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: Area, circumference, volume
31-35, area, circumference and volume. Created by Sal Khan.
Want to join the conversation?
- how do you find circumference? still a little lost.(8 votes)
- 2 pi r, you multiply the radius times two and then multiply your product by pi.
use pi as, 3 or 3.14.(6 votes)
- If I have two cubes of the same exact figure, (volume,area,etc.) would one fit inside the other?(5 votes)
- theoretically, yes, but you couldn't take one out because then they would be the same thing. My reasoning is, when you fit two pipes together (one inside the other) and lets say it was a perfect fit down to the atom, then the atoms on the pipe a would be no closer to the other atoms in pipe a than to the atoms in pipe b, because its a perfect fit. Therefore the atoms in pipe a can interact with the atoms in pipe b and they would fuse. Another way to put it, is if you fit those same two pipes together again, then the outside circumfrence of the pipe on the inside would be exactly the same as the inner circumfrence of the pipe on the outside. If its all hard to understand, then blame my curiosity for quantum and theoretical physics.(1 vote)
- Cant you just use the special right triangles theorem/postulate? The one about
45-45-90 triangles, and 30-60-90 triangles?(4 votes)
- yes, yes you can.(4 votes)
- ok i get the problems and such but when i go to take like a test or something i can never recognize what i'm supposed to do to find the answer. for example for the problem where you had to subtract the triangles from the rectangle i paused the video and was thinking that i had to find the missing sides of the triangle tto get the answer... ik fail but i need help(6 votes)
- How do you find the volume of a sphere? I am very much confused.(2 votes)
- This is the formula for the volume of a sphere:
v = 4/3πr^3(4 votes)
- Where can you get more of the CST questions at? I think these questions are great to practice with if you are preparing to join a competitive mathmatical environment like I am. I would really appreciate a copy of a link to these problems. For now, I will just use these videos. Thank you very much, Sal, for posting these competition math and test preparation math problems!(3 votes)
- At6:17I didn't exactly understand what the rhombus part was.(2 votes)
- a rhombus is kinda like a squished square. it has 4 sides that are all the same length. the sides that are across from each-other are parallel. the angles that are across from eachother are congruent.
hope that helped❀(3 votes)
- Question: The diameter of a wheel is 4. How many complete revolutions will the wheel make if it rolls a distance of 100.5pi?(2 votes)
- If the diameter of a wheel is 4, the circumference of the wheel is 4pi.
The wheel will make a complete revolution every time it rolls a distance of 4pi units. Thus, to find how many complte revolutions it makers after rolling 100.5pi units, we must divide 100.5pi by 4pi.
Since you said complete revolutions, we should omit the 1/8 of a revolution.
Thus, a wheel with a diameter of 4 units will make 25 complete revolutions if it rolls 100.5pi units.
I hope this answers your question well.(2 votes)
- Thank you for posting video's it really helps people that need to see the big picture.(2 votes)
- I'm a liitle lost on how he did the equation of the area on number 35 and how he got the answer, little help ??(2 votes)
Area of triangle = (1/2)*base * height
He showed that the triangle was a 30-60-90 triangle.
In a 30-60-90 triangle, the longest side (across from the 90 degree angle) is 2 times the length of the shortest side (across from the 30 degree angle).
In this problem, the longest side of the triangle is 12, so the shortest side =
6. This side is the
baseof the triangle.
Also in a 30-60-90 triangle, the side across from the 60 degree angle has a length which is sqrt(3)*(shortest side)
In this problem, the shortest side is 6, so the side across from the 60 degree angle is sqrt(3)* (6) =
6sqrt(3). This side is the
heightof the triangle.
Area of triangle = (1/2)*base*height
= 18sqrt(3)(1 vote)
We're on problem 31. A sewing club is making a quilt consisting of 25 squares, with each side of the square measuring 30 centimeters. OK, and there's going to be 25 of these. And they're going to be 30 by 30. If the quilt has 5 rows and 5 columns, what is the perimeter of the quilt? OK, so let me draw that out. So it has 5 rows and 5 columns and they're all squares. So the quilt itself is going to be a square. It has 4 rows, one two, three, four, and then five. And then one, two, three, four. So that's a 5 by 5. And each of these squares, their sides are 30 centimeters. So how long is one side of this quilt going to be? It's going to be 30 times 5. So it's going to be 150 centimeters. Same argument, that's going to be 150 centimeters. And that's going to be 150 centimeters. 30 times 5. And this is going to be 150 centimeters. So the perimeter is 150 plus 150 plus 150 plus 150 and that's 600. So that's choice C. 32. The four sides of this figure will be folded up and taped to make a box. Fair enough. What will be the volume of the box? OK, so if we cut this out right now and we folded it along where I'm drawing these green lines, we'll get a box. And they want to say, what's the volume? Well, volume is just the base times the height times the depth. So if I were to fold up this box is going to look something like this. You're going to have the base, which is this base right here. And this is one, one, two, three, four, five by one, two, three, four, five. So it's 5 by 5 base. And then each of the sides are going to be 2 high if I fold this up, it's going to be 2 high like that. It's going to look like that if I folded that side up. This side, when I fold it up, is going to look like this. One, two, three, four, five. That's side when I fold it is going to look like that. And this side when I fold it up is going to look like that. One, two, three, four, five. The big picture, the width is 5, the depth is 5, and the height is 2. So the volume is 5 times 5 is 35 times 2. Which is equal to 50. And that's choice A. Problem 33. Where's 33, I think it's on the next page. OK, let me copy and paste it. I should just copy and paste the whole test. OK, it says a classroom globe has a diameter of 18 inches. If I were to go from the center to the side, it's 18 inches. Which of the following is the approximate surface area. Oh no, sorry, I just drew the radius. It has a diameter of 18 inches. This is 18. Which of the following is an approximate surface area in square inches of the globe? And surface area, they give us the equation, they give it in terms of the radius. So if the diameter is 18, what's the radius? The radius is half of the diameter. So the radius is equal to 9. And we just plug that into here. So the surface area is equal to 4 pi times the radius squared, times 9 squared. That equals 4 times 81 times pi. So it's 324 pi. And they actually multiplied it out. So let's see. Let's see, 324 pi. And if I were to guess this, I mean, look at all the choices. Pi is more than 3. So this value is going to be more than 3 times 324. So it's going to be around 1,000 or a little bit more than 1,000. And the only one that's even close to that is D. But if you wanted to confirm that, you could multiply 324 times 3.14 and that is equal to 1,017.4. All right, next problem. Problem 34. I'll copy and paste 34 and 35 at the same time. There you go. I put this there. All right, ready to do it. The rectangle shown below has a length of 20 meters and width of 10. So this is 10 and this is 20. I just picked that because this looks longer than that. That's a 10 I drew. I know it doesn't look like a 10. If the four triangles are removed from the rectangle as shown, what will be the area of the remaining figure? So what's the are before I remove them? It's 20 times 10. That's the area of the whole rectangle. So it's 200. And then how much area am I removing? So each of these triangles, what's its area? It's base times height times 1/2. That's the area of a triangle. Because if you just did base times height, you'd be figuring out the area of this little rectangle there. So the area of this is 4 times 4 is 16 times 1/2, which is 8. This is going to be 8, going to be 8, going to be 8. So we're removing four 8's from this area. So we're removing 32. So minus 32. And that's 168. So that's choice C. Problem 35. If RSTW is a rhombus, so rhombus tells us that all the sides are equal and they're parallel. What is the area of WXT. So this right here. OK, so this is something that you may or may not have learned about a rhombus already. But its diagonals actually intersect at a perpendicular line. And let me see what else we can see. This is 60 degrees, so this is 30 degrees. Let's see what we can get from this. This is 12. Then this is 12. OK, I see where they're going with this. So if this is 90 degrees, this is 90 degrees. This is a rhombus, so all the sides are the same. If this is 60, this is 90, this has to be 30 degrees. And then you could actually make a very strong argument that these are similar triangles. Whatever length this is, that's the same length, because this is a parallelogram and the diagonals bisect each other. This side is equal to that side. That side is equal to this side. So these are congruent triangles. So this is also going to be 60 degrees. This is going to be 30. But if you have a 60, let me do it in another color, if you have a 60, 60, 60 triangle, all of the angles are 60 degrees, you're dealing with an equilateral triangle. So that tells you that all the sides are the same. So if this side is 12, that side is 12, this side right here also has to be 12. If that whole side is 12, what's this length? We already know that in a parallelogram the diagonals bisect each other. So this length is 6. And this length is 6. Fair enough. And let's see, if each of these lengths are 6, can we figure out what this height is equal to? Because if we know the base and the height, we're ready to figure out the area of a triangle. So let's see if we can use the Pythagorean Theorem. If we called this x, we could say x squared plus 6 squared plus 36 is equal to 12 squared, is equal to 144. And we you could say that x squared is equal to, what's 144 minus 36, that's 108. Let's see, 108. x squared is equal to 108. x is equal to the square root of 108. And I can simplify that more because 9 goes into 108 12 times. Let me do that. So x is equal to the square root of 9 times 12, that's 108. So that's equal to the square root of 9 times the square root of 12. That's equal to 3 times the square root of 12. Square root of 12 is the same thing as the square root of 3 times the square root of 4. Square root of 4 is 2. So that's 2 times 3 is 6 square root of 3. This is 36 times 3. We could have said this is equal to the square root of 36 times the square root of 3. All right, so 6 square roots of 3. That's this side. So what's the area of this triangle right there? It's 1/2 times this base times 6 times 6 square roots of 3. So that's 1/2 times 6 is 3, times, 6 square roots of 3 is 18 square roots of 3. Now that's just this triangle. This triangle is congruent to this triangle, so it will have the same area. And you can make the same argument that all of these triangles are congruent. So the area of this entire rhombus is going to be 4 times this. Is that what they wanted? Oh no, they wanted of the area of WXD. That's what we just figured out. They didn't want the area of the whole rhombus. They want just the area of this triangle right there, which we just figured out, which is 18 square roots of 3. I'm trying to think if there's a simpler way of doing this. There might be some formula for the equation for the area of a rhombus that I've forgotten in my memory. But we were able to re-prove it. And that's always better, to come from basic principles. Anyway, I'll see you in the next video.