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Boa tarde Curiosa!
Solução!
Vamos usar as relações métricas no triângulo retângulo.
Vamos primeiro calcular o valor de b.
![(15)^{2}=(8)^{2}+b^{2}\\\\\
225=81+b^{2}\\\\\
225-81=b^{2} \\\\\\
144=b^{2}\\\\\
b= \sqrt{144}\\\\\
b=12](https://tex.z-dn.net/?f=%2815%29%5E%7B2%7D%3D%288%29%5E%7B2%7D%2Bb%5E%7B2%7D%5C%5C%5C%5C%5C%0A225%3D81%2Bb%5E%7B2%7D%5C%5C%5C%5C%5C%0A225-81%3Db%5E%7B2%7D+%5C%5C%5C%5C%5C%5C%0A144%3Db%5E%7B2%7D%5C%5C%5C%5C%5C%0Ab%3D+%5Csqrt%7B144%7D%5C%5C%5C%5C%5C%0Ab%3D12%0A%0A++++++ "(15)^{2}=(8)^{2}+b^{2}\\\\\
225=81+b^{2}\\\\\
225-81=b^{2} \\\\\\
144=b^{2}\\\\\
b= \sqrt{144}\\\\\
b=12")
Calculando o valor de x.
a=15
b=12
c=8
![b^{2}=a.x\\\\\
(12)^{2}=15m\\\\\\
144=15m\\\\\\
m= \frac{144}{15}\\\\\\
m=9,6cm](https://tex.z-dn.net/?f=b%5E%7B2%7D%3Da.x%5C%5C%5C%5C%5C%0A%2812%29%5E%7B2%7D%3D15m%5C%5C%5C%5C%5C%5C%0A144%3D15m%5C%5C%5C%5C%5C%5C%0Am%3D+%5Cfrac%7B144%7D%7B15%7D%5C%5C%5C%5C%5C%5C%0Am%3D9%2C6cm+%C2%A0+ "b^{2}=a.x\\\\\
(12)^{2}=15m\\\\\\
144=15m\\\\\\
m= \frac{144}{15}\\\\\\
m=9,6cm")
Calculo do valor de y.
![a.y=b.c\\\\\
15.y=12.8\\\\\
15y=96\\\\\\
y= \dfrac{96}{15}\\\\\\
y=6,4cm](https://tex.z-dn.net/?f=a.y%3Db.c%5C%5C%5C%5C%5C%0A15.y%3D12.8%5C%5C%5C%5C%5C%0A15y%3D96%5C%5C%5C%5C%5C%5C%0Ay%3D+%5Cdfrac%7B96%7D%7B15%7D%5C%5C%5C%5C%5C%5C+%0Ay%3D6%2C4cm "a.y=b.c\\\\\
15.y=12.8\\\\\
15y=96\\\\\\
y= \dfrac{96}{15}\\\\\\
y=6,4cm")
Boa tarde!
Bons estudos!
Solução!
Vamos usar as relações métricas no triângulo retângulo.
Vamos primeiro calcular o valor de b.
Calculando o valor de x.
a=15
b=12
c=8
Calculo do valor de y.
Boa tarde!
Bons estudos!
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