# Manhattan GMAT Challenge Problem of the Week – 2 Nov 2010

by Manhattan Prep, Nov 2, 2010

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## 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?

(A) [pmath](d^2-p)/3[/pmath]

(B) [pmath](2d^2-p)/2[/pmath]

(C) [pmath](p-d^2)/2[/pmath]

(D) [pmath](12d^2-p^2)/8[/pmath]

(E) [pmath](p^2-4d^2)/8[/pmath]

Lets take an algebraic approach. The first step is to realize that we should create variables for the simplest things about this rectangle: its length and its width. The reason is that we can express all of these secondary features (diagonal, perimeter, and area) in terms of length and width. Then we can look for a relationship between these expressions.

We can call the length of rectangle a and its width b. Now we can write equations relating a and b to the diagonal, perimeter, and area, respectively:

Diagonal: [pmath]a^2 + b^2 = d^2[/pmath], by the Pythagorean Theorem

Perimeter: [pmath]2(a + b) = p[/pmath]

Area: Area = ab = ? (in terms of p and d)

So we are looking to manipulate the first two equations to isolate ab on one side. The expression involving p and d on the other side will be our answer.

To avoid square roots, lets square the second equation (for p) and see what we get.

[pmath]2(a + b) = p[/pmath]

[pmath]4(a + b)^2 = p^2[/pmath]

[pmath]4(a^2 +2ab + b^2) = p^2[/pmath]

[pmath]4a^2 +8ab + 4b^2 = p^2[/pmath]

Comparing this equation to the first equation (for d), we hopefully notice that the [pmath]a^2[/pmath] and [pmath]b^2[/pmath] terms can be made to line up and cancel. We multiply the first equation by 4, to begin with.

[pmath]a^2 + b^2 = d^2[/pmath]

[pmath]4a^2 + 4b^2 = 4d^2[/pmath]

Now we can line up and subtract:

[pmath]4a^2 +8ab + 4b^2 = p^2[/pmath]

-[pmath][4a^2 + 0 + 4b^2 = 4d^2][/pmath]

yields

[pmath]8ab=p^2-4d^2[/pmath]

Now just divide by 8 to isolate ab:

[pmath]ab=(p^2-4d^2)/8[/pmath]