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# In the xy-coordinate system, rectangle ABCD is inscribed within a circ

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In the xy-coordinate system, rectangle ABCD is inscribed within a circ  [#permalink]

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24 Mar 2011, 22:59
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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

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Re: In the xy-coordinate system, rectangle ABCD is inscribed wit  [#permalink]

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14 Jul 2013, 22:53
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skamal7 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

If you are able to get base as 10 please explain how did you find that out?

x^2 + y^2 = 25 is the equation of the circle centered at the origin and radius of $$\sqrt{25}=5$$ (check here for more: math-coordinate-geometry-87652.html):
Attachment:

Inscibed rectangle.png [ 8.51 KiB | Viewed 12052 times ]
AC = diameter of the circle = diagonal of the rectangle = 2*radius = 10.

B lies on y = 3x + 15 as well as on x^2 + y^2 = 25. Solve to get the y-coordinate of B: y=0 (discard) or y=3. Now, since y-coordinate of B is 3, then the area of triangle ABC is 1/2*10*3=15, this the area of the rectangle ABCD is twice of that: area=2*15=30.

Hope it's clear.
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Re: In the xy-coordinate system, rectangle ABCD is inscribed within a circ  [#permalink]

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25 Mar 2011, 00:29
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AnkitK wrote:
in the xy axis ,rectangle ABCD is inscribed within a circle having equation x^2 +y^2 =25 .Line segment AC is a diagonal of rectangle and lies on the x axis.Vertex B lies in quadrant 2 and vertex D lies in quadrant 4.If side BC lies on line y=3x+15 ,what is area of rectangle?

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

Please see attached image for visual clarification.

Let's see what has been provided in the question.

Circle's equation: $$x^2+y^2=5^2$$
Circle has a radius of 5 and is centered at (0,0)

AC is the diagonal of the rectangle and lies on x-axis; means AC=10

B lies in 2nd quadrant and D lies in 4th quadrant. See the image.

BC lies on line "y=3x+15". B and C are two vertices of the rectangle. We can find B and C if we find the solutions for x and y for both line and circle. Line y=3x+15 must intersect the circle on two points giving us the vertices B and C. These two points can be found by solving the simultaneous equations for the circle and the line.

Line: $$y=3x+15$$ ------ 1
Circle: $$y^2+x^2=25$$ ------ 2

Substituting 1 in 2:
$$(3x+15)^2+x^2=25$$
$$9x^2+225+90x+x^2=25$$
$$10x^2+200+90x=0$$
$$x^2+9x+20=0$$
$$x^2+9x+20=0$$
$$(x+5)(x+4)=0$$
x=-5 and x=-4

if x=-5; y= 3x+15 = 3*(-5)+15=0
if x=-4; y= 3x+15 = 3*(-4)+15=3

We found the vertices B and C now; B(-4,3) and C(-5,0)

Length of BC;
Distance between two points $$(x_1,y_1) and (x_2,y_2)$$ is found using following formula:
$$BC = sqrt{(y_2-y_1)^2+(x_2-x_1)^2}$$

Distance between B(-4,3) and C(-5,0)
$$BC = sqrt{(0-3)^2+(-5-(-4))^2} = sqrt{9+1} = sqrt{10}$$

We now know $$BC=sqrt{10} & AC=10$$

We can find AB; $$\triangle{ABC}$$ is a right angled triangle with hypotenuse as AC. We can use Pythagoras theorem to find AB

$$(AC)^2=(AB)^2+(BC)^2$$
$$10^2=(AB)^2+(\sqrt{10})^2$$
$$100=(AB)^2+10$$
$$(AB)^2=100-10=90$$
$$AB=\sqrt{90}$$

Area of the rectangle ABCD
$$BC*AB = \sqrt{10}*\sqrt{90} = \sqrt{900} = 30$$

Ans: "B"
Attachments

inscribed_rectangle.PNG [ 14.63 KiB | Viewed 10804 times ]

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Re: In the xy-coordinate system, rectangle ABCD is inscribed within a circ  [#permalink]

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Updated on: 25 Mar 2011, 00:08
3
First find the intercept of the line and circle.
x^2 + y^2 = 25 ----1)
y = 3x + 15------2)

Hence x^2 + (3x+15)^2 = 25
x = -5, -4
Since we are given B is in 2nd quadrant its coordinate is (-4,3) i.e. (-4)^2 + 3^2 = 25
Since we are given D is in 4th quadrant its coordinate is (4,-3) i.e. 4^2 + (-3)^2 = 25

We already know the points C (-5,0) and A (5,0).

b = Length BC = B(-4,3) to C(-5,0) = sqrt(1^2 + 3^2) = sqrt(10)
l = Length BA = B(-4,3) to A(5,0) = sqrt(9^2 + 3^2) = sqrt(90)

Area of the rectangle = b * l = sqrt(10) * sqrt(90) = sqrt(900) = 30. Answer B.

PS : I still want someone to verify all the above statements.

AnkitK wrote:
in the xy axis ,rectangle ABCD is inscribed within a circle having equation x^2 +y^2 =25 .Line segment AC is a diagonal of rectangle and lies on the x axis.Vertex B lies in quadrant 2 and vertex D lies in quadrant 4.If side BC lies on line y=3x+15 ,what is area of rectangle?

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

Originally posted by gmat1220 on 24 Mar 2011, 23:22.
Last edited by gmat1220 on 25 Mar 2011, 00:08, edited 2 times in total.
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Re: In the xy-coordinate system, rectangle ABCD is inscribed within a circ  [#permalink]

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25 Mar 2011, 04:52
1
fluke wrote:
AnkitK wrote:
in the xy axis ,rectangle ABCD is inscribed within a circle having equation x^2 +y^2 =25 .Line segment AC is a diagonal of rectangle and lies on the x axis.Vertex B lies in quadrant 2 and vertex D lies in quadrant 4.If side BC lies on line y=3x+15 ,what is area of rectangle?

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

Please see attached image for visual clarification.

Let's see what has been provided in the question.

Circle's equation: $$x^2+y^2=5^2$$
Circle has a radius of 5 and is centered at (0,0)

AC is the diagonal of the rectangle and lies on x-axis; means AC=10

B lies in 2nd quadrant and D lies in 4th quadrant. See the image.

BC lies on line "y=3x+15". B and C are two vertices of the rectangle. We can find B and C if we find the solutions for x and y for both line and circle. Line y=3x+15 must intersect the circle on two points giving us the vertices B and C. These two points can be found by solving the simultaneous equations for the circle and the line.

Line: $$y=3x+15$$ ------ 1
Circle: $$y^2+x^2=25$$ ------ 2

Substituting 1 in 2:
$$(3x+15)^2+x^2=25$$
$$9x^2+225+90x+x^2=25$$
$$10x^2+200+90x=0$$
$$x^2+9x+20=0$$
$$x^2+9x+20=0$$
$$(x+5)(x+4)=0$$
x=-5 and x=-4

if x=-5; y= 3x+15 = 3*(-5)+15=0
if x=-4; y= 3x+15 = 3*(-4)+15=3

We found the vertices B and C now; B(-4,3) and C(-5,0)

Length of BC;
Distance between two points $$(x_1,y_1) and (x_2,y_2)$$ is found using following formula:
$$BC = sqrt{(y_2-y_1)^2+(x_2-x_1)^2}$$

Distance between B(-4,3) and C(-5,0)
$$BC = sqrt{(0-3)^2+(-5-(-4))^2} = sqrt{9+1} = sqrt{10}$$

We now know $$BC=sqrt{10} & AC=10$$

We can find AB; $$\triangle{ABC}$$ is a right angled triangle with hypotenuse as AC. We can use Pythagoras theorem to find AB

$$(AC)^2=(AB)^2+(BC)^2$$
$$10^2=(AB)^2+(\sqrt{10})^2$$
$$100=(AB)^2+10$$
$$(AB)^2=100-10=90$$
$$AB=\sqrt{90}$$

Area of the rectangle ABCD
$$BC*AB = \sqrt{10}*\sqrt{90} = \sqrt{900} = 30$$

Ans: "B"

Btw, once we solve for the coordinates of B, we can also see that the y-coordinate of B is the height of triangle BCA which has base as 10 and hence area of triangle BCA is 1/2*3*10 = 15 and area of rectangle is twice of that, so area is 30.
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Re: In the xy-coordinate system, rectangle ABCD is inscribed within a circ  [#permalink]

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25 Mar 2011, 05:03
beyondgmatscore wrote:
~fluke - whats a quick way to plot and label such figures and attach them to one's posts? This can be really useful while discussing co-ordinate geometry problems.

It comes handy indeed!!! I use "The Geometer's Sketchpad".

http://www.themathlab.com/toolbox/geometry%20stuff/geosketch.htm
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Re: In the xy-coordinate system, rectangle ABCD is inscribed within a circ  [#permalink]

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25 Mar 2011, 19:26
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That is a great solution fluke and good catch beyondgmatscore.

I would like to tell you guys my first instinct when I looked at the question. I started drawing the diagram while reading the question and saw that BC has equation y = 3x + 15. What I said to myself was that slope of BC is 3 which means that if x changes by 1, y changes by 3. So if x becomes -4, y becomes 3. Since the radius of the circle is 5 so OB is 5 and it was clear that the point (-4, 3) lies on the circle and is B since the line intersects with the circle on only 2 points... So I didn't have to solve for the line and the circle. If you start thinking logically, you will be surprised how many many things suddenly fall into place in GMAT.
Of course, a valid concern is that what if co-ordinates of B were not integrals... Since the area was an integer, I was more inclined towards integers as B's co-ordinates. Also, as I have said before, GMAT is considerate that way - they give you numbers that fall beautifully in place. They are testing your aptitude and innovative ability, not calculation skills.
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In the xy-coordinate system, rectangle ABCD is inscribed wit  [#permalink]

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14 Jul 2013, 22:27
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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

If you are able to get base as 10 please explain how did you find that out?
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Re: In the xy-coordinate system, rectangle ABCD is inscribed wit  [#permalink]

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14 Jul 2013, 23:05
Bunuel
Can you tell me how did you get the base of the triangle as 10?

My thinking was since ABC is a right angled triangle Diamater of the triangle will be the hypotenuse. Here Diameter of circle(hpontenuse of triangle) will be 10.SInce the hypotenuse of triangle ABC is 10 i thought the base and height should be AB and BC.
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Re: In the xy-coordinate system, rectangle ABCD is inscribed wit  [#permalink]

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14 Jul 2013, 23:07
skamal7 wrote:
Bunuel
Can you tell me how did you get the base of the triangle as 10?

My thinking was since ABC is a right angled triangle Diamater of the triangle will be the hypotenuse. Here Diameter of circle(hpontenuse of triangle) will be 10.SInce the hypotenuse of triangle ABC is 10 i thought the base and height should be AB and BC.

It's explained in the solution above:
AC = diameter of the circle = diagonal of the rectangle = 2*radius = 10.

Does this make sense?
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Re: In the xy-coordinate system, rectangle ABCD is inscribed wit  [#permalink]

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14 Jul 2013, 23:12
I am not understanding how the diagonal of the rectangle 10 becomes base of the triangle ABC.. Hope my question is clear to you?
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Re: In the xy-coordinate system, rectangle ABCD is inscribed wit  [#permalink]

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14 Jul 2013, 23:15
skamal7 wrote:
I am not understanding how the diagonal of the rectangle 10 becomes base of the triangle ABC.. Hope my question is clear to you?

Your question does not make much sense...

The diagonal of ABCD is AC, the base of ABC is AC... Refer to the diagram in my post.
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Re: In the xy-coordinate system, rectangle ABCD is inscribed wit  [#permalink]

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12 Sep 2013, 03:24
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ghanashyam250 wrote:
In the question, it is stated that side BC lies in the line Y=3X+15, but in the solution you are taking about point B. am I missing something ? Please clarify...

please solve the equation .. as we know the diagonals intersect X axis .. so it means the value of y = 0. so take equation of the circle

x^2 + y^2 = 25.
(as y = 0 at x axis) so
x^2 = 25 => x = +-5 .. so one point on x axis is +5 and the other is -5. the distance between 5 and -5 is 10. Hope its clear now.
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Re: In the xy-coordinate system, rectangle ABCD is inscribed wit  [#permalink]

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12 Sep 2013, 03:26
1
Bunuel wrote:
skamal7 wrote:
B lies on y = 3x + 15 as well as on x^2 + y^2 = 25. Solve to get the y-coordinate of B: y=0 (discard) or y=3. Now, since y-coordinate of B is 3, then the area of triangle ABC is 1/2*10*3=15, this the area of the rectangle ABCD is twice of that: area=2*15=30.

Hope it's clear.

Hello Bunuel, how did you get y = 0 , 3 ??
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Re: In the xy-coordinate system, rectangle ABCD is inscribed wit  [#permalink]

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12 Sep 2013, 03:40
muzammil wrote:
Bunuel wrote:
skamal7 wrote:
B lies on y = 3x + 15 as well as on x^2 + y^2 = 25. Solve to get the y-coordinate of B: y=0 (discard) or y=3. Now, since y-coordinate of B is 3, then the area of triangle ABC is 1/2*10*3=15, this the area of the rectangle ABCD is twice of that: area=2*15=30.

Hope it's clear.

Hello Bunuel, how did you get y = 0 , 3 ??

Solve for y: y = 3x + 15 and x^2 + y^2 = 25.
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Re: In the xy-coordinate system, rectangle ABCD is inscribed within a circ  [#permalink]

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16 Dec 2014, 23:50
circle has a radius = 5.
lets get x and y intercept = y=15 and x=-5. ok x=-5 that mean side BC is a line y = 3x+15. Now since BC is one side of the rectangle and AC is diameter then we know that angleB is right angle. So we can see that ABC is a 3-4-5 triangle. Co-ordinates of B (-4,3). Now since its a rectangle and diagonal bisect it into right traingle then area of rectangle = 2*0.5*10*3 (y=3 is basically height of triangle ABC) = 30.

AnkitK 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

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Re: In the xy-coordinate system, rectangle ABCD is inscribed within a circ  [#permalink]

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17 Dec 2014, 16:49
I took the same approach as fluke. My only question is why do we assume that the diameter is the full diameter of the circle? why cant it be -4,0 to 4,0?
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In the xy-coordinate system, rectangle ABCD is inscribed within a circ  [#permalink]

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17 Dec 2014, 22:57
bankerboy30 wrote:
I took the same approach as fluke. My only question is why do we assume that the diameter is the full diameter of the circle? why cant it be -4,0 to 4,0?

The circle is centered at (0, 0) with radius 5. A and C are two points lying on the circle on x axis (when a rectangle is inscribed in a circle, all its 4 vertices must lie on the circle so A and C both lie on the circle). So AC is a line whose both end points lie on the circle and it passes through the center of the circle. Such a line is the diameter of the circle.
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Re: In the xy-coordinate system, rectangle ABCD is inscribed within a circ  [#permalink]

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06 Apr 2015, 03:29
I dont thing this question is from gmatprep.

how to solvve the equation 3x+15= y and x^+y^2=25,

it is not easy to solve.

a small change in the math question leads to a quite different question with much higher diffuculty , which no one can sovle in the test room, for 2 minutes.
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Re: In the xy-coordinate system, rectangle ABCD is inscribed within a circ  [#permalink]

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21 Apr 2015, 17:17
Hello Karishma and Bunuel,

I would kindly ask on your opinion or where Im going wrong in my reasoning.

So in order to do this question fast i tried to come up with a shortcut. I thought since the radius is 5 => the dimater is 10 and in the same time it will be a hypothenuse of the triangle ABC which is a right triangle. So having that in mind I was thinking to use the 3 - 4 -5 right triangle and the sides of the triangle would be 3*2=6 and 4*2=8, or the area of the rectangle to be 6*8=48 which is not correct.

please correct me why I cant use this 3-4-5 property here

Thanks soooo muchh
Re: In the xy-coordinate system, rectangle ABCD is inscribed within a circ &nbs [#permalink] 21 Apr 2015, 17:17

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