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If ABCD is a square, what are the coordinates of C?
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30 Nov 2018, 18:04
30 Nov 2018, 18:04
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If ABCD is a square, what are the coordinates of C?
GRE exam - If ABCD is a square, what are the coordinates of C .jpg [ 12.34 KiB | Viewed 9378 times ]
A. \((\sqrt{3}, \sqrt{3} )\)
B. \((\sqrt{3}, 1+ \sqrt{3})\)
C. \((2 \sqrt{3}, \sqrt{3} )\)
D. \(( 1+ \sqrt{3}, \sqrt{3} )\)
E. \((\sqrt{3}, 2 \sqrt{3} )\)
Attachment:
GRE exam - If ABCD is a square, what are the coordinates of C .jpg [ 12.34 KiB | Viewed 9378 times ]
A. \((\sqrt{3}, \sqrt{3} )\)
B. \((\sqrt{3}, 1+ \sqrt{3})\)
C. \((2 \sqrt{3}, \sqrt{3} )\)
D. \(( 1+ \sqrt{3}, \sqrt{3} )\)
E. \((\sqrt{3}, 2 \sqrt{3} )\)
Re: If ABCD is a square, what are the coordinates of C?
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06 Dec 2018, 04:31
06 Dec 2018, 04:31
2
It's 30-60-90 triangle
The ratio of sides will be 1:\sqrt{3}:2
So if we take origin O
OA=1
OB= \sqrt{3}
AD=2
Draw a perpendicular from C onto x-axis, and take the intersection point E
Again we have a 30-60-90 triangle where
EC=\sqrt{3}
ED=1
Hence
OE= 1 + \sqrt{3}
EC=\sqrt{3}
Coordinates of C are ( 1+ \sqrt{3} , \sqrt{3} )
It's D
The ratio of sides will be 1:\sqrt{3}:2
So if we take origin O
OA=1
OB= \sqrt{3}
AD=2
Draw a perpendicular from C onto x-axis, and take the intersection point E
Again we have a 30-60-90 triangle where
EC=\sqrt{3}
ED=1
Hence
OE= 1 + \sqrt{3}
EC=\sqrt{3}
Coordinates of C are ( 1+ \sqrt{3} , \sqrt{3} )
It's D
Re: If ABCD is a square, what are the coordinates of C?
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30 Apr 2020, 09:24
30 Apr 2020, 09:24
1
Since the side opposite 30 is length 1, so the hypotenuse is twice the side opposite to 30 which is 1*2 = 2.
And the side opposite 60 is root(3)/2 times the hypotenuse so 2 * root(3) / 2 = root(3).
Now, we know the square is titled at an angle so the point C is slightly to the right of bottom tip of the square. So X coordinate is greater than root(3).
Eliminate A, B and D which have root(3) as X-coordinate.
We need to check how far the X coordinate is. If the square were tilted at 45 degrees on X axis, then left half and the right half would be equivalent and the X coordinate would be twice root(3). But we know the square is tilted at a less than 45 degrees on the X axis (30 degrees given), so X coordinate is greater than root(3) and lesser than 2root(3). So eliminate option C.
D is the answer (1+root(3)).
And the side opposite 60 is root(3)/2 times the hypotenuse so 2 * root(3) / 2 = root(3).
Now, we know the square is titled at an angle so the point C is slightly to the right of bottom tip of the square. So X coordinate is greater than root(3).
Eliminate A, B and D which have root(3) as X-coordinate.
We need to check how far the X coordinate is. If the square were tilted at 45 degrees on X axis, then left half and the right half would be equivalent and the X coordinate would be twice root(3). But we know the square is tilted at a less than 45 degrees on the X axis (30 degrees given), so X coordinate is greater than root(3) and lesser than 2root(3). So eliminate option C.
D is the answer (1+root(3)).
