Dear sir ,

I would appreciate if anyone can show me the solution to the below question.

thanks

Kingman

What is the value of the angles A+B+C+D+E+F+G in the beolw figure ?

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- Apr 4th 2011, 03:50 AMkingmansum of interior angles of a 7-point star.
Dear sir ,

I would appreciate if anyone can show me the solution to the below question.

thanks

Kingman

What is the value of the angles A+B+C+D+E+F+G in the beolw figure ? - Apr 4th 2011, 06:44 AMSoroban
Hello, kingman!

I have a very primitive solution.

Quote:

Place a pencil on , the eraser at , the point at

Rotate the pencil about so that the eraser is at .

Rotate the pencil about so that the point is at

Rotate the pencil about so that the eraser is at

Rotate the pencil about so that the point is at

Rotate the pencil about so that the eraser is at

Rotate the pencil about so that the point is at

Rotate the pencil about so that the eraser is at

We find that the pencil has undergone rotations.

Therefore: .

- Apr 4th 2011, 07:01 AMmathaddict
Assuming that the star is uniform, you can try this method.

Lets begin with the triangle with angle A as one of its angle, label the two remaining angles with**1**and**2**(your preference). Then, label the opposite angles in the neighbouring triangle as well. Introduce some new angles, and repeat the process until you have all the angles in the 7 exterior triangles labelled. Now, label the heptagon inside, ie 180 -**1**, 180 -**2**....

A +**1**+**2**= 180

B +**2**+**3**= 180

Do for all the rest and get 7 equations and sum them all up:

A+...+G + 2 (**1**+**2**+ ... +**7**) =1260

Then set up another equation by summing up all the angles in the inner heptagon.

Yeah, just follow these boring steps if it comes out in a test, else go for Soroban's more interesting way of solving.(Wink) - Apr 4th 2011, 07:21 AMkingman
Thanks very much Soroban for the graphical approach to the question but I wonder whether it can done in the similar fashion iwhen solving the above problem if we take a 5 -point star instead and taking advantage of fact that the exterior angle of a triangle equals to sum of the interior opposite angles.

Thanks

kingman - Apr 4th 2011, 09:02 AMbjhopper
another method

mark intersection of GE and DF, H.StraigtenAGEF to form a rectangle.Let AGD be equilateral and GHD isosceles 30-30-120. Revised figure contains 4 90's and 3 60's= 540 - Apr 4th 2011, 06:54 PMkingman
thanks very much for the solution but would appreciate very much if you can draw out how the final figure looks like and how a rectangle can be formed by straigtening AGEF .Finally can you explain how AGD becomes equilateral and GHD becomes isosceles .

thanks - Apr 4th 2011, 09:10 PMbjhoppersum of interior angles
Hi kingman,

if you look at your diagram with my added pointH you can see that BGHD is shaped like akite and ACEF can be straitened to a rectanglewhich I will call akite with tails at F and E.THe rearrangement does not change the sum of the interior angles but makes it easy to find the sum.Draw the kite as previously described. Extend GH and DH .Draw AC parallel to GD.Drop perpendiculars from A and C meeting DH extended at Fand GH extended at E. Count the angles AC must be equal to FE

bjh - Apr 4th 2011, 10:41 PMkingman
Thanks very much for the response and would appreciate if you can use Paintbrush and make a sketch

of the diagram you have just described. Really expeiencing difficulty in visually something without a diagram.

thanks