Data Sufficiency

What is the area of rectangle ?

1. Diagonal is twice as long as side
2. Diagonal is 0.1 meters longer than side

goyalsau wrote:What is the area of rectangle ?

1. Diagonal is twice as long as side
2. Diagonal is 0.1 meters longer than side

Are you sure that its rectangle and not square??
it doesnt mention which side.

1. Diagonal is twice as long as side

Assuming, its still rectangle, and that the diagonal is twice the shorter side,

d^2 = l^2 + b^2
no info about the absolute lengths. hence Area cant be found.

2. Diagonal is 0.1 meters longer than side
d = 0.1 + b
still no absolute values... hence INSUFF

Combining:
form stmt1, we get a relationship between L and B
from stmt2 also we get a relationship between L and B.

2 eqtns, 2 variable. Hence L and B can be found, hence the area.

pick C.

I feel that the question might be missing something.

Whats OA?

kvcpk wrote:
Are you sure that its rectangle and not square??

It is a rectangle,

kvcpk wrote: it doesnt mention which side.

1. Diagonal is twice as long as side

Assuming, its still rectangle, and that the diagonal is twice the shorter side,

Can you please explain that why it is the shorter side not the longer side,
It can we D = 2l or D = 2b

kvcpk wrote: 2. Diagonal is 0.1 meters longer than side
d = 0.1 + b
still no absolute values... hence INSUFF

It can we d = 0.1 + l and will consider d = 0.1 + b

kvcpk wrote: Combining:
form stmt1, we get a relationship between L and B
from stmt2 also we get a relationship between L and B.

2 eqtns, 2 variable. Hence L and B can be found, hence the area.

I feel that the question might be missing something.

Now we have 4 different equations. Question is like this, and the OA is C, Can you please explain where I am wrong in the above reasoning.

My approach:

To calculate the rectangle area, we need to get its sides' measures. Let's say the two sides are: a and b, and the diagonal is d. a,b,d form a right triangle.
As a and b can be substitute for each other, I'll consider the relationship between a and d only.
For example
d is 5, one side of the rectangle is 4, what is the area of it?. If a = 4 then b = 3, if a = 3 then b = 4, however, the product a*b is unchanged. So I just take a = 4, the a = 3 is the reversal version ( when b takes the position of a to be 4).

1. d^2 = a^2 + b^2,
and d = 2a
We have no measure here, just the ratio. So we can only figure out the ratio db with such given information.
Insuff.

2. d^2 = a^2 + b^2
and d = a + 0.1
Plug in d = a+0.1 into the equation:
(a + 0.1)^2 = a^2 + b^2 => 1 equation, 2 unknowns, we can only get a relationship between a and b
or a = f(b) or b = f(a)
Insuff.

1&2.
Given information:
d = 2a
d = a +0.1 or d = b + 0.1

d = 2a gives us the shape of the triangle, the ratio among a,b,d is fixed.
The shape has been fixed, now we can zoom up or zoom down the triangle to let:

either d = a + 0.1
or d = b + 0.1

Thus, we have two different sets of value, so the area of the rectangle can have 2 different possible values

To make it clearer take this example:

The ratio of d to one side of the triangle is 5/3,
d is 2 units longer than 1 one side, what is the area of the triangle?

You may first think of : 5,3,4 as a set of value ( 5 - 3 = 2)
However, 10,6,8 also a possible set of value ( 10 - 8 = 2)
The first area = 3*4 /2 = 6
The second = 6*8/2 = 24
Now, you get what I mean.
1&2 is also insuff.

Pick E.

Please cite the source and OA's explanation for C.

IMO C.

Let D be the diagonal of rectangle
B be the width (< Length)
L be the length

so B<L
and D^2 = B^2 + L^2

1:Diagonal is twice as long as side

Say if you take D = 2L then
D^2 = B^2 + L^2
=> (2L)^2 = B^2 + L^2
=> 4L^2 = B^2 + L^2
=> 3L^2 = B^2 Thus B seems greater than L. But we have taken B<L (Not possible)

So D has to be multiple of B not L.

Thus 3B^2 = L^2

So Area of rectangle is not found (Not Sufficient)

2. Diagonal is 0.1 meters longer than side
D = 0.1 + L
So Area of rectangle is not found (Not Sufficient)

Combine,

D = 0.1 + L => 2B = 0.1 + L => L = 2B - 0.1
Put in 3B^2 = (2B - 0.1)^2
We can solve this and get B = 0.32 (approx)
and L = 0.54(approx)
and D = 0.6 (approx)

Thus Both sufficient.

goyalsau wrote:

kvcpk wrote: it doesnt mention which side.

1. Diagonal is twice as long as side

Assuming, its still rectangle, and that the diagonal is twice the shorter side,

Can you please explain that why it is the shorter side not the longer side,
It can we D = 2l or D = 2b

Let D be diagonal, B be width and L be length
So B<L
Now if you take D = 2L then
D^2 = B^2 + L^2
=> (2L)^2 = B^2 + L^2
=> 4L^2 = B^2 + L^2
=> 3L^2 = B^2 Thus B > L. But we have taken B<L (Not possible)

So D has to be multiple of B not L.

sorry for the delayed response. I think Shovan has answered your queries in clear terms.
Let me know if you still have any questions.

Cheers!!

goyalsau wrote:What is the area of rectangle ?

1. Diagonal is twice as long as side
2. Diagonal is 0.1 meters longer than side

The question is definitely missing some words - neither the stem nor the statements are grammatically correct. Since every word can make a difference, it's imperative that you reproduce questions verbatim.

For example, we don't know if statements (1) and (2) reference the same side of the rectangle. If they do, then the answer is (C); if they don't, then the answer is (E).

Let's call the diagonal "d", the long side "l" and the short side "w".

Q: what's l*w?

We're asked to solve for a measurement and we haven't been given any numbers. So, to solve we need at least 1 actual number. We also have 2 variables and 0 equations.

(1) no actual numbers provided, therefore automatically insufficient.

(2) Either d = l + 0.1 OR d = w + .01. Either way, only 1 equation for our 2 unknowns - insufficient.

Combined:

From (1), we know that d = 2w.

(Let's think logically about triangles. We know that each side has to be smaller than the sum of the other two sides. If the diagonal were double the length of the longer side, then we'd have d > l + w, which is impossible. Accordingly, d must be double w, not l.)

So, applying thy pythagorean theorem and substituting 2w for d, we know that:

(2w)^2 = l^2 + w^2

4w^2 = l^2 + w^2

3w^2 = l^2

(Or, if you recognize common ratios, you see that we have an x:xroot3:2x or 30/60/90 triangle.)

From (2), we know that either:

d = l + 0.1

OR

d = w + 0.1

we can substitute d = 2w into each equation:

2w = l + 0.1

OR

2w = w + 0.1

However, since each equation (combined with 3w^2 = l^2) will give us a different answer for l*w, we don't have enough information to answer the question.

If, on the other hand, we knew that the two statements were referring to the same side of the rectangle, we'd have two equations and two unknowns and could solve.