Chapter 8.3, Problem 15WE

To determine

To specify: the value of x for which the triangle with given sides is acute

Answer to Problem 15WE

The value of x for the given condition is $0<x<12$

Explanation of Solution

Given Information: A triangle with sides x,x+4 and 20 with 20 being the longest side.

Formula used:

Any triangle is said to be acute, if all its three angles are less than 90⁰. This implies that, it does not satisfy the Pythagoras theorem. This can be expressed for an acute triangle as,

  $a^2+b^2<c^2$

Calculation:

For an acute triangle with given sides x,x+4 and 20, 20 being the longest side, the relation between sides can be expressed as,

  $\begin{array}{l}{x}^2+{\left(x+4\right)}^2<{20}^2\\ x^2+x^2+8x+16<400\\ 2x^2+8x+16<400\\ x^2+4x+8<200\end{array}$

This implies ,

  $x^2+4x-192<0$

Solving the equation and expressing the above inequality in terms of its factors of the quadratic polynomial,

  $\begin{array}{l}\left(x+16\right)\left(x-12\right)<0\\ x<-16,x<12\end{array}$

Since x is one of the sides of the triangle, a negative value cannot be considered. Thus, the value of x for the triangle to be acute can be summarized as,

  $0<x<12$