Draw one diagonal of the parallelogram and find the area of the two triangles formed. If you add them, what do you get? Alternatively, you can cut and paste a section of the parallelogram to make a rectangle.
See solution.
Practice makes perfect
Let's begin by considering a parallelogram with base b and height h.
We will show that the area of this parallelogram is equal to the area of a rectangle with the same base and height. That is, A= b h. To do that, we will divide the parallelogram into two triangles by drawing a diagonal.
The area of each of the two triangles above is equal to 12 b h. Since the area of the parallelogram is equal to the sum of the areas of the triangles, we can write the equation below.
A = b h/2 + b h/2
⇓
A = b h
That way, we've shown that the area of the parallelogram is equal to the area of a rectangle with the same base and height.
Alternative Solution
Alternative Solution
Let's show another way to prove that the area of a parallelogram equals the area of a rectangle with the same base and height. This way can be divided into two disjoint cases, depending on the shape of the rectangle.
Cutting Off Method
Let's consider the parallelogram with base b and height h shown below.
Next, let's draw a second triangle on the right-hand side of the parallelogram.
Due to the parallelism between the sides, we have that the two triangles marked above are congruent. Now, we will cut off the left-hand side triangle and translate it to the right-hand side of the parallelogram.
As we can see, we obtained a rectangle with base b and height h. In consequence, the area of the original parallelogram is equal to the area of a rectangle with the same base and height.
A = b h
Drawing Auxiliary Triangles
Let's consider the parallelogram with base b and height h shown below.
Next, we will draw a line from B perpendicular to DC and mark some points.
We will repeat the same process, but this time we will draw the line from A.
By the Side-Angle-Side (SAS) Congruence Postulate, we get that △ BFC ≅ △ AGD, so they have the same area. Additionally, we can rewrite each of these areas in terms of the areas labeled in the diagram.
A(â–ł BFC)_(A_2+A_3) = A(â–ł AGD)_(A_4+A_3)
From the above, we obtain that A_2+A_3 = A_4+A_3 which implies that A_2=A_4. In consequence, A_1+A_2 = A_1+A_4.
A_1+A_2^(Area ParallelogramABCD) = A_1+A_4_(Area RectangleABFG)
Since the area of the rectangle ABFG is equal to b h, we obtain that the area of the parallelogram is also b h. This proves what we wanted to show.
