If The Diagonal of Rectangle Z is d, And The Perimeter of Rectangle Z is p GMAT Problem Solving

Question: 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?

1. (d^2 – p)/3
2. (2d^2 – p)/2
3. (p – d^2)/2
4. (12d^2 – p^2)/8
5. (p^2 – 4d^2)/8

‘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?’ - is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book “Official Guide for GMAT Reviews”. To solve GMAT Problem Solving examiners measure how well the candidates make analytical and logical approaches to solve numerical problems. In this section, candidates have to evaluate and interpret data from given graphical representation. In this section, mostly one finds out mathematical questions. Five answer choices are given for each GMAT Problem solving question.

Solution and Explanation:

Approach Solution 1:

There is only one approach to solve this problem.
It is given in the question that the diagonal of a rectangle Z is d and the perimeter of rectangle Z is p. It is asked to find out the area of the rectangle in terms of d and p.
Let the side of the rectangle be x and y.
Given that the diagonal of the rectangle is d
Which means,
\(x^2\)+ \(y^2\)= \(d^2\)—--(1)
We are also given that the perimeter of the rectangle is p.
Perimeter of a rectangle = 2*(l+b)
Here l = x and b = y
We get perimeter = 2 *(x+y) = 2x + 2y
This is equal to p
P = 2x + 2y —--- (2)
Since now the area of the rectangle is = l*b
Area = xy
From equation 2,
Squaring both sides,
\(p^2\)= 4\((x+y)^2\) = 4(\(x^2\)+\(y^2\)+ 2xy)
Putting the value of (\(x^2\)+\(y^2\)) from (1)
\(p^2\)= 4(\(d^2\)+ 2xy)
We know that area = xy
4(\(d^2\)+ 2xy) = \(p^2\)
2xy = \(p^2\)/4 - \(d^2\)
xy = \(p^2\)/ 8 - \(d^2\)/2
Area = \(p^2\)/ 8 - \(d^2\)/2 = (\(p^2\)-4\(d^2\))/8
Therefore the correct answer will be E.
Correct Answer: E

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