Online Course Discussion Forum

Math IIB

 
 
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Re: Math IIB
by David Reynoso - Monday, September 30, 2019, 11:27 AM
 

Hi Andy. I'll reply to all of your posts in here. Since homework is still not due, for homework problems I'll just give you a hint. Also, the class you are enrolled in is II-A not II-B.

  • 2.14: In the diagram you should have a hexagon and a pentagon that share a side ($EF$). There are two possibilities: the pentagon is inside the hexagon or the pentagon is outside the hexagon. This will give you the two possible values of $\angle GAF$ ($\alpha$ and $\beta$). To find $\angle GAF$ recall what the measure of the interior angles of a regular pentagon and hexagon are; you also want to use the fact that $\triangle GAF$ is isosceles.
  • 2.17: You have a good idea there. Just be careful checking which factors work as a solution to the problem.
  • 2.22: Try to look at the area of $\triangle AED$ in terms of the triangles (or portions of triangles) that were used to construct hexagon $ABCDEF$. What do you notice about this triangles? How many triangles make for the area of $\triangle $AED$?
  • 2.26: Look at the sides of the quadrilaterals $PDCE$, $PFBD$, and $PEAF$. What can you say about them? (Note they should have at least a pair of equal sides and a pair of parallel sides.) Compare also with the solution of problem 2.6.
  • 2.3: Here's a diagram for the problem


Squares have angles of $90^\circ$ and equilateral trianlges have angles of $60^\circ$. Thus so far the angles at vertex $A$ add up to $90 + 90 + 60 = 240$ degrees. We know that the angles of all $4$ regular polygons add to $360^\circ$, so the fourth unknown polygon must have an interior angle measure of $360 - 240 - 120$ degrees. The only regular polygon with interior angle measure of $120$ degrees is a hexagon, so that must be the missing polygon.