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In the xy-plane, ABCD is a rectangle inscribed in a circle O with
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05 Sep 2021, 23:02
05 Sep 2021, 23:02
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In the xy-plane, ABCD is a rectangle inscribed in a circle O.png [ 7.9 KiB | Viewed 2464 times ]
In the xy-plane, ABCD is a rectangle inscribed in a circle O with equation \(x^2 + y^2 = 25\). Also, side BC lies on line \(y = 3x + 15\).
Quantity A |
Quantity B |
Area of rectangle ABCD |
30 |
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Re: In the xy-plane, ABCD is a rectangle inscribed in a circle O with
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07 Sep 2021, 05:03
07 Sep 2021, 05:03
1
Can anyone give a solution to this question?
I could only find the diagonal i.e. 10. Both the equations are serving the same purpose to get the x coordinated of C and A (-5,0) and (5,0) respectively.
I could only find the diagonal i.e. 10. Both the equations are serving the same purpose to get the x coordinated of C and A (-5,0) and (5,0) respectively.
Re: In the xy-plane, ABCD is a rectangle inscribed in a circle O with
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08 Sep 2021, 02:36
08 Sep 2021, 02:36
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Since ABCD is a rectangle and AC its diagonal, we know that the diagonal divides the rectangle into two equal parts.
Basically, Area of Rectangle ABCD is twice the area of Traingle ABC or Triangle ADC. Thus we can proceed to find the AREA of triangle ABC.
For that we need the height of the triangle which is nothing but the Y co-ordinate of Point B.
Here, we must realize that Point B is point of intersection of Line y=3x+15 and Circle x^2 + y^2 = 25.The co-ordinates of point B satisfy both the equations.
Thus we can substitute X = (y-15)/3 in the equation of the circle to get the Y value of point of intersection.
After solving the resulting equation we get value of Y as 3. Thus we now have the height of triangle ABC.
So Area of Triangle ABC = 0.5*10*3= 15.
Thus Area of Rectangle(ABCD) = 15*2 = 30
Hence option C
Basically, Area of Rectangle ABCD is twice the area of Traingle ABC or Triangle ADC. Thus we can proceed to find the AREA of triangle ABC.
For that we need the height of the triangle which is nothing but the Y co-ordinate of Point B.
Here, we must realize that Point B is point of intersection of Line y=3x+15 and Circle x^2 + y^2 = 25.The co-ordinates of point B satisfy both the equations.
Thus we can substitute X = (y-15)/3 in the equation of the circle to get the Y value of point of intersection.
After solving the resulting equation we get value of Y as 3. Thus we now have the height of triangle ABC.
So Area of Triangle ABC = 0.5*10*3= 15.
Thus Area of Rectangle(ABCD) = 15*2 = 30
Hence option C
