Olá, Lilian.
Pelo Teorema de Pitágoras, temos:
[tex]a^2=(\overline{QM})^2+(\overline{MR})^2=2^2+4^2=4+16 \Rightarrow a=\sqrt{20}=\sqrt{2^2\cdot5} \\\\ \Rightarrow \boxed{a=2\sqrt5} \\\\ b^2=(\overline{PN})^2+(\overline{RN})^2=8^2+4^2=64+16=80 \Rightarrow b=\sqrt{2^4\cdot 5}\\\\ \Rightarrow b=2^2\sqrt5 \Rightarrow \boxed{b=4\sqrt5}\\\\ c^2=a^2+b^2=4\cdot 5+16\cdot 5=5(4+16)=5\cdot20 \Rightarrow c=\sqrt{100}\\\\ \Rightarrow \boxed{c=10}[/tex]
Perímetro P do trapézio MNPQ:
[tex]P=\overline{QM}+\overline{MR}+\overline{RN}+\overline{NP}+\overline{PQ}=2+4+4+8+10 \Rightarrow \\\\ \boxed{P=28}[/tex]
