GMAT Quantitative - Calculating an angle in a polygon

What is the measure of one exterior angle of a regular twenty-four sided polygon?

$15^{\circ }$

$18^{\circ }$

$20^{\circ }$

$24^{\circ }$

$25^{\circ }$

Explanation

The sum of the measures of the exterior angles of any polygon, one at each vertex, is $360^{\circ }$ . Since a regular polygon with twenty-four sides has twenty-four congruent angles, and therefore, congruent exterior angles, just divide:

$360 \div 24 = 15$

Which of the following figures would have exterior angles none of whose degree measures is an integer?

A regular polygon with eighty sides.

A regular polygon with forty-five sides.

A regular polygon with thirty sides.

A regular polygon with twenty-four sides.

A regular polygon with ninety sides.

Explanation

The sum of the degree measures of any polygon is $360^{\circ }$ . A regular polygon with sides has exterior angles of degree measure $\left (\frac{360}{N} \right )^{\circ }$ . For this to be an integer, 360 must be divisible by .

We can test each of our choices to see which one fails this test.

$360 \div 24 = 15$

$360 \div 30 = 12$

$360 \div 45 = 8$

$360 \div 80 = 4.5$

$360 \div 90 = 4$

Only the eighty-sided regular polygon fails this test, making this the correct choice.

The measures of the angles of a pentagon are: $x ^{\circ },\left ( x+23 \right ) {^\circ},\left ( x+23 \right ) {^\circ},\left ( x+33 \right ) {^\circ},\left ( x+41 \right ) {^\circ}$

What is equal to?

Explanation

The degree measures of the interior angles of a pentagon total $180 \left ( 5-2\right ) ^{\circ } = 540^{\circ }$ , so

$x + \left ( x+23 \right ) +\left ( x+23 \right ) +\left ( x+33 \right ) +\left ( x+41 \right ) = 540$

$5x \div 5= 420 \div 5$

Polygons_1

The above diagram shows a regular pentagon and a regular hexagon sharing a side. Give $m\angle AVB$ .

$132^{\circ }$

$168^{\circ }$

$120^{\circ }$

$144^{\circ }$

$150^{\circ }$

Explanation

This can more easily be explained if the shared side is extended in one direction, and the new angles labeled.

Polygons_2

$\angle1$ and $\angle 2$ are exterior angles of the regular polygons. Also, the measures of the exterior angles of any polygon, one at each vertex, total $360^{\circ }$ . Therefore,

$m \angle1 = \frac{360}{5} = 72^{\circ }$

$m \angle2 = \frac{360}{6} = 60^{\circ }$

Add the measures of the angles to get $m\angle AVB$ :

$m\angle AVB = m\angle 1 + m\angle 2 = 72 + 60 = 132^{\circ }$

What is the arithmetic mean of the measures of the angles of a nonagon (a nine-sided polygon)?

$140^{\circ }$

$120^{\circ }$

$108^{\circ }$

$135^{\circ }$

The question cannot be answered without knowing the measures of the individual angles.

Explanation

The sum of the measures of the nine angles of any nonagon is calculated as follows:

$180 \cdot \left ( 9-2\right ) =180 \cdot 7= 1,260^{\circ }$

Divide this number by nine to get the arithmetic mean of the measures:

$1,260\div 9 = 140 ^{\circ }$

What is the median of the measures of the angles of a nonagon (a nine-sided polygon)?

The question cannot be answered without knowing the measures of the individual angles.

$140^{\circ }$

$120^{\circ }$

$108^{\circ }$

$135^{\circ }$

Explanation

The sum of the measures of the nine angles of any nonagon is calculated as follows:

$180 \cdot \left ( 9-2\right ) =180 \cdot 7= 1,260^{\circ }$

The median of an odd quantity of numbers is the number that falls in the center position when they are arranged in ascending order; for nine numbers, it will be the fifth-highest number. We now need to show that we need to know the actual numbers in order to find the median.

Case 1: Each angle measures $140 ^{\circ }$ .

The set is $\left { 140,140,140,140,140,140,140,140,140\right }$ and the median is 140.

Case 2: Eight of the angles measure $139 ^{\circ }$ and one of them measures $148 ^{\circ }$ .

The set is $\left {}139, 139, 139, 139, 139, 139, 139, 139, 148 \right }$ and the median is 139.

In both cases, the sum of the angle measures is 1,260, but the medians differ between the two.

Thingy

The above diagram shows a regular pentagon and a regular hexagon sharing a side. What is the measure of $\angle ABC$ ?

$12 ^{\circ }$

$18 ^{\circ }$

$15 ^{\circ }$

$10 ^{\circ }$

$20 ^{\circ }$

Explanation

The measure of each interior angle of a regular pentagon is

$\frac{180 (5-2)}{5} = \frac{180 (3)}{5} = 108^{\circ }$

The measure of each interior angle of a regular hexagon is

$\frac{180 (6-2)}{6} = \frac{180 (4)}{6} = 120^{\circ }$

The measure of $\angle ABC$ is the difference of the two, or $12 ^{\circ }$ .

Which of the following cannot be the measure of an exterior angle of a regular polygon?

$16^{\circ }$

$15^{\circ }$

$9^{\circ }$

$6^{\circ }$

Each of the given choices can be the measure of an exterior angle of a regular polygon.

Explanation

The sum of the measures of the exterior angles of any polygon, one per vertex, is $360^{\circ }$ . In a regular polygon of sides , then all of these exterior angles are congruent, each measuring $\frac{360^{\circ }}{N}$ .

If is the measure of one of these angles, then $x = \frac{360^{\circ }}{N}$ , or, equivalently, $N = \frac{360^{\circ }}{x}$ . Therefore, for to be a possible measure of an exterior angle, it must divide evenly into 360. We divide each in turn:

$360 \div 6 = 60$

$360 \div 9 = 40$

$360 \div 15 = 24$

$360 \div 16 = 22.5$

Since 16 is the only one of the choices that does not divide evenly into 360, it cannot be the measure of an exterior angle of a regular polygon.

The angles of a pentagon measure $x^{\circ }, \left (x + 10 \right )^{\circ }, \left (x + 10 \right )^{\circ }, \left (x + 15 \right )^{\circ }, \left (x + 25 \right )^{\circ }$ .

Evaluate .

This pentagon cannot exist

Explanation

The sum of the degree measures of the angles of a (five-sided) pentagon is $180 (5 -2 ) ^{\circ } = 540^{\circ }$ , so we can set up and solve the equation: