Ensino Superior ⇒ Geometria Euclidiana Espacial Tópico resolvido

[Idocrase]

As dimensões de um paralelepípedo retângulo estão em progressão aritmética, uma diagonal mede [tex3]2\sqrt{14}[/tex3] cm e sua área total vale 88 cm². Quais são as dimensões do paralelepípedo?

Alguém me ajuda nesse exercício por favor?

[petras]

Idocrase,
[tex3]\mathtt{a=a_1=n, \\
b = a_ 2=n+r, \\
c=a_3=n+2r\\
d^2=(a)^2+(b)^2+(c)^2\\
(2√14)^2=n^2++(n+r)^2+(n+2r)^2 \implies \underline{56-5r^2=3n^2+6nr} (I)\\
S_t=2.(ab+ac+bc)\\
88=2.(ab+ac+bc)\implies 44=ab+ac+bc\\
44=n.(n+r)+n.(n+2r)+(n+r).(n+2r) \implies \underline{44-2r^2=3n^2+6nr} (II)\\
I=II: 56-5r^2=44-2r^2 \implies \underline{r = \pm 2}\\
\underline{r=-2}: 56-5(-2)^2 = 3n^2+6n(-2)\\
3n^2-12n-36=0 \implies \cancel{n = -2} : n=6\\
\therefore a = 6, b = 4, c =2\\
\underline{r=2}:\\
56-5(2)^2=3n^2+6n(2) \\
3n^2+12n-36=0 \implies \cancel{n =-6}, n =2 \\
\therefore a= 2, b = 4, c = 6\\

\therefore \boxed{2,4,6}\color{green}\checkmark}
[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\mathtt{a=a_1=n, \\
b = a_ 2=n+r, \\
c=a_3=n+2r\\
d^2=(a)^2+(b)^2+(c)^2\\
(2√14)^2=n^2++(n+r)^2+(n+2r)^2 \implies \underline{56-5r^2=3n^2+6nr} (I)\\
S_t=2.(ab+ac+bc)\\
88=2.(ab+ac+bc)\implies 44=ab+ac+bc\\
44=n.(n+r)+n.(n+2r)+(n+r).(n+2r) \implies \underline{44-2r^2=3n^2+6nr} (II)\\
I=II: 56-5r^2=44-2r^2 \implies \underline{r = \pm 2}\\
\underline{r=-2}: 56-5(-2)^2 = 3n^2+6n(-2)\\
3n^2-12n-36=0 \implies \cancel{n = -2} : n=6\\
\therefore a = 6, b = 4, c =2\\
\underline{r=2}:\\
56-5(2)^2=3n^2+6n(2) \\
3n^2+12n-36=0 \implies \cancel{n =-6}, n =2 \\
\therefore a= 2, b = 4, c = 6\\

\therefore \boxed{2,4,6}\color{green}\checkmark}
[/tex3]

[Idocrase]

Idocrase,
[tex3]\mathtt{a=a_1=n, \\
b = a_ 2=n+r, \\
c=a_3=n+2r\\
d^2=(a)^2+(b)^2+(c)^2\\
(2√14)^2=n^2++(n+r)^2+(n+2r)^2 \implies \underline{56-5r^2=3n^2+6nr} (I)\\
S_t=2.(ab+ac+bc)\\
88=2.(ab+ac+bc)\implies 44=ab+ac+bc\\
44=n.(n+r)+n.(n+2r)+(n+r).(n+2r) \implies \underline{44-2r^2=3n^2+6nr} (II)\\
I=II: 56-5r^2=44-2r^2 \implies \underline{r = \pm 2}\\
\underline{r=-2}: 56-5(-2)^2 = 3n^2+6n(-2)\\
3n^2-12n-36=0 \implies \cancel{n = -2} : n=6\\
\therefore a = 6, b = 4, c =2\\
\underline{r=2}:\\
56-5(2)^2=3n^2+6n(2) \\
3n^2+12n-36=0 \implies \cancel{n =-6}, n =2 \\
\therefore a= 2, b = 4, c = 6\\

\therefore \boxed{2,4,6}\color{green}\checkmark}
[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\mathtt{a=a_1=n, \\
b = a_ 2=n+r, \\
c=a_3=n+2r\\
d^2=(a)^2+(b)^2+(c)^2\\
(2√14)^2=n^2++(n+r)^2+(n+2r)^2 \implies \underline{56-5r^2=3n^2+6nr} (I)\\
S_t=2.(ab+ac+bc)\\
88=2.(ab+ac+bc)\implies 44=ab+ac+bc\\
44=n.(n+r)+n.(n+2r)+(n+r).(n+2r) \implies \underline{44-2r^2=3n^2+6nr} (II)\\
I=II: 56-5r^2=44-2r^2 \implies \underline{r = \pm 2}\\
\underline{r=-2}: 56-5(-2)^2 = 3n^2+6n(-2)\\
3n^2-12n-36=0 \implies \cancel{n = -2} : n=6\\
\therefore a = 6, b = 4, c =2\\
\underline{r=2}:\\
56-5(2)^2=3n^2+6n(2) \\
3n^2+12n-36=0 \implies \cancel{n =-6}, n =2 \\
\therefore a= 2, b = 4, c = 6\\

\therefore \boxed{2,4,6}\color{green}\checkmark}
[/tex3]

Entendi, muito obrigado.