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In the xy-coordinate system, rectangle ABCD is inscribed

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Joined: 13 Apr 2006
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In the xy-coordinate system, rectangle ABCD is inscribed [#permalink]

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04 May 2006, 09:06
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In the xy-coordinate system, rectangle ABCD is inscribed within a circle having the equation x^2 + y^2=2t. 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?

Can you please help?! A thorough explanation would be great b/c I am totally lost with this.
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05 May 2006, 04:19
The first thing to notice is that the Circle has its center as (0,0) which makes our job simpler.

Once you have the figure right , you will see that point 'C' is on the x-axis => it is of the form (...,0).

Now, 'C' also lies on the line y = 3x + 15 , so 'C' is ( -5 , 0 )
since 0 = 3(-5) + 15.

This means the radius of our circle is 5 , and so
x^2 + y^2 = 25 = 2t. => t = 12.5

Since 'C' is (-5,0) , 'A' should be (5,0). [ AC lies on the x-axis ].

Now to find 'B' , we substitute , y as 3x + 15 in
the eqn x^2 + y^2 = 25.

x^2 + (3x + 15)^2 = 25 ..solving this , we get
x = -5 or -4 .

Since -5 refers to 'C', then point 'B' must be
( -4 , 3 ). [ we got 3 this way 3 = 3(-4) + 15. ]

So, point B is 3 units above the x-axis. This means,
The triangle ABC has area as
0.5 * base * height = 0.5 * 10 * 3 = 15.

Now, a diagonal divides a rectangle into two equal parts ,

so the area of the rectangle is
2 * 15 = 30 units.
[#permalink] 05 May 2006, 04:19
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In the xy-coordinate system, rectangle ABCD is inscribed

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