Question

determine a área do triângulo retângulo cujos cada catetos medem ( x + 3) cm e (5x-7) cm e cuja hipotenusa mede (2x + 4) cm​

Answer 1

${(2x + 4)}^{2} = {(x + 3)}^{2} + {(5x - 7)}^{2} \\ 4 {x}^{2} + 16x + 16 = {x}^{2} + 6x + 9 \\ + 25 {x}^{2} - 70x + 49$

$4 {x}^{2} - {x}^{2} - 25 {x}^{2} \\ + 16x - 6x + 70x = 49 + 9 - 16 \\ - 22 {x}^{2} + 80x = 42 \\ 22 {x}^{2} - 80x + 42 = 0 \div 2$

$11 {x}^{2}- 40x + 21 = 0$

$\Delta = {b}^{2} - 4ac \\ \Delta = {( - 40)}^{2} - 4.11.21 \\ \Delta = 1600 - 924 \\ \Delta = 676$

$x = \frac{-b±\sqrt{\Delta} }{2a} \\ x = \frac{ - ( - 40)± \sqrt{676} }{2.11} \\ x = \frac{40±26}{22}$

$x' = \frac{40 + 26}{22} = \frac{66}{22} = 3 \\ x'' = \frac{40 - 26}{22} = \frac{14 \div 2}{22 \div 2} = \frac{7}{11}$