Learn how to find the area of a quadrilateral by splitting it into two triangles. Use the formula for triangle area (1/2 base times height) for each triangle. Add the areas together for the total quadrilateral area. It's as easy as pie...or should I say, as easy as area! Created by Sal Khan.
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- it would have been much faster is he just did 8*5=40/2 which equals 20 and then 4*5=20/2 whish is 10 then add them which is 30(11 votes)
- why the height for second triangle is 5 :/(2 votes)
- The height is perpendicular to the base, so you would have to extend the line segment of the base which is 4 at least to the point of vertex on top left. Then the same 5 on bottom would be the perpendicular distance to this vertex. Think of the two bases as being parallel, distance between them will remain constant.(14 votes)
- It will be much easier and quick if you make a parallelogram and a triangle out of the shape first and then it is just 20 + 10 = 30 unit(5 votes)
- Shouldn't you be doing this for a trapezium? (We call that in Hong Kong):
- well some people only learned the area of the triangle so you just split it up into 2 triangles(5 votes)
- Me: Accidentally clicks on someone elses boi
Also me: "NOOOOOOOOOOO😭" *Video restarts🥲*(3 votes)
- Would the formula 1/2x(b1 + b2)h work for composite figures(2 votes)
- [Instructor] What we're going to try to do in this video is find the area of this figure. And we can see it's a quadrilateral. It has 1, 2, 3, 4 sides. And we know that this side and this side that they're parallel to each other. You can see that they both form right angles with this dotted line. So pause this video and see if you can find the area. All right. Now, if you had a little bit of trouble with that I'll give you a hint. What if we were to take this quadrilateral and divide it into two triangles? So let me do this in a color that you are likely to see. So if I were to draw a line like this it now divides the quadrilateral into two triangles. If I were to take this triangle right over here I could take it out and reorient it, so it looks something like this, where the base has length 8, and then the height right over here, the height this has length 5. So that would be that triangle. And then, this triangle over here if you were to take it out and reorient it a little bit it could look like this, where the base is 4, and the triangle looks something like, looks something like this. So the base is 4, and then the height is going to be 5. So this height right over here, this height, we notice this is a right angle. So from here to here, which is the same thing as from here to right over here, we know that this is 5. So that's my fairly big hint to you, if you know how to find the area of a triangle. The area of a triangle we know is 1/2 base times height. So the area of this one right over here is gonna be 1/2 times 8 times 5. And the area of this one over here is going to be 1/2 times the base, which is 4 times the height, which is 5. And we could evaluate each of these, or we could just add them together. That the area of the entire thing is going to be 1/2 times this base right over here which is 8, times the height, which is 5, plus 1/2 times this side, you could do that as the other base, times 4 times that same height, times 5. And obviously, you could just evaluate this or we could see some interesting things about it. We could express this as well if we were to factor out 1/2 and the 5 here. This could be rewritten as 1/2, 1/2 times 8 plus 4, 8 plus 4. And then, all of that, all of that times 5 right over here. And so, another way you could think about it is the average of the length of these two bases. You could view this as base 1 and base 2, you multiply that times the height and you have the area of this quadrilateral. Well, what's that going to be? 8 plus 4 is 12. 1/2 times 12, this is all going to be 6. 6 times 5 is going to be equal to 30 square units. You could have figured it out here too. 1/2 times 8 is 4, times 5 is 20. And then, this would've been 1/2 times 4 is 2, times 5 is 10. 20 plus 10 is 30 square units once again.