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1. Two regular pentagons and a regular decagon, all with the same side length, can completely surround a point, as shown. A square, a regular pentagon, and a regular n-gon, all with the same side length, also completely surround a point. Find n. 

2. In the diagram below, each side of convex quadrilateral ABCD is trisected. (For example, AP=PQ=QB). The area of convex quadrilateral ABCD is 180. Find the area of the shaded hexagon. 

3. Let ABCDEF be a convex hexagon. Let A'B'C'D'E'F' be the centroids of triangles FAB, ABC, BCD, CDE, DEF, EFA, respectively.
(a) Show that every pair of opposite sides in hexagon A'B'C'D'E'F' (namely A'B' and D'E', and B'C' and E'F', and C'D' and F'A) are parallel and equal in length.
(b) Show that triangles A'C'E' and B'D'F' have equal areas.

 

Sorry if that's a lot! I got 24 for #1, which was wrong, and I don't know what to do now. If anyone could help and provide a full explanation with the solution to the problems, it would be really appreciated, thank you!

 Jul 3, 2020
edited by Otterstar  Jul 3, 2020
 #1
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1)  A square, a regular pentagon, and a regular n-gon surround a point.

 

     The number of degrees in each vertex angle of a regular n-gon is:  (n - 2) · 180o / n

 

     We know that each vertex angle of a square has 90o.

 

     Each vertex angle of a regular pentagon has  (5 - 2) · 180o / 5  =  108o.

 

     There are 360o around each point.

 

     Subtracting the vertex angle of a square and the vertex angle of a regular pentagon,

     we are left with:  360o - 90o - 108o  =  162o.

 

     Now we use the formula:   (n - 2) · 180o / n  =  162o  to find the value of n.

                  (n - 2) · 180o  =  162o · n

               180o · n - 360o  =  162o · n

          180o · n - 162o · n  =  360o 

                            18o · n  =  360o 

                                     n  =  20

 Jul 3, 2020
 #2
avatar+157 
+1

Thank you! Does anyone know how to do #2 and 3?

Otterstar  Jul 4, 2020
 #3
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2) 

Triangle(BPS) is similar to Triangle(BAC)     --->     Area(BPS) / Area(BAC)  =  BP2 / BA2  =  22 / 32  =  4/9

     Consequently, Area(APSC)  =  5/9 · Area(ABC)

 

Similarly, Area(DWT) / Area(DAC)  =  4/9     --->     Area(WACT)  =  5/9 · Area(DAC)

 

Therefore, Area(APSCTW)  =  5/9 · Area(ABCD)     --->     Area(APSCTW)  =  5/9 · 180  =  20

 Jul 5, 2020

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