In a rectangle, if the length is increased by 3 metres and breadth is decreased by 4 metres the area of the rectangle is reduced by 67 square metres. If length is reduced by 1 metre and breadth is increased by 4 metres, the area is increased by 89 sq. metres. Find the dimensions of the rectangle.
Last updated date: 21st Mar 2023
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Answer
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Hint : Create equations on the basis of given conditions.
Let the length of the rectangle be $x$ metres and the breadth be $y$ metres.
Area of the rectangle $ = length \times breadth = x \times y = xy$ sq. metres
From the given information, we have,
\[(x + 3) \times (y - 4) = xy - 67\]
& \[(x - 1) \times (y + 4) = xy + 89 \\\]
$
(x + 3) \times (y - 4) = xy - 67 \\
= > xy - 4x + 3y - 12 = xy - 67 \\
= > 4x - 3y = 55 \\
= > 4x = 3y + 55....(i) \\
$
Also,
$
(x - 1) \times (y + 4) = xy + 89 \\
= > xy + 4x - y - 4 = xy + 89 \\
= > 4x - y = 93....(ii) \\
$
Substituting equation (i) in equation (ii), we get,
$
4x - y = 93 \\
= > 3y + 55 - y = 93 \\
= > 2y = 38 \\
= > y = 19 \\
$
Substituting $y = 19$ in equation (i), we get,
$
4x = 3y + 55 \\
= > 4x = 3(19) + 55 \\
= > 4x = 112 \\
= > x = 28 \\
$
Therefore, the length of the rectangle $ = x = 28$ metres.
breadth of rectangle $ = y = 19$ metres.
Note – In this problem, first let assume the breadth and length of the rectangle then follow the given conditions which gives us two equations. On solving these two equations, we get the final answer.
Let the length of the rectangle be $x$ metres and the breadth be $y$ metres.
Area of the rectangle $ = length \times breadth = x \times y = xy$ sq. metres
From the given information, we have,
\[(x + 3) \times (y - 4) = xy - 67\]
& \[(x - 1) \times (y + 4) = xy + 89 \\\]
$
(x + 3) \times (y - 4) = xy - 67 \\
= > xy - 4x + 3y - 12 = xy - 67 \\
= > 4x - 3y = 55 \\
= > 4x = 3y + 55....(i) \\
$
Also,
$
(x - 1) \times (y + 4) = xy + 89 \\
= > xy + 4x - y - 4 = xy + 89 \\
= > 4x - y = 93....(ii) \\
$
Substituting equation (i) in equation (ii), we get,
$
4x - y = 93 \\
= > 3y + 55 - y = 93 \\
= > 2y = 38 \\
= > y = 19 \\
$
Substituting $y = 19$ in equation (i), we get,
$
4x = 3y + 55 \\
= > 4x = 3(19) + 55 \\
= > 4x = 112 \\
= > x = 28 \\
$
Therefore, the length of the rectangle $ = x = 28$ metres.
breadth of rectangle $ = y = 19$ metres.
Note – In this problem, first let assume the breadth and length of the rectangle then follow the given conditions which gives us two equations. On solving these two equations, we get the final answer.
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