How are we supposed to do that?

Geometry Level 3

In a polygon, the greatest angle is 110 degrees and all the angles are distinct measures (in degrees). Find the maximum number of sides that the polygon can have.


The answer is 5.

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3 solutions

Rohit Ner
Mar 29, 2015

Regular polygon is the ideal polygon which has all angles equal and greatest. If we add a side to the regular polygon , the greatest angle changes. Hence regular polygon is the polygon with maximum no. of sides. possible. According to exterior angle theorem , For a polygon of n sides, n(180-angle)=360, 70n=360, n=5.14, But no of sides is supposed to be integral Hence n=5.

It is not mentioned that it is a regular polygon... and "There is nothing to be assumed."

Sakshi Jain - 6 years, 2 months ago

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I will tell you my solution. Let one angle be 110° and the sum of all angles will be 180n-360° so the sum of remaining angles will be 180n-470°. Now all angles are less than 110° so their average will also be less than 110° so \frac{180n-470°}{n-1}<110°. So 180n-470°<110n-110°, 70n<360, n<5.143. And n should be a positive integer so greates positive value of n which is less than 5.143 is 5. Hence n=5

Kushagra Sahni - 5 years, 11 months ago

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Got it! Thanks :)

Sakshi Jain - 5 years, 8 months ago

First we notice that in a hexagon, the average angle is 120 degrees, but in the question the greatest angle is 110, so a hexagon won't work. Next, we check the pentagon, with an average degree measure of 108. Quickly, we realize that 5 possible angles in a pentagon are 106, 107, 108, 109, and 110, which are all distinct. So we have the answer 5 \boxed{5}

Lucas Maia
Sep 11, 2015

I use this desaquality to solve the problem because the polygon have the biggest size we use that the sum of the angles descreacing one by one is bigger than the sum of angles of the polygon and we can find the biggest n that is exactly 5. (N-2)180=<110 +109 +...+111-n N^2 +139n-720=<0 N<=5 Now using the exemple 110,109,108,107,106 that is 540 we can prove that we can create this polygon

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