Question

3) (UNISC INV/2014) Os lados de um losango medem 4 cm e um de seus
ângulos 60o. As medidas da diagonal menor e da diagonal maior do losango medem, respectivamente

a) 2 cm e 2√3 cm.
b) 2√3 cm e 4 cm.
c) 2√3 cm e 4√3 cm.
d) 4 cm e 4√3 cm.
e) 4 cm e 8 cm.

Desenvolva

Answer 1

Explicação passo-a-passo:

• Diagonal menor

Pela lei dos cossenos:

$\sf d^2=4^2+4^2-2\cdot4\cdot4\cdot cos~60^{\circ}$

$\sf d^2=4^2+4^2-2\cdot4\cdot4\cdot\dfrac{1}{2}$

$\sf d^2=16+16-\dfrac{32}{2}$

$\sf d^2=32-16$

$\sf d^2=16$

$\sf d=\sqrt{16}$

$\sf \red{d=4~cm}$

• Diagonal maior

Os ângulos obtusos desse losango medem $\sf 180^{\circ}-60^{\circ}=120^{\circ}$

Pela lei dos cossenos:

$\sf D^2=4^2+4^2-2\cdot4\cdot4\cdot cos~120^{\circ}$

$\sf D^2=4^2+4^2-2\cdot4\cdot4\cdot\Big(-\dfrac{1}{2}\Big)$

$\sf D^2=16+16+\dfrac{32}{2}$

$\sf D^2=32+16$

$\sf D^2=48$

$\sf D=\sqrt{48}$

$\sf \red{D=4\sqrt{3}~cm}$

Letra D

Answer 2

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$\tt{\underline{C\acute{a}lculo~da~Diagonal~maior~}}\\\sf{D^2=4^2+4^2-2\cdot4\cdot4\cdot cos~120^\circ}\\\sf{D^2=16+16-\diagup\!\!\!2\cdot16\cdot\left(-\dfrac{1}{\diagup\!\!\!2}\right)}\\\sf{D^2=32+16}\\\sf{D^2=48}\\\sf{D=\sqrt{48}}\\\sf{D=\sqrt{16\cdot3}}\\\sf{D=4\sqrt{3}~cm}$

$\tt{\underline{C\acute{a}lculo~da~diagonal~menor}}\\\sf{d^2=4^2+4^2-2\cdot4\cdot4\cdot cos~60^\circ}\\\sf{d^2=16+16-\diagup\!\!\!2\cdot16\cdot\left(\dfrac{1}{\diagup\!\!\!2}\right)}\\\sf{d^2=16+\diagdown\!\!\!\!\!16-\diagdown\!\!\!\!\!16}\\\sf{d^2=16}\\\sf{d=\sqrt{16}}\\\sf{d=4~cm}$