Geometry - GMAT Quantitative

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Question

The measures of the acute angles of a right triangle are $x^{\circ }$ and $y^{\circ }$ . Also,

.

Evaluate .

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Answer

The measures of the acute angles of a right triangle have sum $90^{\circ }$ , so

Along with , a system of linear equations can be formed and solved as follows:

$\underline{x+y = 90}$

$2x \div 2 = 105 \div 2$

$x = 52 \frac{1}{2}$

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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

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

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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

Give the -intercept(s) of the graph of the equation

$y = 3x^{2} -2x - 16$

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Answer

Set

$y = 3x^{2} -2x - 16$

$3x^{2} -2x - 16 = 0$

Using the -method, we look to split the middle term of the quadratic expression into two terms. We are looking for two integers whose sum is and whose product is $3 \left ( -16 \right ) = -48$ ; these numbers are .

$3x^{2} -8x + 6x - 16 = 0$

$\left (3x^{2} -8x \right )+ \left ( 6x - 16 \right ) = 0$

$x \left (3x -8 \right )+ 2 \left (3 x - 8 \right ) = 0$

$\left ( x + 2 \right )\left (3x -8 \right ) = 0$

Set each linear binomial to 0 and solve:

or

$3x\div 3 = 8 \div 3$

$x = \frac{8}{3}$

There are two -intercepts - $(-2,0), \left ( \frac{8}{3}, 0 \right )$

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Question

Which of the following lines are perpendicular to $y=\frac{5}{6}x+10$ ?

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Answer

In order for a line to be perpendicular to another line defined by the equation , the slope of line must be a negative reciprocal of the slope of line . Since line 's slope is in the slope-intercept equation above, line 's slope would therefore be $-\frac{1}{m}$ .

Since in this instance the slope $m=\frac{5}{6}$ , $-\frac{1}{m}=-\frac{1}{\frac{5}{6}}=-\frac{6}{5}$ . Two of the above answers have this as their slope, so therefore that is the answer to our question.

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Question

Given a right triangle $\bigtriangleup ABC$ with right angle $\angle B$ , what is the measure of $\angle A$ ?

Statement 1: $m \angle A - m \angle C = 10^{\circ }$

Statement 2: $4 \cdot m \angle A = 5 \cdot m \angle C$

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Answer

Let be the measure of $\angle A$ . The sum of the measures of the acute angles of a right triangle, $\angle A$ and $\angle C$ , is $90^{\circ }$ , so

$m \angle A + m \angle C = 90^{\circ }$

$X + m \angle C = 90^{\circ }$

$X + m \angle C -X = 90^{\circ } -X$

$m \angle C = 90^{\circ } -X$

Assume Statement 1 alone. This can be rewritten:

$m \angle A - m \angle C = 10^{\circ }$

$X - (90^{\circ } -X) = 10^{\circ }$

$2 X - 90^{\circ }= 10^{\circ }$

$2 X - 90^{\circ }+ 90^{\circ } = 10^{\circ } + 90^{\circ }$

$2 X = 10 0^{\circ }$

$2 X \div 2 = 10 0^{\circ } \div 2$

$X = 50^{\circ }$

Assume Statement 2 alone. This can be rewritten:

$4 \cdot m \angle A = 5 \cdot m \angle C$

$4 X = 5 (90^{\circ }-X)$

$4 X = 450 ^{\circ }-5X$

$4 X+ 5X = 450 ^{\circ }-5X + 5X$

$9X = 450 ^{\circ }$

$9X\div 9 = 450 ^{\circ } \div 9$

$X = 50^{\circ }$

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Question

Find the slope of a line that is perpendicular to the line running through the points and .

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Answer

To find the slope of the line running through and , we use the following equation:

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{8-4}{5-7}=\frac{4}{-2}=-2$

The slope of any line perpendicular to the given line would have a slope that is the negative reciprocal of , or $-\frac{1}{m}$ . Therefore, $-\frac{1}{-2}=\frac{1}{2}$

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Question

Which of the following lines is perpendicular to $y=\frac{3}{4}x+16$ ?

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Answer

Given a line defined by the equation with a slope of , any line perpendicular to would have a slope that is the negative reciprocal of , $-\frac{1}{m}$ . Given our equation $y=\frac{3}{4}x+16$ , we know that $m=\frac{3}{4}$ and that $-\frac{1}{m}=-\frac{1}{\frac{3}{4}}=-\frac{4}{3}$ .

The only answer choice with this slope is $y=-\frac{4}{3}x+12$ .

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Question

Which of the following lines is perpendicular to

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Answer

Given a line defined by the equation with a slope of , any line perpendicular to would have a slope that is the negative reciprocal of , $-\frac{1}{m}$ . Given our equation , we know that and that $-\frac{1}{m}=-\frac{1}{5}$ .

There are two answer choices with this slope, $y=-\frac{1}{5}x-4$ and $y=-\frac{1}{5}x+7$ .

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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

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. 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

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

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

Consider the Circle :

(Figure not drawn to scale.)

If Circle represents the bottom of a silo, what is the area of the base of the silo in square meters?

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Answer

The area of circle is found by the following equation:

$A=\pi r^2$

where is our radius, which is .

So,

$A=\pi*(15:m)^2=225 \pi:m^2$

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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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