The area of a rectangle is numerically equal to twice its perimeter and both length and breadth are integers. How many such rectangles are possible?
Answers
Only one rectangle will be possible.
Step-by-step explanation:
Given: The area of a rectangle is numerically equal to twice its perimeter and both length and breadth are integers.
Find: How many such rectangles are possible?
Solution:
Let l be the length and b be the breadth of the rectangle. l and b are integers.
Area = l * b
Perimeter = 2(l+b)
Given l*b = 2*2(l+b)
lb = 4l + 4b
1 = 4/b + 4/l
4/l = 1 - 4/b
4 = l( b - 4 )/ b
4 = l - 4l/b
b = l-4l / 4
Breadth of the rectangle = (length - 4length)/4
Only one rectangle will be possible.
Given : The area of a rectangle is numerically equal to twice its perimeter and both length and breadth are integers
To find : How many such rectangles are possible
Solution:
Length Say Length = L
& Breadth = B
Perimeter = 2(L + B)
Area = LB
LB = 2 * 2(L + B)
=> LB = 4L + 4B
=> 4L = LB - 4B
=> 4L = B ( L - 4)
=> B = 4L/(L - 4) or L = 4B/(B - 4)
if L < 4 as then B will be - ve
=> L > 4 & B > 4
L = 5
=> B = 20
Perimeter = 50 Area = 100
L = 6
=> B = 12
Perimeter = 36 Area = 72
L = 8
=> B = 8
Perimeter = 32 Area = 64
Hence 3 Rectangle Possibles
( 5 , 20) & ( 6 , 12) & ( 8 , 8 )
Where area of a rectangle is numerically equal to twice its perimeter
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