Question

Va rog mult, doar pct a) , am nevoie și de desen

Answer 1

Răspuns:

Explicație pas cu pas:

a}

Construim DA si EB inaltimi

A trapezului = 24√3

A trapezului  = (Baza mare +baza mica)x inaltimea /2   =>$\displaystyle 24\sqrt{3} = \frac{(8\sqrt{2}+4\sqrt{2})\times h }{2} \\\\2\times 24\sqrt{3} =(8\sqrt{2} +4\sqrt{2} )\times h\\\\ 48\sqrt{3} = 12\sqrt{2} \times h \\ \\ \displaystyle h = \frac{\not48\sqrt{3} }{\not12\sqrt{2} } \\ \\ h = \frac{4\sqrt{3} }{\sqrt{2} } \\\\h = \frac{\not4\sqrt{6} }{\not2} \\\\h= 2\sqrt{6}$

$\displaystyle tg=\frac{cateta opusa}{cateta alaturata}$

in ΔEBF aplicam tangenta de F=>

$\displaystyle tg F=\frac{EB}{FB} =\frac{2\sqrt{6} }{2\sqrt{2} } =\frac{\sqrt{6} }{\sqrt{2} } =\frac{\sqrt{12} }{2} =\frac{2\sqrt{3} }{2} =\sqrt{3}$

tangenta de 60° = √3  ⇒ ∠F = 60°

b) Aplicam Pitagora in ΔEBF ⇒ EF²=EB²+BF²  ⇒ EF²=(2√6)²+(2√2)² ⇒ EF²=24+8 ⇒ EF²=32 ⇒ EF=√32 ⇒ EF=4√2

Perimetrul defg = DE+EF+FG+GD=4√2+4√2+8√2+4√2=20√2

c

c)

Aria demg=DA×GM=2√6 × 4√2 = 8√12=16√3

M mij  FG => GM=FM = 8√2/2=4√2

BF=2√2 ⇒ MB=MF-BF=4√2-2√2=2√2

MF=4√2 DAR EF=4√2 si ∠F=60° ⇒ ΔEMF echilateral ⇒EM=4√2