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I'm currently studying for the SAT. I've found a question that I can't seem to figure out.

I'm sure there is some logical postulate or assumption that is supposed to be made here. Here is the exact problem:

enter image description here

I don't really care for an answer, I would rather know steps on how to solve this. I'm trying to be prepared for all types of questions on the SAT.

Thanks!

EDIT:

Thanks so much for the help guys! I've figured it out and wanted to explain it in detail for anybody who wanted to know:

SOLUTION SPOILER:
1. The figure displayed is a non-regular quadrilateral.
2. Because we know that, we know that the interior angles of the shape are 360 because of the formula (n-2)180.
3. I then created two statements. x+y=80 and x+y+z where z is the top two full angles (above both x and z).
4. I then simply solved for z (the total of the two angles above x and y).
5. I found z to equal 280. When split between the two angles it represented, I determined that each angle in the shape was equal to 140 degrees.
6. Because each angle is congruent (the shape is regular) we now know the measurement of every angle.
7. I then plugged this into the interior angle formula: (n-2)180=n140.
8. After solving for n, you learn that the number of sides is 9 :)

Hope this helps!

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I'm disappointed that the problem doesn't explicitly say that the uncovered portion of the polygon is a quadrilateral. In standardized tests, never go by the diagram alone; always refer to the wording of the problem. –  chharvey Jan 27 '12 at 4:24
    
@TestSubject528491 Do you know of another way to prove that there are 9 sides? –  Xan Jan 27 '12 at 10:51
    
Not that I'm aware of. IMO, this is a poorly worded problem. The only way to complete it successfully is to assume the diagram is correct and that the uncovered part has 4 sides. Though, if it had 3 sides, the top vertex would be 100 degrees, which you know can't be the interior angle of a regular polygon (extra credit: why not?;) and if it had 5 sides or more, you wouldn't have enough information to determine the answer. So I guess you could deductively reason that the uncovered part must have 4 sides. –  chharvey Jan 31 '12 at 0:38
    
Self Correction: the uncovered portion could have any even number of sides. Try it with 6. I end up with a whole polygon with 18 sides. –  chharvey Jan 31 '12 at 1:07

3 Answers 3

up vote 4 down vote accepted

Hints to walk through this problem:

  1. In the visible grey quadrilateral, what do the top two angles add up to?
  2. How big is each angle of the partly hidden regular polygon?
  3. How many sides does the regular polygon have?
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This is a cute problem.

The basic idea is to know how to find the sum of the interiors angles of a given convex polygon.

The way to do that is to join each vertex to a point inside, and use the fact that the angles at the point sum to 360, and the angles of a triangle sum to 180.

If you also know that the interior angles are all equal, then given the sum of interior angle and one of the interior angles, you can calculate the number of sides.

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Hints: The sum of the measures of the interior angles of an $n$-sided convex polygon is $(n-2)*180^\circ$. So, if the polygon is regular, each interior angle has measure $180^\circ-{360^\circ\over n}$. (You could also use the fact that the sum of the exterior angles of a convex polygon is $360^\circ$. An "exterior angle" is the angle formed by a line coinciding with a side and the "next" side.)

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