Question

2- construa um triângulo retangulo abc que tenha como vértices A (8,3), B (-4,8) e C (-4,3)

Answer 1

Explicação passo-a-passo:

2)

a)

• Cateto AC

$\sf \overline{AC}=\sqrt{(x_A-x_C)^2+(y_A-y_C)^2}$

$\sf \overline{AC}=\sqrt{(8+4)^2+(3-3)^2}$

$\sf \overline{AC}=\sqrt{12^2+0^2}$

$\sf \overline{AC}=\sqrt{144+0}$

$\sf \overline{AC}=\sqrt{144}$

$\sf \overline{AC}=12$

• Cateto BC

$\sf \overline{BC}=\sqrt{(x_B-x_C)^2+(y_B-y_C)^2}$

$\sf \overline{BC}=\sqrt{(-4+4)^2+(8-3)^2}$

$\sf \overline{BC}=\sqrt{0^2+5^2}$

$\sf \overline{BC}=\sqrt{0+25}$

$\sf \overline{BC}=\sqrt{25}$

$\sf \overline{BC}=5$

Os catetos medem 12 e 5

b)

Seja $\sf x$ a medida da hipotenusa

Pelo Teorema de Pitágoras:

$\sf x^2=12^2+5^2$

$\sf x^2=144+25$

$\sf x^2=169$

$\sf x=\sqrt{169}$

$\sf x=13$

A hipotenusa mede 13

3)

a)

• Cateto AC

$\sf \overline{AC}=\sqrt{(x_A-x_C)^2+(y_A-y_C)^2}$

$\sf \overline{AC}=\sqrt{(-3+3)^2+(7-1)^2}$

$\sf \overline{AC}=\sqrt{0^2+6^2}$

$\sf \overline{AC}=\sqrt{0+36}$

$\sf \overline{AC}=\sqrt{36}$

$\sf \overline{AC}=6$

• Cateto BC

$\sf \overline{BC}=\sqrt{(x_B-x_C)^2+(y_B-y_C)^2}$

$\sf \overline{BC}=\sqrt{(5+3)^2+(1-1)^2}$

$\sf \overline{BC}=\sqrt{8^2+0^2}$

$\sf \overline{BC}=\sqrt{64+0}$

$\sf \overline{BC}=\sqrt{64}$

$\sf \overline{BC}=8$

Os catetos medem 6 e 8

b)

Seja $\sf x$ a medida da hipotenusa

Pelo Teorema de Pitágoras:

$\sf x^2=6^2+8^2$

$\sf x^2=36+64$

$\sf x^2=100$

$\sf x=\sqrt{100}$

$\sf x=10$

A hipotenusa mede 10

4)

a)

• Cateto AC

$\sf \overline{AC}=\sqrt{(x_A-x_C)^2+(y_A-y_C)^2}$

$\sf \overline{AC}=\sqrt{(-3-3)^2+(5-5)^2}$

$\sf \overline{AC}=\sqrt{6^2+0^2}$

$\sf \overline{AC}=\sqrt{36+0}$

$\sf \overline{AC}=\sqrt{36}$

$\sf \overline{AC}=6$

• Cateto BC

$\sf \overline{BC}=\sqrt{(x_B-x_C)^2+(y_B-y_C)^2}$

$\sf \overline{BC}=\sqrt{(3-3)^2+(13-5)^2}$

$\sf \overline{BC}=\sqrt{0^2+8^2}$

$\sf \overline{BC}=\sqrt{0+64}$

$\sf \overline{BC}=\sqrt{64}$

$\sf \overline{BC}=8$

Os catetos medem 6 e 8

b)

Seja $\sf x$ a medida da hipotenusa

Pelo Teorema de Pitágoras:

$\sf x^2=6^2+8^2$

$\sf x^2=36+64$

$\sf x^2=100$

$\sf x=\sqrt{100}$

$\sf x=10$

A hipotenusa mede 10