20.6: Área de paralelogramos sólidos
- Page ID
- 114509
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Dejar\(\square ABCD\) ser un paralelogramo en el plano euclidiano,\(a=AB\) y\(h\) ser la distancia entre las líneas\((AB)\) y\((CD)\). Entonces
\(\text{area }(\blacksquare ABCD)=a\cdot h.\)
- Prueba
-
Dejar\(A'\) y\(B'\) denotar los puntos del pie de\(A\) y\(B\) en la línea\((CD)\).
Tenga en cuenta que\(ABB'A'\) es un rectángulo con lados\(a\) y\(h\). Por Teorema 20.5.1,
\[\text{area }(\blacksquare ABB'A')=h\cdot a.\]
Sin pérdida de generalidad, podemos suponer que\(\blacksquare ABCA'\) contiene\(\blacksquare ABCD\) y\(\blacksquare ABB'A'\). En este caso\(\blacksquare ABCA'\) admite dos subdivisiones:
\(\blacksquare ABCA'=\blacksquare ABCD\cup\blacktriangle AA'D=\blacksquare ABB'A'\cup\blacksquare BB'C.\)
\[\begin{aligned} \text{area }( \blacksquare ABCD)&+\text{area }(\blacktriangle AA'D)= \\ &= \text{area }(\blacksquare ABB'A')+ \text{area } (\blacktriangle BB'C). \end{aligned}\]
Tenga en cuenta que
En efecto, dado que los\(ABB'A'\) cuadriláteros y\(ABCD\) son paralelogramos, por Lemma 7.5.1, tenemos eso\(AA'=BB'\),\(AD=BC\), y\(DC=AB=A'B'\). De ello se deduce que\(A'D=B'C\). Aplicando la condición de congruencia SSS, obtenemos 20.6.3.
En particular,
\[\text{area }(\blacktriangle BB'C)=\text{area } (\blacktriangle AA'D). \]
Restar 20.6.4 de 20.4.2, obtenemos que
\[\text{area } (\blacksquare ABCD)=\text{area }(\blacksquare ABB'D).\]
Queda por aplicar 20.6.1.
Supongamos\(\square ABCD\) y\(\square AB'C'D'\) son dos paralelogramos tales que\(B'\in[BC]\) y\(D\in [C'D']\). Demostrar que
\(\text{area }(\blacksquare ABCD)=\text{area }(\blacksquare AB'C'D').\)
- Pista
-
Supongamos que\(E\) denota el punto de intersección de las líneas\((BC)\) y\((C'D')\).
Utilice Proposición\(\PageIndex{1}\) para probar las dos identidades siguientes:
\(\begin{array} {l} {\text{area } (\blacksquare AB'ED) = \text{area } (\blacksquare ABCD),} \\ {\text{area } (\blacksquare AB'ED) = \text{area } (\blacksquare AB'C'D')} \end{array}\)