6.5: El Área de un Trapezoide

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En la Figura$\PageIndex{1}$,$b_1$ y$b_2$ son las bases del trapecio$ABCD$ y$h$ es la altura o altitud. La fórmula para el área se da en el siguiente teorema:

Screen Shot 2020-12-28 a las 11.40.42 AM.png — Figura$\PageIndex{1}$: Trapezoide$ABCD$ con bases$b_1$ y$b_2$ y altura$h$.

Teorema$\PageIndex{1}$

El área de un trapecio es igual a la mitad del producto de su altura y la suma de sus bases.

\[A = \dfrac{1}{2} h(b_1 + b_2)\]

Prueba: En Figura$\PageIndex{1}$ dibujar$BD$ (ver Figura$\PageIndex{2}$). Tenga en cuenta que$CD = b_2$ es la base y$BF = h$ es la altura de$\triangle BCD$. Área de trapecio$ABCD =$ Área de$\triangle ABD$ + Área de$\triangle BCD$ =$\dfrac{1}{2} b_1h + \dfrac{1}{2} b_2 h = \dfrac{1}{2} h(b_1 + b_2)$.

$Screen Shot 2020-12-28 a las 11.46.50 AM.png$
Figura$\PageIndex{2}$: Dibujar$BD$. $CD$es la base y$BF$ es la altura de$\triangle BCD$.

Ejemplo$\PageIndex{1}$

Encuentra la zona:

$Screen Shot 2020-12-28 a las 11.42.45 AM.png$

Solución

$A = \dfrac{1}{2} h(b_1 + b_2)= \dfrac{1}{2} (6)(28 + 16) = \dfrac{1}{2} (6)(44) = 132$.

Respuesta:$A = 132$.

Ejemplo$\PageIndex{2}$

Encuentra el área y perímetro:

$Screen Shot 2020-12-28 a las 11.48.05 AM.png$

Solución

Dibujar alturas$DE$ y$CF$ (Figura$\PageIndex{3}$). $\triangle ADE$es$30^{\circ}-60^{\circ}-90^{\circ}$ triángulo. Entonces$AE$ = pierna corta =$\dfrac{1}{2}$ hipotenusa =$\dfrac{1}{2}(10) = 5$, y$DE$ = pierna larga = (pierna corta) ($\sqrt{3}$) =$5\sqrt{3}$. $CDEF$es un rectángulo así$EF = CD = 10$. Por lo tanto$BF = AB - EF = 22 -10 - 5 = 7$. Vamos$x = BC$.

Screen Shot 2020-12-28 a las 11.50.59 AM.png — Figura$\PageIndex{3}$: Dibujar alturas$DE$ y$CF$.

\[\begin{array} {rcl} {\text{CF}^2 + \text{BF}^2} & = & {\text{BC}^2} \\ {(5\sqrt{3})^2 + 7^2} & = & {x^2} \\ {75 + 49} & = & {x^2} \\ {124} & = & {x^2} \\ {x} & = & {\sqrt{124} = \sqrt{4} \sqrt{31} = 2\sqrt{31}} \end{array}\]

Área =$\dfrac{1}{2} h(b_1 + b_2) = \dfrac{1}{2} (5\sqrt{3}) (22 + 10) = \dfrac{1}{2} (5\sqrt{3})(32) = 80\sqrt{3}$.

Perímetro =$22 + 10 + 10 + 2\sqrt{31} = 42 + 2\sqrt{31}$.

Respuesta:$A = 80\sqrt{3}$,$P = 42 + 2\sqrt{31}$.