Question on Polygons - Interior Angles | Geometry

Today, let’s take a look at a polygons question which tests our grasp of statistics too! This will be seen commonly in GMAT – a question testing multiple topics. That is why we do not suggest you to skip any topic during your prep.

Question: The measures of the interior angles in a polygon are consecutive odd integers. The largest angle measures 153 degrees. How many sides does this polygon have?

A) 8

B) 9

C) 10

D) 11

E) 12

Solution:

The interior angles are: 153, 151, 149, 147 … and so on.

Now there are two ways to approach this question – one which is straight forward but uses a good bit of algebra so is time consuming, and another which makes you think but doesn’t take much time. You can guess which one we are going to focus on! But before we do that let’s take a quick look at the algebraic solution too.

Method 1: The Algebra

The sum of the interior angles of this irregular polygon $= 153 + 151 + 149 + … (153 - 2(n-1)) = (n - 2)\times 180$

If there are $n$ sides, there are $n$ interior angles. The second largest angle will be $153 - 2\times 1$ . The third largest will be $153 - 2\times2$ . The smallest will be $153 - 2\times(n-1)$ . This is an arithmetic progression.

Sum of all terms $= \frac{First\ term\ +\ Last\ term}{2} \times n = \frac{153 + 153 - 2(n-1))}{2} \times n$

Equating, we get $\frac{(153 + 153 - 2(n-1))}{2} \times n = (n - 2)\times180$

Solving this you get, $n = 10$

But let’s figure out a solution without going through this painful calculation, because as you know on the GMAT time matters, and there’s usually a shortcut for those who deeply understand these concepts!

Method 2: Capitalize on what you know

Angles of the polygon: $153, 151, 149, 147, 145, 143, 141, … , (153 - 2(n-1))$

The average of these angles must be equal to the measure of each interior angle of a regular polygon with $n$ sides since the sum of all angles is the same in both the cases.

Measure of each interior angle of $n$ sided regular polygon $= \frac{Sum\ of\ all\ angles}{n} = \frac{(n-2)\times180}{n}$

Using the options:

Measure of each interior angle of 8 sided regular polygon $= 180\times \frac{6}{8} = 135$ degrees

Measure of each interior angle of 9 sided regular polygon $= 180\times \frac{7}{9} = 140$ degrees

Measure of each interior angle of 10 sided regular polygon $= 180\times \frac{8}{10} = 144$ degrees

Measure of each interior angle of 11 sided regular polygon $= 180\times \frac{9}{11} = 147$ degrees approx.
and so on…

Notice that the average of the given angles can be 144 if there are 10 angles.

The average cannot be higher than 144 i.e. 147 since that will give us only 7 sides (153, 151, 149, 147, 145, 143, 141 – the average is 147 is this case). But the regular polygon with interior angle measure of 147 has 11 sides. Similarly, the average cannot be less than 144 i.e. 140 either because that will give us many more sides than the required 9.

Hence, the polygon must have 10 sides.

Answer (C).

Interesting, eh? Well, it will be when you understand method 2 well and can do it intuitively!

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Geometry – Question on Polygons – Interior Angles

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