Since the equation of the circle x^2+y^2=25, it represents a circle with centre at the origin and radius 5.
The explaination about the rectagle suggests that one of the diagonals is the diameter of the circle. Therefore, the two diametrically opposite points on the X axis would be (-5,0) and (5,0) the length of the diameter being 10.
Now, to find out about the location of point B. Since it lies on the circle, it should satisfy the equation x^2+y^2=25. Moreover, since it lies on the line y=3x+15, it should satisfy this equation as well. Solving these two equations, we get the possible values of B to be (-5, 0) and (-4, 3). Since B lies in the II quadrant, we take B to be (-4, 3).
Area of the rectangle ABCD = 2 * area of the triangle ABC
Area of triangle ABC = 1/2 * base * height = 1/2 * 10 * 3 = 15
=> Area of the rectangle = 2 * 15 = 30
Edit 1: calcuation mistake.
Edit 2: Attached a file displaying the problem pictorially.
File comment: Figure for explaination
PS_Geometry.doc [25.5 KiB]
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