Question

calcule o valor de X nas figuras:

Answer 1

a) Esses traços indicam que esses lados têm a mesma medida.

Pelo Teorema de Pitágoras:

$y^2+y^2=(4\sqrt{2})^2 \iff 2y^2=32 \iff y^2=\dfrac{32}{2}$

$y^2=16 \iff y=\sqrt{16} \iff \boxed{y=4}$

Novamente, pelo Teorema de Pitágoras:

$y^2+(y+x)^2=(4\sqrt{5})^2 \iff 4^2+(4+x)^2=80$

$16+16+8x+x^2=80 \iff x^2+8x+32-80=0 \iff x^2+8x-48=0$

$\Delta=8^2-4\cdot1\cdot(-48)=64+192=256$

$x=\dfrac{-8\pm\sqrt{256}}{2\cdot1}=\dfrac{-8+16}{2}=\dfrac{8}{2}=4$

$\boxed{x=4}$

e) No triângulo azul: $y^2=30^2+(50-x)^2$

No triângulo verde: $y^2=40^2+x^2$

Igualando essas equações: $(y^2=y^2)$

$30^2+(50-x)^2=40^2+x^2 \iff 900+2500-100x+x^2=1600+x^2$

$3400-100x+x^2=1600+x^2 \iff 3400-1600+x^2-x^2=100x$

$100x=1800 \iff x=\dfrac{1800}{100} \iff \boxed{x=18}$


f) $d^2=9^2+12^2 \iff d^2=81+144 \iff d^2=225$

$d=\sqrt{225} \iff d=15$

Utilizando a relação métrica $ah=bc$:

$a=15 \ \ \ \ \ h=x \ \ \ \ \ b=9 \ \ \ \ \ c=12$

$15x=9\cdot12 \iff 15x=108$

$x=\dfrac{108}{15} \iff x=\dfrac{36}{5} \iff \boxed{x=7,2}$.