In the xy-coordinate system, rectangle ABCD is inscribed within a circle having the equation x^2+y^2=25 . Line segment AC is a diagonal of the rectangle and lies on the x-axis. Vertex B lies in quadrant II and vertex D lies in quadrant IV. If side BC lies on line y=3x+15 , what is the area of rectangle ABCD?
(A) 15
(B) 30
(C) 40
(D) 45
(E) 50
x^2+y^2=25 (circle of radius 5 with the center at the origin)
Line segment AC lies on the x axis, vertex B lies in quadrant II and vertex D lies in quadrant IV. Drawing a picture of the rectangle in the circle we can determine that:
A = (5,0) and C = (-5,0)
The rectangle can be viewed as two triangles (ABC and ADC) with CA as the base. The base dimension is already known (10), and the height has to be determined in order to calculate the area of the triangle. Summing the areas of the two equal triangles gives the area of the rectangle.
If side BC lies on line Y=3x+15 , Then side BA, which is perpendicular to BC, has a slope that’s the negative reciprocal of BC’s slope (i.e. Y=(-x/3 + y-intercept).
Since A is on line BA and we know A’s values, plug in the values of A (5,0) and solve for the y-intercept.
0=-5/3 + y-intercept
Y-intercept = 5/3
So side BA lies on line Y = -X/3 + 5/3
Since point B is where BC and BA meet they must share the same coordinates. Equate the two Y equations and solve for X:
3x + 15 = -X/3 + 5/3
X= -4
So Y=3 (which is the height)
Calculate the area of triangle ABC
5*3=15
Multiplying by 2 since there are 2 triangles gives the area of the rectangle 15*2 = 30 (B)
(A) 15
(B) 30
(C) 40
(D) 45
(E) 50
x^2+y^2=25 (circle of radius 5 with the center at the origin)
Line segment AC lies on the x axis, vertex B lies in quadrant II and vertex D lies in quadrant IV. Drawing a picture of the rectangle in the circle we can determine that:
A = (5,0) and C = (-5,0)
The rectangle can be viewed as two triangles (ABC and ADC) with CA as the base. The base dimension is already known (10), and the height has to be determined in order to calculate the area of the triangle. Summing the areas of the two equal triangles gives the area of the rectangle.
If side BC lies on line Y=3x+15 , Then side BA, which is perpendicular to BC, has a slope that’s the negative reciprocal of BC’s slope (i.e. Y=(-x/3 + y-intercept).
Since A is on line BA and we know A’s values, plug in the values of A (5,0) and solve for the y-intercept.
0=-5/3 + y-intercept
Y-intercept = 5/3
So side BA lies on line Y = -X/3 + 5/3
Since point B is where BC and BA meet they must share the same coordinates. Equate the two Y equations and solve for X:
3x + 15 = -X/3 + 5/3
X= -4
So Y=3 (which is the height)
Calculate the area of triangle ABC
5*3=15
Multiplying by 2 since there are 2 triangles gives the area of the rectangle 15*2 = 30 (B)
