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# In the figure above, the coordinates of points P and Q are (0, 1) and

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Math Expert
Joined: 02 Sep 2009
Posts: 56300
In the figure above, the coordinates of points P and Q are (0, 1) and  [#permalink]

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13 Dec 2017, 21:03
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Difficulty:

25% (medium)

Question Stats:

80% (00:32) correct 20% (00:45) wrong based on 33 sessions

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In the figure above, the coordinates of points P and Q are (0, 1) and (1, 0) respectively. What is the area of square PQRS?

(A) 1
(B) √2
(C) 2
(D) 2√2
(E) 4

Attachment:

2017-12-12_2125_002.png [ 7.47 KiB | Viewed 671 times ]

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Posts: 3087
In the figure above, the coordinates of points P and Q are (0, 1) and  [#permalink]

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14 Dec 2017, 12:20
Bunuel wrote:

In the figure above, the coordinates of points P and Q are (0, 1) and (1, 0) respectively. What is the area of square PQRS?

(A) 1
(B) √2
(C) 2
(D) 2√2
(E) 4

Attachment:
2017-12-12_2125_002.png

Let origin be O. Find side length of the square from ∆ OPQ. Its hypotenuse = side of square

∆ OPQ is a right isosceles triangle
Side OP = OQ = 1
∠ at vertex O = 90°
(180° - 90°) = 90° remain
Angles opposite equal sides are equal:
∠ OPQ = ∠ OQP = x : (2x = 90), x = 45°

∆ OPQ has side lengths in ratio $$x : x : x\sqrt{2}$$
x = 1
Hypotenuse* = $$x\sqrt{2}=1\sqrt{2}=\sqrt{2}$$

Area of square = $$s^2=(\sqrt{2})^2= 2$$

*OR Pythagorean theorem
$$1^2 + 1^2 = h^2$$
$$2 = h^2$$
$$h = \sqrt{2}$$

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In the figure above, the coordinates of points P and Q are (0, 1) and   [#permalink] 14 Dec 2017, 12:20
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