GMAT prep Q4

Question

Hi,

Can someone help with this question ?

Thanks in advance.

jaynayak · Answer

<PoQ = 90 degress and OP = OQ (radius) hence triangle OPQ must be an icoscelles rigth triangle.
Hence PQ = OPsqr2
We know OP = 2
Hence PQ = 2Sqr2

Now lets take s = 1

Then t = sqr3
PQ^2 = (-sqr3 - 1)^2 +(1-sqr3)^2
 = 8
 = 2sqr2

hence proved.
Answer s = 1

shoonya · Answer

consider a point R along the y axis where the semicircle intersects the y axis.

we know that the radius is 2.

if angle <ROP is k, k can be determined as

cos(k) = 1/2

=> k = 60 degrees.

so angle ROQ is 90-60=30 degrees.

s is the x co-ordinate of Q.


cos(90-30) = s/2
s = 2*cos60 = 1.

rajiv_av · Answer

the triangle is an right angle isoceles triangle. so isnt the answer root 3

jaynayak · Answer

No you are taking the picture as you are seeing it. 
Take a graph and draw a right angled isosceles triangle..... the points can be anywhere as long as the slope is -1 .

gmatmathguru · Answer

There are several other ways to solve the problem

Method I:

Note that the radius of the circule = OP = OQ = sqrt(3+1) = 2 units.

therefore, s^2+t^2 = 4 ...equation 1.

given, OP and OQ are perpendicular, the product of the slopes should be -1. Hence, -1/sqrt(3) * t/s = -1 or t = s * sqrt(3) or t^2 = s^2 * 3 ..equation 2.

Substitute t^2 = 3* s^2 into equation 1 to get s = 1.

Method II:

The angle in a semi-circle is 180 degrees.

Draw a perpendicular from P and Q onto X axis and call the intersection points with x-axis P1 and Q1 respectively. You should have two right triangles P1OP and Q1OQ.

Note that angle P1OP = 30 degrees because tan30 = 1/sqrt(3).

which means, Q1OQ = 180-90-30=60 degrees.

because tan60 = sqrt(3), s = 1, and t = sqrt(3).

-mathguru

gmatmathguru · Answer

There are several other ways to solve the problem

Method I:

Note that the radius of the circule = OP = OQ = sqrt(3+1) = 2 units.

therefore, s^2+t^2 = 4 ...equation 1.

given, OP and OQ are perpendicular, the product of the slopes should be -1. Hence, -1/sqrt(3) * t/s = -1 or t = s * sqrt(3) or t^2 = s^2 * 3 ..equation 2.

Substitute t^2 = 3* s^2 into equation 1 to get s = 1.

Method II:

The angle in a semi-circle is 180 degrees.

Draw a perpendicular from P and Q onto X axis and call the intersection points with x-axis P1 and Q1 respectively. You should have two right triangles P1OP and Q1OQ.

Note that angle P1OP = 30 degrees because tan30 = 1/sqrt(3).

which means, Q1OQ = 180-90-30=60 degrees.

because tan60 = sqrt(3), s = 1, and t = sqrt(3).

-mathguru

GMAT prep Q4

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