So, by doing 6*3 and ADDING 2*3, Sal now had not only the area of the trapezoid (middle + 2 triangles) but also had an additional "middle + 2 triangles". The area of a figure that looked like this would be 6 times 3. In Area 3, the triangle area part of the Trapezoid is exactly one half of Area 3. Then, in ADDITION to that area, he also multiplied 2 times 3 to get a second rectangular area that fits exactly over the middle part of the trapezoid. πβπβ = 2π΄ is true for any rhombus with diagonals πβ, πβ and area π΄, so in order to find the lengths of the diagonals we need more information.
Aligned with most state standardsCreate an account. That's why he then divided by 2. That is a good question! Well, now we'd be finding the area of a rectangle that has a width of 2 and a height of 3.
A rhombus as an area of 72 ft and the product of the diagonals is. So what would we get if we multiplied this long base 6 times the height 3? Now, what would happen if we went with 2 times 3? So it completely makes sense that the area of the trapezoid, this entire area right over here, should really just be the average. Access Thousands of Skills. 6 plus 2 is 8, times 3 is 24, divided by 2 is 12. A width of 4 would look something like this. Why it has to be (6+2). So that is this rectangle right over here. I'll try to explain and hope this explanation isn't too confusing!
Multiply each of those times the height, and then you could take the average of them. This is 18 plus 6, over 2. What is the formula for a trapezoid? How do you discover the area of different trapezoids? Either way, you will get the same answer. Or you could say, hey, let's take the average of the two base lengths and multiply that by 3. Let's call them Area 1, Area 2 and Area 3 from left to right.
Of the Trapezoid is equal to Area 2 as well as the area of the smaller rectangle. Well, that would be the area of a rectangle that is 6 units wide and 3 units high. Maybe it should be exactly halfway in between, because when you look at the area difference between the two rectangles-- and let me color that in. Adding the 2 areas leads to double counting, so we take one half of the sum of smaller rectangle and Area 2. Now let's actually just calculate it. So these are all equivalent statements. If you take the average of these two lengths, 6 plus 2 over 2 is 4. Now, the trapezoid is clearly less than that, but let's just go with the thought experiment. 6th grade (Eureka Math/EngageNY). So let's just think through it. So what do we get if we multiply 6 times 3?
So you could imagine that being this rectangle right over here. Area of a trapezoid is found with the formula, A=(a+b)/2 x h. Learn how to use the formula to find area of trapezoids. Okay I understand it, but I feel like it would be easier if you would just divide the trapezoid in 2 with a vertical line going in the middle. Sal first of all multiplied 6 times 3 to get a rectangular area that covered not only the trapezoid (its middle plus its 2 triangles), but also included 2 extra triangles that weren't part of the trapezoid. 6 plus 2 divided by 2 is 4, times 3 is 12. Well, that would be a rectangle like this that is exactly halfway in between the areas of the small and the large rectangle. You could view it as-- well, let's just add up the two base lengths, multiply that times the height, and then divide by 2.
So that's the 2 times 3 rectangle. That is 24/2, or 12. You're more likely to remember the explanation that you find easier. In other words, he created an extra area that overlays part of the 6 times 3 area.
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