AnkitK 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
(Please refer to the attachment)The equation of a circle given in the form
X^2 + Y^2 =25 indicates that the circle has a radius of rand that its
center is at the origin (0,0) of the xy-coordinate system. Therefore, we know that the circle with the equation
X^2 + Y^2 = 25will have a radius of 5 and its center at (0,0).
If a rectangle is inscribed in a circle, the diameter of the circle must be a diagonal of the rectangle (if you try
inscribing a rectangle in a circle, you will see that it is impossible to do so unless the diagonal of the rectangle
is the diameter of the circle). So diagonal AC of rectangle ABCD is the diameter of the circle and must have
length 10 (remember, the radius of the circle is 5). It also cuts the rectangle into two right triangles of equal
area. If we find the area of one of these triangles and multiply it by 2, we can find the area of the whole
rectangle.
We could calculate the area of right triangle ABC we had the base and height. We already know that the
base of the triangle, AC, has length 10. So we need to find the height.
The height will be the distance from the x-axis to vertex B. We need to find the coordinate of point B in order
to find the height. Since the circle intersects triangle ABCD at point B, the coordinates of point B will satisfy
the equation of the circle
X^2 + Y^2 =25 . Point B also lies on
Y = 3X + 15 the line , so the coordinates of point
Bwill satisfy that equation as well.
Since the values of x and y are the same in both equations and since
Y = 3x + 5 , we can substitute (3x+ 15)
for yin the equation X^2 + Y^2 = 25 and solve for x:
( See attachment for solution)
So the two possible values of x are -4 and -5. Therefore, the two points where the circle and line intersect
(points Band C) have x-coordinates -4 and -5, respectively. Since the x-coordinate of point C is -5 (it has
coordinates (-5, 0)), the x-coordinate of point B must be -4. We can plug this into the equation Y = 3X + 5
and solve for the y-coordinate of point B:
Y = 3(-4) + 15
Y = -12 + 15
Y = 3So the coordinates of point Bare (-4, 3) and the distance from the x-axis to point Bis 3, making the height of
triangle ABC equal to 3. We can now find the area of triangle ABC:
Area = 1/2 * Base * Height
= 1/2 * 10 * 3
=15
The area of rectangle ABCD will be twice the area of triangle ABC. So if the area of triangle ABC is 15, the
area of rectangle ABCD is
(2)(15) = 30. Correct Answer B
Attachments
equation.JPG [ 18.41 KiB | Viewed 10355 times ]
File comment: Main Figure
Rectangle_in_Circle.JPG [ 25.11 KiB | Viewed 10365 times ]