Area of trapezoid on the coordinate plane | High School Math | Khan Academy

So we have a trapezoid here on the coordinate plane, and what we want to do is find the area of this trapezoid just given this diagram. Like always, pause this video and see if you can figure it out.

Well, we know how to figure out the area of a trapezoid. We have videos where we derive this formula, but the area of a trapezoid, just put simply, is equal to the average of the lengths of the bases. We could say base 1 plus base 2 times the height.

So what are our bases here, and what is going to be our height over here? Well, we could call Base 1, we could call that segment CL, so it would be the length of segment CL right over here. I'll do that in magenta. That is going to be Base 1. Base 2, that could—let me do that in a different color—Base 2 would be the length of segment O, or B2, would be the length of seg OW right over there.

And then our height, our height H, well, that would just be an altitude, and they did one in a dotted line here. Notice it intersects B— the base one, I guess you could say segment CL— at a right angle here. This would be the height. So if we know the lengths of each of these, if we know each of these values which are the lengths of these segments, then we can evaluate the area of this actual trapezoid.

Once again, if this is completely unfamiliar to you or if you're curious, we have multiple videos talking about the proofs or how we came up with this formula. You can even break down a trapezoid into two triangles and a rectangle, which is one way to think about it.

But anyway, let's see how we could figure this out. The first one is: what is B1 going to be? B1 is the length of segment CL, and you could say, "Well look, we know what the coordinates of these points are." You could say, "Let's use the distance formula."

You could say, “Well, the distance formula is just an application of the Pythagorean theorem.” So this is just going to be our square root of our change in x squared. Our change in x is going to be this right over here. Notice we're going from x = -4 to x = 8 as we go from C to L, so our change in x is equal to 8 - -4, which is equal to 12.

And our change in y, we’re going from y = 1 to y = 5, so we could say our change in y is equal to 5 - -1, which of course is equal to 6. You see that here: 1, 2, 3, 4, 5, 6.

And the segment that we care about—its length that we care about—that’s just the hypotenuse of this right triangle that has one side 12 and one side 6. So the length of that hypotenuse, from the Pythagorean theorem, and as I mentioned, the distance formula is just an application of the Pythagorean theorem, this is going to be our change in x squared, 12 squared, plus our change in y squared, so plus 6 squared.

And this is going to be equal to 144 + 36. So the square root of 144 + 36 is 180, which is equal to—let's see—180 is 36 * 5, so that is 6 square roots of 5. Let me not skip some steps, so this is a square root of 36 * 5 which is equal to the square root of 36, which is 6, so 6 square root of 5.

Now, let's figure out B2. So B2, once again: change in x is the square root of change in x squared plus change in y squared. Well, let's see. If we're going from— we could set up a right triangle if you like— like this to figure those things out. So our change in x: we're going from x = -2 to x = 4, so our change in x is 6.

Our change in y, we are going from y = 5 to y = 8, so our change in y is equal to 3. So just applying the Pythagorean theorem to find the length of the hypotenuse here, it's going to be the square root of change in x squared, 6 squared, plus change in y squared, 3 squared.

Which is going to be equal to 36 + 9, which is 45. So square root of 45, which is equal to the square root of 9 * 5, which is equal to 3 square roots of 5. Now we only have one left to figure out. We have to figure out H; we have to figure out the length of H.

So H is going to be equal to— what is our—if we're going from W to N? Our change in x is 2. Change in x is equal to 2; we're going from x = 4 to x = 6. If you want to do that purely numerically, you would say, "Okay, our end point—our x value is 6; our starting point—our x value is 4, 6 - 4 is 2." You see that visually here.

So it's going to be the square root of 2 squared plus—let me write that radical a little bit better—so it's the square root of our change in x squared plus our change in y squared. Our change in y is -4. Change in y is -4, but we're going to square it, so it's going to become a positive 16.

So this is going to be equal to the square root of 4 + 16, square root of 20, which is equal to the square root of 4 * 5, which is equal to 2 * the square root of 5. It's nice that the square root of 5 keeps popping up.

And so now we just substitute into our original expression. The area of our trapezoid is going to be (12 * (6√5 + 3√5)) * (2√5). Let me close that parenthesis. So let's see how we can simplify this.

So (6√5 + 3√5) is 9√5. Let's see, the 1/2 * the 2, those cancel out to just be 1. So we're left with 9√5 * √5. Well, √5 * √5 is just going to be 5. So this is equal to 9 * 5, which is equal to 45 square units or units squared.