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The figure shows the graph of the equation y =k–x^2, where k

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The figure shows the graph of the equation y =k–x^2, where k [#permalink] New post 29 Oct 2017, 01:27
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The figure shows the graph of the equation \(y = k – x^2\), where k is a constant. If the area of triangle ABC is \(\frac{1}{8}\), what is the value of k?

Give your answer to the nearest 0.01


[Reveal] Spoiler: OA
0.25
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Re: The figure shows the graph of the equation y =k–x^2, where k [#permalink] New post 31 Oct 2017, 00:55
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In order to find the coordinates of the vertices of the triangle, let's exploit the equation of the parabola.

To find the coordinates of points B and C, let's evaluate the parabola at y = 0, i.e. \(0 = -x^2+k\) which leads to \(x=\pm\sqrt{k}\).

Then, vertices A has the same coordinates of the vertex of the parabola, i.e. evaluating at x = 0, we get \(y=k\).

Then, we can impose an equation for the area of the triangle as follows \(\frac{BC*OC}{2} = \frac{2\sqrt{k}*k}{2} =\frac{1}{8}\). Solving for k, we get \(k\sqrt{k}=\sqrt{k^3}=\frac{1}{8}\), i.e. k = 1/4 = 0.25.
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Re: GRE Math Challenge #2 (170 level question) [#permalink] New post 20 Sep 2018, 04:29
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Tough question.

The first part of the solution:

Let the origin be O.

The area of the triangle ABC is 1/8.
The area of the triangle ABC = the area of the triangle AOB*2
The area of the triangle AOB = (distance between Origin and B * distance between Origin and A)/2
The area of the triangle ABC = ((distance between Origin and B * distance between Origin and A)/2)*2 =>
The area of the triangle ABC = distance between Origin and B * distance between Origin and A = 1/8

The second part of the solution:

Point A is the point when x=0
If x=0, then y=k-0 => y=k
So, we now know that the distance between the origin and the point A is k.

Points B and C are points when y=0
0=k-x^2 => x^2 = k => x = +- sqrt(k)

So, we now know that the distance between the origin and the point B is sqrt(k)

The third part of the solution:
k*sqrt(k) = 1/8
k^3 = 1/64
k = 1/sqrt(8)
k = 0.35

Kudos if you liked this solution :)
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The figure shows the graph of the equation y = k – x^2, [#permalink] New post 11 Dec 2018, 14:11
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The figure shows the graph of the equation \(y = k – x^2\), where k is a constant. If the area of triangle ABC is \(\frac{1}{8}\), what is the value of k?

Give your answer to the nearest 0.01


[Reveal] Spoiler: OA
0.25

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Re: The figure shows the graph of the equation y = k – x^2, [#permalink] New post 16 Dec 2018, 04:48
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Carcass wrote:
Attachment:
#GREpracticequestion The figure shows the graph of the equation.jpg


The figure shows the graph of the equation \(y = k – x^2\), where k is a constant. If the area of triangle ABC is \(\frac{1}{8}\), what is the value of k?

Give your answer to the nearest 0.01


[Reveal] Spoiler: OA
0.25



Explanation::

Given

\(y = k - x^2\)

Now considering the y - axis and x = 0

Therefore, \(y = k - 0\)

or \(y = k\)

So the point A co-ordinate =\(k,0\)

Similarly, for point c, \(y =0\)

then,\(y = k -x^2\)

or\(0 = k - x^2\)

or\(x = \sqrt{k}\)

therefore point c , co-ordinate will be = \(- \sqrt{k},0\)

and point B, will be \(\sqrt{k},0\)

Now the area of \(\triangle ABC = \frac{1}{8}\)

or \(\frac{1}{2} * 2\sqrt{k}* k = \frac{1}{8}( base = \sqrt{k} + \sqrt{k}) = 2\sqrt{k}\) and height = point A and the distance = k)

or \(k = \frac{1}{4} = 0.25\)
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Re: The figure shows the graph of the equation y = k – x^2, [#permalink] New post 06 Jan 2019, 18:46
pranab01 wrote:
height = point A and the distance = k)

How? Please explain.
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Re: The figure shows the graph of the equation y = k – x^2, [#permalink] New post 06 Jan 2019, 19:34
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AE wrote:
pranab01 wrote:
height = point A and the distance = k)

How? Please explain.


Both x-axis and y-axis are perpendicular to each other, so the line from origin to A is the altitude of the triangle...
And we know that when x=0, y=k, so coordinates of point A are (0,k)..

Thus the height of the triangle is k
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Re: The figure shows the graph of the equation y =k–x^2, where k [#permalink] New post 07 Jan 2019, 05:54
IlCreatore wrote:

Then, we can impose an equation for the area of the triangle as follows \(\frac{BC*OC}{2} = \frac{2\sqrt{k}*k}{2} =\frac{1}{8}\). Solving for k, we get \(k\sqrt{k}=\sqrt{k^3}=\frac{1}{8}\), i.e. k = 1/4 = 0.25.


Is it OC or AC?
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Re: The figure shows the graph of the equation y = k – x^2, [#permalink] New post 07 Jan 2019, 06:00
chetan2u wrote:
AE wrote:
pranab01 wrote:
height = point A and the distance = k)

How? Please explain.


Both x-axis and y-axis are perpendicular to each other, so the line from origin to A is the altitude of the triangle...
And we know that when x=0, y=k, so coordinates of point A are (0,k)..

Thus the height of the triangle is k

Now I got It. My confusion was how the Point A is (0,k). Now it is clear.
In equation if x=0, what is y, and here, y is k.
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Re: The figure shows the graph of the equation y =k–x^2, where k [#permalink] New post 04 Nov 2019, 21:22
Magoosh prefers it as Very Hard Question not medium.

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Re: The figure shows the graph of the equation y =k–x^2, where k [#permalink] New post 14 Dec 2019, 09:20
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Re: The figure shows the graph of the equation y =k–x^2, where k   [#permalink] 14 Dec 2019, 09:20
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