# Thread: Area between two curves?

1. ## Area between two curves?

How do I calculate the area between these two curves?

Thanks guys and girls,

John

2. ## Re: Area between two curves?

I will need for information before I can answer this question. There are no "curves" in the question. I will need some equations also. Thanks

3. ## Re: Area between two curves?

I don't have equations for the lines/curves sadly. Is it not possible to calculate the area without them?

4. ## Re: Area between two curves?

The horizontal line forms a right triangle. The area of a triangle is $\dfrac{1}{2}bh$ where $b$ is the length of the base and $h$ is the height. In this case, it appears the base is 69 units and the height is 5,000 units.

There are then three areas you need to subtract from the area of the large triangle. You have two line segments to consider. The left line segment forms a right triangle with base 45 units and height 3,150 units. To the right of that is a rectangle with length 69-45 = 24 units and width 3,150 units. Above that is a right triangle with base 24 units and height 5,000-3,150 = 1,850 units.

Edit: The picture changed and no longer has minutes listed as the units for the x-axis. Also, the rightmost number changed from 60 to 69.

5. ## Re: Area between two curves?

Another way to do this: The figure is itself a triangle so has area (1/2)bh where b is the length of a "base" (any on of the sides) and h is the "height" (the distance from that "base" to the opposite vertex- measured along a perpendicular to the "base". It is simplest to take the "base" as the long line from (0, 0) to (69, 5000). It's length is $\sqrt{69^2+ 5000^2}$. Finding the height is a little harder. The slope of the base is 5000/69 so any line perpendicular it has slope -69/5000. The line with slope -69/5000 through (45, 3150) has equation y= (-69/5000)(x- 45)+ 3150. That intersects the line y= (5000/69)x where (-69/5000)(x- 45)+ 3150= (5000/69)x. Solve for x and y had find the distance between that point and (41, 3150).