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The length of the tangent drawn from a point P to a circle o

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by gmatter2012 » Tue Jul 10, 2012 6:12 am
I know this looks easy , how to calculate square root(42.25)

because : 6^2 + 2.5^2 = H^2 using Pythagoras theorem so H = square root( 36+ 6.25) = sqrroot( 42.25)

answer would of course be H - 2.5
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by Anurag@Gurome » Tue Jul 10, 2012 6:23 am
gmatter2012 wrote:The length of the tangent drawn from a point P to a circle of radius 2.5 cm is 6 cm. What is the shortest distance from point P to the circle?
Image
We need to find the length of PD.

Now, OP = OD = 2.5 cm and PT = 6 cm
And, OP² = (OT² + PT²)
---> (OD + PD)² = (OT² + PT²)
---> (OD² + 2*OD*PD + PD²) = (OT² + PT²)
---> (2.5² + 2*2.5*PD + PD²) = (2.5² + 6²)
---> (PD² + 5PD) = 6²
---> (PD² + 5PD - 36) = 0
---> (PD - 4)(PD + 9) = 0

As length of a line segment cannot be negative, PD must be equal to 4

The correct answer is E.
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by Anurag@Gurome » Tue Jul 10, 2012 6:28 am
gmatter2012 wrote:I know this looks easy , how to calculate square root(42.25)

because : 6^2 + 2.5^2 = H^2 using Pythagoras theorem so H = square root( 36+ 6.25) = sqrroot( 42.25)

answer would of course be H - 2.5
You don't need to find the exact value if you proceed this way.

--> √36 < √42.25 < √49
--> 6 < √42.25 < 7
--> (6 - 2.5) < (√42.25 - 2.5) < (7 - 2.5)
--> 3.5 < (√42.25 - 2.5) < 4.5

Only option that lies between 3.5 and 4.5 is option E.
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by gmat_and_me » Tue Jul 10, 2012 6:29 am
Well, let's just remove the decimal for now. What is asked is root of 4225.
We know that this value falls between 3600 (60^2) and 4900(70^2). Assuming
that GMAT provides us with good numbers, the one and only number whose
square can result in 5 in the units place is 65. So the root of 42.25 has to be
6.5.

There is an easier way to find square of a number ending with 5. The square
will always end in 25. The most significant number will be equal to the square
of the rest of the digits added with the rest of the digit. For e.g.: 15^2. Result
will end in 25. What is left is 1. Square of 1 + 1 = 2. So square = 225. What
about 65. Result will end with 25. The most significant digits will be 6^2 + 6 = 42.
So the result is 4225.

HTH
gmatter2012 wrote:I know this looks easy , how to calculate square root(42.25)

because : 6^2 + 2.5^2 = H^2 using Pythagoras theorem so H = square root( 36+ 6.25) = sqrroot( 42.25)

answer would of course be H - 2.5
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by gmatter2012 » Tue Jul 10, 2012 6:51 am
gmatter2012 wrote:I know this looks easy , how to calculate square root(42.25)

because : 6^2 + 2.5^2 = H^2 using Pythagoras theorem so H = square root( 36+ 6.25) = sqrroot( 42.25)

answer would of course be H - 2.5
Anurag@Gurome wrote:
You don't need to find the exact value if you proceed this way.

--> √36 < √42.25 < √49
--> 6 < √42.25 < 7
--> (6 - 2.5) < (√42.25 - 2.5) < (7 - 2.5)
--> 3.5 < (√42.25 - 2.5) < 4.5

Only option that lies between 3.5 and 4.5 is option E.
This is really smart !
gmat_and_me wrote: Well, let's just remove the decimal for now. What is asked is root of 4225.
We know that this value falls between 3600 (60^2) and 4900(70^2). Assuming
that GMAT provides us with good numbers, the one and only number whose
square can result in 5 in the units place is 65. So the root of 42.25 has to be
6.5.

There is an easier way to find square of a number ending with 5. The square
will always end in 25. The most significant number will be equal to the square
of the rest of the digits added with the rest of the digit. For e.g.: 15^2. Result
will end in 25. What is left is 1. Square of 1 + 1 = 2. So square = 225. What
about 65. Result will end with 25. The most significant digits will be 6^2 + 6 = 42.
So the result is 4225.
Thank you.
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