Explanation
Let's see what has been provided in the question.
Circle's equation:
x2+y2=52 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:
y2+x2=25 ------ 2
Substituting 1 in 2:
(3x+15)2+x2=259x2+225+90x+x2=2510x2+200+90x=0x2+9x+20=0x2+9x+20=0(x+5)(x+4)=0x=-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
(x1,y1)and(x2,y2) is found using following formula:
BC=sqrt(y2−y1)2+(x2−x1)2Distance between B(-4,3) and C(-5,0)
BC=sqrt(0−3)2+(−5−(−4))2=sqrt9+1=sqrt10We now know
BC=sqrt{10} & AC=10We can find AB;
△ABC is a right angled triangle with hypotenuse as AC. We can use Pythagoras theorem to find AB
(AC)2=(AB)2+(BC)2102=(AB)2+(√10)2100=(AB)2+10(AB)2=100−10=90AB=√90Area of the rectangle ABCD
BC∗AB=√10∗√90=√900=30Ans: "B"