The diagonal of a rectangular paper is 10 cm. The length and width of the paper are changed. What is the diagonal of the new rectangle?
(1) The new area is ½ the original area
(2) The new perimeter is 3/4th the original perimeter
DS Rectangle question
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is it E?
let l and b are the initial length and breadth
l^2+b^2=100
let l' and b' are the new length and breadth
1) l'b'=lb/2
not sufficient. We cannot find l'^2+b'^2
2)2(l' +b')=3/4*2*(l+b)
l'+b'=3/4*(l+b). not sufficient to find l'^2 +b'^2
combined
l'^2+b'^2=(l'+b')^2  2 l'b'
=9/16(l+b)^2 lb
=9/16(l^2+b^2+2lb) lb
=9/16(100+2lb) lb
225/4+ lb/8
still not sufficient
let l and b are the initial length and breadth
l^2+b^2=100
let l' and b' are the new length and breadth
1) l'b'=lb/2
not sufficient. We cannot find l'^2+b'^2
2)2(l' +b')=3/4*2*(l+b)
l'+b'=3/4*(l+b). not sufficient to find l'^2 +b'^2
combined
l'^2+b'^2=(l'+b')^2  2 l'b'
=9/16(l+b)^2 lb
=9/16(l^2+b^2+2lb) lb
=9/16(100+2lb) lb
225/4+ lb/8
still not sufficient
The powers of two are bloody impolite!!

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OA is Cern5231 wrote:The diagonal of a rectangular paper is 10 cm. The length and width of the paper are changed. What is the diagonal of the new rectangle?
(1) The new area is ½ the original area
(2) The new perimeter is 3/4th the original perimeter
special triangles(306090)>1:sqrt(3):2
from given info: 10=diagonal and length =5sqrt(3) and width=5
statement 1)
NA=1/2 OA
NA=1/2 x 5 sqrt(3) x 5 = 25sqrt(3)/2
new lengthx new width=25sqrt(3)/2
but in order to calculate new diagonal we need length and new width
insufficient
statement 2)
NP=3/4OP
OP=2*5sqrt(3) + 2*5=10sqrt(3)+10
NP=3/4*10sqrt(3)+10
2*new length + 2*new width=3/4*10sqrt(3)+10
but in order new diagonal we need new length and new width
insufficient
combined
system of equations/substitute method >solvable
you got this man!

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 Posts: 752
 Joined: 17 May 2009
 Location: Tokyo
 Thanked: 81 times
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shah,shahdevine wrote:OA is Cern5231 wrote:The diagonal of a rectangular paper is 10 cm. The length and width of the paper are changed. What is the diagonal of the new rectangle?
(1) The new area is ½ the original area
(2) The new perimeter is 3/4th the original perimeter
special triangles(306090)>1:sqrt(3):2
from given info: 10=diagonal and length =5sqrt(3) and width=5
statement 1)
NA=1/2 OA
NA=1/2 x 5 sqrt(3) x 5 = 25sqrt(3)/2
new lengthx new width=25sqrt(3)/2
but in order to calculate new diagonal we need length and new width
insufficient
statement 2)
NP=3/4OP
OP=2*5sqrt(3) + 2*5=10sqrt(3)+10
NP=3/4*10sqrt(3)+10
2*new length + 2*new width=3/4*10sqrt(3)+10
but in order new diagonal we need new length and new width
insufficient
combined
system of equations/substitute method >solvable
you got this man!
OA might be C.
but I don't agree with your method. You cannot assume a certain rectangle here because IMO that question is based on this fact that "there can be a lot of rectangles with diagonal 10".
for example l=6,b=8
The powers of two are bloody impolite!!
how do u find the highlighted statement?shahdevine wrote:OA is Cern5231 wrote:The diagonal of a rectangular paper is 10 cm. The length and width of the paper are changed. What is the diagonal of the new rectangle?
(1) The new area is ½ the original area
(2) The new perimeter is 3/4th the original perimeter
special triangles(306090)>1:sqrt(3):2
from given info: 10=diagonal and length =5sqrt(3) and width=5
statement 1)
NA=1/2 OA
NA=1/2 x 5 sqrt(3) x 5 = 25sqrt(3)/2
new lengthx new width=25sqrt(3)/2
but in order to calculate new diagonal we need length and new width
insufficient
statement 2)
NP=3/4OP
OP=2*5sqrt(3) + 2*5=10sqrt(3)+10
NP=3/4*10sqrt(3)+10
2*new length + 2*new width=3/4*10sqrt(3)+10
but in order new diagonal we need new length and new width
insufficient
combined
system of equations/substitute method >solvable
you got this man!

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 Posts: 197
 Joined: 18 May 2008
 Thanked: 12 times
special triangles occur whenever you have a right triangle. A right triangle with a 30° angle or 60° angle must be a 30° 60° 90° special right triangle and corresponds to length ratios of 1:sqrt(3):2. However, that said it might have been wrong for me to assume that the diagonal cuts the rectangle into a 306090 ratio, even although there is triangle is a right triangle.real2008 wrote:how do u find the highlighted statement?shahdevine wrote:OA is Cern5231 wrote:The diagonal of a rectangular paper is 10 cm. The length and width of the paper are changed. What is the diagonal of the new rectangle?
(1) The new area is ½ the original area
(2) The new perimeter is 3/4th the original perimeter
special triangles(306090)>1:sqrt(3):2
from given info: 10=diagonal and length =5sqrt(3) and width=5
statement 1)
NA=1/2 OA
NA=1/2 x 5 sqrt(3) x 5 = 25sqrt(3)/2
new lengthx new width=25sqrt(3)/2
but in order to calculate new diagonal we need length and new width
insufficient
statement 2)
NP=3/4OP
OP=2*5sqrt(3) + 2*5=10sqrt(3)+10
NP=3/4*10sqrt(3)+10
2*new length + 2*new width=3/4*10sqrt(3)+10
but in order new diagonal we need new length and new width
insufficient
combined
system of equations/substitute method >solvable
you got this man!