PQRS is a quadrilateral whose diagonals are perpendicular to each othe

Since PQRS quadrilateral has two diagonals perpendicular to each other, four right angled triangles are formed right angled at point of intersection O.
Now \(PQ^2 = PO^2 + OQ^2\)
\(16^2 = PO^2 + OQ^2\) - -- -- Eqn. 1

\(QR^2 = QO^2 + OR^2\)
\(12^2 = QO^2 + OR^2\) - -- -- Eqn. 2

\(RS^2 = RO^2 + OS^2\)
\(20^2 = RO^2 + OS^2\) - -- -- Eqn. 3

\(SP^2 = SO^2 + OP^2\) ?

Adding Eqn. 1, 2 and 3
\(16^2 + 12^2 + 20^2 = PO^2 + OQ^2 + QO^2 + OR^2 + RO^2 + OS^2\)
\(256 + 144 + 400 = PO^2 + OS^2 + 2(OQ^2 + OR^2)\)
\(256 + 144 + 400 = PO^2 + OS^2 + 2*144\)
\(256 + 144 + 400 - 2*144 = PO^2 + OS^2 \)
\(PO^2 + OS^2 = 256 + 400 -144 = 512\)
\(PO^2 + OS^2 = 2^9 = (2^4 * 2^{\frac{1}{2}})^2\)

\(PO^2 + OS^2 = 16\sqrt{2}\)

Answer C.

Bunuel How do you know which vertex corresponds to PQRS? Wouldn't the answer change depending on this? For example for PQ = 16, if P and Q were on diagonally opposite to each other versus being adjacent to each other.
Tagging VeritasKarishma as well

You can trust the relative ordering of points. So, PQRS means that PQ, QR, RS and SQ are the edges of the quadrilateral.

Hi Bunuel
I solved it the same way as above. But this will take some time to solve in a real exam right ? Is there a better way to solve this quickly ?

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.