Why the Area of a Rhombus is Half the Product of its Diagonals

A rhombus is a parallelogram with four congruent sides. Since it is a parallelogram, it has also all the properties of a parallelogram. One of these properties is that the diagonals bisect each other. That is, they divide each other into two equal parts.

Another property of a rhombus is that the diagonals are perpendicular. So, summarizing all the properties above, if we have rhombus ABCD, then,

\overline{AB} \cong \overline{BC} \cong \overline{CD} \cong \overline{DA}.

and

\overline{AC} \perp \overline{BD}.

 From the figure below, it is clear that if we divide the parallelogram into two triangles, ABC and ACD,  \overline{BE} and \overline{DE} are their altitudes because they are perpendicular to \overline{AC}.

area of parallelogram

To find the area of parallelogram ABCD, first we find the area of triangle ABC, and then multiply it by 2. This is because triangle ABC is congruent to triangle CDA by SSS congruence (can you see why?).

Now, we know that the area of a triangle is 1/2 the product of the base and the height, so

area of triangle ABC = \frac{1}{2} (AC)(BE).

But we also know that BE is half of BD, so

Area of ABC = \frac{1}{2}(AC) (\frac{1}{2} BD) = \frac{1}{4} (AC)(BD).

Now, to get the area of ABCD, we multiply the area of ABC by 2. That is,

Area of parallelogram ABCD = 2 \left (\displaystyle \frac{1}{4}(AC)(BD) \right ) = \frac{1}{2} (AC)(BD)

This is what we want to prove: the area of parallelogram ABCD is half the product of diagonals AC and BD.

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