Question

61. Na figura abcd é um trapézio de bases AB e BC determine as medidas dos ângulos internos desse trapézio sabendo que AD//BE

62. Na figura abcd é um trapézio retangular e AB = BC calcule as medidas de x e y

Answer 1

$\large\begin{array}{l} \textsf{Para melhor compreens\~ao da resposta, veja os anexos.}\\\\ \textsf{61) O quadril\'atero ADEB \'e um paralelogramo. Os \^angulos}\\\textsf{da base s\~ao suplementares, ou seja, a soma \'e igual a }180^\circ.\\\\\textsf{Sendo assim,}\\\\ \mathsf{A\widehat{D}E+D\widehat{E}B=180^\circ}\\\\ \mathsf{A\widehat{D}E+123^\circ=180^\circ}\\\\ \mathsf{A\widehat{D}E=180^\circ-123^\circ}\\\\ \boxed{\begin{array}{c}\mathsf{A\widehat{D}E=57^\circ} \end{array}} \end{array}$


$\large\begin{array}{l} \textsf{Em todo paralelogramo, os \^angulos opostos s\~ao congruentes,}\\\textsf{ou seja, possuem a mesma medida:}\\\\ \mathsf{B\widehat{A}D=D\widehat{E}B}\\\\ \boxed{\begin{array}{c}\mathsf{B\widehat{A}D=123^\circ} \end{array}}\\\\\\ \mathsf{E\widehat{B}A=A\widehat{D}E}\\\\ \boxed{\begin{array}{c}\mathsf{E\widehat{B}A=57^\circ} \end{array}}\end{array}$


$\large\begin{array}{l} \textsf{Observe que }\mathsf{D\widehat{E}B}\textsf{ \'e um \^angulo externo ao tri\^angulo BEC.}\\\textsf{Ent\~ao, a medida de }\mathsf{D\widehat{E}B}\textsf{ \'e igual \`a soma dos \^angulos internos }\\\textsf{de BEC n\~ao-adjacentes a ele:}\\\\ \mathsf{D\widehat{E}B=E\widehat{B}C+B\widehat{C}E}\\\\ \mathsf{123^\circ=90^\circ+B\widehat{C}E}\\\\ \mathsf{B\widehat{C}E=123^\circ-90^\circ}\\\\ \boxed{\begin{array}{c}\mathsf{B\widehat{C}E=33^\circ} \end{array}} \end{array}$


$\large\begin{array}{l} \textsf{Por \'ultimo, encontramos o \^angulo }\mathsf{C\widehat{B}A:}\\\\ \mathsf{C\widehat{B}A=C\widehat{B}E+E\widehat{B}A}\\\\ \mathsf{C\widehat{B}A=90^\circ+57^\circ}\\\\ \boxed{\begin{array}{c}\mathsf{C\widehat{B}A=147^\circ} \end{array}} \end{array}$

_________

$\large\begin{array}{l} \textsf{62) Como os segmentos }\overline{\mathsf{AB}}\textsf{ e }\overline{\mathsf{BC}}\textsf{ s\~ao congrumentes, ent\~ao o tri\^angulo}\\\textsf{ABC \'e is\'osceles. Dessa forma, os \^angulos da base s\~ao congruentes:}\\\\ \mathsf{B\widehat{A}C=B\widehat{C}A}\\\\ \mathsf{B\widehat{A}C=x}\\\\\\ \textsf{Os \^angulos }\mathsf{B\widehat{A}C}\textsf{ e }\mathsf{C\widehat{A}D}\textsf{ s\~ao complementares:}\\\\ \mathsf{B\widehat{A}C+C\widehat{A}D=90^\circ}\\\\ \mathsf{x+72^\circ=90^\circ}\\\\ \mathsf{x=90^\circ-72^\circ}\\\\ \boxed{\begin{array}{c}\mathsf{x=18^\circ} \end{array}} \end{array}$


$\large\begin{array}{l} \textsf{A soma dos \^angulos internos do tri\^angulo ABC \'e }\mathsf{180^\circ:}\\\\ \mathsf{x+y+x=180^\circ}\\\\ \mathsf{y+2x=180^\circ}\\\\ \mathsf{y=180^\circ-2x}\\\\ \mathsf{y=180^\circ-2\cdot 18^\circ}\\\\ \mathsf{y=180^\circ-36^\circ}\\\\ \boxed{\begin{array}{c}\mathsf{y=144^\circ} \end{array}} \end{array}$


$\large\begin{array}{l} \textsf{D\'uvidas? Comente.}\\\\\\ \textsf{Bons estudos!} \end{array}$