A rectangle with sides (2m1) cm X (2n1) cm is divided in to
squares by drawing parallel lines in such a way that distance between two
consecutive parallel lines is 1 cm.
The
number of rectangles that can be drawn on the figure such that both the length
and breadth are in odd cm is:
(a) 4^{m+n+1}
(b) mn(m+1)(n+1) (c) m^{2}n^{2}
(d) (m+n+1)^{2}
Solution follows
here:
Solution:
For this problem, we go in Reverse Engineering Method. We go
from answer options.
We take diagrams with smaller
dimensions and find out the possible number of rectangles and then check out
with the answer options.
As (2m1) and (2n1) are odd numbers,
we need to take examples such that topmost horizontal line and rightmost
vertical lines are tagged with odd numbers.
Simplest example:
1


0

1

2

3

In this case, 2m1 = 3 => m=2; 2n1 = 1 => n=1
Rectangles with oddcm dimensions:
1cmX1cm => 3 numbers
3cmX1cm => 1 number
∴Number of rectangles with oddcm
dimensions = 4
From all the multiple choices the only possible option is (c)
=> m^{2}n^{2} = 2^{2}1^{2} = 4
So answer is (c) only.
But let us repeat the exercise with one more example:
1


0

1

2

3

4

5

In this case, 2m1 = 5 => m=3; 2n1 = 1 => n=1
Rectangles with oddcm dimensions:
1cm X 1cm => 5 numbers
3cm X 1cm => 3 numbers
5cm X 1cm => 1 number
∴Total number of rectangles
with oddcm dimensions = 5+3+1 = 9
Even for this example, from all the multiple choices the only
possible option is (c)
=> m^{2}n^{2} = 3^{2}1^{2} =
9
Answer(c)
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