Geometry - GMAT Quantitative

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Question

What is the area of a rectangle given the length of and width of ?

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Answer

To find the area of a rectangle, you must use the following formula:

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Question

Find the perimeter of a rectangle whose width is and length is .

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Answer

To solve, simply use the formula for the perimeter of a rectangle:

$P=2(w+l)=2(7+6)=2\cdot13=26$

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Question

Determine whether and $y+8=-\frac{2}{3}x +2$ are parallel lines.

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Answer

Parallel lines have the same slope. Therefore, we need to find the slope once both equations are in slope intercept form :

$y=-\frac{2}{3}x+\frac{10}{3}$

$m=-\frac{2}{3}$

$y+8=-\frac{2}{3}x +2$

$y=-\frac{2}{3}x+2-8$

$y=-\frac{2}{3}x-6$

$m=-\frac{2}{3}$

The lines are parallel because the slopes are the same.

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Question

What polynomial represents the area of a rectangle with length and width ?

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Answer

The area of a rectangle is the product of the length and the width. The expression can be multplied by noting that this is the product of the sum and the difference of the same two terms; its product is the difference of the squares of the terms, or

$(A+4)(A-4) = A^{2}-4^{2}= A^{2}-16$

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Question

What is the length of the diagonal of a cube if its side length is ?

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Answer

The diagonal of a cube extends from one of its corners diagonally through the cube to the opposite corner, so it can be thought of as the hypotenuse of a right triangle formed by the height of the cube and the diagonal of its base. First we must find the diagonal of the base, which will be the same as the diagonal of any face of the cube, by applying the Pythagorean theorem:

$c^2=18\rightarrow c=\sqrt{18}=3\sqrt{2}$

Now that we know the length of the diagonal of any face on the cube, we can use the Pythagorean theorem again with this length and the height of the cube, whose hypotenuse is the length of the diagonal for the cube:

$(3\sqrt{2})^2+3^2=d^2$

$d^2=27\rightarrow d=\sqrt{27}=3\sqrt{3}$

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Question

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

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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 $\dpi{100} \small (0,-4)\ and\ (-1,-7)$ .

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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 $\overrightarrow{OE}$ .

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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?

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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?

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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 $m \angle A$

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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 $\angle ABC$ ?

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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 $m\angle AVB$ .

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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 $m\angle 1$ .

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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?

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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 $m \angle A$ .

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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)?

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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

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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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 $m \angle APN$ ?

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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 .

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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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