DS Rectangle question

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

ern5231 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

OA is C

special triangles(30-60-90)-->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!

shahdevine wrote:

ern5231 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

OA is C

special triangles(30-60-90)-->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!

shah,
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

I am still not clear with the explanation. Can any of the instructors help me with this?

shahdevine wrote:

ern5231 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

OA is C

special triangles(30-60-90)-->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!

how do u find the highlighted statement?

real2008 wrote:

shahdevine wrote:

ern5231 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

OA is C

special triangles(30-60-90)-->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!

how do u find the highlighted statement?

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 30-60-90 ratio, even although there is triangle is a right triangle.