
# 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:

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

\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

$\triangle AA'D\cong \triangle BB'C.$

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:

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