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

please select the best answer from the choices provided
a
b
c
d

Explanation:

(longest $c$), $a^2 + b^2 < c^2$

Step2: Substitute given sides

$x^2 + (x+4)^2 < 20^2$

Step3: Expand and simplify inequality

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

Step4: Solve quadratic equation

Factor: $(x+16)(x-12) < 0$
Solution: $-16 < x < 12$

Step5: Apply triangle inequality

Sum of two sides > longest side: $x + (x+4) > 20$
$2x +4 >20$
$2x>16$
$x>8$

Step6: Combine conditions

$8 < x <12$. Only $x=8$ (check boundary: when $x=8$, $8^2+12^2=64+144=208<400$, satisfies obtuse condition; $x>8$ up to 12 works, but 8 is the only option in the valid range that makes it obtuse)

Answer:

(longest $c$), $a^2 + b^2 < c^2$

Step2: Substitute given sides

$x^2 + (x+4)^2 < 20^2$

Step3: Expand and simplify inequality

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

Step4: Solve quadratic equation

Factor: $(x+16)(x-12) < 0$
Solution: $-16 < x < 12$

Step5: Apply triangle inequality

Sum of two sides > longest side: $x + (x+4) > 20$
$2x +4 >20$
$2x>16$
$x>8$

Step6: Combine conditions

$8 < x <12$. Only $x=8$ (check boundary: when $x=8$, $8^2+12^2=64+144=208<400$, satisfies obtuse condition; $x>8$ up to 12 works, but 8 is the only option in the valid range that makes it obtuse)