Hi All,
Please help in answering the attached question from GMAT Prep.
Answer options are:
1. 1/2
2. 1
3. square root 2
4. square root 3
5. (square root 2)/ 2
Thanks!
Co-ordinate geometry
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slope of PO = a/b
slope of QO = t/s
PO and QO is perpendicular so [a/b]*[t/s]=-1, so t/s = -b/a
In this case, b = √3, a = -1 , so t/s = [-√3]/[-1] = √3/1, so s =1
Alternatively
from the figure we can see that OP=OQ = radius of the semicircle. Also Angle POQ = 90degree
thus PQ^2 = OP^2 + OQ^2-------eqn1
using distance formula (O is origin)
OP = sqrt [(1-0)^2 + (sqrt -3 -0)^2] = sqrt [4] = 2 = OQ
also OQ = sqrt [ (s-0)^2 + (t-0)^2] = sqrt[s^2 + t^2]
OQ^2 = s^2+t^2 =4 ----eqn2
now PQ = 2sqrt 2 when we put values of OP and OQ in eqn 1
using distance formula length of PQ = sqrt [ (s+ sqrt3)^2 + (t-1)^2]
PQ^2 = (s+sqrt3)^2 + (t-1)^2 = 8--------eqn3
soving eqn 2 and 3 we get sqrt 3 s = t.Equating this in eqn 2 we get s=1
slope of QO = t/s
PO and QO is perpendicular so [a/b]*[t/s]=-1, so t/s = -b/a
In this case, b = √3, a = -1 , so t/s = [-√3]/[-1] = √3/1, so s =1
Alternatively
from the figure we can see that OP=OQ = radius of the semicircle. Also Angle POQ = 90degree
thus PQ^2 = OP^2 + OQ^2-------eqn1
using distance formula (O is origin)
OP = sqrt [(1-0)^2 + (sqrt -3 -0)^2] = sqrt [4] = 2 = OQ
also OQ = sqrt [ (s-0)^2 + (t-0)^2] = sqrt[s^2 + t^2]
OQ^2 = s^2+t^2 =4 ----eqn2
now PQ = 2sqrt 2 when we put values of OP and OQ in eqn 1
using distance formula length of PQ = sqrt [ (s+ sqrt3)^2 + (t-1)^2]
PQ^2 = (s+sqrt3)^2 + (t-1)^2 = 8--------eqn3
soving eqn 2 and 3 we get sqrt 3 s = t.Equating this in eqn 2 we get s=1
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Last edited by Brent@GMATPrepNow on Thu Apr 19, 2018 1:34 pm, edited 1 time in total.
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Hi DevB,
When dealing with geometry/graphing questions, there are certain patterns you should be on the lookout for:
1) A diagonal line on a graph IS the hypotenuse of a right triangle. You can draw that triangle and figure out its base and height. From there, you have a variety of triangle formulas that might be helpful involving the question.
2) The GMAT has a variety of standard formulas and patterns that it will test you on. When you come across a triangle, chances are it fits some type of pattern. Look to use the Pythagorean Theorem, and look for the special right triangles (30/60/90, 45/45/90, 3/4/5, 5/12/13).
3) Weird numbers almost always point to a pattern. Notice the (root3) in this question? I wonder what that might refer to in a right triangle....?
GMAT assassins aren't born, they're made,
Rich
When dealing with geometry/graphing questions, there are certain patterns you should be on the lookout for:
1) A diagonal line on a graph IS the hypotenuse of a right triangle. You can draw that triangle and figure out its base and height. From there, you have a variety of triangle formulas that might be helpful involving the question.
2) The GMAT has a variety of standard formulas and patterns that it will test you on. When you come across a triangle, chances are it fits some type of pattern. Look to use the Pythagorean Theorem, and look for the special right triangles (30/60/90, 45/45/90, 3/4/5, 5/12/13).
3) Weird numbers almost always point to a pattern. Notice the (root3) in this question? I wonder what that might refer to in a right triangle....?
GMAT assassins aren't born, they're made,
Rich
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Please explain this part
In this case, b = √3, a = -1
In this case, b = √3, a = -1
abhasjha wrote:slope of PO = a/b
slope of QO = t/s
PO and QO is perpendicular so [a/b]*[t/s]=-1, so t/s = -b/a
In this case, b = √3, a = -1 , so t/s = [-√3]/[-1] = √3/1, so s =1
Alternatively
from the figure we can see that OP=OQ = radius of the semicircle. Also Angle POQ = 90degree
thus PQ^2 = OP^2 + OQ^2-------eqn1
using distance formula (O is origin)
OP = sqrt [(1-0)^2 + (sqrt -3 -0)^2] = sqrt [4] = 2 = OQ
also OQ = sqrt [ (s-0)^2 + (t-0)^2] = sqrt[s^2 + t^2]
OQ^2 = s^2+t^2 =4 ----eqn2
now PQ = 2sqrt 2 when we put values of OP and OQ in eqn 1
using distance formula length of PQ = sqrt [ (s+ sqrt3)^2 + (t-1)^2]
PQ^2 = (s+sqrt3)^2 + (t-1)^2 = 8--------eqn3
soving eqn 2 and 3 we get sqrt 3 s = t.Equating this in eqn 2 we get s=1
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He's using similar triangles here. Each triangle has sides of length 1 and √3, which, applied to coordinate plane, gives (-√3, 1) for the first and (1, √3) for the second.Gurpreet singh wrote:Please explain this part
In this case, b = √3, a = -1
H Brent,Here's one approach:
Can you explain why you assumed in your solution that the second triangle is 30:60:90? like the traingle on the left?
I know the r=2 but this does not mean that the second triangle is the same as the triangle to the left.
Wham am I missing here?
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The three angles along the x-axis must sum to 180.Zoser wrote:H Brent,Here's one approach:
Can you explain why you assumed in your solution that the second triangle is 30:60:90? like the traingle on the left?
I know the r=2 but this does not mean that the second triangle is the same as the triangle to the left.
Wham am I missing here?
In Brent's solution, the leftmost angle along the x-axis = 30.
The figure indicates that the middle angle along the x-axis = 90.
Thus, the rightmost angle along the x-axis = 180-30-90 = 60.
Result:
The triangle on the right is also a 30-60-90 triangle.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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I dont know why I keep missing those basic concepts while answering Gmat questions! Even though, I know them but they just do not come accross my mind while solving the question!The three angles along the x-axis must sum to 180.
In Brent's solution, the leftmost angle along the x-axis = 30.
The figure indicates that the middle angle along the x-axis = 90.
Thus, the rightmost angle along the x-axis = 180-30-90 = 60.
Result:
The triangle on the right is also a 30-60-90 triangle.
Many thanks!