Question

A dog sits at a corner of a square with side length 44 meters. the dog runs 10 meters along a diagonal toward the opposite corner. it stops, makes a 90 degrees right turn and runs 5 more meters. a scientist measures the shortest distance between the dog and each side of the square. what is the average of these four distances in meters?

Answer 1

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.

1. The shortest distance of B from PQ = AC-AD
2. The shortest distance of B from SR = 44-(AC-AD)
3. The shortest distance of B from SP = AC+AD
4. The shortest distance of B from RQ = 44-(AC+AD)

So, the average of the shortest distances of B from each side is $\frac{(AC-AD)+44-(AC-AD)+(AC+AD)+44-(AC+AD)}{4}=\frac{44+44}{4}=22$

Hence, the average of the shortest distance of B from each side is 22 m

Learn more about average here-

brainly.com/question/24057012

#SPJ10