Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Apr 2610:00 AM EDT
-11:00 AM EDT
The Target Test Prep course represents a quantum leap forward in GMAT preparation, a radical reinterpretation of the way that students should study. Try before you buy with a 5-day, full-access trial of the course for FREE!
Apr 2711:00 AM EDT
-12:00 PM EDT
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors
Apr 2808:00 AM PDT
-11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.
24 Dec 2019, 03:26
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
50% (02:56) correct
50%
(02:49)
wrong
based on 84
sessions
History
Date
Time
Result
Not Attempted Yet
PQRS is a quadrilateral whose diagonals are perpendicular to each other. If PQ=16 cm, QR=12 cm and RS=20 cm, what is the value of PS.
(A) \(8\sqrt{2}\)
(B) \(12\sqrt{2}\)
(C) \(16\sqrt{2}\)
(D) \(20\sqrt{2}\)
(E) \(24\sqrt{2}\)
(A) \(8\sqrt{2}\)
(B) \(12\sqrt{2}\)
(C) \(16\sqrt{2}\)
(D) \(20\sqrt{2}\)
(E) \(24\sqrt{2}\)
24 Dec 2019, 06:48
Kudos
Bookmarks
Bunuel
From the above figure,
We have 4 right angled triangles PTQ, QTR, RTS & STP
Let QT = a, PT = b, RT = c & ST = d
--> \(16^2 = a^2 + b^2\) ....... (1)
--> \(12^2 = a^2 + c^2\) ....... (2)
--> \(20^2 = c^2 + d^2\) ....... (3)
--> \(x^2 = b^2 + d^2\) ....... (4)
(1) + (3)
--> \(16^2 + 20^2 = a^2 + b^2 + c^2 + d^2\)
&
(2) + (4)
--> \(12^2 + x^2 = a^2 + b^2 + c^2 + d^2\)
--> \(16^2 + 20^2 = 12^2 + x^2\)
--> \(256 + 400 = 144 + x^2\)
--> \(x^2 = 656 - 144 = 512\)
--> \(x = \sqrt{512} = \sqrt{256*2}\)
--> \(x = 16\sqrt{2}\)
IMO Option C
Attachments
WhatsApp Image 2019-12-24 at 7.08.54 PM.jpeg [ 96.12 KiB | Viewed 17709 times ]
25 Dec 2019, 10:41
Kudos
Bookmarks
Bunuel
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.
