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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
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
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:
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?
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.
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