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The figure shows the graph of the equation y =k–x^2, where k [#permalink]
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Question Stats: 57% (03:56) correct 42% (02:54) wrong based on 26 sessions
Attachment: Coordinate Geometry - area- NE question.png [ 10.06 KiB | Viewed 11623 times ]

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?

[Reveal] Spoiler: OA
0.25 Director Joined: 03 Sep 2017
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Re: The figure shows the graph of the equation y =k–x^2, where k [#permalink]
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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. Intern Joined: 27 Aug 2018
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Re: GRE Math Challenge #2 (170 level question) [#permalink]
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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

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The figure shows the graph of the equation y = k – x^2, [#permalink]
Expert's post
Attachment: #GREpracticequestion The figure shows the graph of the equation.jpg [ 11.01 KiB | Viewed 8258 times ]

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?

[Reveal] Spoiler: OA
0.25

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Re: The figure shows the graph of the equation y = k – x^2, [#permalink]
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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?

[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]
pranab01 wrote:
height = point A and the distance = k)

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Re: The figure shows the graph of the equation y = k – x^2, [#permalink]
Expert's post
AE wrote:
pranab01 wrote:
height = point A and the distance = k)

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]
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?
Director Joined: 09 Nov 2018
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Re: The figure shows the graph of the equation y = k – x^2, [#permalink]
chetan2u wrote:
AE wrote:
pranab01 wrote:
height = point A and the distance = k)

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.
Director  Joined: 22 Jun 2019
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Re: The figure shows the graph of the equation y =k–x^2, where k [#permalink]
Magoosh prefers it as Very Hard Question not medium.

Attachment: Screenshot from 2019-11-05 11-22-37.png [ 119.29 KiB | Viewed 6155 times ]

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Re: The figure shows the graph of the equation y =k–x^2, where k [#permalink]
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