Question

what will be the value, when substituted for x, that would represent an acute triangle with side lengths x, x + 4, and 20? assume that the longest side of the triangle is of length 20 units. 12 14 8 10

Explanation:

Step1: Set triangle inequality rule

For sides $x$, $x+4$, $20$ (20 is longest):
$x + (x+4) > 20$

Step2: Solve inequality for $x$

$2x + 4 > 20 \implies 2x > 16 \implies x > 8$

Step3: Use acute triangle condition

For acute triangles, $a^2 + b^2 > c^2$ (c is longest side):
$x^2 + (x+4)^2 > 20^2$

Step4: Expand and simplify the inequality

$x^2 + x^2 + 8x + 16 > 400$
$2x^2 + 8x - 384 > 0$
$x^2 + 4x - 192 > 0$

Step5: Solve quadratic equation

Factor: $(x+16)(x-12) > 0$
Since $x>0$, solution is $x > 12$

Step6: Verify upper bound (20 is longest)

$x+4 < 20 \implies x < 16$
So $12 < x < 16$, only 14 fits.

Answer:

14