Question

Estou com duvida, me ajudem por favor
Eh a 14

Answer 1

x+y+10=24
x+y=24-10
x+y=14
x=14-y

hipotenusa⇒10cm
cateto⇒y
cateto⇒14-y

$a^2=b^2+c^2 \\ 10^2=y^2+(14-y)^2 \\ 100=y^2+196-28y+y^2 \\ \\ 2y^2-28y+196-100=0 \\ 2y^2-28y+96=0$

a=2
b=-28
c=96
Δ=b²-4ac
Δ=(-28)²-4(2)(96)
Δ=784-768
Δ=16
√Δ=√16=±4

$y= \frac{-b\pm \sqrt{delta} }{2a} = \frac{-(-28)\pm \sqrt{16} }{2.2} = \frac{28\pm4}{4} \\ \\ y'= \frac{28+4}{4} = \frac{32}{4} =8 \\ \\ y"= \frac{28-4}{4} = \frac{24}{4} =6$

cateto⇒y=8cm
cateto⇒14-y=14-8=6cm