Question

Determine a menor altura de um triangulo cujos lados medem 4 m, 5 m e 6 m.

Answer 1

h²+x²=5² ⇒ h²+x²=25 ⇒ h²=25-x²
h²+(6-x)²=4² ⇒ h²+36-12x+x²=16 ⇒ h²-12x+x²=20 ⇒ h²= -x²+12 - 20

h²=h²
-x²+12x-20 = 25-x²
12x=45
x=45/12
x=15/4

h²-x²=25
(15/4)²+h²=25
h²=25-225/16
h=5√7/4

 

Answer 2

$S = \frac{a+b+c}{2} = \frac{4+5+6}{2} = \frac{15}{2} = 7,5 \\ \\ A= \sqrt{s(s-a)(s-b)(s-c)} \\ \\ A= \sqrt{7,5(7,5-4)(7,5-5)(7,5-6)} \\ \\ A= \sqrt{7,5*3,5*2,5*1,5 } \\ \\ A= \sqrt[]{98,4375} \\ \\ A=9,9... \\ \\ A=10 \\ \\ A= \frac{b*h}{2} \\ \\ 10= \frac{6*h}{2} \\ \\ 6h=20 \\ h= \frac{20}{6} \\ \\ h=3,3m$