Question

Fie dreptunghiul ABCD cu AB=6 cm si BC=4cm.Perpendiculara in C pe AC intersecteaza dreapta AD in M.Aflati lungimea segmentului AM.

Answer 1

Aplic teorema lui Pitagora in triunghiul ABC-dreptunghic in B:
$AB^{2}+ BC^{2}= AC^{2}=>AC= \sqrt{AB^{2}+BC^{2}}= \sqrt{36+16}= \sqrt{52}= \\ = 2\sqrt{13} (cm).$

Fie {E}=MC ∩ AB.

Aplic teorema inaltimii in ΔACE-dreptunghic in C:
$BC^{2}=AB*BE=>BE= \frac{BC^{2}}{AB}= \frac{16}{6}= \frac{8}{3}(cm).$

Aplic teorema lui Pitagora in ΔBEC-dreptunghic in B:
$BC^{2}+ BE^{2}= CE^{2} =>CE= \sqrt{ BC^{2}+BE^{2} } = \sqrt{16+ \frac{64}{9} }= \sqrt{ \frac{208}{9} } = \\ = \frac{4 \sqrt{3} }{3} (cm).$

Aplic teorema inaltimii in ΔMAE-dreptunghic in A:
$AC^{2}=MC*CE=>MC= \frac{AC^{2}}{CE}= \frac{52}{ \frac{4 \sqrt{3} }{3} }= \frac{52*3}{ 4\sqrt{3} }= \frac{39}{ \sqrt{3} }= \frac{39 \sqrt{3} }{3}= \\ 13 \sqrt{3}(cm).$

ME=MC+CE=$13\sqrt{3}+ \frac{4 \sqrt{3} }{3}= \frac{43 \sqrt{3} }{3} (cm).$

Aplic teorema lui Pitagora in ΔMAE-dreptunghic in A:
$AM^{2}+ AE^{2}= ME^{2}=>AM= \sqrt{ME^{2}-AE^{2}}= \sqrt{ \frac{1849}{3}- \frac{676}{9} }= \\ = \sqrt{ \frac{4871}{9} }= \frac{ \sqrt{4871} }{3}(cm).$