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# PQRS is a quadrilateral whose diagonals are perpendicular to each othe

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Math Expert
Joined: 02 Sep 2009
Posts: 64222
PQRS is a quadrilateral whose diagonals are perpendicular to each othe  [#permalink]

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24 Dec 2019, 02:26
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Difficulty:

85% (hard)

Question Stats:

37% (02:12) correct 63% (02:50) wrong based on 26 sessions

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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) $$24\sqrt{2}$$

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Re: PQRS is a quadrilateral whose diagonals are perpendicular to each othe  [#permalink]

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24 Dec 2019, 05:48
Bunuel wrote:
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) $$24\sqrt{2}$$

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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PQRS is a quadrilateral whose diagonals are perpendicular to each othe  [#permalink]

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25 Dec 2019, 09:41
Bunuel wrote:
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) $$24\sqrt{2}$$

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

$$PO^2 + OS^2 = 16\sqrt{2}$$

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PQRS is a quadrilateral whose diagonals are perpendicular to each othe   [#permalink] 25 Dec 2019, 09:41