Answer:
[tex] (x+11)^2 = (x+3)^2 +16^2[/tex]
And if we solve this equation for x we got:
[tex] x^2 +22x +121 = x^2 +6x +9 +256[/tex]
We can cancel [tex]x^2[/tex] in both sides and we have this:
[tex] 22x -6x= 256+9-121 =144[/tex]
And then we got:
[tex] 16 x= 144[/tex]
[tex] x =\frac{144}{16}= 9[/tex]
And then the length of the sides are 9+11= 20 m for the hypothenuse, 16 for the adjacent side and 9+3 = 12m for the last side.
Lenght of the smaller unknown side: 12m
Lenght of the larger unknown side: 20m
Step-by-step explanation:
For this case we have a right triangle and we can use the Pythagoras Theorem and using the info given by the triangle we can set up the following equation:
[tex] (x+11)^2 = (x+3)^2 +16^2[/tex]
And if we solve this equation for x we got:
[tex] x^2 +22x +121 = x^2 +6x +9 +256[/tex]
We can cancel [tex]x^2[/tex] in both sides and we have this:
[tex] 22x -6x= 256+9-121 =144[/tex]
And then we got:
[tex] 16 x= 144[/tex]
[tex] x =\frac{144}{16}= 9[/tex]
And then the length of the sides are 9+11= 20 m for the hypothenuse, 16 for the adjacent side and 9+3 = 12m for the last side side.
Lenght of the smaller unknown side: 12m
Lenght of the larger unknown side: 20m
