Verified Answer :

dalendrk

a, b, c - the lengths of the sides of the triangle
and a ≤ b ≤ c
then:
a + b > c and if the triangle is an acute triangle then a² + b² > c².

\(1^o\\5\leq10\leq x\\\\\left\{\begin{array}{ccc}5+10 \ \textgreater \ x\\5^2+10^2 \ \textgreater \ x^2\end{array}\right\to\left\{\begin{array}{ccc}x \ \textless \ 15\\ x^2 \ \textless \ 125\end{array}\right\to\left\{\begin{array}{ccc}x \ \textless \ 15\\ x \ \textless \ \sqrt{125}\approx11.1\end{array}\right\\\\\boxed{x=11}\\\\2^o\\5\leq x\leq10\\\\\left\{\begin{array}{ccc}5+x \ \textgreater \ 10\\ 5^2+x^2 \ \textgreater \ 10^2\end{array}\right\to\left\{\begin{array}{ccc}x \ \textgreater \ 5\\ x^2 \ \textgreater \ 75\end{array}\right\to\left\{\begin{array}{ccc}x \ \textgreater \ 5\\ x \ \textgreater \ \sqrt{75}\approx8.7\end{array}\right\\\boxed{x=9}\)

\(3^o\\x\leq5\leq10\\\\\left\{\begin{array}{ccc}x +5 \ \textgreater \ 10\\ x^2+5^2 \ \textgreater \ 10^2\end{array}\right\to\left\{\begin{array}{ccc}x \ \textgreater \ 5\\ x^2 \ \textgreater \ 75\end{array}\right\to\left\{\begin{array}{ccc}x \ \textgreater \ 5\\ x \ \textgreater \ \sqrt{75}\approx8.7\end{array}\right\\\boxed{x\in\O}\\\\Answer:\boxed{x=9\ or\ x=11}\to your\ answer:\boxed{\boxed{x=11}}\)

olemakpadu

If given the two sides of a triangle, the third side, x, must be greater than the difference between the given two sides.

The third side, x must also be less than the sum of the given two sides.

Given  5, and 10.

The third side, x > (10 - 5)            x > 5

The third side, x < (10 + 5)            x < 15

x > 5  and    x < 15

5 < x < 15

The third side x, is between 5 and 15.  It could be any 6, 7, 8, 9, 10, 11,...., 14

But since the question stated that the triangle is acute, x can be like 11.

You can use Cosine Rule to check the angles of triangle, 5, 10, 11. You would discover all the angles are acute.

Answer is option D.

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