$ $12$ $15$ $24$

$2$ $2$ $2$ $2$ $2$

$l = 3 b + 2$ $8$ $14$ $17$ $26$

Points: $(2, 8), (4, 14), (5, 17), (8, 26)$

Both the lines are intersecting. The solution is the point that is common to both the lines. Here we find it to be $(4, 14)$ . We can give the solution to be $b = 4, l = 14$ .

Verification:

$2 (l + b) = 36 . . . . . (1)$

$2 (14 + 4) = 36$

$2 \times 18 = 36$

$36 = 36$ true

$l = 3 b + 2 . . . . . (2)$

$14 = 3 (4) + 2$

$14 = 12 + 2$

$14 = 14$ true

Tamil Nadu Board Solutions for Chapter: Algebra, Exercise 47: Solved Example

Tamil Nadu Board Mathematics Solutions for Exercise - Tamil Nadu Board Solutions for Chapter: Algebra, Exercise 47: Solved Example

Questions from Tamil Nadu Board Solutions for Chapter: Algebra, Exercise 47: Solved Example with Hints & Solutions

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