In the xy-coordinate system, rectangle ABCD is inscribed within a circ

x^2 + y^2 = 25 is the equation of the circle centered at the origin and radius of \(\sqrt{25}=5\) (check here for more: math-coordinate-geometry-87652.html):AC = diameter of the circle = diagonal of the rectangle = 2*radius = 10.

B lies on y = 3x + 15 as well as on x^2 + y^2 = 25. Solve to get the y-coordinate of B: y=0 (discard) or y=3. Now, since y-coordinate of B is 3, then the area of triangle ABC is 1/2*10*3=15, this the area of the rectangle ABCD is twice of that: area=2*15=30.

Answer: B.

Hope it's clear.
_________________

Please see attached image for visual clarification.

Let's see what has been provided in the question.

Circle's equation: \(x^2+y^2=5^2\)
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: \(y^2+x^2=25\) ------ 2

Substituting 1 in 2:
\((3x+15)^2+x^2=25\)
\(9x^2+225+90x+x^2=25\)
\(10x^2+200+90x=0\)
\(x^2+9x+20=0\)
\(x^2+9x+20=0\)
\((x+5)(x+4)=0\)
x=-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 \((x_1,y_1) and (x_2,y_2)\) is found using following formula:
\(BC = sqrt{(y_2-y_1)^2+(x_2-x_1)^2}\)

Distance between B(-4,3) and C(-5,0)
\(BC = sqrt{(0-3)^2+(-5-(-4))^2} = sqrt{9+1} = sqrt{10}\)

We now know \(BC=sqrt{10} & AC=10\)

We can find AB; \(\triangle{ABC}\) is a right angled triangle with hypotenuse as AC. We can use Pythagoras theorem to find AB

\((AC)^2=(AB)^2+(BC)^2\)
\(10^2=(AB)^2+(\sqrt{10})^2\)
\(100=(AB)^2+10\)
\((AB)^2=100-10=90\)
\(AB=\sqrt{90}\)

Area of the rectangle ABCD
\(BC*AB = \sqrt{10}*\sqrt{90} = \sqrt{900} = 30\)

Ans: "B"

That is a great solution fluke and good catch beyondgmatscore.

I would like to tell you guys my first instinct when I looked at the question. I started drawing the diagram while reading the question and saw that BC has equation y = 3x + 15. What I said to myself was that slope of BC is 3 which means that if x changes by 1, y changes by 3. So if x becomes -4, y becomes 3. Since the radius of the circle is 5 so OB is 5 and it was clear that the point (-4, 3) lies on the circle and is B since the line intersects with the circle on only 2 points... So I didn't have to solve for the line and the circle. If you start thinking logically, you will be surprised how many many things suddenly fall into place in GMAT.
Of course, a valid concern is that what if co-ordinates of B were not integrals... Since the area was an integer, I was more inclined towards integers as B's co-ordinates. Also, as I have said before, GMAT is considerate that way - they give you numbers that fall beautifully in place. They are testing your aptitude and innovative ability, not calculation skills.
_________________

Slope is the change in y when you change x by 1 unit. If slope is 3, it means the y coordinate will increase by 3 for every unit increase in x coordinate. And also, y coordinate will decrease by 3 for every 1 unit decrease in x coordinate.

For more on this, check: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2016/0 ... line-gmat/

You are given a point (-5, 0).
So if x increases by 1 unit and becomes -5 + 1 = -4, then y increases by 3 units i.e. it becomes 0 + 3 = 3.

That is how we get the point (-4. 3)
_________________

First find the intercept of the line and circle.
x^2 + y^2 = 25 ----1)
y = 3x + 15------2)

Hence x^2 + (3x+15)^2 = 25
x = -5, -4
Since we are given B is in 2nd quadrant its coordinate is (-4,3) i.e. (-4)^2 + 3^2 = 25
Since we are given D is in 4th quadrant its coordinate is (4,-3) i.e. 4^2 + (-3)^2 = 25

We already know the points C (-5,0) and A (5,0).

b = Length BC = B(-4,3) to C(-5,0) = sqrt(1^2 + 3^2) = sqrt(10)
l = Length BA = B(-4,3) to A(5,0) = sqrt(9^2 + 3^2) = sqrt(90)

Area of the rectangle = b * l = sqrt(10) * sqrt(90) = sqrt(900) = 30. Answer B.

PS : I still want someone to verify all the above statements.

Btw, once we solve for the coordinates of B, we can also see that the y-coordinate of B is the height of triangle BCA which has base as 10 and hence area of triangle BCA is 1/2*3*10 = 15 and area of rectangle is twice of that, so area is 30.

The distance of any point lying on the circle from (0, 0) must be 5 (since the radius of the circle is 5).
(-4, 3) is at a distance of 5 from (0, 0) but (-3, 6) is not (even though it lies on the line). It will not lie on the circle.
_________________

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?

Point B
x^2 + (3x+15)^2 = 25
x^2 + 9x^2 + 90x +225 = 25
10x^2 + 90x + 200 = 0
x^2 + 9x + 20 = 0
(x+5)(x+4) = 0
x = -5 or x = -4
(x,y) = {(-5,0),(-4,3)}
Since point B lies in II quadrant
(x,y) = (-4,3)
Area of ABC = 1/2 * 10 * 3 = 15
Area of ABCD = 15*2 = 30

IMO B

The circle is centered at (0, 0) with radius 5. A and C are two points lying on the circle on x axis (when a rectangle is inscribed in a circle, all its 4 vertices must lie on the circle so A and C both lie on the circle). So AC is a line whose both end points lie on the circle and it passes through the center of the circle. Such a line is the diameter of the circle.
_________________

You are already given in the question that the rectangle is "INSCRIBED" in the circle. So all vertices of the rectangle have to lie on the circle, not inside.
_________________

Rectangles with same diagonals can have different area.
_________________

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.


_________________

I took the same approach as fluke. My only question is why do we assume that the diameter is the full diameter of the circle? why cant it be -4,0 to 4,0?

DO I have to remember the formular x^2 + y ^2= r^2

this formular never appear in og books, and so, this question is weid.

I do belive it is from gmatprep but it should be at 51 level.

gmat normally dose not require us to remember formular.

The question gives you the equation of a circle. So it is absolutely acceptable. If GMAC gives you a particular formula in the question itself, it is not expecting you to remember it. For example, they might ask you the volume of an irregular figure but give you the formula to find the volume.
That said, it does help sometimes to know these formulas. You will be able to solve the question even without it but it might take longer.
_________________

Hi,

Please can anyone explain that how can we be sure that all the vertices of rectangle actually touch some point on the circle ??

I can draw a smaller rectangle than the ones drawn in figures in discussion above (where the vertices would not touch the circle)

Hi, silly question, but please can anyone correct me here since i often wonder why area of rectangle can't be calculated by 1/2 (diagonal)square, in which case 1/2* (10)square = 50. Why can we use the equation for square but not rectangle, since the diagonals are congruent in both cases.

Thanks

(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 = 25
will 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 = 3

So 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

While I do understand that slope 3 means y/x = 3/1, How did you reach at this conclusion ?