Question

Pe planul triunghiului dreptunghic ABC, cu m(BAC)=90,se construieste perpendiculara MC,MC=24 cm. Stiine ca AB=30 cm si AC=40 cm,calculati distanta de la M la inaltimea AD, unde AD perpendicular pe BC,D apartine (BC) si distanta de la M la latura AB. Va rog ajutor am incercat de multe ori dar nu imi iese. Multumesc anticipat!

Answer 1

MC ⊥ (ABC)  |
CD ⊥ AD       | => MC ⊥ AD => d(M,AD)=MD
AD ⊂ (ABC)  |

In Δ ABC , $BC^{2}= AC^{2}+ AB^{2}$
[tex] BC^{2}= 40^{2}+ 30^{2} = 1600+900 = 2 500 =\ \textgreater \ BC= \sqrt{2 500}=\ \textgreater \ BC= 50[/tex]cm

AD-inaltime in ΔABC => AD=$\frac{AC*AB}{BC}= \frac{40*30}{50}= \frac{1200}{50}= 24$cm

In ΔACD ,$CD^{2}= AC^{2}- AD^{2}$
[tex] CD^{2}= 40^{2} - 24^{2} =1 600- 576 =1024 =\ \textgreater \ CD= \sqrt{1024}=\ \textgreater \ CD= 32[/tex] cm

In ΔMCD , [tex] MD^{2}= MC^{2}+ CD^{2} = 24^{2}+ 32^{2} =576+ 1024 = 1600 [/tex] => MD=$\sqrt{1600}=40cm$

b)

MC⊥(ABC)  |
CA⊥AB       | => MA ⊥ AB => d(M,AB)=MA
AB⊂(ABC)  |

In ΔMCA , $MA^{2}= MC^{2}+ CA^{2}$
[tex] MA^{2}= 24^{2} + 40^{2} =576 + 1600 =MA= \sqrt{2176} = 8 \sqrt{34}cm [/tex]