Question

Ajutoor la acest exercitiu

Answer 1

$a - cateta \: 1 \\ b - cateta \: 2 \\ c - ipotenuza$

$a = b - 4$

${a}^{2} + {b}^{2} = {c}^{2} \:(teorema\: lui\:Pitagora)$

${(b - 4)}^{2} + {b}^{2} = {20}^{2}$

${b}^{2} - 8b + 16 + {b}^{2} = 400$

${b}^{2} + {b}^{2} - 8b + 16 - 400 = 0$

$2 {b}^{2} - 8b - 384 = 0 \: | \div 2$

${b}^{2} - 4b - 192 = 0$

$x = 1$

$y = - 4$

$z = - 192$

$\Delta = {y}^{2} - 4xz$

$\Delta = {( - 4)}^{2} - 4 \times 1 \times ( - 192)$

$\Delta = 16 + 768$

$\Delta = 784$

$b_{1,2}=\frac{-y\pm\sqrt{\Delta}}{2x}$

$b_{1,2}=\frac{-( - 4)\pm\sqrt{784}}{2 \times 1}$

$b_{1,2}=\frac{4\pm28}{2 }$

$b_{1}=\frac{4 + 28}{2 }$

$b_{1}=\frac{32}{2 }$

$b_{1}=16$

$b_{2}=\frac{4 - 28}{2 }$

$b_{2}= - \frac{24}{2 }$

$b_{2}= - 12 \: nu \: convine$

$= > b = 16 \: cm$

$a = b - 4$

$a = 16 - 4$

$a = 12$

${a}^{2} + {b}^{2} = {c}^{2}$

${12}^{2} + {16}^{2} = {20}^{2}$

$144 + 256 = 400$

$400 = 400 \: (A)$