# Find the value of k

• Sep 19th 2008, 09:09 PM
fabxx
Find the value of k
(page 350 #9)
Sorry for the sloppy graph. The outter line should be a curve concave down. And inside the curve is a triangle.

The question: The figure above shows the graph of y=k-x^2, where k is constant. If the area of triangle ABC is 64, what is the value of k? I don't get why the correct answer is 16. Thanks in advance
• Sep 19th 2008, 09:26 PM
mr fantastic
Quote:

Originally Posted by fabxx
(page 350 #9)
Sorry for the sloppy graph. The outter line should be a curve concave down. And inside the curve is a triangle.

The question: The figure above shows the graph of y=k-x^2, where k is constant. If the area of triangle ABC is 64, what is the value of k? I don't get why the correct answer is 16. Thanks in advance

Are points A, B and C meant to coincide with the vertices of the triangle?
• Sep 19th 2008, 10:35 PM
fabxx
yes they are. sorry for the sloppy graph. A, C should be the points for both the parabola and the triangle.
• Sep 19th 2008, 10:44 PM
Moo
Okay
Quote:

Originally Posted by fabxx
yes they are. sorry for the sloppy graph. A, C should be the points for both the parabola and the triangle.

Note that k is positive.
A and C are the x intercepts of the curve y=k-x².
This means that y=0 for these points.
Thus x=+ or - sqrt(k)
The distance AB is then 2sqrt(k).

Can you see what the altitude of the triangle from point B is ? It's the y-ordinate of B, which is the y intercept of the curve. This means that its x is 0. So what is its y ?

Then remember that area of triangle = (1/2)*AB*altitude.

Solve for k !
• Sep 19th 2008, 10:44 PM
mr fantastic
Quote:

Originally Posted by fabxx
(page 350 #9)
Sorry for the sloppy graph. The outter line should be a curve concave down. And inside the curve is a triangle.

The question: The figure above shows the graph of y=k-x^2, where k is constant. If the area of triangle ABC is 64, what is the value of k? I don't get why the correct answer is 16. Thanks in advance

The x-intercepts of the parabola are $\displaystyle x = \pm \sqrt{k}$ and the y-coordinate of the turning point is $\displaystyle y = k$. So you have a triangle of baselength $\displaystyle 2\sqrt{k}$ and height $\displaystyle k$.

Therefore solve $\displaystyle k \sqrt{k} = 64 \Rightarrow k^{3/2} = 64$ for k.
• Sep 20th 2008, 01:52 AM
fabxx
Quote:

Originally Posted by Moo
Okay
A and B are the x intercepts of the curve y=k-x².

shouldn't the x intercepts of the curve be A and C?

Also another question is why do you make y=0?

• Sep 20th 2008, 02:05 AM
fabxx
Quote:

Originally Posted by mr fantastic

Therefore solve $\displaystyle k \sqrt{k} = 64 \Rightarrow k^{3/2} = 64$ for k.

I don't get how you got from $\displaystyle k \sqrt{k} = 64 to \Rightarrow k^{3/2} = 64$ and to the final k=16.
Can you please show me the steps?

• Sep 20th 2008, 02:07 AM
Moo
Quote:

Originally Posted by fabxx
shouldn't the x intercepts of the curve be A and C?

Also another question is why do you make y=0?

Excuse me, I put B instead of C. Now I've edited.

The x intercepts are the points intercept points of the curve with the x-axis. But any point on the x-axis has a null ordinate. This is why y=0 for these points.

Quote:

Originally Posted by fabxx
I don't get how you got from $\displaystyle k \sqrt{k} = 64 to \Rightarrow k^{3/2} = 64$ and to the final k=16.
Can you please show me the steps?

$\displaystyle \sqrt{k}=k^{1/2}$

and remember a rule of exponents : $\displaystyle a^b a^c=a^{b+c}$
• Sep 20th 2008, 02:58 AM
fabxx
Quote:

Originally Posted by moo
$\displaystyle \Rightarrow k^{3/2} = 64$ for k.

how do you get from k^(3/2) to k=16?

Thanks again!! sorry for the trouble.
• Sep 20th 2008, 03:10 AM
mr fantastic
Quote:

Originally Posted by fabxx
how do you get from k^(3/2) to k=16?

Thanks again!! sorry for the trouble.

$\displaystyle k^{3/2} = (k^{1/2})^3 = (\sqrt{k})^3 = 64$

Cube root both sides: $\displaystyle \sqrt{k} = 4$.

Square both sides: $\displaystyle k = 16$.
• Aug 24th 2009, 01:16 PM
mtax
How come you cannot use the 30, 60, 90 triangle method?