Question

Ad si dc direct proportionale cu 0.1(6) si 0,375

Answer 1

   
[tex]\texttt{Enuntul problemei:} \\ \text{intr-un triunghi dreptunghic lungimea ipotenuzei este egala } \\ \text{cu 20,8 dm iar lungimiile proiectiilor catetelor sunt direct proportionale} \\ \text{cu 0,1(6) si 0,375. Aflati lungimea inaltimii corespunzatoare ipotenuzei.} \\ \\ Notat: \\ \Delta ABC ~cu~ \ \textless \ A=90^o \\ Ipotenuza~BC \\ Catetele AB~ si ~AC ~si ~ inaltimea AD ~unde ~D \in BC\\ Se ~cere: \\ AD = ?[/tex]

[tex]\displaystyle Rezolvare: \\ \frac{BD}{0,1(6)}= \frac{CD}{0,375} ~\Longleftrightarrow~ \frac{BD}{ \frac{16-1}{90}}= \frac{CD}{ \frac{375}{1000}} ~\Longleftrightarrow~ \frac{BD}{ \frac{1}{6}}= \frac{CD}{ \frac{3}{8}} =k \\ BD+CD=20,8 \\ --- \\ BD= \frac{1}{6}k ~~si~~CD=\frac{3}{8}k \\ \\ BD+CD= \frac{1}{6}k+ \frac{3}{8}k= 20,8 \\ \\ \frac{4k+9k}{24}=20,8 \\ \\ 13k= 24 \times 20,8 \\ k= \frac{24 \times 20,8}{13}= 24 \times 1,6 = 38,4 [/tex]

[tex]\displaystyle \\ BD= \frac{1}{6}k = \frac{1}{6} \times 38,4 = \frac{38,4}{6}= \boxed{6,4 ~dm} \\ \\ CD=\frac{3}{8}k= \frac{3}{8}\times 38,4= \frac{3 \times 38,4}{8}=3 \times 4,8 = \boxed{14,4 ~dm} \\ \\ \text{Aflam catetele folosind teorema catetei:}\\ \text{Patratul catetei = prod. dintre ipotenuza si proiectia ei pe ipotenuza} \\ AB = \sqrt{ BD \times BC}= \sqrt{ 6,4 \times 20,8}= \sqrt{\frac{64 \times 208}{100}}= \sqrt{\frac{64 \times 16 \times 13}{100}}= [/tex]

[tex]\displaystyle =\frac{8 \times 4 \times \sqrt{13}}{10}= \boxed{3,2\sqrt{13}~dm} \\ \\ AC= \sqrt{ CD \times BC}= \sqrt{ 14,4 \times 20,8}= \sqrt{\frac{144 \times 208}{100}}= \\ \\ \sqrt{\frac{144 \times 16 \times 13}{100}}=\frac{12 \times 4 \times \sqrt{13}}{10}=\boxed{4,8\sqrt{13}~dm} \\ \\ \text{Calculam inaltimea AD :} \\ \\ AD = \frac{AB \times AC}{BC} = \frac{3,2\sqrt{13} \times 4,8\sqrt{13}}{20,8} = \frac{3,2 \times 4,8 \times 13}{20,8}= [/tex]

[tex]\displaystyle \\ = \frac{3,2 \times 3 \times 1,6 \times 13}{20,8}=\frac{3,2 \times 3 \times 20,8}{20,8}=3,2 \times 3 = \boxed{\boxed{9,6 ~dm }}[/tex]