Geometry - GMAT Quantitative

Card 0 of 6760

Question

In the above figure, give the union of $\overrightarrow{GE}$ and $\overrightarrow{OE}$ .

Answer

$\overrightarrow{OE}$ can be seen to be completely contained in $\overrightarrow{GE}$ - that is, $\overrightarrow{OE}\subseteq \overrightarrow{GE}$ . The union of a set and its subset is the containing set, so the correct response is $\overrightarrow{GE}$ .

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Question

Find the equation of the line that is perpendicular to the line connecting the points $\dpi{100} \small (0,-4)\ and\ (-1,-7)$ .

Answer

Lines are perpendicular if their slopes are negative reciprocals of each other. First we need to find the slope of the line in the question stem.

$slope = \frac{rise}{run} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{-7 + 4}{-1 - 0} = \frac{-3}{-1} = 3$

The negative reciprocal of 3 is $\dpi{100} \small -\frac{1}{3}$ , so our answer will have a slope of $\dpi{100} \small -\frac{1}{3}$ . Let's go through the answer choices and see.

$\dpi{100} \small y=3x-1$ : This line is of the form $\dpi{100} \small y=mx+b$ , where $\dpi{100} \small m$ is the slope. The slope is 3, so this line is parallel, not perpendicular, to our line in question.

$\dpi{100} \small y=-4x+8$ : The slope here is $\dpi{100} \small -4$ , also wrong.

$\dpi{100} \small y=\frac{x}{3}+1$ : The slope of this line is $\dpi{100} \small \frac{1}{3}$ . This is the reciprocal, but not the negative reciprocal, so this is also incorrect.

The line between the points $\dpi{100} \small (3,0)\ and\ (-3,2)$ : $\dpi{100} \small slope = \frac{2}{(-3-3)}=\frac{2}{-6}=-\frac{1}{3}$ .

This is the correct answer! Let's check the last answer choice as well.

The line between points $\dpi{100} \small (0,0)\ and\ (2,2)$ : $\dpi{100} \small slope = \frac{2}{2}=1$ , which is incorrect.

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Question

In the above figure, give the intersection of $\overrightarrow{GE}$ and $\overrightarrow{OE}$ .

Answer

$\overrightarrow{OE}$ can be seen to be completely contained in $\overrightarrow{GE}$ - that is, $\overrightarrow{OE}\subseteq \overrightarrow{GE}$ . The intersection of a set and its subset is the subset, so the correct response is $\overrightarrow{OE}$ .

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Question

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

Answer

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:

$360 \div 24 = 15$

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Question

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

Answer

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.

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Question

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 $m \angle A$

Answer

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

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

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

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

$m \angle A = 140$

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Question

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

Answer

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 $12 ^{\circ }$ .

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Question

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

Answer

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 $360^{\circ }$ . Therefore,

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

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

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

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

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Question

Note: Figure NOT drawn to scale

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

Answer

Each angle of a square measures $90^{\circ }$ ; each angle of a regular pentagon measures $180 \cdot3 \div 5 = 108^{\circ }$ . To get $m\angle 1$ , subtract:

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

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Question

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

Answer

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.

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Question

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 $m \angle A$ .

Answer

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 $\angle A$ .

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Question

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

Answer

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:

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

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Question

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

Answer

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.

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Question

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

Answer

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:

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

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Question

Note: Figure NOT drawn to scale.

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

Answer

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

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

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

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

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Question

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 .

Answer

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:

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

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

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Question

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?

Answer

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

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

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

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Question

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

Answer

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.

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

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Question

What is the maximum possible area of a quadrilateral with a perimeter of 48?

Answer

A quadrilateral with the maximum area, given a specific perimeter, is a square. Since $Area=side^{2}$ and a square has four equal sides, the max area is

$(\frac{48}{4})^{2} =12^{2}=144$

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Question

$\textup{QWERT}$ is a pentagon with two sets of congruent sides and one side that is longer than all the others.

The smallest pair of congruent sides are 5 inches long each.

The other two congruent sides are 1.5 times bigger than the smallest sides.

The last side is twice the length of the smallest sides.

What is the perimeter of $\textup{QWERT}$ ?

Answer

A pentagon is a 5 sided shape. We are given that two sides are 5 inches each.

Side 1 = 5inches

Side 2 = 5 inches

The next two sides are each 1.5 times bigger than the smallest two sides.

$\small 5*1.5=7.5$

Side 3 =Side 4= 7.5 inches

The last side is twice the size of the smallest side,

Side 5 =10 inches

Add them all up for our perimeter:

5+5+7.5+7.5+10=35 inches long

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