1wel it makes no sense naa......
A has double area
B has double perimeter
so its obvious tht there are no solutions......
Q) Two rectangles are given.The Area of the 1st rectangle is double the area of the 2nd rectangle.The Perimeter of 2nd rectangle is double the perimeter of the 1st rectangle.Then Find the Sides of the 2 rectangles?
Plz give the answer with solution!!!!
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9 Answers
62Q) Two rectangles are given.The Area of the 1st rectangle is double the area of the 2nd rectangle.The Perimeter of 2nd rectangle is double the perimeter of the 1st rectangle.Then Find the Sides of the 2 rectangles?
ab=2cd
a+b=2(c+d)
a2+b2+2ab = 4c2+4d2+8cd
substract 4ab on left and 8 cd on the right side, we get
a2+b2-2ab = 4c2+4d2
(a-b)= 2 √c2+d2
Thus, a=(c+d)+√c2+d2
and b=(c+d)-√c2+d2
for different values of c and d.
that is all I guess!
1The Area of the 1st rectangle is double the area of the 2nd rectangle.The Perimeter of 2nd rectangle is double the perimeter of the 1st rectangle.
bhaiyya.....check ur ans again
62Q) Two rectangles are given.The Area of the 1st rectangle is double the area of the 2nd rectangle.The Perimeter of 2nd rectangle is double the perimeter of the 1st rectangle.Then Find the Sides of the 2 rectangles?
2ab=cd
a+b=2(c+d)
a2+b2+2ab = 4c2+4d2+8cd
substract 4ab on left and 2 cd on the right side, we get
a2+b2-2ab = 4c2+4d2+6cd
(a-b)= 2 √c2+d2+1.5cd
Thus, a=(c+d)+√c2+d2+1.5cd
and b=(c+d)-√c2+d2+1.5cd
Hence infinite solutions given by the above :)
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1hmmm.........hww.....der is no solution bhaiyya........1/2 can also be der naaa...........
62what does 1/2 can be der also!!
do you have an example?
May be I am wrong!
9yes its got to have infinite sols as the scale can be adjusted proportionately always
i think Q demands c d in terms of a b