Question

The sides of a triangle have lengths of x, x + 4, and 20. If the longest side is 20, which of the following values of x would make the triangle acute?

a. 8
b. 10
c. 12
d. 14

Answer 1

The answer is 14, which is D.

Explanation:

Answer 2

If c is the longest side and a and b are the shorter sides, then the triangle is acute if
`c^2\ \textless \ a^2+b^2`.

The longest side is 20, the shorter sides are x and x+4.


`20^2 \ \textless \ x^2+(x+4)^2 \\ 400\ \textless \ x^2+x^2+8x+16 \\ 400\ \textless \ 2x^2+8x+16 \ \ \ \ \ \ \ \ \ \ \ \ \ |-400 \\ 0\ \textless \ 2x^2+8x-384 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |/ 2 \\ 0\ \textless \ x^2+4x-192 \\ 0\ \textless \ x^2+16x-12x-192 \\ 0\ \textless \ x(x+16)-12(x+16) \\ 0\ \textless \ (x-12)(x+16)`

The zeros are -16 and 12, the coefficient of x² is positive so the parabola opens upwards. The values greater than zero are in the interval
`x \in (-\infty, -16) \cup (12, \infty)`.

The sides of the triangle must be positive numbers, so
`x\ \textgreater \ 12`.
There's one number that satisfies this inequality: 14.

The answer is D.