We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website.
Please click on "Accept cookies" if you agree to the setting of cookies. Cookies that do not require consent remain unaffected by this, see
cookie policy and privacy policy.
DECLINE COOKIES

s is a writing problem; please explain your answers in complete sentences.

(a) A convex, 11-sided polygon can have at most how many obtuse interior angles?

(b) A convex, 11-sided polygon can have at most how many acute interior angles?

Note: Convex means that each interior angle measure is less than 180.

Hint(s):

Try drawing 11-gons of different shapes on paper. Remember that they have to be convex for this problem -- you can't have interior angles bigger than 180 degrees.

For part (b), think about the sum of the interior angles of an 11-gon. How many acute angles can you have before the remaining angles are forced to add up to more than they can possibly add up to?

Guest Aug 1, 2017

#1**+2 **

a)

Let's think about this problem. Before we do, though, we must understand a few things.

An obtuse angle is defined by an angle wherein its measure follows the rule \(90^{\circ} . In other words, an angle that is obtuse is larger than 90 degrees and is less than 180 degrees.

Another thing to consider is the combined number of degrees of the interior angles of an 11-gon (known as a hendecagon). To do this, there is a formula for it. It is the following:

\(S=180(n-2)\)

Let S = sum of all interior angles in polygon

Let n = number of sides of the polygon.

An 11-gon has 11 sides total, so let's plug that into the formula:

\(S=180(n-2)\) | Solve for S by plugging in the number of sides in the polygon, 11. |

\(S=180(11-2)\) | Simplify inside the parentheses first; 11-2=9. |

\(S=180*9=1620^{\circ}\) | |

I will assume that the angle I will use is 90 degrees. Divide the sum of the interior angles of degrees by 90 and see what you get. \(\frac{1620}{90}=18\). Of course, I can't have right angles as that does not fit the definition of an obtuse angle, so that means that there can only be 17 obtuse angles.

But wait! An 11-gon only has 11 sides and subsequently 11 angles, which means that every angle can be obtuse!

B)

I actually found b easier than A) but that is OK.

There can only be, at most, 3 acute angles in a convex polygon. Only 3. This is because the exterior angle of an acute angle is greater than 90 degrees, and a 4th angle would make the exterior angle measure passed the limit of 360 degrees. Does this make sense?

TheXSquaredFactor Aug 1, 2017