Question

Find an integer, x, such that 5, 10, and x represent the lengths of the sides of an acute triangle.

Answer 1

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 $a,b,\text{ and } c$ where $c$ is the largest then $c^2$.

Now if the triangle is acute (with the assumption x is the largest) then:

$x^2$

$x^2$

$x^2$

This implies that $-\sqrt{125}$ with the condition that x>10 since we assumed it largest so the actual restriction on x is: $10$

($\sqrt{125} \approx 11.18$)

So this includes 11 and not 6.

Now if the triangle is acute (with the assumption x is not the largest) then:

$10^2$

$100$

$75$

$x^2>75$

This means that $x\sqrt{75}$ with condition x is less than 10 since we are assuming x is not the largest.

($\sqrt{75} \approx 8.66$)

So this mean that x would have to be included between $\sqrt{75}$ 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.

$11^2$

$121$

$121$ is true and 11 is a number between 5 and 15.