Get reliable answers to your questions at Westonci.ca, where our knowledgeable community is always ready to help. Discover precise answers to your questions from a wide range of experts on our user-friendly Q&A platform. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
Sagot :
To determine the possible lengths for the third side [tex]\( x \)[/tex] of a triangle with given sides 200 units and 300 units, we need to use the triangle inequality theorem. The triangle inequality theorem states that the sum of the lengths of any two sides of a triangle must be greater than the length of the remaining side.
Given the sides [tex]\( a = 200 \)[/tex] units and [tex]\( b = 300 \)[/tex] units, we consider the following inequalities:
1. [tex]\( a + b > x \)[/tex]
2. [tex]\( a + x > b \)[/tex]
3. [tex]\( b + x > a \)[/tex]
Now, we can simplify these inequalities:
1. Since [tex]\( a = 200 \)[/tex] and [tex]\( b = 300 \)[/tex]:
[tex]\[ 200 + 300 > x \implies 500 > x \][/tex]
2. Since [tex]\( a = 200 \)[/tex] and [tex]\( b = 300 \)[/tex]:
[tex]\[ 200 + x > 300 \implies x > 300 - 200 \implies x > 100 \][/tex]
3. Since [tex]\( a = 200 \)[/tex] and [tex]\( b = 300 \)[/tex]:
[tex]\[ 300 + x > 200 \][/tex]
The third inequality [tex]\( 300 + x > 200 \)[/tex] is always true for any positive [tex]\( x \)[/tex] as long as [tex]\( x \)[/tex] satisfies the first two inequalities. So, we can put that aside.
Combining the simplified inequalities from steps 1 and 2 gives us the compound inequality:
[tex]\[ 100 < x < 500 \][/tex]
Therefore, the range of possible lengths for the third side, [tex]\( x \)[/tex], is:
[tex]\[ 100 < x < 500 \][/tex]
Given the sides [tex]\( a = 200 \)[/tex] units and [tex]\( b = 300 \)[/tex] units, we consider the following inequalities:
1. [tex]\( a + b > x \)[/tex]
2. [tex]\( a + x > b \)[/tex]
3. [tex]\( b + x > a \)[/tex]
Now, we can simplify these inequalities:
1. Since [tex]\( a = 200 \)[/tex] and [tex]\( b = 300 \)[/tex]:
[tex]\[ 200 + 300 > x \implies 500 > x \][/tex]
2. Since [tex]\( a = 200 \)[/tex] and [tex]\( b = 300 \)[/tex]:
[tex]\[ 200 + x > 300 \implies x > 300 - 200 \implies x > 100 \][/tex]
3. Since [tex]\( a = 200 \)[/tex] and [tex]\( b = 300 \)[/tex]:
[tex]\[ 300 + x > 200 \][/tex]
The third inequality [tex]\( 300 + x > 200 \)[/tex] is always true for any positive [tex]\( x \)[/tex] as long as [tex]\( x \)[/tex] satisfies the first two inequalities. So, we can put that aside.
Combining the simplified inequalities from steps 1 and 2 gives us the compound inequality:
[tex]\[ 100 < x < 500 \][/tex]
Therefore, the range of possible lengths for the third side, [tex]\( x \)[/tex], is:
[tex]\[ 100 < x < 500 \][/tex]
We appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.