Respuesta :

Let's draw the triangle to better understand the problem:

Let,

x = the length of the third side of the triangle.

Let's recall a property of the sides of a triangle:

"The length of one side of a triangle must always be lesser than the sum of its other two sides"

Let's check each choice to verify which will satisfy the criteria:

a.) x = 5

[tex]\text{ Side 1: 5 }<\text{ 5 + 12 }\rightarrow\text{ 5 }<\text{ 17 ; satisfies the criteria}[/tex][tex]\text{ Side 2: 5 }<\text{ 5 + 12 }\rightarrow\text{ 5 }<\text{ 17 ; satisfies the criteria}[/tex][tex]\text{ Side 3: 13 }<\text{ 5 + 5 }\rightarrow\text{ 12 }<\text{ 10 ; does not satisfies the criteria}[/tex]

Therefore, x =5 is not applicable.

b.) x = 11

[tex]\text{ Side 1: 11 }<\text{ 5 + 12 }\rightarrow\text{ 11 }<\text{ 17 ; satisfies the criteria}[/tex][tex]\text{ Side 2: 5 }<\text{ 11 + 12 }\rightarrow\text{ 5 }<\text{ 23 ; satisfies the criteria}[/tex][tex]\text{ Side 3: 12 }<\text{ 11 + 5 }\rightarrow\text{ 12 }<\text{ 16 ; satisfies the criteria}[/tex]

Therefore, x = 11 is applicable.

c.) x = 19

[tex]\text{ Side 2: 19 }<\text{ 5 + 12 }\rightarrow\text{ 19 }<\text{ 17 ; does not satisfies the criteria}[/tex]

Therefore, x =19 is not applicable.

d.) x = 9

[tex]\text{ Side 1: 9 }<\text{ 5 + 12 }\rightarrow\text{ 9 }<\text{ 17 ; satisfies the criteria}[/tex][tex]\text{ Side 2: 5 }<\text{ 9 + 12 }\rightarrow\text{ 5 }<\text{ 21 ; satisfies the criteria}[/tex][tex]\text{ Side 3: 12 }<\text{ 9 + 5 }\rightarrow\text{ 12 }<\text{ 14; satisfies the criteria}[/tex]

Therefore, x = 9 is applicable.

e.) x = 7

[tex]\text{ Side 1: 7 }<\text{ 5 + 12 }\rightarrow\text{ 7 }<\text{ 17 ; satisfies the criteria}[/tex][tex]\text{ Side 2: 5 }<\text{ 7 + 12 }\rightarrow\text{ 5 }<\text{ 19 ; satisfies the criteria}[/tex][tex]\text{ Side 3: 12 }<\text{ 7 + 5 }\rightarrow\text{ 12 }<\text{ 12 ; does not satisfies the criteria}[/tex]

Therefore, x = 7 is not applicable.

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