Question

Se considera dreptunghiul MNPQ. Daca MN=8cm si MP=10cm , aflati: a) Perimemetrul lui MNPQ si b) d(N,MP)

Answer 1

   
[tex]Avem: \texttt{Dreptunghiul MNPQ in care:} \\ \texttt{MP = 10 cm = diagonala lui MNPQ si ipotenuza } \Delta NMP \\ \texttt{MN = 8 cm = lungimea lui MNPQ si cateta } \Delta NMP \\ \\ \texttt{a) Calculam NP = latimea lui MNPQ si cateta } \Delta NMP \\ \\ NP = \sqrt{MP^2 - MN^2}= \sqrt{10^2 - 8^2}= \sqrt{100 - 64}=\sqrt{36}=\boxed{6~cm} \\ \\ \texttt{Calculam perimetrul dreptunghiului MNPQ.} \\ \\ P= 2(L + l) = 2(MN + NP) = 2(8 + 6) =2 \times 14 = \boxed{28 ~cm} [/tex]


[tex] \displaystyle \\ \texttt{b) Calculam d(N, MP)} \\ \texttt{d(N, MP) este inaltimea ipotenuzei in }\Delta MNP \\ \\ d(N, MP) = \frac{\texttt{Produsul catetelor}}{\texttt{Ipotenuza}} \\ \\ d(N, MP) = \frac{MN \times NP}{MP}=\frac{8 \times 6}{10}= \frac{48}{10}= \boxed{\frac{24}{5}~cm} = \boxed{4,8 ~cm}[/tex]