area of parallelogram proof?

Answer 1

A parallelogram is a four-sided plane figure that has two sets of opposite parallel sides. Every parallelogram is a polygon, and more specifically a quadrilateral. Special cases of a parallelogram are the rhombus, in which all four sides are of equal length, the rectangle, in which the two sets of opposing, parallel sides are perpendicular to each other, and the square, in which all four sides are of equal length and the two sets of opposing, parallel sides are perpendicular to each other. In any parallelogram, the diagonals bisect each other, i.e, they cut each other in half.

The parallelogram law distinguishes Hilbert spaces from other Banach spaces.

It is possible to create a tessellation with any parallelogram.

The three-dimensional counterpart of a parallelogram is a parallelepiped.

The area of a parallelogram can be seen as twice the area of a triangle created by one of its diagonals. The area of a parallelogram can be found by using the formula (Base multiplied by Height equals area). The area can also be computed as the magnitude of the vector cross product of two of its non-parallel sides.

Answer 2

Which formula for area of a parallelogram are you trying to prove? If it's base times height, you can conduct a geometric proof. Suppose the parallelogram has a base of b and a height of h. Cut the parallelogram into a rectangle and two congruent triangles; the two triangles are 180 degrees rotated from each other on opposite sides of the rectangle. Each triangle has a base of a and a height of h, so the area of each is 0.5*ah, and the area of both together is ah. The rectangle has a base of (b - a) and a height of h, so its area is bh - ah. Sume the areas of the rectangle and both triangles, and you get bh, which is what was to be proved.

Answer 3

evaluate a parallelogram as a rectangle with all aspects straight away{this does not happen with regards to parallelogram.NOW,from one in each and every of its vertices draw a straight away line to the alternative area this is the perpendicular to that area.it is going to likely be its vertical top.we already be responsive to that each and all of the climate of a rectangle are perpendicular to a minimum of one yet another making an attitude of ninety degree in any respect of its vertices.the area of the rectangle is length*breadth the place length is the backside and breadth will become its vertical top.hence THE formula FOR RECTANGLE IS BASE *top.evaluate the comparable FOR A PARALLELOGRAM.you will get YOUR answer.hence resembling THE CASE OF A RECTANGLE,IF WE MARK VERTICAL top FOR A PARALLELOGRAM,ITS formula additionally will grow to be BASE*top. thank you

Answer 4

AREA OF A PARALLELOGRAM.
Let ABCD be a parallelogram. Clearly, the diagonal BD divides this parallelogram into two equal triangles.


Area of parallelogram ABCD= 2 * (area of ▲ABD)
=2 * (½ * AB * DE) sq units
= (AB * DE) sq units
= (BASE*HEIGHT) sq units
Hence, area of parallelogram = (b * h) sq units



AREA OF A QUADRILATERAL.
Let us consider a quadrilateral ABCD whose one diagonal, say AC, and the lengths of the perpendiculars to Ac from the opposite vertices B and D are given.
Let BE┴AC and DF┴AC.
Then, the area of quadrilateral ABCD=area of ▲ABC + area of ▲ACD
= ½ * AC * BE + ½ * AC * DF
= ½ *AC *(BE+DF) sq units
Hence, area of quadrilateral = ½ *diagonal*(sum of heights) sq units

Answer 5

TO PROVE: Area of //gm = base. height ______
Given: /\ /
Area of triangle = 1/2 . base. height / \ /
/ \ /
/____\ /
a diagonal f a //gm divides it in2 2 congruent traingles.
Area of each tri. = 1/2 . base. height
hence, area f //gm = 2. 1/2 . base. height

Answer 6

The proof is explained on this page:

http://www.cut-the-knot.org/Curriculum/Geometry/AreaOfParallelogram.shtml

Answer 7

Easy, Isn't it.. everyone has explained

Answer 8

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