Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

20.6: Area of solid parallelograms

  • Page ID
    23715
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)

    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

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

    截屏2021-03-03 上午9.01.52.png

    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

    \[\triangle AA'D\cong \triangle BB'C.\] 

    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

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

    截屏2021-03-03 上午9.02.53.png

    Hint

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

    截屏2021-03-03 上午9.09.22.png

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

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