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### Course: 7th grade foundations (Eureka Math/EngageNY) > Unit 6

Lesson 3: Topic D: Foundations- Intro to nets of polyhedra
- Nets of polyhedra
- Surface area of a box using nets
- Surface area using nets
- Surface area of a box (cuboid)
- Surface area
- Area of a parallelogram
- Area of parallelograms
- Area of parallelograms
- Area of a triangle
- Area of triangles
- Area of triangles
- Area of composite shapes
- Area of composite shapes
- Area of a quadrilateral on a grid
- Areas of shapes on grids

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# Area of a quadrilateral on a grid

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

## Want to join the conversation?

- If 2 triangles have the same base, will they have the same area?(2 votes)
- No. If the height differs then they can't have the same area(9 votes)

- What if it is a quadrilateral but it has a semi circle on top of it? What would you do then(3 votes)
- 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.(2 votes)

- 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.(4 votes)
- 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(0 votes)

- Ho do you find the area of this crazy shape ?(2 votes)
- 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)(1 vote)

- I do not understand , is their another way to solve a simalar problem?(1 vote)
- 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.(2 votes)

- At5:30you gave the formula of how to find the area of a quadrilateral but it was quite confusing. Could you please explain it again?(1 vote)
- 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)(2 votes)

- What if the projecting triangle wasn't 1/2(0 votes)
- 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.(4 votes)

- 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?(1 vote)
- 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.(2 votes)

- At2:05, isn't the triangle right below it slanted a bit? If it is, this would make the whole problem wrong.(2 votes)
- we don't care about the right angle right now, we cut that angle in half to make the bottom triangle's angle(0 votes)

- The diagonal of a quadrilateral is 36cm remaining opposite vertices are 10 centimetre and 12 centimetre find the area of the field(1 vote)
- 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.(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.