If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

### Course: Geometry (all content)>Unit 7

Lesson 6: Area of shapes on grids

# 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?
• No. If the height differs then they can't have the same area
• 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.
• 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.
• 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
• Ho do you find the area of this crazy shape ?
• 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.
• At you 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)
• 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.
• 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.