mirzohidjon wrote:
Regular hexagon ABCDEF has a perimeter of 36. O is the center of the hexagon and of circle O. Circles A, B, C, D, E, and F have centers at A, B, C, D, E, and F, respectively. If each circle is tangent to the two circles adjacent to it and to circle O, what is the area of the shaded region (inside the hexagon but outside the circles)?
A. \(108-18\pi\)
B. \(54\sqrt{3}-9\pi\)
C. \(54\sqrt{3}-18\pi\)
D. \(108-27\pi\)
E. \(54sqrt{3}-27\pi\)
Guys,
I did not understand why we can not form two trapezoids of this hexagon, and solve through it. The official solution says, that we have to form 6 equiaterial triangles in the hexagon and form through it. Forming trapezoids, I got 108 as a square of the hexagon.
Please. elaborate
You can do this by calculating the area of the trapezoid as well.
First of all the angle = 120 degrees. Non shaded area will consist of the areas of 6*120 degrees sectors plus one whole circle = 3 whole circles = \(3\pi{r^2}=27\pi\), (as \(r=3\), half of the side of the hexagon).
Now if we want to proceed with the trapezoid areas. We would have two trapezoids. Bases would be \(a=6\) (the smallest base) and \(b=12\) (the largest base). \(Area=\frac{a+b}{2}*h\). The height of the trapezoid can be calculated as the side in 30-60-90 right triangle
. Height would be the opposite side to the 60 degrees angle (half of 120 degrees). In 30-60-90 right triangle the ratio of the sides is \(1/\sqrt{3}/2\), the height would correspond with \sqrt{3} and the hypotenuse (which on the other hand would be the side of the hexagon=6) would correspond with 2. so we'll have \(2/\sqrt{3}=\frac{6}{h}\) --> \(h=3\sqrt{3}\). Area of the trapezoid would be \(Area=\frac{a+b}{2}*h=\frac{6+12}{2}*3\sqrt{3}=27\sqrt{3}\). As we have two trapezoids the whole area would be \(2*27\sqrt{3}=54\sqrt{3}\).
Area of the shaded region \(54\sqrt{3}-27\pi\).
Answer: E.
1) Regarding your method above, it's still a little blurry as to how you calculated the area of the trapezoid, namely the height. Can you please elaborate a little?
2) The way I went about calculating the area of the trapezoid was 2 traingles plus a rectangle. To find the area of the triangle, I used pyt theorem -- Square Root of 72 was the hypotenuse of the triangle and the length of the rectangle. So I had (2*1/2*6*6) + (Root 72 * 6) = 36 + 36root 2. Why is that wrong? Doesn't that encapsulate the total hexagon?