Finding area of a triangle using a parabola and another triangle

[Joshbr32]

Hi! I was given a math question today to solve and present to the class tomorrow morning but I don't really understand how to do it. Here it is:

I would appreciate a step by step solutions as I would also like to know how to solve it rather than just finding the answer. Thank you!

 

[Dr.Peterson]

Hi! I was given a math question today to solve and present to the class tomorrow morning but I don't really understand how to do it. Here it is:

I would appreciate a step by step solutions as I would also like to know how to solve it rather than just finding the answer. Thank you!

We're not here to give you a ready-made answer, even if you can learn from it; we want to help you solve it. Did you read the posting guidelines (at the top of each forum)?

I'd like to know what topic you are learning and what techniques you think you might use here. There are probably many ways to solve this.

My first thoughts are that you can find the location of A from the given area, and then write an equation for the parabola using the two zeros and the y-intercept. Then you can find the location of the vertex, and use one of several methods to find the area of DBC. But it is possible that you have learned some special fact about parabolas that could be used as a shortcut.

 

[Joshbr32]

We're not here to give you a ready-made answer, even if you can learn from it; we want to help you solve it. Did you read the posting guidelines (at the top of each forum)?

I'd like to know what topic you are learning and what techniques you think you might use here. There are probably many ways to solve this.

My first thoughts are that you can find the location of A from the given area, and then write an equation for the parabola using the two zeros and the y-intercept. Then you can find the location of the vertex, and use one of several methods to find the area of DBC. But it is possible that you have learned some special fact about parabolas that could be used as a shortcut.

I replied before but it is not showing up so I will again.

I have not learned anything about this in particular, which is why I am not sure what to do. But so far I have used the formula Base = 2*Area/Height to find the base which is 2. Therefore point A is at (2,0). I then found a possible equation for the parabola using the two zeros and the y-intercept, which is
y = (-1/2)*(x-2)*(x-4). From the I subbed in 3 as that is the x position of the vertex to get the y position which is 1/2, so the location of the vertex is (3, 1/2). But I do not know how to continue to find the area

 

[mmm4444bot]

I replied before but it is not showing up so I will again.

You were told twice when you joined (by registration e-mail and by private message) that your first three posts need to wait for approval.

:idea: Maybe part of your problem is that you're not paying enough attention.

Did you read the forum guidelines? Please do not create duplicate posts.

 

[Dr.Peterson]

I replied before but it is not showing up so I will again.

I have not learned anything about this in particular, which is why I am not sure what to do. But so far I have used the formula Base = 2*Area/Height to find the base which is 2. Therefore point A is at (2,0). I then found a possible equation for the parabola using the two zeros and the y-intercept, which is
y = (-1/2)*(x-2)*(x-4). From the I subbed in 3 as that is the x position of the vertex to get the y position which is 1/2, so the location of the vertex is (3, 1/2). But I do not know how to continue to find the area

Very good. Now you just need the area of the triangle. As I said, there are several ways you might do it; you are presumably expected either to use a method you have been taught, or to be creative. One method is a formula, sometimes called the shoelace formula (you can look it up) to find the area of any polygon given the coordinates of its vertices.

Here's one way: draw a vertical line through D and find its intersection E with BC. Then add the areas of triangles BDE and CDE.

Here's another: draw a horizontal line through D, and call its intersection with the y axis F. Combine the areas of triangles BFD and BOC, and trapezoid OFDC.

 

[mmm4444bot]

Another way would be finding the lengths of each side of triangle DBC, using the Distance Formula. Then apply Heron's Formula. :cool:

 

[stapel]

I have not learned anything about this in particular, which is why I am not sure what to do. But so far I have used the formula Base = 2*Area/Height to find the base which is 2. Therefore point A is at (2,0). I then found a possible equation for the parabola using the two zeros and the y-intercept, which is y = (-1/2)*(x-2)*(x-4). From the I subbed in 3 as that is the x position of the vertex to get the y position which is 1/2, so the location of the vertex is (3, 1/2). But I do not know how to continue to find the area

What is the slope of the line through BC? What then is the perpendicular slope?

What then is the equation of the perpendicular to BC, passing through D? Where does this perpendicular intersect BC? What is the distance between this intersection point (let's call it "E") and D? So what then is the height of the triangle BDC?

What is the length of BC? So what then is the base of the triangle BDC?

So what is the area?