SOLUTION: Two regular quadrilateral vinyl tiles each of 1 ft sides overlap each other such that the overlapping region is a regular octagon. What is the area of the overlapping region?
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Question 1116886: Two regular quadrilateral vinyl tiles each of 1 ft sides overlap each other such that the overlapping region is a regular octagon. What is the area of the overlapping region?
The answer is 119.29 squared inches. Need solution.
Found 2 solutions by greenestamps, solver91311:
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!
You can view the overlapping region as one of the two tiles with its corners cut off at 45 degree angles.
If s is the side length of the octagon, then each of those corner triangles is a 45-45-90 right triangle with hypotenuse and legs .
The length of one side of a tile is then .
But the length of an edge of each tile is 12 inches. So
The four triangles cut off can be rearranged into a square with side length s. So the area of the four triangles cut off the tile is
And then the area of the overlapping region (the whole tile minus the four corners) is
square inches.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
A quadrilateral is, as the name implies, a four-sided polygon. A polygon is regular if and only if all of its sides are of equal measure and all of its interior angles are of equal measure. Being a typically lazy mathematician, I shall refer henceforth to such figures as "squares".
In order to overlap two squares such that the overlap is a regular octagon, one must make the centers of the two squares coincident and then rotate one of the squares through an angle of 45 degrees relative to the other square. Note that the center of a square is defined as the intersection of its two diagonals.
See diagram
Since a diagonal of the magenta square passes through the center of the blue square (on the line 
), and the measure of the diagonal of a square with sides that measure 1 is 
, the distance from the center of the two squares to point A is 
. Hence, given that the lower left corner of the blue square was placed at the origin, the abscissa of point A must be 
and the ordinate is on the line 
, and the coordinates of point A are as illustrated.
Points B and C are similarly fixed; verification left as an exercise for the student.
Using the two-point form of an equation of a straight line, we can derive an equation for the line that contains segment 
, thus:
}{0.5\ -\ \frac{\sqrt{2}}{2}\ -\ 0.5}(x\ -\ (0.5\ -\ \frac{\sqrt{2}}{2})))
Or, more simply:
Since the left side of the blue square is coincident with the 
axis, the point of intersection, 
, of segment and the left side of the blue square is )
as illustrated.
I leave it as another exercise for the student to verify the coordinates of the intersection, 
, between segment 
and the left side of the blue square.
Since the abscissa's of the two points of intersection are equal, the distance between the two points, and therefore the measure of one side of the regular octagon is simply the difference between the two ordinates:
\ =\ \sqrt{2}\ -\ 1)
The formula for the area of a regular octagon given the measure of one side is:
a^2)
Where 
is the measure of one side of the octagon.
For this problem:
\(\sqrt{2}\ -\ 1\)^2)
I leave you to your own arithmetic. Note that the above calculation will provide the area in square feet. You will need to multiply the result by the number of square inches in a square foot (namely 144) in order to achieve the answer that you desire.
Further note: Since the only measurement given is the measure of a side of one of the squares, and that measurement is given to the nearest whole foot, it is utterly inappropriate to express your answer to a precision greater than a whole number. The correct answer to this problem is NOT 119.29 square inches, it is actually 1 square foot. It is never correct to provide a result of a calculation involving measurements to a greater precision than the least precise given measurement.
There is good reason for this rule. When someone says that a thing measures 1 foot, they are only promising that the true measurement is in the interval 
feet. If your instructor wanted greater precision than the nearest whole square foot, then s/he should have posed the problem as consisting of two regular quadrilaterals each measuring 12.00 inches on a side. Only in that case would it be appropriate to claim the area is 119.29 square inches.
John
My calculator said it, I believe it, that settles it
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