Question 1203593: George took out a piece of paper and placed it on his desk. He then drew a polygon. All the angles in the polygon were obtuse. What type of polygon could could he have drawn?
A. quadrilateral
B. rectangle
C. triangle
D. hexagon (the correct answer)
I need to know how to determine this answer (hexagon).
Found 3 solutions by math_helper, math_tutor2020, greenestamps:
Answer by math_helper(2461)   (Show Source):
You can put this solution on YOUR website!
Using formal Algebra, you can use the sum of interior angles formula:
S = (n - 2)*180
where S = sum of interior angles
n = number of sides
[ You can remember this formula by noting that n-2 is the number
of non-overlapping triangles you can draw by connecting verticies
of an n-sided polygon. Each triangle having 180 degrees. ]
We can compute the average for each of the proposed polygons:
A. quadrilateral: 4 sides --> Sum = (4-2)*180 = 360
Since we are told each angle is obtuse, looking at 360/4 = 90, there
must be some angle that is less than or equal to 90 degrees (allowing
for irregular quadrilaterals). Therefore, this can not be the correct
answer (not all angles can be obtuse).
B. Rectangle: each angle of a rectangle is 90 degrees, clearly not the correct answer. This is already covered by A above as well.
C. Triangle: 3 sides --> Sum = (3-2)*180 = 180 degrees, at most one angle can be obtuse in a triangle, so clearly not the answer.
D. Hexagon: 6 sides: Sum = (6-2)*180 = 720 degrees. Since 720/6 = 120, George must have drawn a hexagon (it doesn't have to be a regular hexagon, BUT it can't be any arbitrary hexagon either, as he must've drawn it such that every angle was obtuse).
In summary, the hexagon shape is the only one possible from the choices given.
Hope this helps you!
Answer by math_tutor2020(3817)  (Show Source):
You can put this solution on YOUR website!
An angle is obtuse when the angle is between 90 degrees and 180 degrees excluding each endpoint.
90 < angle < 180
Quadruple all sides of that inequality to get 360 < 4*angle < 720
If a quadrilateral could have all 4 obtuse angles, then the angles would sum to a value anywhere between 360 and 720, excluding those endpoints.
But recall that the sum of the interior angles of any quadrilateral is always 360 degrees.
It comes from this formula:
S = 180(n-2)
where,
S = sum of interior angles
n = number of sides
When n = 4 it leads to S = 360
We can see that 360 < 4*angle < 720 wouldn't be possible due to S = 360.
This contradiction would allow us to rule out choice A.
Quadrilaterals cannot have all angles that are obtuse.
The most obtuse angles a quadrilateral can have is 3.
Choice B is ruled out because all four angles of any rectangle are always 90 degrees.
Use the formula
S = 180(n-2)
to plug in n = 3 and you should get S = 180
The sum of all three interior angles of any triangle is always 180 degrees.
Start from 90 < angle < 180 and triple each side to get 270 < 3*angle < 540
Unfortunately 180 is not in this interval, so we cannot have a triangle with all angles obtuse.
The most obtuse angles a triangle can have is 1.
If one angle was obtuse, then the other two angles must be acute. The other two acute angles must be in the interval 0 < angle < 90.
So we can rule out choice C as well.
If n = 6, then S = 180*(n-2) = 180*(6-2) = 720 is the sum of all six angles of a hexagon.
Also if all 6 angles of a hexagon are obtuse, then 90 < angle < 180 leads to 540 < 6*angle < 1080
The 720 is indeed between 540 and 1080, which means it is possible to have a hexagon with all obtuse angles.
In fact, any regular hexagon will have each interior angle of 720/6 = 120 degrees
Bonus Questions:- What is the smallest positive integer value of n such that 90 < 180(n-2)/n < 180 is true?
- Can a quadrilateral have exactly 3 right angles, where the 4th angle is not 90 degrees?
Answer by greenestamps(13203)  (Show Source):
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