Question

which value from the list below, when substituted for x, 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.
8
10
12
14

Explanation:

Step1: Use triangle inequality first

For a triangle with sides $x$, $x+4$, 20 (20 is longest), the sum of the two shorter sides must exceed the longest side:
$x + (x+4) > 20$
Simplify: $2x + 4 > 20 \implies 2x > 16 \implies x > 8$
Also, since 20 is the longest side, $x+4 \leq 20 \implies x \leq 16$

Step2: Use acute triangle condition

For an acute triangle with longest side $c$, the Pythagorean theorem generalization gives $a^2 + b^2 > c^2$. Here $a=x$, $b=x+4$, $c=20$:
$x^2 + (x+4)^2 > 20^2$
Expand: $x^2 + x^2 + 8x + 16 > 400$
Simplify: $2x^2 + 8x - 384 > 0$
Divide by 2: $x^2 + 4x - 192 > 0$

Step3: Solve quadratic inequality

Factor the quadratic: $x^2 + 4x - 192 = (x+16)(x-12)$
The inequality $(x+16)(x-12) > 0$ holds when $x > 12$ (since $x>0$ for side length)

Step4: Combine constraints

From Step1: $8 < x \leq 16$; from Step3: $x > 12$. So $12 < x \leq 16$

Answer:

D. 14