Final Project Part IIA:

Quarter Circle Arcs

by

Jenny Johnson

1. First, I created a sketch of GSP of a square with a quarter circle arc from each corner producing a region bordered by four smaller arcs. A screen shot of the GSP figure is shown below. To see the file on GSP, click here.

2. Next, I determined the area of the shaded region of the figure.

It is helpful to see that the figure is symmetric. If we draw a vertical line segment exactly halfway between the vertical sides (through the midpoints of the two horizontal sides), the figure is split into two identical rectangles, the one a reflection of the other.

This vertical line goes through E and G. Similarly, a horizontal line through H and F will cut the square into identical rectangles. Now in order to find the area of the shaded section, we can find the sum of the area of the quadrilateral EFGH and the area of the remaining four shaded sectors. Let us start by finding the area of the quadrilateral EFGH. We can find the length of EG by drawing right triangle DEI.

The length of DI will be s/2 since it is exactly half of the square side s (which is also the radius of the circle centered at D). DE = s since it is the radius of the circle. Thus, we can figure out the angle θ (x in the picture) using trigonometry.

Now that we know the angles and two of the sides of the triangle DEI, we can find the side EI using trigonometry.

And, since we know JI is length s, we can find length JE by subtracting JI – EI. Since the length of JE is the same as the length of GI (by symmetry), we can then subtract the length of JE (GI) from the length of EI to find the length of EG. This algebra is shown below.

Now that we know the length of EG, let us look at the quadrilateral EFGH. By symmetry, we can see the EG = HF (if we did all the calculations for HF that we just did for EG, we would get the same length). Thus, the diagonals of the quadrilateral are congruent. This means the quadrilateral must be an isosceles trapezoid, a rectangle, or a square.

Also, HE = EF = FG = GH by symmetry as well. So, all sides of the quadrilateral EFGH are congruent. This means it must be a square because trapezoids don’t have all four sides congruent. Since this is a square, we know that angles EFG, FGH, GHE, and HEF are right angles and the diagonals bisect the angles at the vertices. Thus, triangle EFG is a triangle with angles measure 45-45-90. This means we can again use some simple geometry to find the side of the square g. Then the area of the square will just be g².

Now that we’ve found the area of the square EFGH, we can calculate the area of the four shaded sectors that are part of the shaded region, but outside the square. To do this, we will create triangle EDF. If we find the area of the entire sector of the circle (shaded in orange)

and subtract the area of triangle EDF (orange),

then we will have the area of the section we want.

So, first we want to find the angle measure of EDF. We know EDC is 60⁰ (shown above) and that ADF is 60⁰ (by the same argument we made for angle EDC), so by basic arithmetic, angle EDF = 30⁰. We can then find the area of the whole sector of the circle (remember r = s).

We can also calculate the area of the triangle fairly quickly using the following formula.

We now have the area of the small sector in the orange-shaded region in the picture below.

By symmetry, we see that the area of the four shaded regions we need (shaded in blue below) are congruent. Thus, we can multiply the area we got for the orange-shaded region above by 4 to get the area of all 4.

Finally, we can add the area of these four blue sectors to the area we found for quadrilateral EFGH, and we will have the area of the entire yellow shaded region.

Thus, the total area of the original yellow shaded region is .

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