Find the area of a regular 12-gon inscribed in a unit circle.


Shown below is a regular octagon. Each of the four red regions has area 12. What is the area of the blue region?


Isosceles triangle OPQ has legs OP=OQ, base PQ=2, and angle POQ=45 degrees. Find the distance from O to PQ.


ABCDEF is a regular hexagon with area 1. The intersection of triangle ACE and triangle BDF is a smaller hexagon. What is the area of the smaller hexagon?


A, B, C, D, and E are points on a circle of radius 2 in counterclockwise order. We know AB=BC=DE=2 and CD=EA. Find [ABCDE].

Enter your answer in the form x+y sqrt z in simplest radical form.

 Jan 20, 2019

using your other diagram, take the midpoint of AB as F, and draw line segment OF. now you have 15-75-90 triangle, and it has hypotenuse 1.


because sin 15 is (sqrt6-sqrt2)/4 and cos 15 is (sqrt6+sqrt2)/4, you just multiply these values to get 1/4. now you divide by 2 because of triangle area, so it is 1/8. there are 12*2=24 of these triangles, so the answer is 24*1/8 or 3.


for the second one, this means the legs of the right triangles are 2sqrt6, and the hypotenuse is 4sqrt3. the hypotenuse is also the same as the side length of the square, so the answer is 48.


for the third one, i am not sure, but i think you need to use the fact that sin(2x)=2 sin x * cos x.


for the fourth one, triangle BFD trisects line AE with some intuition.


because line AE is sqrt3, the side length of the smaller hexagon is sqrt3/3. using the formula 3s^2sqrt3/2 for the area of a hexagon, you get sqrt3/2.


i will leave the last one for you to do because i do not know how to do it.



 Jan 20, 2019

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