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terricherri
Express answer in exact form. A regular hexagon with sides of 3" is inscribed in a circle. Find the area of a segment formed by a side of the hexagon and the circle. (Hint: remember Corollary 1--the area of an equilateral triangle is 1/4 s2 √3.) HELP CONFUSED
K, it would help if I could draw a picture, but that won't work. Think of a hexagon with all of the same length sides (regular). Now we draw a circle around it. We need to find the area of the space between one of the sides and the outer edge of the circle. The way I might do this is find the area of the hexagon, find the area of the circle, subtract the areas. This would give us the area between all 6 sides and the edge of the circle. Since we only need one side, we would divide that area by 6. The Corollary helps remind us that a hexagon can be broken into 6 equilateral triangles (think the 6 wedges from Trivial Pursuit). With this information, we can find out lots of things. Since they are equilateral triangles, we can find the area of the circle using radius 3. A=pi(r^2) So the area of the circle is simply 9pi. Using things we should know about special triangles, we can take one of the triangles we broke the hexagon into and split it in half vertically. This gives us two 30-60-90 triangles. We will use this to find the height of the triangle which is needed to find the area. We know the base of the 30-60-90 triangle is 1.5 and the hypotinuse is 3, we need to find the other leg using properties of 30-60-90 triangles. That leg would be 1.5(sqrt(3)). Now we have a base of 3 and a height of 1.5(sqrt(3)). using A=1/2(b)(h), we get 2.25(sqrt(3)). This is the area of 1 triangle, we have 6 in our hexagon. Multiply by 6 and get 13.5(sqrt(3)) Now for the EXACT answer, we subtract this answer from our area of the circle. 9pi - 13.5(sqrt(3)) and as i mentioned before we have to divide this by 6. (9pi - 13.5(sqrt(3)))/6 would be an acceptable answer. You could reduce to (1.5pi - 2.25(sqrt(3)). Hope this helps!
mine is 1/2pi- 6inches^2 thats pretty much what the answer has to be in