DarshanGaikwad wrote:
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
Sir, why are you including AC's length to calculate the triangle's area? AC is actually the hypoteneus of the ABC. Also, if you can find AB, then the area of the rectangle is BC * AB = 3 * 9 = 27 which is < B. Answer should be B. Where am I going wrong?
DarshanGaikwad Carcass