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How do you draw a triangle and a parallelogram that each have an area of 20 units?

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2011-02-27 00:01:23
2011-02-27 00:01:23

Do you know the formula for the area of a triangle ?

Pick a base-length and a height for your triangle so that 1/2 (base x height) = 20.

Do you know the formula for the area of a parallelogram ?

Pick a base-length and a height for your parallelogram so that (base x height) = 20.

We're having a problem understanding your difficulty.

Of course, if you don't know the formulas for area . . .

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draw a parallelogram which is not a rectangle.verify that its area is equal to the rectangle on the same base and altitude



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You don't have to do anything at all. As soon as you draw a triangle, it automaticallyhas area. The only trick for you is to figure out how much area it has.



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Right triange with both sides around the 90 degrees = 4 with a right trianle the area = 1/2 * height * width (think of it as half a rectangle)


YES From your start point draw a line 5 units up, from this point draw a line 5 units across, from this point draw a line 5 units down, from this point draw a line 5 units back to the start. You have drawn a square with a total perimeter length of 20 units and a area of 25 square units.


Because - if you draw lines at right-angles from the base of a triangle vertically until they reach the highest point, then draw a horizontal line that connects those points, touching the highest point of the triangle - the area outside the triangle (but inside the resulting quadrilateral) is exactly half the area of the quadrilateral.


The area of qa triangle is always half of the area of a rectangle with the same dimensions




Well, it helps immensely to know the formula for the area of a triangle. The area of a triangle is 1/2 (length of the base) times (height) .Knowing that, all you have to do is draw a triangle in such a way that the length of the base (in cm), multiplied by the height (in cm) is 12 . There are an infinite number of ways to do it, so finding one should not be a problem.


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