Let one side of rectangle be x
then other side becomes (p/2)-x .................(because p= 2* sum of both sides)
Now, using pythagorus theorem,
d^2= x^2 + ((p/2)-x)^2
Now solve for the quadratic equation for x
we get,
x= (p+sqrt(8d^2-p^2))/4 and (p-sqrt(8d^2-p^2))/4

If we take x= (p+sqrt(8d^2-p^2))/4,then P/2-x= (p-sqrt(8d^2-p^2))/4
And if we take x= (p-sqrt(8d^2-p^2))/4, then P/2-x= (p+sqrt(8d^2-p^2))/4
Now area is multiplication of these sides, which is same in both the cases. =(p^2-4d^2)/2

________________________________________

Shahrukh Moin Khan
QA Mentor at Breakspace
https://www.mbabreakspace.com
PGDM: IIM Calcutta
B.Tech IIT Roorkee

Hello Vjesus12.

Let's take a look at your question.

We have the following information: $$diagonal\ =\ d\ \ \ \ \ \ \ \ \ \ \ and\ \ \ \ \ \ \ \ \ \ \ \ \ perimeter\ =\ p\ .$$ And we want to find the area $$A=l\cdot w\ \ $$ where l is the lenght and w its widht.

We also have that: $$d^2=w^2+l^2 \ \ \ \ and\ \ \ \ \ \ \ p=2w+2l = 2(w+l)\ .$$

Now, we can calculate $$\left(w+l\right)^2=w^2+2wl+l^2$$ But, using the previous equations we have $$\left(w+l\right)^2=\left(\frac{p}{2} \right)^2=\frac{p^2}{4}\ $$ and also $$w^2+2wl+l^2=\left(w^2+l^2\right)+2wl=d^2+2wl.$$ Hence, we get $$d^2+2wl=\frac{p^2}{4}\ \ \ \Rightarrow\ \ \ wl=\frac{p^2}{8}-\frac{d^2}{2}=\frac{p^2-4d^2}{8}.$$ So, the correct answer is the option E.

I hope this helps you. <i class="em em-smiley"></i>

VJesus12 wrote: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) (d^2 - p)/3
(B) (2d^2 - p)/2
(C) (p - d^2)/2
(D) (12d^2 - p^2)/8
(E) (p^2 - 4d^2)/8

The OA is the option E.

What is the best way to find the correct answer? Could anyone show me a good approach? Thanks in advance.

Pick some easy numbers for the rectangle and diagonal, determine the area and test the anwers.

Use 3, 4 and 5 ( for the diagonal). So d= 5 and perimeter p is 14 and area is 3x4 = 12.

Try E . (14^2 - 4*5^2)/8 = (196-100)/8 = 96/8 = 12 Success !