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

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

Can anyone elaborate it with a figure..I am not able to draw it..

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.