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Rectangle inscribed in a circle

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Rectangle inscribed in a circle

by surajgarg » Fri Aug 06, 2010 12:25 pm
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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by clock60 » Fri Aug 06, 2010 2:25 pm
i got B-30
reasoning
at first let us find the point where line y=3x+15 intersect with the circle x^2+y^2=5^2 with radiu=5 ( as circle with the center in the origin is given wiith the equation x^2+y^2=R^2)
x^2+(3x+15)^2=25
solving and simplifing
x^2+9x+20=0
x1=-5,x2=-4. we need here x2=-4. ( as -5 already given in the query)
insert x2=-4 in the y=3x+15,
y=-12+15=3
here y=3 is the height of triangle ABC and we can find its area
S=1/2*10(base)*3(height)=15
our rectangle consists of two equal triangles with area =15
so area of rectangle ABCD=2*15=30
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by surajgarg » Fri Aug 06, 2010 8:48 pm
clock60 wrote:i got B-30
reasoning
at first let us find the point where line y=3x+15 intersect with the circle x^2+y^2=5^2 with radiu=5 ( as circle with the center in the origin is given wiith the equation x^2+y^2=R^2)
x^2+(3x+15)^2=25
solving and simplifing
x^2+9x+20=0
x1=-5,x2=-4. we need here x2=-4. ( as -5 already given in the query)
insert x2=-4 in the y=3x+15,
y=-12+15=3
here y=3 is the height of triangle ABC and we can find its area
S=1/2*10(base)*3(height)=15
our rectangle consists of two equal triangles with area =15
so area of rectangle ABCD=2*15=30
Thanks, I overlooked the fact that the line BC will intersect the circle :D
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by ankurmit » Sun Aug 08, 2010 6:02 am
Can anyone elaborate it with a figure..I am not able to draw it..
--------
Ankur mittal
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by zareentaj » Mon Aug 09, 2010 11:16 pm
clock60 wrote:i got B-30
reasoning
at first let us find the point where line y=3x+15 intersect with the circle x^2+y^2=5^2 with radiu=5 ( as circle with the center in the origin is given wiith the equation x^2+y^2=R^2)
x^2+(3x+15)^2=25
solving and simplifing
x^2+9x+20=0
x1=-5,x2=-4. we need here x2=-4. ( as -5 already given in the query)
insert x2=-4 in the y=3x+15,
y=-12+15=3
here y=3 is the height of triangle ABC and we can find its area
S=1/2*10(base)*3(height)=15
our rectangle consists of two equal triangles with area =15
so area of rectangle ABCD=2*15=30
Thanks clock60. for solve this question ,i got idea for this type of questions.
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