What is the area of parallelogram ABCD in square units

My way of solving this if to make four triangles and one rectangle out of this shape.

Use AB, BC, CD, DA to make the triangles.
The left-overs make up a rectangle that has the area of 2x3, which is 6.

Triangle AB and CD are equivalent, they are both (1/2)(1)(3)=1.5 each, which add up to 3.

Triangle AD and BC are equivalent, they are both (1/2)(1)(4)=2 each, which add up to 4.

So, just add up the 6 from the rectangle and the 3 & 4 and you get 13.

The area of that parallelogram is 13.

BladeRunner212 BladeRunner212 · Answer 2 · 2025-04-05T05:40:14-04:00

13 square units

Further explanation

Consider attachment for details.

We make a KLMN rectangle that touches all the vertices of the ABCD parallelogram. Consequently, the ABCD parallelogram is right inside the KLMN rectangle.

Let us take the following strategic steps:

Calculate the area of KLMN.
Calculate the area of the triangles ABL, CDM, ADK, and BCN.
Subtract the area of the KLMN rectangle with the area of all triangles.
The difference in the area above is the area of the ABCD parallelogram.

The Process:

The area of KLMN = 4 x 5 = [tex] \boxed{ \ 20 \ square \ units. \ }[/tex]
The ADK triangle is congruent to the BCN triangle, and each area is [tex]\boxed{ \ \frac{1}{2} \times 4 \times 1 = 2 \ square \ units. \ }[/tex] Thus the total area of ADK and BCN is [tex]\boxed{ \ 2 + 2 = 4 \ square \ units. \ }[/tex]
The ABL triangle is congruent to the CDM triangle, and each area is [tex]\boxed{ \ \frac{1}{2} \times 3 \times 1 = 1.5 \ square \ units \ }.[/tex] Thus, the combined area of ABL and CDM is [tex]\boxed{ \ 1.5 + 1.5 = 3 \ square \ units. \ }[/tex]
Finally, the area of ABCD = 20 - 4 - 3 = 13.

As a result, we get the area of the parallelogram ABCD is 13 square units.

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Keywords: what is the area of parallelogram ABCD, in square units, graph, cartesian coordinates, triangle, rectangular, congruent, touches all the vertices