Question

50 PONTOS PFV AJUDEM
Calcule a área dos triângulos a seguir:

Answer 1

Explicação passo-a-passo:

b)

$\sf S=\sqrt{p\cdot(p-a)\cdot(p-b)\cdot(p-c)}$

$\sf p=\dfrac{a+b+c}{2}$

$\sf p=\dfrac{5+6+7}{2}$

$\sf p=\dfrac{18}{2}$

$\sf p=9$

$\sf S=\sqrt{9\cdot(9-5)\cdot(9-6)\cdot(9-7)}$

$\sf S=\sqrt{9\cdot4\cdot3\cdot2}$

$\sf S=\sqrt{216}$

$\sf \red{S=6\sqrt{6}~cm^2}$

c)

Seja $\sf b$ a medida do lado HI

Pela lei dos cossenos:

$\sf (2\sqrt{19})^2=b^2+6^2-2\cdot b\cdot6\cdot cos~60^{\circ}$

$\sf (2\sqrt{19})^2=b^2+6^2-2\cdot b\cdot6\cdot\dfrac{1}{2}$

$\sf 76=b^2+36-6b$

$\sf b^2-6b+36-76=0$

$\sf b^2-6b-40=0$

$\sf \Delta=(-6)^2-4\cdot1\cdot(-40)$

$\sf \Delta=36+160$

$\sf \Delta=196$

$\sf b=\dfrac{-(-6)\pm\sqrt{196}}{2\cdot1}=\dfrac{6\pm14}{2}$

$\sf b'=\dfrac{6+14}{2}~\Rightarrow~b'=\dfrac{20}{2}~\Rightarrow~b'=10$

$\sf b"=\dfrac{6-14}{2}~\Rightarrow~b"=\dfrac{-8}{2}~\Rightarrow~b"=-4$ (não serve)

Assim, $\sf b=10$

A área do triângulo é:

$\sf S=\dfrac{b\cdot c\cdot sen~\hat{A}}{2}$

$\sf S=\dfrac{10\cdot6\cdot sen~60^{\circ}}{2}$

$\sf S=\dfrac{10\cdot6\cdot\frac{\sqrt{3}}{2}}{2}$

$\sf S=\dfrac{\frac{60\sqrt{3}}{2}}{2}$

$\sf S=\dfrac{30\sqrt{3}}{2}$

$\sf \red{S=15\sqrt{3}~cm^2}$