Geometry - GMAT Quantitative

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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

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

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

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

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

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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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?

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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?

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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?

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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

The perimeter of a regular hexagon is 72 centimeters. To the nearest square centimeter, what is its area?

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Answer

This regular hexagon can be seen as being made up of six equilateral triangles, each formed by a side and two radii; each has sidelength $\small 72\div6=12$ centimeters. The area of one triangle is

$\small \small A = \frac{s^{2} \sqrt{3}}{4} = \frac{12^{2} \sqrt{3}}{4} \approx 62.4$

There are six such triangles, so multiply this by 6:

$\small 62.4 \cdot 6 \approx 374$

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Question

A man wants to design a room such that, looking from above, it appears as a trapezoid with a square attached (shown below). The area of the entire room is to be 100 square meters. The red line shown bisects the dotted line and has a length of 15. How many of the following answers are possible values for the length of one side of the square?

a) 5

b) 6

c) 7

d) 8

Figure is not to scale, but the trapezoidal figure will be similar in dimensions to the one shown.

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Answer

Let denote the length of one side of a square. This is also the top of the trapezoid. Let denote the bottom of the trapezoid. Finally, let be the height of the trapezoid. The area of the trapezoid is then $\frac{x+y}{2}\cdot h$ while the area of the square is $x^{2}$ .

We then have the total area as 100, so: $\frac{x+y}{2}\cdot h + x^{2} = 100$

Now we know that the red line has length 15. is the region of this line that is in the trapezoid. What we notice, however, is that the remainder is precisely the length of one side of a square. So or

Rewriting the previous equation: $\frac{x+y}{2}\cdot (15-x)+ x^{2} = 100$

This is now an equation of 2 variables and we can easily cross out answers by plugging in possible values. What we find is that for $x=5\ and\ 6$ , $y=10\ and\ 8.\overline{2}$ respectively. For we get , which is too small ( must be greater than ). For we get $y=2.\overline{285714}\ \ or\ \frac{16}{7}$ .

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Question

What is the area of a regular hexagon with sidelength 10?

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Answer

A regular hexagon can be seen as a composite of six equilateral triangles, each of whose sidelength is the sidelength of the hexagon:

Each of the triangles has area

$A =\frac{ s^{2}\sqrt{3}}{4}$

Substitute to get $A =\frac{ 10^{2}\sqrt{3}}{4}= \frac{ 100 \sqrt{3}}{4} = 25 \sqrt{3}$

Multiply this by 6: $25 \sqrt{3} \times 6 = 150 \sqrt{3}$ , the area of the hexagon.

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Question

What is the area of a regular octagon with sidelength 10?

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Answer

The area of a regular polygon is equal to one-half the product of its apothem - the perpendicular distance from the center to a side - and its perimeter.

The perimeter of the octagon is $10 \times 8 = 80$

From the diagram below, the apothem of the octagon is .

is one half of the sidelength, or 5. can be seen to be the length of a leg of a $45^{\circ}-45^{\circ}-90^{\circ}$ triangle with hypotenuse 10, or

$\frac{10 }{ \sqrt{ 2}}= \frac{10 \cdot \sqrt{ 2}}{ \sqrt{ 2} \cdot \sqrt{ 2}} = \frac{10 \cdot \sqrt{ 2}}{ 2} = 5\sqrt{ 2}$

This makes the apothem $5 + 5\sqrt{2}$ .

The area is therefore $\frac{1}{2} \cdot 80 \cdot \left ( 5 + 5\sqrt{2} \right ) = 200 + 200 \sqrt{2}$

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Question

What is the area of the figure with vertices ?

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Answer

This figure can be seen as a composite of two simple shapes: the rectangle with vertices , and the triangle with vertices .

The rectangle has length and height , so its area is the product of these dimensions, or $9 \times 10 = 90$ .

The triangle has as its base the length of the horizontal segment connecting and , which is ; its height is the vertical distance from the other vertex to this segment, which is . The area of this triangle is half the product of the base and the height, which is $\frac{1}{2} \times 10 \times 2 = 10$ .

Add the areas of the rectangle and the triangle to get the total area:

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