Hi All,
Could someone explain how to solve these two problems on Coordinate geometry.
Thanks
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2 Coordinate geometry questions - source: GmatPrep
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mathew.tony
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Rahul@gurome
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Solution to first question:
Arc OR is half of diameter which is given by pi*r = pi*9.
If C is the centre, angle PCO is 2*35.
So arc OP is{(2*35)/360}* 2*pi*9 = 7/2 * pi. (since if x is the angle an arc subtends at the centre, then the length of arc is (x/360)*2*pi*r)
Or arc PR is pi*9 - pi*7/2 = pi*11/2.
Since PQ is parallel to OR, angle ORP = angle RPQ.
Or angle RPQ = 35 degree.
Arc QR is{(2*35)/360}*2*pi*9 = 7/2 * pi.
So arc PQ is arc PR - arc QR which is pi*11/2 - pi*7/2 =pi * 2 = 2*pi.
Arc OR is half of diameter which is given by pi*r = pi*9.
If C is the centre, angle PCO is 2*35.
So arc OP is{(2*35)/360}* 2*pi*9 = 7/2 * pi. (since if x is the angle an arc subtends at the centre, then the length of arc is (x/360)*2*pi*r)
Or arc PR is pi*9 - pi*7/2 = pi*11/2.
Since PQ is parallel to OR, angle ORP = angle RPQ.
Or angle RPQ = 35 degree.
Arc QR is{(2*35)/360}*2*pi*9 = 7/2 * pi.
So arc PQ is arc PR - arc QR which is pi*11/2 - pi*7/2 =pi * 2 = 2*pi.
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Solution to second question:
Slope of line OP is y/x = 1/(-sqrt3) = -1/(sqrt3).
Since OQ is perpendicular to OP, its slope is -1/{-1/(sqrt3)} = sqrt3.
So t/s = sqrt3 or t = sqrt(3)*s.
Also radius of this circle is 2.
So sqrt(t^2 + s^2) = 2.
Or sqrt(3*s^2 + s^2) = 2.
Or 2*s = 2.
Or s = 1.
Slope of line OP is y/x = 1/(-sqrt3) = -1/(sqrt3).
Since OQ is perpendicular to OP, its slope is -1/{-1/(sqrt3)} = sqrt3.
So t/s = sqrt3 or t = sqrt(3)*s.
Also radius of this circle is 2.
So sqrt(t^2 + s^2) = 2.
Or sqrt(3*s^2 + s^2) = 2.
Or 2*s = 2.
Or s = 1.
Rahul Lakhani
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Most of the time, GMAT Geometry problems can be solved by more than one ways. That is the beauty of geometry.
I go like this:
Problem 1:
We have following information; radius (9), angle ORP = 35, Lines PQ and OR are parallel. Let C be the center of the circle. So basically we need to find the ∠QCP to find the length of minor arc PQ.
Since lines PQ and OR are parallel, ∠ORP and ∠QPR are same, 35 each.
Tringle PCR is an isosceles triangle and so base angles are same. That means ∠CRP and ∠CPR are 35 each. Similarly for the Triangle QCP is an isosceles triangle and so ∠CPQ is 70 (35+35) and so does ∠CQP. That means ∠QCP is 40.
So length of minor arc PQ (say, x) can be calculated as follows.
x= 40*2*9*pi/360 = 2*pi
Answer is: A
Problem 2:
The length OP or the radius of the semicircle is = SQRT{(sqrt3)^2+1^2} = SQRT{4} = 2
Now consider a triangle by drawing a straight line between P and Q. Since the ∠POQ is 90 and two sides of the triangle OPQ are equal, the distance between O and Q is same as OP = 2, and its base angles are 45 each. Using the 45:45:90 ratio of angle, the length PQ is basically 2*sqrt(2), which is the sum of x-axis distance of P from O [i.e. sqrt(3)] and x-axis distance {s} of Q from O.
In other words, 2*sqrt(2) = sqrt(3) + s.
The distance s (x value) = 2*sqrt(2) - sqrt(3) = 2*1.41 - 1.73 = 1.08
Hence the answer is: B
I am attaching a pdf file with explanation.
Thank You!
I go like this:
Problem 1:
We have following information; radius (9), angle ORP = 35, Lines PQ and OR are parallel. Let C be the center of the circle. So basically we need to find the ∠QCP to find the length of minor arc PQ.
Since lines PQ and OR are parallel, ∠ORP and ∠QPR are same, 35 each.
Tringle PCR is an isosceles triangle and so base angles are same. That means ∠CRP and ∠CPR are 35 each. Similarly for the Triangle QCP is an isosceles triangle and so ∠CPQ is 70 (35+35) and so does ∠CQP. That means ∠QCP is 40.
So length of minor arc PQ (say, x) can be calculated as follows.
x= 40*2*9*pi/360 = 2*pi
Answer is: A
Problem 2:
The length OP or the radius of the semicircle is = SQRT{(sqrt3)^2+1^2} = SQRT{4} = 2
Now consider a triangle by drawing a straight line between P and Q. Since the ∠POQ is 90 and two sides of the triangle OPQ are equal, the distance between O and Q is same as OP = 2, and its base angles are 45 each. Using the 45:45:90 ratio of angle, the length PQ is basically 2*sqrt(2), which is the sum of x-axis distance of P from O [i.e. sqrt(3)] and x-axis distance {s} of Q from O.
In other words, 2*sqrt(2) = sqrt(3) + s.
The distance s (x value) = 2*sqrt(2) - sqrt(3) = 2*1.41 - 1.73 = 1.08
Hence the answer is: B
I am attaching a pdf file with explanation.
Thank You!
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