Calculating an angle in a polygon - GMAT Quantitative

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

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The sum of the measures of the exterior angles of any polygon, one at each vertex, is . 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?

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The sum of the degree measures of any polygon is . 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.

You are given Pentagon such that:

$m \angle A = $\frac{7}{5}$ m \angle B$

and

$\angle B \cong \angle C \cong \angle D \cong \angle E$

Calculate $m \angle A$

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Let be the common measure of $\angle B$ , $\angle C$ , $\angle D$ , and $\angle E$

Then

$m \angle A = $\frac{7}{5}$ x$

The sum of the measures of the angles of a pentagon is $180 (5 - 2) = $540^{\circ }$$ degrees; this translates to the equation

$$\frac{7}{5}$ x + x + x + x + x = 540$

or

$$\frac{27}{5}$ x = 540$

$x = 540 \cdot $\frac{5}{27}$ = 100$

$m \angle A = $\frac{7}{5}$ x = $\frac{7}{5}$ \cdot 100$

$m \angle A = 140$

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

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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 $12 ^{\circ }$ .

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

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

$\angle1$ 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 . 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 }$$

Note: Figure NOT drawn to scale

The figure above shows a square inside a regular pentagon. Give $m\angle 1$ .

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Each angle of a square measures ; each angle of a regular pentagon measures $180 \cdot3 \div 5 = $108^{\circ }$$ . To get $m\angle 1$ , subtract:

$108 - 90 = $18^{\circ }$$ .

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

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The sum of the measures of the exterior angles of any polygon, one per vertex, is . In a regular polygon of sides , then all of these exterior angles are congruent, each measuring .

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.

Note: Figure NOT drawn to scale.

Given:

$\angle A \cong \angle C \cong \angle E$

$\angle B \cong \angle D \cong \angle F$

$m \angle A + $6^{\circ }$ = m \angle B$

Evaluate $m \angle A$ .

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Call the measure of $\angle A$

$m \angle A = m \angle C = m \angle E = x$

$m \angle A + $6^{\circ }$ = m \angle B$ , and $\angle B \cong \angle D \cong \angle F$

so

$m \angle B = m \angle D = m \angle F = x + 6$

The sum of the measures of the angles of a hexagon is $180 (6 - 2) = 720 ^{ \circ }$ , so

$m \angle A + m \angle B + m \angle C + m \angle D + m \angle E + m \angle F = 720$

$\ 6x = 702$

$\ x = 702 \div 6 = 117$ , which is the measure of $\angle A$ .

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

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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:

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

You are given a quadrilateral and a pentagon. What is the mean of the measures of the interior angles of the two polygons?

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The mean of the measures of the four angles of the quadrilateral and the five angles of of the pentagon is their sum divided by 9.

The sum of the measures of the interior angles of any quadrilateral is $180 (4-2) = $360^{\circ }$$ . The sum of the measures of the interior angles of any pentagon is $180 (5-2) = $540^{\circ }$$ .

The sum of the measures of the interior angles of both polygons is therefore $360 + 540 = $900^{\circ }$$ . Divide by 9:

$900 \div 9 = $100^{\circ }$$

Note: Figure NOT drawn to scale.

Given Regular Pentagon . What is $m \angle APN$ ?

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Quadrilateral is a trapezoid, so $m \angle APN + m \angle A = 180 ^{\circ }$ .

$m \angle A = $\frac{180 (5-2 )}{5}$ ^{\circ } = 108 ^{\circ }$ , so

$m \angle APN $+108^{\circ }$ = 180 ^{\circ }$

$m \angle APN $+108^{\circ }$ - $108^{\circ }$ = 180 ^{\circ } - $108^{\circ }$$

$m \angle APN = $72^{\circ }$$

The angles of a pentagon measure .

Evaluate .

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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:

$x+ \left(x + 10 \right )+ \left (x + 10 \right )+ \left(x + 15 \right )+ \left(x + 25 \right ) = 540$

$5x \div 5 = 480 \div 5$

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?

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

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

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

What is the measure of an angle in a regular octagon?

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On octagon has sides. The word regular means that all of the angles are equal. Therefore, we can use the general equation for finding the angle measurement of a regular polygon:

$$\frac{(n-2)180,\textup{degrees}$}{n}$ , where is the number of sides of the polygon.

$$\frac{(8-2)180,\textup{degrees}$}{8}=135,\textup{degrees}$ .

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

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The sum of the measures of the exterior angles of any polygon, one at each vertex, is . 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?

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The sum of the degree measures of any polygon is . 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.

You are given Pentagon such that:

$m \angle A = $\frac{7}{5}$ m \angle B$

and

$\angle B \cong \angle C \cong \angle D \cong \angle E$

Calculate $m \angle A$

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Let be the common measure of $\angle B$ , $\angle C$ , $\angle D$ , and $\angle E$

Then

$m \angle A = $\frac{7}{5}$ x$

The sum of the measures of the angles of a pentagon is $180 (5 - 2) = $540^{\circ }$$ degrees; this translates to the equation

$$\frac{7}{5}$ x + x + x + x + x = 540$

or

$$\frac{27}{5}$ x = 540$

$x = 540 \cdot $\frac{5}{27}$ = 100$

$m \angle A = $\frac{7}{5}$ x = $\frac{7}{5}$ \cdot 100$

$m \angle A = 140$

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

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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 $12 ^{\circ }$ .

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

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

$\angle1$ 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 . 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 }$$

Note: Figure NOT drawn to scale

The figure above shows a square inside a regular pentagon. Give $m\angle 1$ .

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Each angle of a square measures ; each angle of a regular pentagon measures $180 \cdot3 \div 5 = $108^{\circ }$$ . To get $m\angle 1$ , subtract:

$108 - 90 = $18^{\circ }$$ .