What if it is a quadrilateral but it has a semi circle on top of it? What would you do then

Find the area of the quadrilateral, then find the area of a circle with the same radius as the semicircle and divide it by 2.

If 2 triangles have the same base, will they have the same area?

No. If the height differs then they can't have the same area

I do not understand , is their another way to solve a simalar problem?

yes there are many, many ways to solve all problem, it just is up to you to find out which way works the best for you.

At 5:30 you gave the formula of how to find the area of a quadrilateral but it was quite confusing. Could you please explain it again?

Pretty much, with the edges, you form right triangles. Then, you find their area and then add that to the area of the regular quadrilateral. :0)

why did'nt Sal calculate the over all area (base of 6 and height of 5) and then subtract the triangles outside the parallelogram but inside the overall area?

because it is not a parallelogram, the slope of the top line is 1/3 and the slope of the bottom line is 1/5, opposite sides of a parallelogram must be parallel, thus must have the same slope. The same could be done for the left and right lines.

What if the projecting triangle wasn't 1/2

Can you specify what you mean? Are you asking why Khan multiplies all of the triangles areas by 1/2? If so, it's because a triangle is half of a quadrilateral, so if he didn't divide by 2, he would solve for 2 triangles at once which would be incorrect. To find the area of a quadrilateral, you multiply the base times the height (or length X width). If you cut the quadrilateral diagonally, you will get a triangle with the same dimensions but half of the original quadrilateral. So obviously, to find the area of the new triangle, you must divide by 2.

Main content

Geometry (all content)

Course: Geometry (all content) > Unit 7

Lesson 6: Area of shapes on grids

Area of a quadrilateral on a grid

Name: Area of a quadrilateral on a grid
Uploaded: 2013-10-30T16:19:00Z
Description: Learn to break up oddly shaped quadrilaterals into shapes where finding the area is more easily determined.

Google Classroom

Learn to break up oddly shaped quadrilaterals into shapes where finding the area is more easily determined. Created by Sal Khan.

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Caroline
Posted 6 years ago. Direct link to Caroline's post “If 2 triangles have the s...”
If 2 triangles have the same base, will they have the same area?
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(2 votes)
Answer
- bluejay2004
  Posted 6 years ago. Direct link to bluejay2004's post “No. If the height differs...”
  No. If the height differs then they can't have the same area
  Comment on bluejay2004's post “No. If the height differs...”
  (9 votes)
vaidehi desai
Posted 9 years ago. Direct link to vaidehi desai's post “What if it is a quadrilat...”
What if it is a quadrilateral but it has a semi circle on top of it? What would you do then
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(3 votes)
Answer
- Colleen Murray
  Posted 9 years ago. Direct link to Colleen Murray's post “Find the area of the quad...”
  Find the area of the quadrilateral, then find the area of a circle with the same radius as the semicircle and divide it by 2.
  Comment on Colleen Murray's post “Find the area of the quad...”
  (2 votes)
Jyoti Bisht
Posted 7 years ago. Direct link to Jyoti Bisht's post “Find the area of a rhombu...”
Find the area of a rhombus with side 14cm and altitude 5.2cm.if one of its diagonals is 15cm long,find the lenght of its other diagonal.
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(4 votes)
Answer
- BariumLanthanum
  Posted 5 years ago. Direct link to BariumLanthanum's post “In a rhombus, all four tr...”
  In a rhombus, all four triangles created by the diagonals are congruent. the side length, 14 is the hypotenuse of the four triangles. Use the Pythagorean Theorem
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  (0 votes)
Andrea Yahrmatter Gray
Posted 7 years ago. Direct link to Andrea Yahrmatter Gray's post “Ho do you find the area o...”
Ho do you find the area of this crazy shape ?
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(2 votes)
Answer
- James Lansdell
  Posted 7 years ago. Direct link to James Lansdell's post “I solved it by first lear...”
  I solved it by first learning the area of a rectangle that encompassed the whole quadrilateral and then subtracted away the area of the triangles not part of the shape. It seemed more intuitive to me than the video.
  So working out would be: (6*7) - (2.5 + 2.5 + 6 + 4)
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  (1 vote)
Sneha J
Posted 9 years ago. Direct link to Sneha J's post “I do not understand , is ...”
I do not understand , is their another way to solve a simalar problem?
Button navigates to signup pageComment on Sneha J's post “I do not understand , is ...”
(1 vote)
Answer
- Michaelthomps3
  Posted 9 years ago. Direct link to Michaelthomps3's post “yes there are many, many ...”
  yes there are many, many ways to solve all problem, it just is up to you to find out which way works the best for you.
  Button navigates to signup page
  (2 votes)
abhijeet.gite
Posted 9 years ago. Direct link to abhijeet.gite's post “At 5:30 you gave the form...”
At
5:30
you gave the formula of how to find the area of a quadrilateral but it was quite confusing. Could you please explain it again?
Button navigates to signup pageComment on abhijeet.gite's post “At 5:30 you gave the form...”
(1 vote)
Answer
- aernado
  Posted 7 years ago. Direct link to aernado's post “Pretty much, with the edg...”
  Pretty much, with the edges, you form right triangles. Then, you find their area and then add that to the area of the regular quadrilateral. :0)
  Button navigates to signup page
  (2 votes)
Zev Sharma
Posted 9 years ago. Direct link to Zev Sharma's post “What if the projecting t...”
What if the projecting triangle wasn't 1/2
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(0 votes)
Answer
- Daberculosis
  Posted 9 years ago. Direct link to Daberculosis's post “Can you specify what you ...”
  Can you specify what you mean? Are you asking why Khan multiplies all of the triangles areas by 1/2? If so, it's because a triangle is half of a quadrilateral, so if he didn't divide by 2, he would solve for 2 triangles at once which would be incorrect. To find the area of a quadrilateral, you multiply the base times the height (or length X width). If you cut the quadrilateral diagonally, you will get a triangle with the same dimensions but half of the original quadrilateral. So obviously, to find the area of the new triangle, you must divide by 2.
  Button navigates to signup page
  (4 votes)
Mike Snow Carini
Posted 4 years ago. Direct link to Mike Snow Carini's post “why did'nt Sal calculate ...”
why did'nt Sal calculate the over all area (base of 6 and height of 5) and then subtract the triangles outside the parallelogram but inside the overall area?
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(1 vote)
Answer
- David Severin
  Posted 4 years ago. Direct link to David Severin's post “because it is not a paral...”
  because it is not a parallelogram, the slope of the top line is 1/3 and the slope of the bottom line is 1/5, opposite sides of a parallelogram must be parallel, thus must have the same slope. The same could be done for the left and right lines.
  Comment on David Severin's post “because it is not a paral...”
  (2 votes)
Inch
Posted 8 years ago. Direct link to Inch's post “At 2:05, isn't the triang...”
At
2:05
, isn't the triangle right below it slanted a bit? If it is, this would make the whole problem wrong.
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(2 votes)
Answer
- Dylan Wang
  Posted 8 years ago. Direct link to Dylan Wang's post “we don't care about the r...”
  we don't care about the right angle right now, we cut that angle in half to make the bottom triangle's angle
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  (0 votes)
anushreegautam1994
Posted 6 years ago. Direct link to anushreegautam1994's post “The diagonal of a quadril...”
The diagonal of a quadrilateral is 36cm remaining opposite vertices are 10 centimetre and 12 centimetre find the area of the field
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(1 vote)
Answer
- David Severin
  Posted 6 years ago. Direct link to David Severin's post “Please rephrase your ques...”
  Please rephrase your question, vertices are angles and are measured in degrees, not distance units Sides of a quad are measured in centimeters, so are you talking about the two sides that form one of the vertices of the diagonal? I am trying to think about this, but not sure I can figure this out. Will wait for clarification, maybe someone else can help.
  Button navigates to signup page
  (1 vote)

Video transcript

I want you to pause this video and figure out if you can figure out the area of this quadrilateral right over here. And I'll give you a hint. Try to break it up into shapes where it is easier to find the area, especially given the grid of these unit squares that we have right over here. So I assume you gave it a go. Now let's try to do it together. So I'm going to start at points where it would be very easy to measure the dimensions of whatever figures we might break it up at. And so in general, I want to go to any of the whole numbers of these units, so that point right over there. And let's see. I could start to move in this direction. And it looks like I might be able to construct a triangle down here. And it would be tempting to go all the way across, but that wouldn't be too useful because this gets me to halfway through a unit. And I'm just eyeballing it to say halfway through a unit. It might not be exactly halfway through a unit. And so instead, let me see if I can get away with making a triangle just like that. Now, I just did that. So let me try to raise this up to see if I can make another triangle where it would be easy to figure out its dimensions. So once again, I don't want to go all the way up here because now I'm not at a whole unit. Instead, let me take a right and go right over here. And notice, both of these are very easy to figure out its dimensions. This is 1, 2, 3, 4, 5 units long and 1 unit high. This one right over here is 1, 2, 3, 4 units long and 1, 2 units wide. So let's see if we can cover the entire quadrilateral, if we can break it up, I should say, into a bunch of figures like this. So it seems like we have another one just like that. And then I could drop this down, and then we're done. All of these are pretty straightforward to figure out what their dimensions are. This is 5 by 1. This is 4 by 2. This is 1, 2, 3, 4, 5, 6 by 1, 2. And this is 1 by 1, 2, 3, 4, 5. So what is the area of this figure? And of course, we have this center rectangle right over here. Well, a triangle that is 5 units long and 1 unit high, its area is going to be 1/2 times 1 times 5. Or I could write it 1/2 times 1 times 5, depending on what multiplication symbol you are more comfortable with. Well that's just going to be 1/2 times 5, which is going to be equal to 2.5. So that's 2.5 right over there. This one is going to be 1/2 times 4 times 2. Well, that's just going to be 2 times 2 or 4. This one is going to be 1/2 times 2 times 1, 2, 3, 4, 5, 6. Well, 1/2 half times 2 is 1 times 6 is just 6. And then this one's going to be 1/2 times 1 times 1, 2, 3, 4, 5. So once again, the area of this one is going to be 2.5. And then finally, this is a 3 by 4 rectangle. And you could even count the unit squares in here. But it has 12 of those unit square, so it has an area of 12. So if we want to find the total area, we just add all of these together. So 2.5 plus 2.5 is 5, plus 4 is 9, plus 6 is 15, plus 12 is 27. So it has a total area of 27.