20.6: Area of solid parallelograms

Theorem $\PageIndex{1}$

Let $\square ABCD$ be a parallelogram in the Euclidean plane, $a=AB$ and $h$ be the distance between the lines $(AB)$ and $(CD)$. Then

$\text{area }(\blacksquare ABCD)=a\cdot h.$

Proof

Let $A'$ and $B'$ denote the foot points of $A$ and $B$ on the line $(CD)$.

Note that $ABB'A'$ is a rectangle with sides $a$ and $h$. By Theorem 20.5.1,

$$\text{area }(\blacksquare ABB'A')=h\cdot a.$$

Without loss of generality, we may assume that \(\blacksquare ABCA'\) contains \(\blacksquare ABCD\) and \(\blacksquare ABB'A'\). In this case \(\blacksquare ABCA'\) admits two subdivisions:

$\blacksquare ABCA'=\blacksquare ABCD\cup\blacktriangle AA'D=\blacksquare ABB'A'\cup\blacksquare BB'C.$

By Proposition 20.4.2,

$$\begin{aligned} \text{area }( \blacksquare ABCD)&+\text{area }(\blacktriangle AA'D)= \\ &= \text{area }(\blacksquare ABB'A')+ \text{area } (\blacktriangle BB'C). \end{aligned}$$

Note that

Indeed, since the quadrangles $ABB'A'$ and $ABCD$ are parallelograms, by Lemma 7.5.1, we have that $AA'=BB'$, $AD=BC$, and $DC=AB=A'B'$. It follows that $A'D=B'C$. Applying the SSS congruence condition, we get 20.6.3.

In particular,

$$\text{area }(\blacktriangle BB'C)=\text{area } (\blacktriangle AA'D).$$

Subtracting 20.6.4 from 20.4.2, we get that

$$\text{area } (\blacksquare ABCD)=\text{area }(\blacksquare ABB'D).$$

It remains to apply 20.6.1.

Exercise $\PageIndex{1}$

Assume $\square ABCD$ and $\square AB'C'D'$ are two parallelograms such that $B'\in[BC]$ and $D\in [C'D']$. Show that

$\text{area }(\blacksquare ABCD)=\text{area }(\blacksquare AB'C'D').$

Hint

Suppose that $E$ denotes the point of intersection of the lines $(BC)$ and $(C'D')$.

Use Proposition $\PageIndex{1}$ to prove the following two identities:

$\begin{array} {l} {\text{area } (\blacksquare AB'ED) = \text{area } (\blacksquare ABCD),} \\ {\text{area } (\blacksquare AB'ED) = \text{area } (\blacksquare AB'C'D')} \end{array}$