Question

determine as medidas x e y indicadas na figura​

Answer 1

No ∆PQR:

${(3\sqrt{41})}^{2}={(x+10)}^{2}+{12}^{2}\\{(x+10)}^{2}+144=369\\{(x+10)}^{2}=369-144$

${(x+10)}^{2}=225\\x+10=\sqrt{225}\\x+10=15$

$\color{blue}{x=15-10=5}\color{green}{\checkmark}$

No ∆PQS:

${y}^{2}={x}^{2}+{12}^{2}\\{y}^{2}={5}^{2}+{12}\\{y}^{2}=25+144\\{y}^{2}=169$

$y=\sqrt{169}\\ \colorbox{black}{\color{cyan}{y=13}}\color{green}{\checkmark}$