Ensino Fundamental ⇒ Romboide Tópico resolvido

[botelho]

O é centro do romboide ABCE e CD=4, SE=2.Calcule a distância de T a ED.

IMG-20231028-WA0004.jpg (38.97 KiB) Exibido 240 vezes

a)[tex3]\frac{3}{2}[/tex3]
b)[tex3]\frac{5}{2}[/tex3]
c)2
d)[tex3]\frac{4}{3}[/tex3]
e)[tex3]\frac{5}{4}[/tex3]

Resposta

---

a

[petrasMOD]

botelho,

[tex3]tg 75^o = \frac{2}{CS} \implies 2+\sqrt3 = \frac{2}{CS} \therefore CS = 2(2-\sqrt3)\\
DS = CD-CS = 4 - (4-2\sqrt3) = 2\sqrt3\\
\triangle ESD: DE^2 = 2^2+(2\sqrt3)^2 = 16 \therefore DE = 4\implies \angle DES = 60^o \\
\triangle CSE: sen75^o = \frac{2}{CE} \implies \frac{\sqrt6+\sqrt2}{4}=\frac{2}{CE} \therefore CE = 2(\sqrt6-\sqrt2)\\
CL \perp AD(L \in AD)\\\\
\triangle CEL: sen75^=\frac{CL}{CE} \implies CL = 2\\
GM \perp AD (M(\in AD) = \frac{CL}{2} = 1\\
TI \perp AD(I \in AD)\\
MGCL: IT_{(base~média)} = \frac{GM+CL}{2}= \frac{2+1}{2} = \boxed{\frac{3}{2}}





[/tex3]