Side lengths of an acute triangle are $\sqrt{n}$ $sqrtn$ , $\sqrt{n + 1}$ $sqrt(n+1)$ , and $\sqrt{n + 2}$ $sqrt(n+2)$ . How do you find $n$ $n$ ?

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Side lengths of an acute triangle are $\sqrt{n}$ $sqrtn$ , $\sqrt{n + 1}$ $sqrt(n+1)$ , and $\sqrt{n + 2}$ $sqrt(n+2)$ . How do you find $n$ $n$ ?

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Answer 1 · 2018-03-28T12:54:45

If the triangle is a right triangle then square of the largest side is equal to the sum of the squares of smaller sides. But the triangle is acute angled one. So square of the largest side is less than the sum of the squares of smaller sides . Hence

${(\sqrt{n + 2})}^{2} < {(\sqrt{n})}^{2} + {(\sqrt{n + 1})}^{2}$ $(sqrt(n+2))^2< (sqrtn)^2+(sqrt(n+1))^2$

$\Rightarrow n + 2 < n + n + 1$ $=>n+2< n+n+1$

$\Rightarrow n > 1$ $=>n > 1$

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Side lengths of an acute triangle are $\sqrt{n}$ $sqrtn$ , $\sqrt{n + 1}$ $sqrt(n+1)$ , and $\sqrt{n + 2}$ $sqrt(n+2)$ . How do you find $n$ $n$ ?

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Side lengths of an acute triangle are sqrtn√nsqrtn, sqrt(n+1)√n+1sqrt(n+1), and sqrt(n+2)√n+2sqrt(n+2). How do you find nnn?

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Side lengths of an acute triangle are $\sqrt{n}$ $sqrtn$ , $\sqrt{n + 1}$ $sqrt(n+1)$ , and $\sqrt{n + 2}$ $sqrt(n+2)$ . How do you find $n$ $n$ ?