Respuesta :
Answer:
[tex] P = 6+2+ \sqrt{40}=14.32[/tex]
So then the best option on this case is:
A. 14 and 15
Step-by-step explanation:
When we have a right triangle we can use the Pythegorean identity given by:
[tex] Hip^2 = Opp^2 +Adj^2[/tex] (1)
Where Hip represent the hypothenuse. Opp represent the opposite side and Adj the adjacent side.
On this case we have given the hypothenuse assumed [tex] Hip = \sqrt{40}[/tex] because is the only possible reasonable value because [tex] 40\sqrt{40}[/tex] is a too much higher value
We have a ratio provided on this case, let's assume that the ratio is:
[tex] \frac{Opp}{Adj} = 3[/tex]
[tex] Opp = 3 Adj[/tex]
If we rpelace this condition into equation (1) we got:
[tex] (\sqrt{40})^2 = (3Adj)^2 + Adj^2[/tex]
And then we have this:
[tex] 40 = 9 Adj^2 + Adj^2 = 10 Adj^2[/tex]
[tex] Adj = \sqrt{\frac{40}{10}}=2[/tex]
And then the opposite side is:
[tex] Opp = 3 Adj =3* 2= 6[/tex]
The perimeter is defined as the sum of all the sides, we can find the perimeter like this:
[tex] P = Opp + Adj + Hip[/tex]
And replacing the values that we found we got:
[tex] P = 6+2+ \sqrt{40}=14.32[/tex]
So then the best option on this case is:
A. 14 and 15
