20.6: Area of solid parallelograms
- Page ID
- 23715
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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
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.
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:
\(\begin{array} {l} {\text{area } (\blacksquare AB'ED) = \text{area } (\blacksquare ABCD),} \\ {\text{area } (\blacksquare AB'ED) = \text{area } (\blacksquare AB'C'D')} \end{array}\)