Refer to the figure given below while reading the solution.
Suppose the dog reaches position A when traveled 10 m diagonally towards the opposite side.
And then position B when traveled 5 m towards the right turning 90°.
We can observe that APC is a right triangle with legs of equal length AC. And the coordinates of the point A is (AC, AC).
Also we can observe that APB is a right triangle with legs of equal length AD. Then the coordinates of the point D is (AC, AC-AD).
Hence, the coordinates of B will be (AC+AD, AC-AD).
Now, we since we have the coordinates we can calculate the shortest distances of B from each of the sides.
- The shortest distance of B from PQ = AC-AD
- The shortest distance of B from SR = 44-(AC-AD)
- The shortest distance of B from SP = AC+AD
- The shortest distance of B from RQ = 44-(AC+AD)
So, the average of the shortest distances of B from each side is 
Hence, the average of the shortest distance of B from each side is 22 m
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Answer:
Trench warfare was a technique used to fight in World War 1. As they could hide/defend there. Get supplie there. They could even have medical areas there, but the problem was a stalemate as that was such a thing that no one attacks and they stay without moving on in the battle.
276/5 is the answer after multiplying the two
From the given problem, I gather that there are only two groups of forces. These are:
13 lb, 35 lb, resultant force 30 lb
20 lb, 15 lb, resultant force 25 lb
We use the pythagorean theorem to determine if the three forces are grouped correctly, such that the resultant force is the hypotenuse.
Resultant Force = √(a² + b²)
So, for the first group,
30 ? √(13² + 35²)
30 ≠ 37.33
Thus, the first group does not pull at right angles to each other.
For the second group,
25 ? √(20² + 15²)
25 = 25
Thus, the second group does pull at right angles to each other.
The answer in simplest form would be 1/6