Hexagons - SAT Math

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Question

If a triangle has 180 degrees, what is the sum of the interior angles of a regular octagon?

Answer

The sum of the interior angles of a polygon is given by where = number of sides of the polygon. An octagon has 8 sides, so the formula becomes

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Question

Find the sum of all the inner angles in a hexagon.

Answer

To solve, simply use the formula to find the total degrees inside a polygon, where n is the number of vertices.

In this particular case, a hexagon means a shape with six sides and thus six vertices.

Thus,

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Question

Calculate the approximate area a regular hexagon with the following side length:

Answer

How do you find the area of a hexagon?

There are several ways to find the area of a hexagon.

In a regular hexagon, split the figure into triangles.

Find the area of one triangle.

Multiply this value by six.

Alternatively, the area can be found by calculating one-half of the side length times the apothem.

Regular hexagons:

Regular hexagons are interesting polygons. Hexagons are six sided figures and possess the following shape:

In a regular hexagon, all sides equal the same length and all interior angles have the same measure; therefore, we can write the following expression.

$\overline{AB}=\overline{BC}=\overline{CD}=\overline{DE}=\overline{EF}=\overline{FA}$

One of the easiest methods that can be used to find the area of a polygon is to split the figure into triangles. Let's start by splitting the hexagon into six triangles.

In this figure, the center point, , is equidistant from all of the vertices. As a result, the six dotted lines within the hexagon are the same length. Likewise, all of the triangles within the hexagon are congruent by the side-side-side rule: each of the triangle's share two sides inside the hexagon as well as a base side that makes up the perimeter of the hexagon. In a similar fashion, each of the triangles have the same angles. There are $360^{\circ}$ in a circle and the hexagon in our image has separated it into six equal parts; therefore, we can write the following:

$\theta=\frac{360^{\circ}}{6}$

$\theta=60^{\circ$

We also know the following:

$\angle AXB=\angle BXC=\angle CXD=\angle DXE=\angle EXF=\angle FXA=60^{\circ}$

Now, let's look at each of the triangles in the hexagon. We know that each triangle has two two sides that are equal; therefore, each of the base angles of each triangle must be the same. We know that a triangle has $180^{\circ}$ and we can solve for the two base angles of each triangle using this information.

$2x+60^{\circ}=180^{\circ}$

$2x+=120^{\circ}$

$x=60^{\circ}$

Each angle in the triangle equals $60^{\circ}$ . We now know that all the triangles are congruent and equilateral: each triangle has three equal side lengths and three equal angles. Now, we can use this vital information to solve for the hexagon's area. If we find the area of one of the triangles, then we can multiply it by six in order to calculate the area of the entire figure. Let's start by analyzing $\triangle BXC$ . If we draw, an altitude through the triangle, then we find that we create two $30^{\circ}-60^{\circ}-90^{\circ}$ triangles.

Let's solve for the length of this triangle. Remember that in $30^{\circ}-60^{\circ}-90^{\circ}$ triangles, triangles possess side lengths in the following ratio:

$1:2:\sqrt{3}$

Now, we can analyze $\triangle BXC$ using the a substitute variable for side length, .

We know the measure of both the base and height of $\triangle BXC$ and we can solve for its area.

$\textup{Area of }\triangle BXC=\frac{1}{2}\times \textup{base}\times \textup{height}$

$\textup{Area of }\triangle BXC=\frac{1}{2}\times s\times \frac{\sqrt{3}s}{2}$

$\textup{Area of }\triangle BXC=\frac{\sqrt{3}s^2}{4}$

Now, we need to multiply this by six in order to find the area of the entire hexagon.

$\textup{Area of Hexagon}[ABCDEF]=6\times\frac{\sqrt{3}s^2}{4}$

$\textup{Area of Hexagon}[ABCDEF]=\frac{6 \sqrt{3}s^2}{4}$

$\textup{Area of Hexagon}[ABCDEF]=\frac{3\sqrt{3}}{2}\times s^2$

We have solved for the area of a regular hexagon with side length, . If we know the side length of a regular hexagon, then we can solve for the area.

If we are not given a regular hexagon, then we an solve for the area of the hexagon by using the side length(i.e. ) and apothem (i.e. ), which is the length of a line drawn from the center of the polygon to the right angle of any side. This is denoted by the variable in the following figure:

Alternative method:

If we are given the variables and , then we can solve for the area of the hexagon through the following formula:

$A=\frac{1}{2}(P)(a)$

In this equation, is the area, is the perimeter, and is the apothem. We must calculate the perimeter using the side length and the equation $P=\textup{number of sides}\times s$ , where is the side length.

Solution:

In the given problem we know that the side length of a regular hexagon is the following:

Let's substitute this value into the area formula for a regular hexagon and solve.

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

$A=\frac{3\sqrt{3}}{2}\times 6^2$

Simplify.

$A=\frac{108\sqrt{3}}{2}$

$A=54\sqrt{3}$

Round the answer to the nearest whole number.

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Question

An equilateral triangle with side length has one of its vertices at the center of a regular hexagon, and the side opposite that vertex is one of the sides of the hexagon. What is the hexagon's area?

Answer

Because it can be split into two triangles, the area of an equilateral triangle can be expressed as...

$\frac{1}{2}bh = \frac{1}{2}s(\frac{s\sqrt{3}}{2}) = \frac{s^2\sqrt{3}}{4}$

With that in mind, the equilateral triangle in question has area of $\frac{16\sqrt{3}}{4} = 4\sqrt{3}$ .

Now consider that a regular hexagon can be split into six congruent equilateral triangles with a vertex at the center and the side opposite the center as one of the hexagon's sides (a handy way of finding a hexagon's area if you can't use the regular polygon formula requiring an apothem.) Knowing that, our answer is $4\sqrt{3} * 6 = 24\sqrt{3}$ .

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Question

How many diagonals are there in a regular hexagon?

Answer

A diagonal connects two non-consecutive vertices of a polygon. A hexagon has six sides. There are 3 diagonals from a single vertex, and there are 6 vertices on a hexagon, which suggests there would be 18 diagonals in a hexagon. However, we must divide by two as half of the diagonals are common to the same vertices. Thus there are 9 unique diagonals in a hexagon. The formula for the number of diagonals of a polygon is:

$\frac{n\cdot (n-3)}{2}$

where n = the number of sides in the polygon.

Thus a pentagon thas 5 diagonals. An octagon has 20 diagonals.

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Question

How many diagonals are there in a regular hexagon?

Answer

A diagonal is a line segment joining two non-adjacent vertices of a polygon. A regular hexagon has six sides and six vertices. One vertex has three diagonals, so a hexagon would have three diagonals times six vertices, or 18 diagonals. Divide this number by 2 to account for duplicate diagonals between two vertices. The formula for the number of vertices in a polygon is:

$\frac{n\cdot (n-3)}{2}$

where $n=#\ of\ sides$ .

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Question

Hexagon is a regular hexagon with sides of length 10. is the midpoint of $\overline{OH}$ . To the nearest tenth, give the length of the segment $\overline{MX}$ .

Answer

Below is the referenced hexagon, with some additional segments constructed.

Note that the segments $\overline{MX}$ and $\overline{HX}$ have been constructed. Along with $\overline{MH}$ , they form right triangle $\bigtriangleup MHX$ with hypotenuse $\overline{MX}$ .

is the midpoint of $\overline{OH}$ , so

$MH= \frac{1}{2} \cdot OH = \frac{1}{2} \cdot 10 = 5$ .

$\overline{HX}$ has been divided by drawing the perpendicular from to the segment and calling the point of intersection . $\bigtriangleup HYE$ is a 30-60-90 triangle with hypotenuse $\overline{HE}$ , short leg $\overline{EY}$ , and long leg $\overline{HY}$ , so by the 30-60-90 Triangle Theorem,

$EY = \frac{1}{2} \cdot HE = \frac{1}{2} \cdot 10 = 5$

and

$HY = EY \cdot \sqrt{3} \approx 5 \cdot 1.732 \approx 8.660$

For the same reason, , so

$HX = HY + YX \approx 8.660 + 8.660 \approx 17.320$

By the Pythagorean Theorem,

$MX= \sqrt{(MH)^{2}+ (HX)^{2}}$

$\approx \sqrt{(5)^{2}+ (17.320 )^{2}}$

$\approx \sqrt{25 + 300 }$

$\approx \sqrt{325 }$

$\approx 18.0$ when rounded to the nearest tenth.

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Question

The provided image represents a track in the shape of a regular hexagon with perimeter one fourth of a mile.

Teresa starts at Point A and runs clockwise until she gets halfway between Point E and Point F. How far does she run, in feet?

Answer

One mile is equal to 5,280 feet; one fourth of a mile is equal to

$\frac{1}{4} \times 5, 280 = 1,320 \textup{ feet}$

Each of the six congruent sides measures one sixth of this, or

$\frac{1}{6} \times 1,320 = 220\textup{ feet}$

Teresa runs clockwise from Point A to halfway between Point E and Point F, so she runs along four and one half sides, for a total of

$4 \frac{1}{2} \times 220 = 990\textup{ feet}$

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Question

Archimedes High School has an unusual track in that it is shaped like a regular hexagon, as above. Each side of the hexagon measures 264 feet.

Alvin runs at a steady speed of seven miles an hour for twelve minutes, starting at point A and working his way clockwise. When he is finished, which of the following points is he closest to?

Answer

Alvin runs at a rate of seven miles an hour for twelve minutes, or $\frac{12}{60} = \frac{1}{5}$ hours. The distance he runs is equal to his rate multiplied by his time, so, setting $r = 7 , t = \frac{1}{5}$ in this formula:

$d = 7 \cdot \frac{1}{5} = \frac{7}{5}$ miles.

One mile comprises 5,280 feet, so this is equal to

$\frac{7}{5} \cdot 5,280 = 7,392$ feet

Since each side of the track measures 264 feet, this means that Alvin runs

$7,392 \div 264 = 28$ sidelengths.

$28 \div 6 = 4 \textrm{ R }4$ ,

which means that Alvin runs around the track four complete times, plus four more sides of the track. Alvin stops when he is at Point E.

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Question

A circle with circumference $60 \pi$ is inscribed in a regular hexagon. Give the perimeter of the hexagon.

Answer

Below is the figure referenced; note that the hexagon is divided by its diameters, and that an apothem—a perpendicular bisector from the center to one side—has been drawn.

The circle has circumference $C = 60 \pi$ ; its radius, which coincides with the apothem of the hexagon, is the circumference divided by $2\pi$ :

$r = \frac{C}{2\pi} = \frac{60 \pi }{2\pi} = 30$

The hexagon is divided into six equilateral triangles. One, $\bigtriangleup AOB$ , is divided by an apothem of the hexagon $\overline{OM}$ - a radius of the circle - into two 30-60-90 triangles, one of which is $\bigtriangleup AMO$ . Since $\overline{OM}$ has length 30, and it is a long leg of $\bigtriangleup AMO$ , then short leg $\overline{AM}$ has length

$AM = \frac{OM}{\sqrt{3}} = \frac{30}{\sqrt{3}} = \frac{30 \sqrt{3}}{\sqrt{3} \cdot \sqrt{3}} =\frac{30 \sqrt{3}}{3} = 10 \sqrt{3}$

is the midpoint of $\overline{AB}$ , one of the six congruent sides of the hexagon, so

$AB = 2 \cdot AM = 2 \cdot 10 \sqrt{3} = 20\sqrt{3}$ ;

this makes the perimeter of the hexagon six times this, or

$P = 6 \cdot 20 \sqrt{3} = 120 \sqrt{3}$ .

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