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In the xy-coordinate system, rectangle ABCD is inscribed

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In the xy-coordinate system, rectangle ABCD is inscribed

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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

OA B

Source: GMAT Prep

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BTGmoderatorDC wrote:
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
x²+y² = r² is the equation for a circle with its center at the origin and a radius of r.
Thus, x²+y² = 25 is a circle with a center at the origin and a radius of 5.
The information in the prompt yields the following figure:


OB is a radius and thus has a length of 5.
Given the GMAT's love of special triangles, it is likely that triangle BOE is a 3-4-5 triangle, implying that B is located either at (-3, 4) or (-4, 3).
B lies on the line y=3x+15.
If we plug x=-4 into y=3x+15, we get y=3, implying that (-4, 3) lies on y=3x+15.
Thus, B is located at (-4, 3).

Since BE=3, the area of triangle ABC = (1/2)bh = (1/2)(CA)(BE) = (1/2)(10)(3) = 15.
Since the triangle ABC is 1/2 of rectangle ABCD, we get:
ABCD=30.

The correct answer is B.

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BTGmoderatorDC wrote:
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

OA B

Source: GMAT Prep


The equation of the circle x^2 + y^2 = 25 suggests that the given circle has its center at the origin and radius equal to √25 = 5. Thus, AD = 2*5 = 10 = Diameter. Since AD lies on the X-axis, the coordinate of A = (-5, 0) and the coordinate of D = (5, 0).

Now we are given that vertex B lies on the quadrant II and n the circle. Thus, the point must satisfy the equation x^2 + y^2 = 25 and the line y = 3x + 15.

Plugging in the value of y = 3x + 15 in x^2 + y^2 = 25, we get x = -5 or -4; the corresponding value of y are 0 and 3, respectively. Thus, B can be (-5, 0) or (-4, 3). Again, the vertex B lies on the quadrant II, the coordinate of B would be (-4, 3).

Area of the rectangle ABCD = 2 * Area of the triangle ABC
Area of the triangle ABC = 1/2 * (Perpendicular dropped from vertex B on AD) * (Base AD) = 1/2 * (Y-coordinate of vertex B) * 10 = 1/2 * 3 * 10 = 15

Area of the rectangle ABCD = 2 * Area of the triangle ABC = 2 * 15 = 30

The correct answer: B

Hope this helps!

-Jay
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