The area of a rectangle gets reduced by 36 square units if its length is reduced by 6 units and breadth is increased by 2 units. If the length is reduced by 8 units
and the breadth is increased by 5 units the area decreases by 51 square units. Find the dimensions of the rectangle.
Length
units
Breadth =
units
Answers
Let length be 'm'
Let breadth be 'n'
Initial Area = (length)×(breadth) = mn
★ According to condition #1:
New length = m - 6
New breadth = n + 2
New Area = mn - 36
=> (m - 6)(n + 2) = mn - 36
=> mn + 2m - 6n - 12 = mn - 36
=> 2m - 6n = -24
=> m - 3n = -12 ............. ×(-5) ........(a)
=> -5m + 15n = 60 .................(1)
★ According to condition #2:
New length = m - 8
New breadth = n + 5
New area = mn - 51
=> (m - 8)(n + 5) = mn - 51
=> mn + 5m -8n - 40 = mn - 51
=> 5m - 8n = -11 ................(2)
Add (1) and (2):
7n = 49
=>
Put n = 7 in (a),
m - 3(7) = -12
=> m - 21 = -12
=>
Therefore,
Length = m = 9 units
Breadth = n = 7 units
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Let length be 'm'
Let breadth be 'n'
Initial Area = (length)×(breadth) = mn
★ According to condition #1:
New length = m - 6
New breadth = n + 2
New Area = mn - 36
=> (m - 6)(n + 2) = mn - 36
=> mn + 2m - 6n - 12 = mn - 36
=> 2m - 6n = -24
=> m - 3n = -12 ............. ×(-5) ........(a)
=> -5m + 15n = 60 .................(1)
★ According to condition #2:
New length = m - 8
New breadth = n + 5
New area = mn - 51
=> (m - 8)(n + 5) = mn - 51
=> mn + 5m -8n - 40 = mn - 51
=> 5m - 8n = -11 ................(2)
Add (1) and (2):
7n = 49
=>\fbox{ n = 7}
n = 7
Put n = 7 in (a),
m - 3(7) = -12
=> m - 21 = -12
=> \fbox{m = 9}
m = 9
Therefore,
Length = m = 9 units
Breadth = n = 7 units