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24 Dec 2019, 03:26
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
50% (02:56) correct
50%
(02:49)
wrong
based on 84
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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
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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
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25 Dec 2019, 10:41
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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.
