Question

na figura a baixo o valor de x é

Answer 1

1)

${x}^{2} + {(3x)}^{2} = {20}^{2} \\ {x}^{2} + 9 {x}^{2} = 400 \\ 10 {x}^{2} = 400$

${x}^{2} = \frac{400}{10} \\ {x}^{2} = 40 \\ x = \sqrt{40} = \sqrt{4.10} = 2 \sqrt{10}$

2)

$p =40 \\ 4l = 40 \\ l = \frac{40}{4} \\ l = 10$

$d = l \sqrt{2} \\ d = 10 \sqrt{2}$

3)

${x}^{2} + {2}^{2} = {6}^{2} \\ {x}^{2} + 4 = 36 \\ {x}^{2} = 36 - 4 \\ {x}^{2} = 32 \\ x = \sqrt{32} = \sqrt{2.16} = 4 \sqrt{2}$

4)

$x = \sqrt{2} \\ y = \sqrt{98} = \sqrt{49.2} = 7 \sqrt{2} \\ x + y = \sqrt{2} + 7 \sqrt{2} = 8 \sqrt{2}$