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11 Dec 2017, 23:13
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
82% (00:56) correct
18%
(01:04)
wrong
based on 68
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Square PQRS is in a rectangular coordinate plane, and has opposite vertices at points P (4,4) and R (–4,–4). What is the area of square PQRS?
(A) 4
(B) 8
(C) 16
(D) 32
(E) 64
(A) 4
(B) 8
(C) 16
(D) 32
(E) 64
12 Dec 2017, 14:10
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Since the Square PQRS has opposite vertices at points P (4,4) and R (–4,–4), the line PR will
be the length of the diagonal of the square can be found using the below given formula :
Distance between two points = \(\sqrt{(x2-x1)^2 + (y2-y1)^2}\)
*where (x1,y1) and (x2,y2) are the two points.
The length of the diagonal is \(\sqrt{(-4-4)^2 + (-4-4)^2} = \sqrt{64+64} = \sqrt{128} = 8\sqrt{2}\)
The side of the square is \(\frac{diagonal}{\sqrt{2}} = 8\)
Therefore, the area of the square is \(side^2 = 8^2 = 64\)(Option E)
be the length of the diagonal of the square can be found using the below given formula :
Distance between two points = \(\sqrt{(x2-x1)^2 + (y2-y1)^2}\)
*where (x1,y1) and (x2,y2) are the two points.
The length of the diagonal is \(\sqrt{(-4-4)^2 + (-4-4)^2} = \sqrt{64+64} = \sqrt{128} = 8\sqrt{2}\)
The side of the square is \(\frac{diagonal}{\sqrt{2}} = 8\)
Therefore, the area of the square is \(side^2 = 8^2 = 64\)(Option E)
12 Dec 2017, 15:14
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Bunuel
The sides of a square are equal, so use just one pair of x- or y-coordinates to find the length of one side.
Distance between vertices = length of side:
Subtract x- OR y-coordinates
P's x-coordinate is 4. R's x-coordinate is -4.
4 - (-4) = 8 = length of side
Area = \(s^2 = 8^2 = 64\)
Answer E
*Parallel and perpendicular sides result in vertices that
lie on the same horizontal or vertical line.
Hence the distance between two vertices of a square in the xy-plane equals one vertex's x- or y-coordinate minus the other vertex's corresponding coordinate.
