Question

applying the acute triangle inequality theorem
use the drop-down menus to complete the statements.
$4^2$ is $square$ $3^2 + 3^2$.
therefore, $\triangle jkl$ is $square$.
$5^2$ is $square$ $3^2 + 4^2$.
applying the same method, $\triangle abc$ is $square$.

Explanation:

Step1: Calculate $4^2$ and $3^2+3^2$

$4^2=16$, $3^2+3^2=9+9=18$

Step2: Compare $4^2$ and $3^2+3^2$

$16 < 18$, so $4^2 < 3^2+3^2$

Step3: Classify $\triangle JKL$

For a triangle with sides $a,b,c$ (longest $c$), if $c^2 < a^2+b^2$, it is acute.

Step4: Calculate $5^2$ and $3^2+4^2$

$5^2=25$, $3^2+4^2=9+16=25$

Step5: Compare $5^2$ and $3^2+4^2$

$25 = 25$, so $5^2 = 3^2+4^2$

Step6: Classify $\triangle ABC$

If $c^2 = a^2+b^2$, the triangle is right.

Answer:

$4^2$ is $\boldsymbol{less\ than}$ $3^2 + 3^2$.
Therefore, $\triangle JKL$ is $\boldsymbol{an\ acute\ triangle}$.
$5^2$ is $\boldsymbol{equal\ to}$ $3^2 + 4^2$.
Applying the same method, $\triangle ABC$ is $\boldsymbol{a\ right\ triangle}$.