Polygons - GMAT Quantitative

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Calculate the length of the diagonal for a regular pentagon with a side length of .

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A regular pentagon has five diagonals of equal length, each formed by a line going from one vertex of the pentagon to another. We can see that for one of these diagonals, an isosceles triangle is formed where the two equal side lengths between the vertices joined by the diagonal are the other two sides. If we draw a line bisecting the angle between those two sides perpendicular with the diagonal that forms the other side of the triangle, we will have two congruent right triangles whose hypotenuse is the side length, , and whose adjacent angle is half the measure of one interior angle of a pentagon. Using these two values, we can solve for the length of the opposite side, which is half of the diagonal, so we can them multiply the result by to calculate the full length of the diagonal. We start by determining the sum of the interior angles of a pentagon using the following formula, where is the number of sides of the polygon:

So to get the measure of each of the five angles in a pentagon, we divide the result by :

So each interior angle of a regular pentagon has a measure of . As explained earlier, we can find the length of half the diagonal by bisecting this angle to form two right triangles. If the hypotenuse is and the adjacent angle is half of an interior angle, or , then the length of the opposite side will be the hypotenuse times the sine of that angle. This only gives half of the diagonal, however, as there are two of these congruent right triangles, so we multiply the result by and we get the full length of the diagonal of a pentagon as follows:

$D=2(6\sin $54^{\circ}$)=9.7$

The hexagon in the above diagram is regular. If $\overline{AB}$ has length 12, which of the following expressions is equal to the length of $\overline{AD}$ ?

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$\overline{AD}$ is a diameter of the regular hexagon. Examine the diagram below, which shows the hexagon with all three diameters:

Each interior angle of a hexagon measures , so, by symmetry, each base angle of the triangle formed is ; also, each central angle measures one sixth of , or . Each triangle is equilateral, so if , it follows that , and .

The octagon in the above diagram is regular. If $\overline{AB}$ has length 8, which of the following expressions is equal to the length of $\overline{AD}$ ?

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Construct two other diagonals as shown.

Each of the interior angles of a regular octagon have measure , so it can be shown that $\bigtriangleup ABX$ is a 45-45-90 triangle. Its hypotenuse is $\overline{AB}$ , whose length is 8, so, by the 45-45-90 Triangle Theorem, the length of $\overline{AX}$ is 8 divided by $\sqrt{2}$ :

$AX= $\frac{8}{\sqrt{2}$}= $\frac{8\cdot \sqrt{2}$}{$\sqrt{2}$\cdot $\sqrt{2}$} = $\frac{8 \sqrt{2}$}{2} = 4 $\sqrt{2}$$

Likewise, $YD = 4 $\sqrt{2}$$ .

Since Quadrilateral is a rectangle, .

$AD = AX+XY+YD = 4$\sqrt{2}$+ 8 + 4 $\sqrt{2}$= 8+8$\sqrt{2}$$

Note: Figure NOT drawn to scale.

Which of the following statements is true of the length of $\overline{AC}$ ?

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By dividing the figure into rectangles and taking advantage of the fact that opposite sides of rectangles are congruent, we have the following sidelengths:

$\overline{AC}$ is the hypotenuse of a triangle with legs of lengths 8 and 16, so its length can be calculated using the Pythagorean Theorem:

$AC = $$\sqrt{(AB)^{2}$$$+(BC)^{2}$} = $$\sqrt{16^{2}$$$+8^{2}$} = $\sqrt{256+64}$ = $\sqrt{320}$$

The question can now be answered by noting that and .

,

so $\sqrt{320}$ falls between 17 and 18.

Note: Figure NOT drawn to scale.

Refer to the above diagram. What is the length of $\overline{AD}$ in terms of ?

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Extend sides $\overline{CD}$ and $\overline{ED}$ as shown to divide the polygon into three rectangles:

Taking advantage of the fact that opposite sides of a rectangle are congruent, we can find and :

$\bigtriangleup DXA$ is right, so by the Pythagorean Theorem,

$AD = $$\sqrt{(AX)^{2}$$$+(DX)^{2}$}$

$=$\sqrt{\left ( n+4\right $)^{2}$$$+n^{2}$}$

$=$\sqrt{\left ( $n^{2}$$+8n+16\right $)+n^{2}$}$

$=$\sqrt{2 $n^{2}$$+8n+16 }$

Each side of convex Pentagon has length 12. Also, $m \angle A = m \angle B = m \angle D = m\angle E = 105 ^{\circ }$ .

Construct diagonal $\overline{BD}$ . What is its length?

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The measures of the interior angles of a convex pentagon total

$180 ^{\circ } \times (5-2) = 540 ^{\circ }$ ,

so

$m \angle C = 540 ^{\circ } - (m \angle A+m \angle B + m\angle D + m \angle E)$

$= 540 ^{\circ } - (105 ^{\circ }+105 ^{\circ }+105 ^{\circ }+105 ^{\circ })$

$= 540 ^{\circ } - 420 ^{\circ }$

$= $120^{\circ }$$

The pentagon referenced is the one below. Note that the diagonal $\overline{BD}$ , along with congruent sides $\overline{BC}$ and $\overline{CD}$ , form an isosceles triangle $\bigtriangleup BCD$ .

Now construct the altitude from to $\overline{BD}$ :

$\overline{CM}$ bisects $\angle BCD$ and $\overline{BD}$ to form two 30-60-90 triangles. Therefore, $CM = $\frac{1}{2}$ BC = $\frac{1}{2}$ \cdot 12 = 6$ ,

and $BM = CM \cdot $\sqrt{3}$= 6$\sqrt{3}$$ .

$BD = 2 \cdot BM = 2 \cdot 6 $\sqrt{3}$= 12 $\sqrt{3}$$

A pentagon with a perimeter of one mile has three congruent sides; one of the other sides is 100 feet longer than any of those three congruent sides, and the remaining side is 100 feet longer than that fourth side. What is the length of that longest side?

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If each of the five congruent sides has measure , then the other two sides have measures and $\left (x + 100 \right ) +100 = x + 200$ . Add the sides to get the perimeter, which is equal to 5,280 feet, the solve for :

$5x \div 5 = 4,980 \div 5$

Each of the shortest sides is 996 feet long; the longest side is feet long.

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

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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.

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

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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.

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

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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.

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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.

Six regular polygons each have perimeter one mile. The polygons are an equilateral triangle, a square, a pentagon, a hexagon, an octagon, and a decagon (ten-sided polygon). What is the mean of their sidelengths?

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The sidelength of a regular polygon is the perimeter of one mile, or 5,280 feet, divided by the number of sides. Therefore, the sidelengths of the six shapes are:

Triangle: $5,280 \div 3 = 1,760 \textrm{ ft}$

Square: $5,280 \div 4 = 1,320 \textrm{ ft}$

Pentagon: $5,280 \div 5 = 1,056 \textrm{ ft}$

Hexagon: $5,280 \div 6 = 880 \textrm{ ft}$

Octagon: $5,280 \div 8 = 660 \textrm{ ft}$

Decagon: $5,280 \div 10 = 528 \textrm{ ft}$

The mean of the six sidelengths:

$(1,760+1,320+1,056+880+660+528 ) \div 6 = 1,034$ feet

If the shape $\textup{ABCDEF}$ is a regular hexagon with side length of and a perimeter of meters, what is the length of one side?

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Perimeter is the total distance around a shape. Here we have a regular (all sides are equal) hexagon with side distance 2x+6. Multiplying this expression by 6 gives us our perimeter.

$\small 72=6*(2x+6)=12x+36$

$\small 72-36=12x$

$\small 12x=36$

$\small x=3$

But we are not done yet. We need to plug this back in to get the side length

$\small 2(3)+6=9+6=15$

so 15 meters

Alternatively, we could divide 72 by 6 to get 12 meters.

A regular hexagon has perimeter . Which of the following is equal to the length of one side?

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A hexagon has six sides of equal length, so divide the perimeter by 6, which is the product of 2 and 3:

A regular octagon has perimeter . Give the length of one of its sides.

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A regular octagon has eight sides of equal length, so divide the perimeter by 8, or equivalently, :

.

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

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The sum of the measures of the exterior angles of any polygon, one at each vertex, is . Since a regular polygon with twenty-four sides has twenty-four congruent angles, and therefore, congruent exterior angles, just divide:

$360 \div 24 = 15$

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

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The sum of the degree measures of any polygon is . 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.

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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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$

$m \angle A = 140$

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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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 $12 ^{\circ }$ .

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

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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 . 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 }$$