calcule o valor de cada cateto do seguinte triângulo retângulo: hipotenusa 20 cateto x+4 cateto x
Soluções para a tarefa
Resposta:
\begin{gathered} {a}^{2} + {b}^{2} = {c}^{2} \\ {x}^{2} + {(4 + x)}^{2} = {20}^{2} \\ {x}^{2} + {4}^{2} + 2 \times 4 \times x + {x}^{2} = 400 \\ {x}^{2} + 16 + 8x + {x}^{2} = 400 \\ 2 {x}^{2} + 8x - 400 + 16 = 0 \\ (2 {x}^{2} + 8x - 384 = 0) \div 2 \\ {x}^{2} + 4x - 192 = 0 \\ \\ x = \frac{ - b \sqrt{ {b}^{2} - 4 \times a \times c } }{2 \times a} \\ x = \frac{ - 4 \sqrt{ {( - 4)}^{2} - 4 \times 1 \times - 192 } }{2 \times 1} \\ x = \frac{ - 4 \sqrt{16 + 768}}{2}\\ x = \frac{{ - 4 \sqrt{784}}}{2} \\ x = \frac{ - 4 + 28}{2} = \frac{24}{2} = 12 \\ x = \frac{ - 4 - 28}{2} = \frac{ - 32}{2} = - 16 \\ \\ 12 \: ou \: - 16\end{gathered}
a
2
+b
2
=c
2
x
2
+(4+x)
2
=20
2
x
2
+4
2
+2×4×x+x
2
=400
x
2
+16+8x+x
2
=400
2x
2
+8x−400+16=0
(2x
2
+8x−384=0)÷2
x
2
+4x−192=0
x=
2×a
−b
b
2
−4×a×c
x=
2×1
−4
(−4)
2
−4×1×−192
x=
2
−4
16+768
x=
2
−4
784
x=
2
−4+28
=
2
24
=12
x=
2
−4−28
=
2
−32
=−16
12ou−16