Geometry (all content)
- Challenge problems: perimeter & area
- Challenging perimeter problem
- CA Geometry: Deductive reasoning
- CA Geometry: Proof by contradiction
- CA Geometry: More proofs
- CA Geometry: Similar triangles 1
- CA Geometry: More on congruent and similar triangles
- CA Geometry: Triangles and parallelograms
- CA Geometry: Area, pythagorean theorem
- CA Geometry: Area, circumference, volume
- CA Geometry: Pythagorean theorem, area
- CA Geometry: Exterior angles
- CA Geometry: Pythagorean theorem, compass constructions
- CA Geometry: Compass construction
- CA Geometry: Basic trigonometry
- CA Geometry: More trig
- CA Geometry: Circle area chords tangent
- Speed translation
CA Geometry: Pythagorean theorem, area
36-40, Pythagorean Theorem and Area. Created by Sal Khan.
Want to join the conversation?
- I know this is simple, but I just wanted to reassure myself. At around8:27when Sal multiplies the triangles, the formula is b times hieght divided by 2. Now does that only count for that one half of the triangle? Is that why he had to multiply it once more? Also, hypothetically, if I were to encounter a triangle such as this again, would I do the method shown here to find its area?(6 votes)
- yes. because he split the triangles into 2 so he could find the height by using the Pythagorean thm. Then he could use 1/2*b*h.(3 votes)
- I know this is simple, but I just wanted to reassure myself. At around8:27when Sal multiplies the triangles, the formula is b times height divided by 2. Now does that only count for that one half of the triangle? Is that why he had to multiply it once more? Also, hypothetically, if I were to encounter a triangle such as this again, would I do the method shown here to find its area?(4 votes)
The formula for the area of a triangle is base times height. This is because for a right triangle you can make a rectangle with twice the area of the right triangle and area is defined by the base times the height of the right triangle. So because the rectangle is twice the area of the right triangle you divide by two and get the formula for the area of a right triangle(and all other triangles) as 1/2 times the base times the height. He did it twice because there are two right triangles that are congruent.
Hope this helps you!
- Is there a formula for solving an area of a trapezoid?(3 votes)
- That depends on the information that you have.
The area is the average of the two bases × height.(4 votes)
- How does Sal know to use number eight instead of number five in problem number 1.(2 votes)
- Sal was calculating the length of the line left to right, not the height of the other line. He used the x value to find the length, but using the y value would not have given him the length he wanted to find. Hope this helps, I'm not so great at explaining things ;)(7 votes)
- for question 38 cant you just use the pathagreon triple 3,4,5?(2 votes)
- At8:15, How does Sal get 25 roots of 3, divided by 2? I don't get where 2 came from.(2 votes)
- The 2 comes from the equation for the area of a triangle, 1/2 times the base times the height, or [1/2bh]. He divides it by 2 because that is the 1/2 in action. The base is 5 and the height is 5√3. So, the area is (25√3)/2, which could also be simplified to 12.5√3.(1 vote)
- Does this viedo help people or no.Because i have a lot to ask YOU(1 vote)
- It did help me but I do have two or three questions myself.(1 vote)
- At2:25can you combine all the parallelograms to make a big parallelogram?(2 votes)
- on question 36 how did yu know the line was 1/2(1 vote)
- How can you tell when the sides are x and the other problem about the square root of 75 was 25 X 3 how did you do that(1 vote)
We're on problem 36. And it says, what is the area in square units of the trapezoid shown below? So, when you just look at this you're like, OK, trapezoid, do I know the formula for the area of a trapezoid. Then you might get confused and all that. But you say, well, trapezoid I can break that up into a rectangle and a triangle. If were to draw a line right here. Then I've broken up the trapezoid into a rectangle and a triangle. And if I know the dimensions of each of those, I know the area of each of them and then I know the area of the entire thing. So let's see, what's this height right Here Or this width I should say. Well we're going from zero to what? x is equal to 8 here. I just went straight down from x is equal to 8, y is equal to 5. So this dimension is 8. And then we go from x is equal to 8 to x is equal to 12, how far is that? Well, that's going to be 4. So this is 4 and this is 8. Fair enough. And then how high is this rectangle? We're going from y is equal to zero to y is equal to 5, so that's 5. And of course this is 5 as well. So we're done. We're ready to figure out the area. The area of the rectangle part is 8 times 5, that's 40. The area of this triangle is 5 times 4 times 1/2. If we didn't put that 1/2 we would be figuring out the area of this rectangle right there. So 5 times 4 is 20 times 1/2 is 10. So the area of both of these combined is 10 plus 40 is 50. 37. The figure below is a square with 4 congruent parallelograms in side. This looks interesting. What is the area in square units of the shaded portion? So the shaded portion is the whole square minus the area of the parallelogram. So the whole square, that's easy, it's 12. And the height is 12, but since you know it's a square we know the width also has to be 12. So the area of the whole square is 144. If we know the area of one of the parallelograms, we know the area of all of the parallelograms because they are congruent. So let's see if we can figure out the area of one of the parallelograms. So there is actually a formula for the area of a parallelogram, it's actually just the base times the height. And they actually give us that. But let me show you that they give us that because it might not be obvious to you. Let me try to draw it. I'll use my line tool. Nope, that's not the line tool. One side, then go straight like that, come down like that, good enough. OK, now if I look at just this parallelgoram, they tell us that the height here is 3. And I know it's the height because they told me it's a 90 degree angle. And they tell us at the base is 5. And I'm telling you that the area of a parallelogram is just the base times the height is equal to 15. But you shouldn't just take my word for it. That should make intuitive sense to you. And the way to think about it intuitively is imagine if we were to take this part of the parallelogram, and if we were to move it over here. If we were to cut that off and move it over here. Then the parallelogram would look something like this. You'd have the part that we didn't cut off. And then you move the cut-off part over here. And now the dimensions, this base would be 5, and this height would be 3. And the area of this rectangle is 15. And there's no reason why the area of this should be any different than that. We just rearranged its parts. So that's why the area of a parallelogram is just the base times the height. So the area of each of these parallelograms is 15. So the area of all of them combined is 15 times 4, which is 60. So 144 minus 60 is 84. And that's choice B. Problem 38. What is the area in square meters of the trapezoid shown below. So to figure out the area and we could break it up into these rectangles and triangles. To figure out the area of this rectangle we need to know its height. And actually we'll need that to figure out the area of the triangles as well. So what's this height right there? Let's see, we know that this distance is going to be 6. It's a rectangle. If that distance is 6 and both of these are 5, both of these triangles here are going to be congruent. Because this length is equal to this length. This length is equal to this length. And we also we make this angle is equal to that angle. But anyway, let me do it in another color. What's the length of these two green sides? Let's call it x. Well we know that when you add x plus 6 plus x it has to equal 12. The whole top part. So you get x plus x is 2x plus 6 is equal to 12. 2x is equal to 6. x is equal to 3. And you might have been able to solve that in your head. That if that's 6 and these are the same, then both of these are going to be 3. And now we can use that information to figure out this height right there. Because if we just draw this triangle right there, that's 3, that's 5, this is some unknown side, a. You might already recognize, we're going to use the Pythagorean theorem. And this is a very typical type of right triangle. So might already be able to guess what a is. But we'll solve for it. So we know that a squared plus 3 squared is equal to the hypotenuse squared, the side opposite the 90 degree angle. So that's equal to 25. 5 squared is 25. a squared plus 9 is equal to 25. a squared is equal to 16. a is equal to 4. a is equal to 4. And now we're ready to figure out the area. What's the area of the rectangle? 6 times 6, it's 24. What's the area of each of these triangles? 3 times 4 times 1/2. 3 times 4 is 12 times 1/2 is 6. So the area of that triangle is 6. The area of this triangle is 6. So 6 plus 24 plus 6 is 36. B. Problem 39. What is the area in square inches of the triangle below. Interesting. OK, so this is an equilateral triangle, all the sides are equal. And so we could actually say that since these two triangles are symmetric. That's equal to that. And this comes to a general formula for the area of an equilateral triangle. But let's just figure it all out. So this side is going to be 5. And this side is going to be 5. If this is 5 and that's 10, what is this side right here? Let's call it x. Pythagorean theorem. This is the hypotenuse. So x squared plus 5 squared plus 25 is going to be equal to the hypotenuse squared, it's equal to 100. x squared is equal to 100 minus 25, 75. x is equal to the square root of 75. 75 is 25 times 3. So that's equal to the square root of 25 times 3. Which is equal to the square root of 25 times the square root of 3. Which is equal to 5 roots of 3. And now, what's the area of just this right triangle right here? This one on the right side. Well its base is 5, its height is 5 roots of 3. So it's going to be 1/2 times the base, 5, times the height, 5 roots of 3. And that's what? 1/2 times 5 times 5. So it's 25 root 3 over 2 and that's just this triangle right there. Well this triangle's going to have the the exact same area. They are congruent triangles. So the area of the figure is this times 2. So 2 times that is equal to just the 25 root 3. And that's choice B. Next problem, problem 40. The perimeter of two squares are in a ratio of 4:9. What is the ratio between the areas of the two squares? Let me draw two squares. That's one square. Let me draw another square. That's another square. Let's say that the sides of this are x and the sides of this one are y. So they say the perimeters of those two squares are in a ratio of 4:9. So the perimeter of the first square is 4x. x plus x plus x plus x. So the perimeter of the first square is 4x. The perimeter of the second square is 4y. So that's the ratio of the perimeter of the first square to the perimeter of the second square. And then that is equal to 4:9. And they say, what is the ratio between the areas of the two squares? So they want us to figure out the area of the first square is x squared. Base times height, x times x. And the are of the second square is y times y. So they want us to figure out what that is equal to. Well this is x squared over y squared. This is the same thing as x over y squared. So if we can figure out what x over y is equal to, we can just square it and we'll get x squared over y squared. So let's try to do that. So they gave us this. Well this just simplifies. x over y is equal to 4 over 9. So let's substitute that here. So x squared over y squared is equal to x over y squared. Which is equal to 4/9 squared. Which is equal to 16 over 81. Or the ratio of the areas of the two squares is 16:81. Choice D. I think we can fit one more problem in there. Actually no, I'm over 10 minutes. I'll stop right there. See you in the next video.