Question

15. In punctul B se construieşte perpendiculara pe planul triunghiului ABC, eu m va rog ajutatima​

Answer 1

$\it MB\perp(ABC),\ AB\subset (ABC),\ \Rightarrow MB\perp AB \Rightarrow \Delta MBA-dreptunghic, \\ \\ m(\widehat{B})=90^o,\ \stackrel{T.P.}{\Longrightarrow}\ AB^2=AM^2-MB^2 \Rightarrow AB^2=13^2-5^2 \Rightarrow \\ \\ \Rightarrow AB^2=169-25=144=12^2 \Rightarrow AB=12cm\\ \\ \\ \Delta ABC-dreptunghic,\ m(\widehat{A})=90^o,\ \stackrel{T.P.}{\Longrightarrow}\ BC^2=AB^2+AC^2 \Rightarrow \\ \\ \Rightarrow BC^2=12^2+16^2=144+256=400=20^2 \Rightarrow BC=20cm$

$\it MB\perp (ABC),\ BC\subset (ABC)\ \Rightarrow MB\perp BC \Rightarrow \Delta MBC-dreptunghic,\\ \\ m(\widehat{B})=90^o,\ \stackrel{T.P.}{\Longrightarrow}\ MC^2=MB^2+BC^2=5^2+20^2=25+400 \Rightarrow \\ \\ \Rightarrow MC^2=425=25\cdot17=5^2\cdot17 \Rightarrow MC=5\sqrt{17}\ cm$

Answer 2

Răspuns:

Explicație pas cu pas:

AB=√(AM²-BM²)=√(13²-5²)=...=12cm

BC=√AC²-AB²=√(16²+12²)=20cm pt ca unghiul drept este in A

MC=√(BM²+BC²)=√(20²+5²)=...=5√17 cm