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605-655 (Medium)|   Geometry|      
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Bunuel
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In the x-y coordinate plane shown above, point R is at the center of 16 Feb 2016, 01:01
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45% (medium)

Question Stats:

66% (01:51) correct 34% (02:04) wrong based on 124 sessions

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In the x-y coordinate plane shown above, point R is at the center of the quarter-circle arc PQ. What is the x-coordinate of point R if the radius of the quarter circle is 2?

A. \(- \sqrt{3}\)

B. \(- \sqrt{2}\)

C. \(\frac{- \sqrt{2}}{2}\)

D. \(-\frac{1}{2}\)

E. \(\sqrt{2}\)

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In the x-y coordinate plane shown above, point R is at the center of 18 Feb 2016, 05:33
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is it C? Since the inscribed triangle is 45-45-90 and the X and Y coordinates need to be equal?
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In the x-y coordinate plane shown above, point R is at the center of 18 Feb 2016, 06:57
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amtanker
is it C? Since the inscribed triangle is 45-45-90 and the X and Y coordinates need to be equal?
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Hi,
your approach has been correct, but execution has been wrong...

x and y coordinate will be the same and the hypotenuse will be the RADIUS here..
\(x^2+y^2=R^2...\\
x^2+y^2=2^2...\\
or x=y=\sqrt{2}\)

But since teh point is in a quad where x will be -ive..
\(x= - \sqrt{2}\)
so ans will be B and not C..
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In the x-y coordinate plane shown above, point R is at the center of 18 Feb 2016, 07:25
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I understand your solution, but I am confused as to why using the side length formula for the hypotenuse of a 45-45-90 triangle being equal to the shorter side times sqrt(2) gives the answer of C instead of B.

x=short sides
sqrt(2) = hypotenuse

x*sqrt(2) = 2
x = 2 / sqrt (2)

What am I doing wrong in applying this rule that gives me answer C instead of answer B?
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In the x-y coordinate plane shown above, point R is at the center of 18 Feb 2016, 08:15
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amtanker
I understand your solution, but I am confused as to why using the side length formula for the hypotenuse of a 45-45-90 triangle being equal to the shorter side times sqrt(2) gives the answer of C instead of B.

x=short sides
sqrt(2) = hypotenuse

x*sqrt(2) = 2
x = 2 / sqrt (2)

What am I doing wrong in applying this rule that gives me answer C instead of answer B?
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Hi,
You are correct in solution, but be careful in selecting choices....
X=2/√2 = √2..
Whereas C is √2/2, which is not the same as what you have got 2/√2..
Hope its clear now..
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In the x-y coordinate plane shown above, point R is at the center of 02 Mar 2017, 01:06
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Bunuel

In the x-y coordinate plane shown above, point R is at the center of the quarter-circle arc PQ. What is the x-coordinate of point R if the radius of the quarter circle is 2?

A. \(- \sqrt{3}\)

B. \(- \sqrt{2}\)

C. \(\frac{- \sqrt{2}}{2}\)

D. \(-\frac{1}{2}\)

E. \(\sqrt{2}\)

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Attachment:
2016-02-14_1401.png
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Using distance formula, x^2 + y^2=4
since point R lies on the line x=y, 2x^2=4, x^2=2, x=\sqrt{2}
Since it lies in the 3rd quadrant, x=-\sqrt{2}
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In the x-y coordinate plane shown above, point R is at the center of 15 Feb 2018, 11:54
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chetan2u
amtanker
is it C? Since the inscribed triangle is 45-45-90 and the X and Y coordinates need to be equal?

Hi,
your approach has been correct, but execution has been wrong...

x and y coordinate will be the same and the hypotenuse will be the RADIUS here..
\(x^2+y^2=R^2...\\
x^2+y^2=2^2...\\
or x=y=\sqrt{2}\)

But since teh point is in a quad where x will be -ive..
\(x= - \sqrt{2}\)
so ans will be B and not C..
Show more


Hi chetan2u

the question says that radius is JUST 2. why do you write so \(x^2+y^2=2^2\) and not \(x^2+y^2=2\) I thought I should have written so \(1^2+1^2=2\) no ? :?

Also where in the formula do you see a square root .... why do you put 2 under radical sign ? :? \(\sqrt{2}\) ? :?
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In the x-y coordinate plane shown above, point R is at the center of 15 Feb 2018, 21:11
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dave13
chetan2u
amtanker
is it C? Since the inscribed triangle is 45-45-90 and the X and Y coordinates need to be equal?

Hi,
your approach has been correct, but execution has been wrong...

x and y coordinate will be the same and the hypotenuse will be the RADIUS here..
\(x^2+y^2=R^2...\\
x^2+y^2=2^2...\\
or x=y=\sqrt{2}\)

But since teh point is in a quad where x will be -ive..
\(x= - \sqrt{2}\)
so ans will be B and not C..


Hi chetan2u

the question says that radius is JUST 2. why do you write so \(x^2+y^2=2^2\) and not \(x^2+y^2=2\) I thought I should have written so \(1^2+1^2=2\) no ? :?

Also where in the formula do you see a square root .... why do you put 2 under radical sign ? :? \(\sqrt{2}\) ? :?
Show more

hi the formula is x^2+y^2=r^2..

take it as a right angled triange..
two sides are x and y and hypotenuse is radius..
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In the x-y coordinate plane shown above, point R is at the center of 27 Jul 2025, 13:05
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