Area of a Triangle

[mario99]

I need to find the orange area. The dimensions of the triangle legs in the picture are wrong. I drew them like this to make the problem visualized easily. The length of the blue leg is 3 units, black 2 units, and red 1 unit. The function \(\displaystyle y = x^3\) passes through the triangle. The function is not necessary passes through the peak of the triangle as the drawing is wrong.

I don't understand why some of you wants to ban me, especially those who thinks I am not trying to show some efforts.

My try is to set an integral and solve it to find the area, but this would be easy if there are no triangle involved. I have no idea how to merge the function with the triangle to create an integral.

Any help would be appreciated.

 

[AvgStudent]

View attachment 32348

I need to find the orange area. The dimensions of the triangle legs in the picture are wrong. I drew them like this to make the problem visualized easily. The length of the blue leg is 3 units, black 2 units, and red 1 unit. The function \(\displaystyle y = x^3\) passes through the triangle. The function is not necessary passes through the peak of the triangle as the drawing is wrong.

I don't understand why some of you wants to ban me, especially those who thinks I am not trying to show some efforts.

My try is to set an integral and solve it to find the area, but this would be easy if there are no triangle involved. I have no idea how to merge the function with the triangle to create an integral.

Any help would be appreciated.

Hi Mario,
Can you find the intersecting points of the curve and the triangle?

 

[Dr.Peterson]

The length of the blue leg is 3 units, black 2 units, and red 1 unit.

One problem is that there is no such triangle. Try drawing it accurately and you will see why!

Please show us the actual original problem you are trying to solve.

 

[Steven G]

As pointed out, there is no such triangle.
More importantly, once again you chose to show no work. Submitting a diagram that even you say is wrong is not showing any work.
I vote for a ban on this member.

 

[pka]

I take you at your that you want the area in orange.
So draw the vertical line [imath]x=1[/imath].
You will see the orange area divided two areas one of which is a right triangle of area [imath](0.5)(1)(2)[/imath].
The other is the area under the graph of [imath]f(x)=x^3[/imath] from [imath]x=0\text{ to }x=1.[/imath]
[imath]\displaystyle\int_0^1 {{x^3}dx}[/imath]


[imath][/imath][imath][/imath]

 

[mario99]

My bad. The black leg is 4 units, blue 3 units, and red 2 units.

Hi Mario,
Can you find the intersecting points of the curve and the triangle?

No. This is the main issue.

One problem is that there is no such triangle. Try drawing it accurately and you will see why!

Please show us the actual original problem you are trying to solve.

I corrected the mistake!

As pointed out, there is no such triangle.
More importantly, once again you chose to show no work. Submitting a diagram that even you say is wrong is not showing any work.
I vote for a ban on this member.

Me too.

I take you at your that you want the area in orange.
So draw the vertical line [imath]x=1[/imath].
You will see the orange area divided two areas one of which is a right triangle of area [imath](0.5)(1)(2)[/imath].
The other is the area under the graph of [imath]f(x)=x^3[/imath] from [imath]x=0\text{ to }x=1.[/imath]
[imath]\displaystyle\int_0^1 {{x^3}dx}[/imath]


[imath][/imath][imath][/imath]

The problem is that we don't know if \(\displaystyle x^3\) will intersect two vertices of the triangle. We know that it will intersect one vertex at the original, but I don't see if it will also intersect the top vertex. If it will, then your approach can be done successfully.

 

[Dr.Peterson]

My bad. The black leg is 4 units, blue 3 units, and red 2 units.

Have you tried drawing the actual triangle? Here is my drawing based on the new numbers, which make it obvious even from a rough sketch what happens:

​

Now you need to find the equation of line BC, and find the intersection, then integrate.

 

[Deleted member 4993]

You can locate the point the point C, by finding the point of intersection of two circular arcs - one centered at A with radius 2 and the other centered at point B and radius 4.

 

[Steven G]

You can locate the point the point C, by finding the point of intersection of two circular arcs - one centered at A with radius 2 and the other centered at point B and radius 4.

I taught you well.

 

[pka]

If the Peterson drawing is finally correct (I miss the orange) then
The point [imath]C[/imath] is located one of the interactions
of [imath]x^2+y^2=4\text{ and }(x-3)^2 +y^2=16[/imath]

 

[mario99]

Have you tried drawing the actual triangle? Here is my drawing based on the new numbers, which make it obvious even from a rough sketch what happens:

View attachment 32391​

Now you need to find the equation of line BC, and find the intersection, then integrate.

Thanks for the drawing. This must be the correct triangle. I didn't even imagine the triangle will be expanding in Quadrant I and II. I will try to find the equation of the line \(\displaystyle BC\).

You can locate the point the point C, by finding the point of intersection of two circular arcs - one centered at A with radius 2 and the other centered at point B and radius 4.

Thanks to you and to pka who prepared the circle equations (the two circular arcs) to use this idea!

If the Peterson drawing is finally correct (I miss the orange) then
The point [imath]C[/imath] is located one of the interactions
of [imath]x^2+y^2=4\text{ and }(x-3)^2 +y^2=16[/imath]

Thanks pka. I drew these equations online and I got the intersection point as \(\displaystyle (-0.5, 1.936)\). Is there a way to get the exact value of \(\displaystyle y\) as putting 1.936 would be an approximation and we might lose some portion of the required area.

Also, would it be possible to find this point \(\displaystyle (-0.5, 1.936)\), by hand without using an online graphing calculator?

Overwhelming?

 

[pka]

Also, would it be possible to find this point \(\displaystyle (-0.5, 1.936)\), by hand without using an online graphing calculator?

I don't understand why anyone who has to do serious study would not make use of WolframAalpha
It is worth whatever. The answer to your answer is yes:
solve [imath]4-x^2=16-(x-3)^2[/imath] When you find [imath]x[/imath] us to find [imath]y\text{ in }x^2+y^2=4[/imath]

 

[Dr.Peterson]

I will try to find the equation of the line BC.

Also, would it be possible to find this point (−0.5,1.936), by hand without using an online graphing calculator?

Yes. You've even been told how. Have you tried?

Look at pka's equations; just eliminate y between them, and solve for x. Then use one of them to find y.

One you have found point C, you can write the equation of line BC using point-slope.

Overwhelming?

No.

 

[Deleted member 4993]

Overwhelming

Mario, what level of mathematics are you doing? Most of the students - learn about point of intersection of graphs in Algebra II (High school).

 

[mario99]

I don't understand why anyone who has to do serious study would not make use of WolframAalpha
It is worth whatever. The answer to your answer is yes:
solve [imath]4-x^2=16-(x-3)^2[/imath] When you find [imath]x[/imath] us to find [imath]y\text{ in }x^2+y^2=4[/imath]

So \(\displaystyle y = \frac{\sqrt{15}}{2}\)

Yes. You've even been told how. Have you tried?

Look at pka's equations; just eliminate y between them, and solve for x. Then use one of them to find y.

One you have found point C, you can write the equation of line BC using point-slope.

No.

Yeah I tried and got \(\displaystyle y\) and the equation. \(\displaystyle y - \frac{\sqrt{15}}{2} = m(x + \frac{1}{2})\).

Mario, what level of mathematics are you doing? Most of the students - learn about point of intersection of graphs in Algebra II (High school).

University.

 

[Dr.Peterson]

So \(\displaystyle y = \frac{\sqrt{15}}{2}\)


Yeah I tried and got \(\displaystyle y\) and the equation. \(\displaystyle y - \frac{\sqrt{15}}{2} = m(x + \frac{1}{2})\).


University.

That's good. You want the line through [imath]B(3,0)[/imath] and [imath]C(-\frac{1}{2},\frac{\sqrt{15}}{2})[/imath]. What is the slope of this line? That's your m.

(You could use B rather than C in writing the equation, which will be a little easier.)

Then you'll have to find the intersection. Unfortunately, that will mean solving a cubic equation.

Where did the problem come from? You never did show the original, as requested.

 

[mario99]

That's good. You want the line through [imath]B(3,0)[/imath] and [imath]C(-\frac{1}{2},\frac{\sqrt{15}}{2})[/imath]. What is the slope of this line? That's your m.

(You could use B rather than C in writing the equation, which will be a little easier.)

Then you'll have to find the intersection. Unfortunately, that will mean solving a cubic equation.

Where did the problem come from? You never did show the original, as requested.

So this problem cannot be solved because it will involve a cubic equation.

The original problem is this. A triangle with legs 2, 3, and 4 units is placed in a cartesian coordinate with one vertex at the origin connecting the 2 and 3 units legs. If the 3 units leg is the base of the triangle and \(\displaystyle x^3\) passes through the triangle, what is the area bounded by the two larger legs and \(\displaystyle x^3\).

 

[Dr.Peterson]

So this problem cannot be solved because it will involve a cubic equation.

The original problem is this. A triangle with legs 2, 3, and 4 units is placed in a cartesian coordinate with one vertex at the origin connecting the 2 and 3 units legs. If the 3 units leg is the base of the triangle and \(\displaystyle x^3\) passes through the triangle, what is the area bounded by the two larger legs and \(\displaystyle x^3\).

I didn't say it can't be solved; it is possible to find the intersection point (even exactly, though most of us would not want to do that by hand). It just won't involve nice numbers.

Is the context such that it is clear that no numerical approximations are appropriate?

 

[mario99]

Oh, so it can be solved. I don't know if no numerical approximations are appropriate, but I think that if we cannot solve the cubic equation to get an exact solution, a 5 or 6 decimal place approximation for the the intersection point would give us an almost exact area.

 

[mario99]

I did this but the answer was rejected by the teacher.

\(\displaystyle \int_{0}^{1.02926110046} x^3 \ dx \ + \ \int_{1.02926110046}^{3} \frac{-\sqrt{15}}{2(3 + 0.5)}(x + 0.5) + \frac{\sqrt{15}}{2} \ dx\)