Polygons

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GMAT Quantitative › Polygons

Questions 1 - 10

The perimeter of a regular octagon is two kilometers. Give its sidelength in meters.

$250 \textrm{ m}$

$333 \frac{1}{3} \textrm{ m}$

$200\textrm{ m}$

$2,000\textrm{ m}$

$2,500 \textrm{ m}$

Explanation

One kilometer is equal to 1,000 meters, so two kilometers comprise 2,000 meters. A regular octagon has eight sides of equal length, so divide by 8 to get the sidelength: $2,000 \div 8 = 250$ meters.

The perimeter of a regular pentagon is three kilometers. Give its sidelength in meters.

$600 \textrm{ m}$

$60 \textrm{ m}$

$500 \textrm{ m}$

$50 \textrm{ m}$

$375 \textrm{ m}$

Explanation

One kilometer is equal to 1,000 meters, so two kilometers comprise 3,000 meters. A regular pentagon has five sides of equal length, so divide by 5 to get the sidelength: $3,000 \div 5 = 600$ meters.

The perimeter of a regular hexagon is one-half of a mile. Give the sidelength in inches.

$5,280 \textrm{ in}$

$2,640 \textrm{ in}$

$1,760 \textrm{ in}$

$10,560 \textrm{ in}$

$7,920 \textrm{ in}$

Explanation

One mile is 5,280 feet. The perimeter of the hexagon is one-half of this, or 2,640 feet. Since each side of a regular hexagon is congruent, the length of one side is one-sixth of this, or $2,640 \div 6 = 440$ feet.

Multiply by 12 to convert to inches: $440 \times 12 = 5,280$ inches.

What is the area of the figure with vertices ?

Explanation

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:

$\textup{QWERT}$ 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 $\textup{QWERT}$ ?

$35\ \textup{meters}$

$40\ \textup{meters}$

$30\ \textup{meters}$

$50\ \textup{meters}$

$45\ \textup{meters}$

Explanation

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

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

$15^{\circ }$

$18^{\circ }$

$20^{\circ }$

$24^{\circ }$

$25^{\circ }$

Explanation

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:

$360 \div 24 = 15$

The perimeter of a regular pentagon is one-fifth of a mile. Give its sidelength in feet.

$211 \frac{1}{5} \textrm{ ft}$

$422 \frac{2}{5} \textrm{ ft}$

$1,056\textrm{ ft}$

$176\textrm{ ft}$

$352\textrm{ ft}$

Explanation

One mile is 5,280 feet. The perimeter of the pentagon is one-fifth of this, or $5,280 \div 5 = 1,056$ feet. Since each side of a regular pentagon is congruent, the length of one side is one fifth-of this, or $1,056 \div 5 = 211 \frac{1}{5}$ feet.

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

A regular polygon with eighty sides.

A regular polygon with forty-five sides.

A regular polygon with thirty sides.

A regular polygon with twenty-four sides.

A regular polygon with ninety sides.

Explanation

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.

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?

Explanation

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

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

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

Polygons_1

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

$132^{\circ }$

$168^{\circ }$

$120^{\circ }$

$144^{\circ }$

$150^{\circ }$

Explanation

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

Polygons_2

$\angle1$ 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,