Answer:
11
Step-by-step explanation:
First of all, for what values of x can we have a triangle.
By triangle inequality, we have that 10-5<x<10+5.
Simplifying this gives us 5<x<15.
So the answer is either 6 or 11.
An acute triangle with sides [tex]a,b,\text{ and } c[/tex] where [tex]c[/tex] is the largest then [tex]c^2<a^2+b^2[/tex].
Now if the triangle is acute (with the assumption x is the largest) then:
[tex]x^2<5^2+10^2[/tex]
[tex]x^2<25+100[/tex]
[tex]x^2<125[/tex]
This implies that [tex]-\sqrt{125}<x<\sqrt{125}[/tex] with the condition that x>10 since we assumed it largest so the actual restriction on x is: [tex]10<x<\sqrt{125}[/tex]
([tex]\sqrt{125} \approx 11.18[/tex])
So this includes 11 and not 6.
Now if the triangle is acute (with the assumption x is not the largest) then:
[tex]10^2<5^2+x^2[/tex]
[tex]100<25+x^2[/tex]
[tex]75<x^2[/tex]
[tex]x^2>75[/tex]
This means that [tex]x<-\sqrt{75} \text{ or } x>\sqrt{75}[/tex] with condition x is less than 10 since we are assuming x is not the largest.
([tex]\sqrt{75} \approx 8.66[/tex])
So this mean that x would have to be included between [tex]\sqrt{75}[/tex] and 10.
Either way 6 is not included in either of the acute triangle cases.
11 is the only one that satisfies the condition in at least one of the cases.
[tex]11^2<5^2+10^2[/tex]
[tex]121<25+100[/tex]
[tex]121<125[/tex] is true and 11 is a number between 5 and 15.
