Question

Question 6
When Braxton walks from art class to math class, he usually stops at his locker. The distance from his art classroom to his locker is 95 feet, and the distance from his locker to his math classroom is 112 feet. What is the range of possible distances from art class to math class if he takes the hallway and goes directly between the classrooms?



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Answer 1

To find the range of possible distances from Braxton's art class to his math class, we can apply the triangle inequality theorem. In this case, we have the following segments:

1. The distance from art class to the locker: \( a = 95 \) feet
2. The distance from the locker to math class: \( b = 112 \) feet
3. The direct distance from art class to math class: \( c \)

According to the triangle inequality theorem, the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. This gives us two inequalities to consider:

1. \( a + b > c \)
2. \( c + a > b \)
3. \( c + b > a \)

We will solve these inequalities to find the range for \( c \):

1. For \( a + b > c \):
\[
95 + 112 > c \implies 207 > c \implies c < 207
\]

2. For \( c + a > b \):
\[
c + 95 > 112 \implies c > 112 - 95 \implies c > 17
\]

3. For \( c + b > a \):
\[
c + 112 > 95 \implies c > 95 - 112 \implies c > -17
\]
(This inequality is always true since \( c \) must be a positive distance).

Now, combining our valid inequalities, we find:
\[
17 < c < 207
\]

Thus, the range of possible distances from art class to math class is:
\[
(17, 207)
\]