Question

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

A.4
B.5
C.6
D.11

Answer 1

11

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<a^2+b^2`.

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

`x^2<5^2+10^2`

`x^2<25+100`

`x^2<125`

This implies that
`-√(125)<x<√(125)` with the condition that x>10 since we assumed it largest so the actual restriction on x is:
`10<x<√(125)`

(
`√(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<5^2+x^2`

`100<25+x^2`

`75<x^2`

`x^2>75`

This means that
`x<-√(75) \text{ or } x>√(75)` with condition x is less than 10 since we are assuming x is not the largest.

(
`√(75) \approx 8.66`)

So this mean that x would have to be included between
`√(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<5^2+10^2`

`121<25+100`

`121<125` is true and 11 is a number between 5 and 15.