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Co-ordinate Geometry - How to solve this one?

by vaibhavob » Mon Jan 31, 2011 4:36 am
In the xy - coordinate sytem, the rectangle ABCD is inscribed within a circle having the equation x^2 + y^2 = 25. Line segment AC is a diagnol of the rectangle and lies on the x - axis. Vertex B lies in quadrant 2 and vertex in quadrant 4. If side BC lies on line y=3x+15, what is the area of rectangle ABCD?

A.15
B.30
C.40
D.45
E.50

Am able to understand the circle part but not clear on how to use the diagnol information and line BC given?

Any clues with regards to the above would be highly appreciative. sorry for the use of numberal for quadrants rather than romans.

Regards,

Vaibhav[/spoiler][/list]
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by GMATGuruNY » Mon Jan 31, 2011 5:54 am
In the xy-coordinate system, rectangle ABCD is inscribed within a circle having the equation x^2 + y^2 = 25. Line segment AC is a diagonal of the rectangle and lies on the x-axis. Vertex B lies in quadrant II and vertex D lies in quadrant IV. If side BC lies on line y=3x+15, what is the area of rectangle ABCD?

A. 15
B. 30
C. 40
D. 45
E. 50
Image

A GMAT-friendly way to solve:

The formula for a circle is x^2 + y^2 = r^2. Since the equation given for the circle is x^2 + y^2 = 25, r = 5.

Looking at triangle ABC:
b = 2r = 2*5 = 10
h = height
line segment OB = r = 5.

Since all the given values are integers, OB is likely the hypotenuse of a 3-4-5 triangle. Thus, the coordinates of point B are either (-3,4) or (-4,3).

Only point (-4,3) works in y = 3x + 15.

Thus, h=3.
The area of triangle ABC = 1/2bh = 1/2*10*3 = 15.
Since rectangle ABCD = 2*(triangle ABC), the area of ABCD = 2*15 = 30.

The correct answer is B.
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by MAAJ » Mon Jan 31, 2011 9:55 am
GMATGuruNY wrote:
A GMAT-friendly way to solve:

The formula for a circle is x^2 + y^2 = r^2. Since the equation given for the circle is x^2 + y^2 = 25, r = 5.

Looking at triangle ABC:
b = 2r = 2*5 = 10
h = height
line segment OB = r = 5.

Since all the given values are integers, OB is likely the hypotenuse of a 3-4-5 triangle. Thus, the coordinates of point B are either (-3,4) or (-4,3).

Only point (-4,3) works in y = 3x + 15.

Thus, h=3.
The area of triangle ABC = 1/2bh = 1/2*10*3 = 15.
Since rectangle ABCD = 2*(triangle ABC), the area of ABCD = 2*15 = 30.

The correct answer is B.
I'm wondering what other logic can you apply to determine the height, if there is any? couldn't find one.... :cry:
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by Night reader » Mon Jan 31, 2011 10:34 am
MAAJ wrote:
GMATGuruNY wrote:
A GMAT-friendly way to solve:

The formula for a circle is x^2 + y^2 = r^2. Since the equation given for the circle is x^2 + y^2 = 25, r = 5.

Looking at triangle ABC:
b = 2r = 2*5 = 10
h = height
line segment OB = r = 5.

Since all the given values are integers, OB is likely the hypotenuse of a 3-4-5 triangle. Thus, the coordinates of point B are either (-3,4) or (-4,3).

Only point (-4,3) works in y = 3x + 15.

Thus, h=3.
The area of triangle ABC = 1/2bh = 1/2*10*3 = 15.
Since rectangle ABCD = 2*(triangle ABC), the area of ABCD = 2*15 = 30.

The correct answer is B.
I'm wondering what other logic can you apply to determine the height, if there is any? couldn't find one.... :cry:
I can't say too much about height, but...
we know that to be rectangle (not square, as one vertice is quadr. II) one side should not be equal the other side --> a>b OR a<b
we have two right triangles (because they share the same diagonal) and we can induce a^2+b^2=c^2 where c is both the diagonal and hypotenuse; a^+b^2=100
now (a+b)^2-100=2ab; since a>b OR a<b we have either a>Sqrt(2)*5 OR a<Sqrt(2)*5 --> a>7 (a<7), then b<7 (b>7)
plug in sample values --> a=6, b=8 and (6+8)^-100=2(6*8) --> 196-100=2*48 , so we can state that the square of our rectangle is 48 <-- ops it's not in the answer choices. Are we wrong in our solution? I don't think so let BA=8, BC=6 then BA^2 + BC^2 = 100 and Sqrt(100)=10 OR 2r=2*5 and the diagonal (our hypotenuse) is equal to 2r.

Basically, we can say that this problem has two answers, depending on solution methods ...
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by GMATGuruNY » Mon Jan 31, 2011 12:37 pm
GMATGuruNY wrote:
In the xy-coordinate system, rectangle ABCD is inscribed within a circle having the equation x^2 + y^2 = 25. Line segment AC is a diagonal of the rectangle and lies on the x-axis. Vertex B lies in quadrant II and vertex D lies in quadrant IV. If side BC lies on line y=3x+15, what is the area of rectangle ABCD?

A. 15
B. 30
C. 40
D. 45
E. 50
Image

A GMAT-friendly way to solve:

The formula for a circle is x^2 + y^2 = r^2. Since the equation given for the circle is x^2 + y^2 = 25, r = 5.

Looking at triangle ABC:
b = 2r = 2*5 = 10
h = height
line segment OB = r = 5.

Since all the given values are integers, OB is likely the hypotenuse of a 3-4-5 triangle. Thus, the coordinates of point B are either (-3,4) or (-4,3).

Only point (-4,3) works in y = 3x + 15.

Thus, h=3.
The area of triangle ABC = 1/2bh = 1/2*10*3 = 15.
Since rectangle ABCD = 2*(triangle ABC), the area of ABCD = 2*15 = 30.

The correct answer is B.
The traditional way to determine the height of triangle ABC would be to recognize that the coordinates of point B must satisfy both the equation of the circle (x^2 + y^2 = 25) and that of the line (y = 3x+15).

Substituting 3x+15 for y in the equation of the circle, we get:

x^2 + (3x+15)^2 = 25.
x^2 + 9x^2 + 90x + 225 = 25.
10x^2 + 90x + 200 = 0.
x^2 + 9x + 20 = 0.
(x+5)(x+4) = 0.

x= -5 is the x coordinate of point C.
x= -4 is the x coordinate of point B.

To determine the y coordinate of point B, we plug x = -4 into y = 3x+15:
y = 3(-4) + 15 = 3.

Thus, in triangle ABC, h=3.

If I were your math teacher in school, I would encourage you to solve the traditional way. But as your GMAT tutor, I would be disappointed if you solved the traditional way. We get no more points for doing all the messy algebra. The method that I first suggested is much more efficient.
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by GMATGuruNY » Mon Jan 31, 2011 1:14 pm
Night reader wrote:
MAAJ wrote:
GMATGuruNY wrote:
A GMAT-friendly way to solve:

The formula for a circle is x^2 + y^2 = r^2. Since the equation given for the circle is x^2 + y^2 = 25, r = 5.

Looking at triangle ABC:
b = 2r = 2*5 = 10
h = height
line segment OB = r = 5.

Since all the given values are integers, OB is likely the hypotenuse of a 3-4-5 triangle. Thus, the coordinates of point B are either (-3,4) or (-4,3).

Only point (-4,3) works in y = 3x + 15.

Thus, h=3.
The area of triangle ABC = 1/2bh = 1/2*10*3 = 15.
Since rectangle ABCD = 2*(triangle ABC), the area of ABCD = 2*15 = 30.

The correct answer is B.
I'm wondering what other logic can you apply to determine the height, if there is any? couldn't find one.... :cry:
I can't say too much about height, but...
we know that to be rectangle (not square, as one vertice is quadr. II) one side should not be equal the other side --> a>b OR a<b
we have two right triangles (because they share the same diagonal) and we can induce a^2+b^2=c^2 where c is both the diagonal and hypotenuse; a^+b^2=100
now (a+b)^2-100=2ab; since a>b OR a<b we have either a>Sqrt(2)*5 OR a<Sqrt(2)*5 --> a>7 (a<7), then b<7 (b>7)
plug in sample values --> a=6, b=8 and (6+8)^-100=2(6*8) --> 196-100=2*48 , so we can state that the square of our rectangle is 48 <-- ops it's not in the answer choices. Are we wrong in our solution? I don't think so let BA=8, BC=6 then BA^2 + BC^2 = 100 and Sqrt(100)=10 OR 2r=2*5 and the diagonal (our hypotenuse) is equal to 2r.

Basically, we can say that this problem has two answers, depending on solution methods ...
Using your values for BC and AB, point B won't lie on the line y = 3x+15. The correct values for BC and AB will not only satisfy your equation above but also place point B on the line y=3x+15.

Since h=3, we can use the pythagorean theorem to determine that the length of BC = √10 and the length of AB = √90:

Image
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by Night reader » Mon Jan 31, 2011 3:08 pm
@Mitch, y=3x+15 when x=-5, y=0 and I did not break out this condition. BC=6, r=5 --> Sqrt(36-u^2)={h, height of yours ;) }=Sqrt(25-v^2) and u+v=5;

36-25=u^2-v^2, 11=u^2-v^2 OR (u-v)(u+v)=11 --> (u-v)*5=11, u-v=11/5 and u+v=5 --> sum up two statements and get 2u=5+11/5 -->
u=18/5; v=18/5-11/5=7/5

now h=Sqrt[36-(18/5)^2] OR h=Sqrt[25-(7/5)^2] <-- these are the same
h=Sqrt[23.04]~ approximately is 4.8

test for the equation of line y=3x+15 --> y~4.8 and our x has value {-3.6} or u line distance on x-abscess 18/5

Everything has been cleared. The problem has two answers so far :)



GMATGuruNY wrote:
Night reader wrote:
MAAJ wrote:
GMATGuruNY wrote:
A GMAT-friendly way to solve:

The formula for a circle is x^2 + y^2 = r^2. Since the equation given for the circle is x^2 + y^2 = 25, r = 5.

Looking at triangle ABC:
b = 2r = 2*5 = 10
h = height
line segment OB = r = 5.

Since all the given values are integers, OB is likely the hypotenuse of a 3-4-5 triangle. Thus, the coordinates of point B are either (-3,4) or (-4,3).

Only point (-4,3) works in y = 3x + 15.

Thus, h=3.
The area of triangle ABC = 1/2bh = 1/2*10*3 = 15.
Since rectangle ABCD = 2*(triangle ABC), the area of ABCD = 2*15 = 30.

The correct answer is B.
I'm wondering what other logic can you apply to determine the height, if there is any? couldn't find one.... :cry:
I can't say too much about height, but...
we know that to be rectangle (not square, as one vertice is quadr. II) one side should not be equal the other side --> a>b OR a<b
we have two right triangles (because they share the same diagonal) and we can induce a^2+b^2=c^2 where c is both the diagonal and hypotenuse; a^+b^2=100
now (a+b)^2-100=2ab; since a>b OR a<b we have either a>Sqrt(2)*5 OR a<Sqrt(2)*5 --> a>7 (a<7), then b<7 (b>7)
plug in sample values --> a=6, b=8 and (6+8)^-100=2(6*8) --> 196-100=2*48 , so we can state that the square of our rectangle is 48 <-- ops it's not in the answer choices. Are we wrong in our solution? I don't think so let BA=8, BC=6 then BA^2 + BC^2 = 100 and Sqrt(100)=10 OR 2r=2*5 and the diagonal (our hypotenuse) is equal to 2r.

Basically, we can say that this problem has two answers, depending on solution methods ...
Using your values for BC and AB, point B won't lie on the line y = 3x+15. The correct values for BC and AB will not only satisfy your equation above but also place point B on the line y=3x+15.

Since h=3, we can use the pythagorean theorem to determine that the length of BC = √10 and the length of AB = √90:

Image
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by MAAJ » Mon Jan 31, 2011 4:01 pm
D'OH!!!! missed that one!!! thanks!
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by GMATGuruNY » Mon Jan 31, 2011 4:13 pm
@Mitch, y=3x+15 when x=-5, y=0 and I did not break out this condition. BC=6, r=5 --> Sqrt(36-u^2)={h, height of yours ;) }=Sqrt(25-v^2) and u+v=5;

36-25=u^2-v^2, 11=u^2-v^2 OR (u-v)(u+v)=11 --> (u-v)*5=11, u-v=11/5 and u+v=5 --> sum up two statements and get 2u=5+11/5 -->
u=18/5; v=18/5-11/5=7/5

now h=Sqrt[36-(18/5)^2] OR h=Sqrt[25-(7/5)^2] <-- these are the same
h=Sqrt[23.04]~ approximately is 4.8

test for the equation of line y=3x+15 --> y~4.8 and our x has value {-3.6} or u line distance on x-abscess 18/5

Everything has been cleared. The problem has two answers so far :)
Night Reader, I'm not quite able to follow your reasoning. Regardless, the line y = 3x+15 and the circle x^2 + y^2 = 25 intersect at two points only: (-5,0) and (-4,3). It is physically and mathematically impossible for there to be any other points of intersection. There is no other coordinate pair that will satisfy both equations.

While it is possible for the x axis to be the hypotenuse of a 6-8-10 triangle, the coordinates of the vertex at the right angle will not satisfy both y = 3x+15 and x^2 + y^2 = 25.

The picture and the explanation below show why AC=8 and BC=6 are not possible:

Image

Looking at the picture above:
(x+5)^2 + y^2 = 64.
(5-x)^2 + y^2 = 36.

Subtracting the second equation from the first, we get:
(x+5)^2 - (5-x)^2 = 28.
x^2 + 10x + 25 - (25 - 10x + x^2) = 28.
20x = 28
x = 28/20 = 7/5
Thus, in the 6-8-10 triangle, the x coordinate of the height is x= -7/5.

Since point B must lie on the line y=3x+15, we plug x = -7/5 into y=3x+15:
y = 3(-7/5) + 15 = -21/5 + 75/5 = 54/5.

In order for B to lie on the circle, (-7/5, 54/5) must satisfy the equation x^2 + y^2 = 25:
(-7/5)^2 + (54/5)^2 = 25
49/25 + 2916/25 = 25
2965/25 = 25
118.6 = 25
Doesn't work.

Hope this helps!
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by Night reader » Mon Jan 31, 2011 4:24 pm
:( please explain why you arbitrarily assign (x-5) and x different line segments?
you found 7/5, please look up the solution presented above where (x-5) and x are not arbitrarily assigned but evenly spaced, given the vertex condition - placed in quadr.II for B [BC=6] so x is not -7/5 but is -18/5...
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by GMATGuruNY » Mon Jan 31, 2011 4:29 pm
Night reader wrote::( please explain why you arbitrarily assign (x-5) and x different line segments?
you found 7/5, please look up the solution presented above where (x-5) and x are not arbitrarily assigned but evenly spaced, given the condition of vertex placed in quadr.II for B [BC=6] so x is not -7/5 but is -18/5...
The assignments follow the conditions given in the problem. The center of the circle is (0,0) and the radius of the circle is 5. So the two segments to the left of the y axis are x and 5-x.
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by Night reader » Mon Jan 31, 2011 4:51 pm
you really don't follow?

64-(5-x+5)^2=36-x^2
x=18/5

arbitrary assignment of x and 5-x :(

x in our equations and x of the equation line are not the same parameters; so you seem making arbitrary assignment here
instead of x and 5-x we could have 'bob' and '5-bob' :)
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by GMATGuruNY » Mon Jan 31, 2011 5:00 pm
Night reader wrote:you really don't follow?

64-(5-x+5)^2=36-x^2
x=18/5

arbitrary assignment of x and 5-x :(

x in our equations and x of the equation line are not the same parameters; so you seem making arbitrary assignment here

Image
Ah, now I understand.

In my drawing, x is distance from the y axis to the height. Thus, -x is the x coordinate of point B.

In your drawing, 5-x is the distance from the y axis to the height. Thus, x-5 will be the x coordinate of point B. Once you've determined that x=18/5, then the x coordinate of point B = x-5 = 18/5 - 5 = 18/5 - 25/5 = -7/5.

This is the same result that I got. As explained above, this x value will not satisfy both y = 3x+15 and x^2 + y^2 = 25. Hence, ABC cannot be a 6-8-10 triangle.

Is the situation now clear?
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by Night reader » Mon Jan 31, 2011 5:18 pm
and the equation of line y=3x+15 and the equation of circle x^2+y^2=25 will cross in point y, agree --> (3x+15)^2=25-x^2
9x^2+90x+225-25+x^2=0, 10x^2+90x+200=0 OR x^2+9x+20=0, (x+5)(x+4)=0, x{-4;-5}

x can't be -18/5 neither -7/5

as the result we have not quite inscribed rectangle

thanks Mitch!
GMATGuruNY wrote:
Night reader wrote:you really don't follow?

64-(5-x+5)^2=36-x^2
x=18/5

arbitrary assignment of x and 5-x :(

x in our equations and x of the equation line are not the same parameters; so you seem making arbitrary assignment here

Image
Ah, now I understand. Using your assignments, once you've determined that x=18/5, then the x coordinate of point B = 5-x = 5 - 18/5 = -7/5. This is the same result that I got. As explained above, this x value will not satisfy both y = 3x+15 and x^2 + y^2 = 25. Hence, ABC cannot be a 6-8-10 triangle.

Is the situation now clear?
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by san03 » Tue Feb 15, 2011 7:18 am
I am using the same figure as provided earlier.
equation of BC is : y=3x+15
hence equation of AB will be y= -x/3 +C; as AB and BC are perpendicular . c is a constant
AB will satisfy pt A (5,0) hence quick substitution gives

equation of AB as y=-x/3 +5/3

solve equation of AB and BC to get Y value of B as 3.

area of ABCD is 2*{1/2*3*10}=30
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