Question

points a, b, and c, form a triangle. the distance between point a and point b is 15 yards. the distance between point b and point c is 25 yards. pete walks directly from point a to point c, without passing through point b. what is the direct distance from a to c?
how far would pete walk if he went from a to b to c? \\( \square \\) yards
the direct distance from a to c is more than \\( \square \\) yards.
the inequality \\( w < \square \\) represents the distance, \\( w \\), that pete might save by taking the direct path.

Explanation:

Step1: Calculate the distance from A to B to C

The distance from A to B is 15 yards and from B to C is 25 yards.
$15 + 25$

Step2: Apply the triangle - inequality theorem for the second blank

In a triangle, the sum of the lengths of any two sides must be greater than the length of the third side. Let \(AC=x\), \(AB = 15\), \(BC=25\). Then \(x+15>25\) (also \(x + 25>15\) which is always true for positive \(x\) and \(15 + 25>x\)). From \(x+15>25\), we get \(x>25 - 15\).
\(25-15\)

Step3: Find the maximum value of the saving

The saving \(w=(15 + 25)-x\). Since \(x>10\) and \(x<40\) (from \(15 + 25>x\)), the maximum value of \(w\) occurs when \(x\) is minimized. But if we consider the non - degenerate triangle case, when \(x\) approaches 10 (but \(x>10\)), the maximum value of \(w\) is when \(x = 10+\epsilon\) (\(\epsilon\to0\)). The maximum value of \(w\) is based on the upper - bound of \(x\). The maximum value of \(w=(15 + 25)-10\) (using the fact that \(x>10\) and we want to find the upper - bound for \(w\) considering the non - degenerate triangle. The sum \(AB + BC=15 + 25\) and the lower - bound of \(AC\) is \(10\) (not inclusive). The maximum value of \(w\) is when \(AC\) is minimized (approaching 10). But if we consider the strict triangle - inequality for non - degenerate triangles, the maximum value of \(w\) is \(15+25 - 10\)

Answer:

• For the first blank: \(15 + 25=\boldsymbol{40}\) yards.
• For the second blank: \(25-15=\boldsymbol{10}\) yards.
• For the third blank: \(15 + 25-10=\boldsymbol{30}\) (since \(w=(15 + 25)-x\) and \(x>10\), the maximum value of \(w\) approaches \(30\) as \(x\) approaches \(10\)).