Here's one approach:
In the figure above, points P and Q lie on the circle with center O. What is the value of S?
a) 1/2
b) 1
c) √2
d) √3
e) (√2)/2
So, s = 1
Answer: B
Cheers,
Brent
Geometry - P, Q and a circle
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Brent@GMATPrepNow
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So, s = 1
Answer: B
Cheers,
Brent
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nikhilgmat31
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Hi Brent
as POQ is 90, I am taking it as right angle triangle & trying to solve using Pythagoras
Radius = OP = root (-root 3 ^2 + 1 ^2 ) = root 4 = 2
Radius = OQ = root (s^2 + t ^2 )
PQ is hypotenuse and distance between PQ is root [ (s + root 3)^2 + (t-1)^2) ]
By Pythagoras
root [ (s + root 3)^2 + (t-1)^2) ] = 2^2 + (s^2 + t ^2 )
But the equation got complex.. Please suggest how to solve it simple way.
as POQ is 90, I am taking it as right angle triangle & trying to solve using Pythagoras
Radius = OP = root (-root 3 ^2 + 1 ^2 ) = root 4 = 2
Radius = OQ = root (s^2 + t ^2 )
PQ is hypotenuse and distance between PQ is root [ (s + root 3)^2 + (t-1)^2) ]
By Pythagoras
root [ (s + root 3)^2 + (t-1)^2) ] = 2^2 + (s^2 + t ^2 )
But the equation got complex.. Please suggest how to solve it simple way.
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theCEO
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1. If two lines are at 90degrees to each other, it means they are perpendicular to each other.
In the figure above, points P and Q lie on the circle with center O. The angle POQ = 90. What is the value of S?
a) 1/2
b) 1
c) √2
d) √3
e) (√2)/2
2. Perpendicular lines have negative inverse slopes.
example line A and B are perpendicular if there slopes are (1/2) and (-2/1) respectively.
slope of OP = rise/run = 1/-sqrt(3)= -1/sqrt(3)
slope of OQ = rise/run = sqrt(3)/1
since s = x value = run; s=1
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nikhilgmat31
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theCEO wrote:1. If two lines are at 90degrees to each other, it means they are perpendicular to each other.
In the figure above, points P and Q lie on the circle with center O. The angle POQ = 90. What is the value of S?
a) 1/2
b) 1
c) √2
d) √3
e) (√2)/2
2. Perpendicular lines have negative inverse slopes.
example line A and B are perpendicular if there slopes are (1/2) and (-2/1) respectively.
slope of OP = rise/run = 1/-sqrt(3)= -1/sqrt(3)
slope of OQ = rise/run = sqrt(3)/1
since s = x value = run; s=1
I agree with you as slope of OQ = sqrt(3)/1
but it is (t-0)/(s-0) = sqrt(3)/1
are you just equating the numerators & denominators ?
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theCEO
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Correct - since both lines pass through the origin, the slopes can be found from each point's x and y values.nikhilgmat31 wrote:theCEO wrote:1. If two lines are at 90degrees to each other, it means they are perpendicular to each other.
In the figure above, points P and Q lie on the circle with center O. The angle POQ = 90. What is the value of S?
a) 1/2
b) 1
c) √2
d) √3
e) (√2)/2
2. Perpendicular lines have negative inverse slopes.
example line A and B are perpendicular if there slopes are (1/2) and (-2/1) respectively.
slope of OP = rise/run = 1/-sqrt(3)= -1/sqrt(3)
slope of OQ = rise/run = sqrt(3)/1
since s = x value = run; s=1
I agree with you as slope of OQ = sqrt(3)/1
but it is (t-0)/(s-0) = sqrt(3)/1
are you just equating the numerators & denominators ?
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[email protected]
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Hi All,
Brent's approach to solving this question is exactly how I would have done it, so I won't rehash that explanation. Instead I'll point out the patterns and "hints" that were in the original drawing (and that you should be on the lookout for when dealing with similar questions).
1) Any time you see a diagonal line segment on a graph, you can draw a right triangle using that line as the hypoteneuse.
2) Right triangles on the GMAT are almost always pattern-based (angles or sides; at the very least, the Pythagorean Theorem). Here, we have a 1 and a √3, so we should be thinking 30/60/90.
3) Radii have the same length no matter what direction they go in.
4) Lines "add up" to 180 degrees.
GMAT assassins aren't born, they're made,
Rich
Brent's approach to solving this question is exactly how I would have done it, so I won't rehash that explanation. Instead I'll point out the patterns and "hints" that were in the original drawing (and that you should be on the lookout for when dealing with similar questions).
1) Any time you see a diagonal line segment on a graph, you can draw a right triangle using that line as the hypoteneuse.
2) Right triangles on the GMAT are almost always pattern-based (angles or sides; at the very least, the Pythagorean Theorem). Here, we have a 1 and a √3, so we should be thinking 30/60/90.
3) Radii have the same length no matter what direction they go in.
4) Lines "add up" to 180 degrees.
GMAT assassins aren't born, they're made,
Rich
Last edited by [email protected] on Wed Dec 09, 2015 10:01 am, edited 1 time in total.
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nikhilgmat31
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[email protected] wrote:Hi All,
Brent's approach to solving this question is exactly how I wouldn't have done it, so I won't rehash that explanation. Instead I'll point out the patterns and "hints" that were in the original drawing (and that you should be on the lookout for when dealing with similar questions).
1) Any time you see a diagonal line segment on a graph, you can draw a right triangle using that line as the hypoteneuse.
2) Right triangles on the GMAT are almost always pattern-based (angles or sides; at the very least, the Pythagorean Theorem). Here, we have a 1 and a √3, so we should be thinking 30/60/90.
3) Radii have the same length no matter what direction they go in.
4) Lines "add up" to 180 degrees.
GMAT assassins aren't born, they're made,
Rich
Hi Rich,
How We can utilize 30:60:90 rule here we don't know which side makes angle of 30 or 60 with hypotenuse.
Please explain.
concept from Brent was fine, But the last step is little tricky & hard to absorb.
Thanks
Nikhil
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Jim@StratusPrep
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If you drop a vertical line from P to the x-axis to a point called A, then POA is the 30 degree angle because PA will equal 1 and AO will equal sqrt 3.
Really all you have to do is flip the coordinates and account for positive values to get (1, √3)
Really all you have to do is flip the coordinates and account for positive values to get (1, √3)
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[email protected]
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Hi nikhilgmat31,
Here's how you can figure out which angle is which...
1) Draw a right triangle.
2) Label the two LEGS 1 and (root3).
3) Using your knowledge of special right triangles, you should be able to determine that you have a 30/60/90 right triangle AND you should be able to determine which angle is 30 and which is 60. (hint: think about the side lengths - how do they relate to one another? how do the angles relate to one another?)
GMAT assassins aren't born, they're made,
Rich
Here's how you can figure out which angle is which...
1) Draw a right triangle.
2) Label the two LEGS 1 and (root3).
3) Using your knowledge of special right triangles, you should be able to determine that you have a 30/60/90 right triangle AND you should be able to determine which angle is 30 and which is 60. (hint: think about the side lengths - how do they relate to one another? how do the angles relate to one another?)
GMAT assassins aren't born, they're made,
Rich
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nikhilgmat31
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yes Rich, But here both the legs are 2 only which is radius of circle.
found it using distance formula P(-root 3,1) to (0,0) which is origin of cirlce.
so distance of (s,t) to (0,0) is also 2
found it using distance formula P(-root 3,1) to (0,0) which is origin of cirlce.
so distance of (s,t) to (0,0) is also 2
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[email protected]
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Hi nikhilgmat31,
When referring to a right triangle, the word "LEGS" refers to the two 'short' sides (NOT the hypotenuse).
In this prompt, we can use the co-ordinate (-root3, 1) to create a right triangle - it will have a base of root3 and a height of 1. With those two LEGS, you can determine the measures of the two non-right angles.
GMAT assassins aren't born, they're made,
Rich
When referring to a right triangle, the word "LEGS" refers to the two 'short' sides (NOT the hypotenuse).
In this prompt, we can use the co-ordinate (-root3, 1) to create a right triangle - it will have a base of root3 and a height of 1. With those two LEGS, you can determine the measures of the two non-right angles.
GMAT assassins aren't born, they're made,
Rich
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Another neat (and quick) approach:
Since <POQ = 90°, the two other angles (call them <POR and <POS) must sum to 90° as well.
Since P has coordinates (-√3, 1), a triangle with PO as the hypotenuse is a 30-60-90. Hence <POR = 30°.
Thus <POS = 60°, and a triangle with QO as the hypotenuse is also 30-60-90, but since the "base" angle is now 60°, we have reversed coordinates: the x-value is 1 and the y-value is √3.
Since <POQ = 90°, the two other angles (call them <POR and <POS) must sum to 90° as well.
Since P has coordinates (-√3, 1), a triangle with PO as the hypotenuse is a 30-60-90. Hence <POR = 30°.
Thus <POS = 60°, and a triangle with QO as the hypotenuse is also 30-60-90, but since the "base" angle is now 60°, we have reversed coordinates: the x-value is 1 and the y-value is √3.
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nikhilgmat31
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I agree Rich, two LEGS refers to sides of triangle.[email protected] wrote:Hi nikhilgmat31,
When referring to a right triangle, the word "LEGS" refers to the two 'short' sides (NOT the hypotenuse).
In this prompt, we can use the co-ordinate (-root3, 1) to create a right triangle - it will have a base of root3 and a height of 1. With those two LEGS, you can determine the measures of the two non-right angles.
GMAT assassins aren't born, they're made,
Rich
my only concern is Why we didn't take 45:45:90 ratio. since the sides of triangle will be of same length since both equals the radius of triangle or distance of P or Q from origin (0,0)
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Hi nikhilgmat31,
In the triangle that we're discussing, ONLY the hypotenuse = radius. The two legs of this right triangle have a base of (root3) and a height of 1; since those two lengths are NOT equal, we CANNOT have a 45/45/90 right triangle.
GMAT assassins aren't born, they're made,
Rich
In the triangle that we're discussing, ONLY the hypotenuse = radius. The two legs of this right triangle have a base of (root3) and a height of 1; since those two lengths are NOT equal, we CANNOT have a 45/45/90 right triangle.
GMAT assassins aren't born, they're made,
Rich
