What is the area of the rectangular region above?

What is the area of the rectangular region above?

\(area=lw=?\)

(1) l + w = 6. Not sufficient to get the value of \(lw\).
(2) d^2 = 20 --> \(l^2 +w^2 = 20\). Not sufficient to get the value of \(lw\).

(1)+(2) Square (1): \(l^2+2lw+w^2=36\), as from (2) \(l^2 +w^2 = 20\) then \(2lw+20=36\) --> \(lw=8\). Sufficient.

Answer: C.



Usually the diagonal does not divide a rectangle into two 45-45-90 triangles (it'll be correct only for squares, so when l=w).

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(1)
\(l+w = 6\)
\((l+w)^2 = l^2+w^2+2w*l=36\)
\(w*l=\frac{36-l^2-w^2}{2}\)
Not Sufficient.

(2)
\(d^2=l^2+w^2=20\)
Not Sufficient.

Combining both;
\(w*l=\frac{36-(l^2+w^2)}{2}\)
\(w*l=\frac{36-20}{2}\)
\(w*l=\frac{16}{2}\)
\(w*l=8\)(Area is w*l)

Sufficient.

Ans: "C"

Need to find area = l*w
1. l+w=6 --> sqrn both sides --> l^2 + w^2 + 2wl=36 -eqn(a)
INSUFFICIENT
2. d^2=20 --> l^2+w^2=d^2=20 -eqn(b)
INSUFFICIENT

Consider eqn (a) and (b)
20 +2wl=36 --> Solve for wl

Ans -> C

What is the area of the rectangular region above?

(1) l + w = 6
(2) d^2 = 20

Sol:
St1: l+w=6----> If l=5,w=1 then Area is 5 but if l=4 and w=2 then area is 8...Since we have 2 answers so not St is not sufficient
A and D ruled out

d^2=20 or l^2+w^2=d^2 or l^2+w^2=20
Clearly not sufficient

Combining we get l+w=6, squaring both sides we get

l^2+w^2+2lw=36 or 2lw=16 or lw=8

Ans is C

600 level may be okay
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No, that's not correct. Usually the diagonal does not divide a rectangle into two 45-45-90 triangles (it'll be correct only for squares, so when l=w).
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and equations ensures a solution.


What is the area of the rectangular region above?

(1) l + w = 6. Not sufficient to get the value of .
(2) d^2 = 20

In the original condition we have 2 variables for the rectangle(width, length) and thus we need 2 variables to match the number of variables and equations. Since there is 1 each in 1) and 2), C is likely the answer.

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Thanks!
Actually, can we not use the pythagorean theorem and 45-45-90 triangle rule to get l and w using (2)?

i think B would be sufficient ! with the help of 90:45:45 triangle ,as we already know 1 side we can find the rest.
plz correct me if i m wrong
thank you

Hi all,

I understood why it's not a 45-45-90 triangle. However I chose B, assuming it was a 30-60-90 triangle.

45-45-90 triangles can only be assumed when l=w.

Is there a similar rule for the 30-60-90 triangle or how can I eliminate the possibility that this is a 30-60-90 triangle here?


Thanks

We cannot say it's a 30-60-90 triangle for the same reason we cannot say it's a 45-45-90: we know only the diagonal (hypotenuse) which is NOT enough to determine the other sides.

As for your other question:


This is one of the 'standard' triangles you should be able to recognize on sight. A fact you should commit to memory is: the sides are always in the ratio \(1 : \sqrt{3}: 2\).

Please check for more here: math-triangles-87197.html

Hope it helps.
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Bunuel, I screwed up and assumed I was to use the 30:60:90 triangle rule for statement #2 (since it's a rectangle).

How am I supposed to know that rule won't work for this question? Is it because d^2 = 20 -> d= 4 square root 5, but not 4 square root 3?

Knowing only the length of a diagonal of a rectangle does not allow to find its adjacent angles. You can squeeze or stretch a rectangle with any length of a diagonal along one of its sides to get any adjacent lengths from 0 to 90 not inclusive.
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Bunuel why are you squaring statement one l + w = 6 ?? the formula for area of rectangle is L*W , isnt it

Hello Dave

I think Bunuel is doing this because he sees two equations from two statements: one equation says that l + w = 6 and the other says that l^2 + w^2 = 20. But he has to find l*w. So how to find l*w from l+w and l^2 + w^2. Immediately the formula which comes to mind is:
(l+w)^2 = l^2 + w^2 + 2*l*w.
We can substitute the values of l+w and l^2 + w^2 in the above formula and we can get the value of l*w. Hope this is clear.

Here is how i would solve this problem:

We are interested in the area of this rectangular region, which is L x W= ?

Statement 1/

L+W= 6 is clearly not sufficient!

If L= 5 and W= 1 , then the area would be 5 x 1= 5, on the other side

if L=4 and W=2, then the area would be 4 x 2= 8,

Thus, since we get different values for the area of this rectangular region, statement 1 is clearly Not Sufficient!


Statement 2/

D^2=20 is clearly not sufficient. The diagonal of a rectangle can be found using Pythagorean Theorem, but we need atleast 2 sides to find the third one and since statement 2 gives us just one side, hence Not sufficient.


Statement 1 and 2 together, are sufficient and here is why:

From statement 1: for L+W=6, we had two scenarios: 5+1=6 or 4+2=6.

From statement 2: D^2=20.


So if we apply the Pythagorean Theorem for the diagonal of the rectangle, which is D, then we come to the conclusion that from the 2 different scenarios we had from statement 1, inwhich L+W=6, just one of them fits the Pythagorean Theorem for the diagonal of the rectangle, and that is when L=4 and W=2, which gives L+W=6.

Pythagorean Theorem:

a^2+b^2=c^2 (C stands for the diagonal, which is D according statement 2), and lets check if that is the case.

4^2 + 2^2 = D^2

16+4=20, thus gives 20 for D^2.

Thus, the area L * W=4 * 2= 8

Both statements together are Sufficient! C

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