A regular hexagon and an equilateral triangle are both inscribed in the same circle so that the hexagon and the triangle share three vertices. The radius of the circle is 8 units. What is the area of the region between the two polygons?Make sure you explain your answer - answers that include just a number will not be considered correct! I know the answer already - what I want to know is how you figured it out.
Annie says: Solutions
Pretty good job this week, though most of you insisted on doing more work than necessary.
What I mean by doing too much work is this: one way, and the most popular way, to do this problem is to find the area of the hexagon and subtract the area of the triangle. That sure makes sense, but it's twice as much work as you need to do! If you draw a good picture and look carefully and think for a second before you start doing the calculations, you'll notice that the hexagon is two of the triangles, so all you have to do is find the area of the triangle, or the area of something that equals the triangle.
Less work means less chance to make a mistake, and believe me, we all make mistakes. So look for those sorts of things when you are doing problems like this.
There were a number of well-explained solutions this week, including the ones highlighted below. Tetsuya Matsuguchi from Centennial High School provides a nice picture with a clear explanation. Kim Dao from Highland Park Senior High School draws a nice picture, and then uses some trig to figure it out. I have to say, though, that Giscard Pongnon from Wingate High School submitted my favorite solution, simply because he did it just the way I do - draw a picture, figure out what you're looking for, draw some separate triangles, figure out their areas, and voilà! You're done.
Tetsuya Matsuguchi Grade: 11 School: Centennial High School, Boise, IdahoHexagon ABCDEF and triangle ACE are inscribed in the circle with center O and radius r, which is 8 units. Draw segments connecting the center O with each of the vertices of the hexagon. This will divide the center in six equal angles of 60 degrees. The triangle ABO is an equilateral triangle, because AO = BO and angle AOB is 60 degrees. So AB = 8. Because ABCDEF is a regular hexagon, AB = BC = 8, and therefore triangle ABC is an isosceles triangle with the vertex measuring 120 degrees. Thus angle BAC must be 30 degrees. Now let the intersection of segments BO and AC be M. Angle B measures 60 degrees (triangle ABC is equilateral), so triangle ABM is a 30-60-90 triangle. BM must be 1/2 of the hypotenuse AB, which is 4, and AM must be MB times the square root of 3, which is 4(sqrt(3)). The area of the triangle AMB, then, is 4 * 4(sqrt(3) * 1/2, or 8(sqrt(3)) Notice that triangle ABM is congruent to triangle BCM, and triangle ABC is congruent to triangle CDE and EFA. So, the area of the region between the two polygons must be 6 * (8(sqrt(3))), or 48(sqrt(3)).
Kim Dao Grade: 12 School: Highland Park Senior High School, St. Paul, Minnesota We know that an equilateral triangle has 60 degree angles, so if we cut it into three equal triangles, we will have a 30 degree angle for each triangle. We want to find the area of the region between the two polygons. We can choose any of these 30 degree angles of the triangles, because they are all the same, equal and congruent to each other. I chose triangle AGC. According to the information we have from this triangle, we know angle GAH = 30 degrees and AG = 8 units; however, we don't know AH or GH. We know that angle AHG is 90 degrees, so if we want to find AH and GH, we first think about to use cosine and sine, because we know its angle = 30 degrees and its side = 8. cos 30 degrees = leg adjacent/hypotenuse = AH/8 sin 30 degrees = leg opposite/hypotenuse = GH/8 AH = 4 times radical 3 and GH = 4. The area of triangle AGC = 16 times radical 3, and the area of triangle ABC is 3(16 times radical 3) = 48 times radical 3. The area of the region between the two polygon is also the area of triangle ABC = 48 times radical 3.
Giscard Pognon Grade: 12 School: George Wingate High School, Brooklyn, New York We remove the right triangle to find 1/2 the base of the figure labeled triangle I.Figure I is shown by itself. The base is 8\/3 and the height is half the radius 4. We use the formula to find the area of the triangle. Area I = 1/2 bxh = 1/2 (8\/3 x 4) = 16\/3 Area I = Area II = Area III Therefore the area between the two polygons is 48\/3 units^2.
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