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
Value of S in Half Circle
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So, s = 1
Answer: B
Cheers,
Brent
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Hi bml1105,
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 10:02 am, edited 1 time in total.
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sanjoy18
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Without drawing...Here is my approach
since OQ=OP (radius)
s^2 +t^2= 4 ------------(1)
since OQ is perpendicular to OP
(t/s)*(- 1/sqrt(3))=-1
=> t=sqrt(3) ...(2)
solving 1 and 2
4s^2=4
S= 1/-1
since S is in 1st quardent then S=1
since OQ=OP (radius)
s^2 +t^2= 4 ------------(1)
since OQ is perpendicular to OP
(t/s)*(- 1/sqrt(3))=-1
=> t=sqrt(3) ...(2)
solving 1 and 2
4s^2=4
S= 1/-1
since S is in 1st quardent then S=1
