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Bunuel
zisis
If the diagonal of rectangle Z is d, and the perimeter of rectangle Z is p, what is the area of rectangle Z, in terms of d and p?
(A) (d2 – p)/3
(B) (2d2 – p)/2
(C) (p – d2)/2
(D) (12d2 – p2)/8
(E) (p2 – 4d2)/8


NO OA - once provided i will update

IMO (A)

Let the sides of rectangle be \(x\) and \(y\).

Given: \(d^2=x^2+y^2\) and \(p=2(x+y)\). Question: \(area=xy=?\)

Square \(p\) --> \(p^2=4(x^2+2xy+y^2)\) --> substitute \(x^2+y^2\) by \(d^2\) --> \(p^2=4(d^2+2xy)\) --> \(p^2-4d^2=8xy\) --> \(area=xy=\frac{p^2-4d^2}{8}\).

Answer: E.

excellent explanation ! kudos
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Equation 1: \(d = \sqrt{L^2+W^2}\)
Equation 2: \(p = 2(L+W)\)

Combine 1 and 2:

\(d^2 = L^2+W^2\)
\(p^2 = 4 ( L^2 + 2LW + W^2)\)
\(p^2 = 4L^2 + 8LW + 4W^2\)
\(p^2 = 4d^2 + 8LW\)

Remember that A = LW and \frac{p}{2}=L+W

\(A = \frac{p^2-4d^2}{8}\)

Answer: E
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if we take a rectangle of 2x1

area A = 2x1 = 2
perimeter P = 2x2 + 2x1 = 6
the diagonal D = sqrt(1²+2²) = sqrt(5)

for ansewer (B) (2d^2 – p)/2

(2D² - P) = 2*sqrt(5)² - 6 = 4 ==> 4/2 = 2 ==> which is the area A

the same reasoning goes for answer E

am i doing something wrong? because B and E are correct for me
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I just used real numbers to solve this question. Pick a rectangle with sides 3 and 4. Diagonal would be 5 so would = d. Perimeter = 14, Area = 12.

Only E works.
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if we take a rectangle of 2x1

area A = 2x1 = 2
perimeter P = 2x2 + 2x1 = 6
the diagonal D = sqrt(1²+2²) = sqrt(5)

for ansewer (B) (2d^2 – p)/2

(2D² - P) = 2*sqrt(5)² - 6 = 4 ==> 4/2 = 2 ==> which is the area A

the same reasoning goes for answer E

am i doing something wrong? because B and E are correct for me

When you take some numbers as you did here (say, distinct sides are 2 and 1 so p = 6, d = ...etc), it is possible that 2 or more options give you the correct answer. In that case, you need to take another set of numbers and put the values in only those two options. Hopefully, only one of them will give you the correct answer.
Also, when picking numbers, try to pick those which need minimum effort. e.g. I would like to pick the sides as 3 and 4 because I know that the diagonal in that case will be 5. No messy square roots to handle.
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Refer diagram below:

Let one side of the rectangle = a

Other side \(= \frac{p}{2} - a\)

Require to find \(Area = a (\frac{p}{2} - a) = \frac{ap}{2} - a^2\)

\(d^2 = a^2 + (\frac{p}{2} - a)^2\)

\(d^2 = 2a^2 + \frac{p^2}{4} - ap\)

\(ap - 2a^2 = \frac{p^2}{4} - d^2\)

\(\frac{ap}{2} - a^2 = \frac{p^2 - 4d^2}{8}\)

Answer = E
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the question rotating on the formula:
(x+y)^2=x^2+y^2+2xy

just solve acc and get answer E
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I made it into a 3-4-5 triangle:
diagonal=5
side = 3,4 so P = 14 and target Area is 3*4=12

Then plug into answers, E) comes out to be (196-100)/8 = 96/8 = 12
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Could be solved by just assuming the rectangle is a square with sides of 2 or 3?
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Is this a 650 level question?
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Is this a 650 level question?

You can check difficulty level of a question along with the stats on it in the first post. For this question Difficulty = 600-700 Level. The difficulty level of a question is calculated automatically based on the timer stats from the users which attempted the question.
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Hello from the GMAT Club BumpBot!

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