How to find the area of a hexagon - ACT Math

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Question

A single hexagonal cell of a honeycomb is two centimeters in diameter.

What’s the area of the cell to the nearest tenth of a centimeter?

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 problem we are told that the honeycomb is two centimeters in diameter. In order to solve the problem we need to divide the diameter by two. This is because the radius of this diameter equals the interior side length of the equilateral triangles in the honeycomb. Lets find the side length of the regular hexagon/honeycomb.

$\textup{Diameter}=2\times \textup{ radius}$

Substitute and solve.

$2\textup{ cm}=2\times \textup{radius}$

$\textup{radius}=\frac{2\textup{ cm}}{2}$

$\textup{radius}=1\textup{ cm}$

We know the following information.

$\textup{radius}=\textup{side length}$

As a result, we can write the following:

$s=1\textup{ cm}$

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

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

$A=\frac{3\sqrt{3}}{2}\times (1\textup{ cm})^2$

Simplify.

$A=\frac{3\sqrt{3}}{2} \textup{ cm}^2$

Solve.

$A=2.59\textup{ cm}^2$

Round to the nearest tenth of a centimeter.

$A=2.6\textup{ cm}^2$

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Question

A hexagon is made up of 6 congruent equilateral triangles. Each equilateral triangle has a length of 8 units. What is the area in square units of the hexagon?

Answer

First, let's draw out the hexagon.

Because the hexagon is made up of 6 equilateral triangles, to find the area of the hexagon, we will first find the area of each equilateral triangle then multiply it by 6.

Using the Pythagorean Theorem, we find that the height of each equilateral triangle is $4\sqrt3$ .

The area of the triangle is then

$\frac{8 \times 4\sqrt3}{2}=16\sqrt3$

Multiply this value by 6 to find the area of the hexagon.

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Question

What is the area of a regular hexagon with a side length of ?

Answer

This question is asking about the area of a regular hexagon that looks like this:

Now, you could proceed by noticing that the hexagon can be divided into little equilateral triangles:

By use of the properties of isosceles and triangles, you could compute that the area of one of these little triangles is:

$\frac{\sqrt{3}}{4}s^2$ , where is the side length. Since there are of these triangles, you can multiply this by to get the area of the regular hexagon:

$6\frac{\sqrt{3}}{4}s^2=\frac{3\sqrt{3}}{2}s^2$

It is likely easiest merely to memorize the aforementioned equation for the area of an equilateral triangle. From this, you can derive the hexagon area equation mentioned above. Using this equation and our data, we know:

$Area=\frac{3\sqrt{3}}{2}6^2=\frac{3\sqrt{3}}{2}*36=54\sqrt{3}$

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Question

What is the area of a regular hexagon with a perimeter of ?

Answer

A hexagon has sides. A regular polygon is one that has sides that are of equal length. Therefore, if the side length of our polygon is taken to be , we know:

, or

This question is asking about the area of a regular hexagon that looks like this:

Now, you could proceed by noticing that the hexagon can be divided into little equilateral triangles:

By use of the properties of isosceles and triangles, you could compute that the area of one of these little triangles is:

$\frac{\sqrt{3}}{4}s^2$ , where is the side length. Since there are of these triangles, you can multiply this by to get the area of the regular hexagon:

$6\frac{\sqrt{3}}{4}s^2=\frac{3\sqrt{3}}{2}s^2$

It is likely easiest merely to memorize the aforementioned equation for the area of an equilateral triangle. From this, you can derive the hexagon area equation mentioned above. Using this equation and our data, we know:

$Area=\frac{3\sqrt{3}}{2}12^2=\frac{3\sqrt{3}}{2}*144=216\sqrt{3}:units^2$

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Question

The figure above is a regular hexagon. is the center of the figure. The line drawn is perpendicular to the side.

What is the area of the figure above?

Answer

You can redraw the figure given to notice the little equilateral triangle that is formed within the hexagon. Since a hexagon can have the degrees of its internal rotation divided up evenly, the central angle is degrees. The two angles formed with the sides also are degrees. Thus, you could draw:

Now, the is located on the side that is the same as $\sqrt{3}$ on your standard triangle. The base of the little triangle formed here is on the standard triangle. Let's call our unknown value .

We know, then, that:

$\frac{5}{\sqrt{3}}=\frac{x}{1}$

Another way to write $\frac{5}{\sqrt{3}}$ is:

$\frac{5}{\sqrt{3}}\frac{\sqrt{3}}{\sqrt{3}}=\frac{5\sqrt{3}}{3}$

Now, there are several ways you could proceed from here. Notice that there are of those little triangles in the hexagon. Since you know that the are of a triangle is:

$A = \frac{1}{2}BH$

and for your data...

$B=\frac{5\sqrt{3}}{3}$

The area of the whole figure is:

$12 (\frac{1}{2}5\frac{5\sqrt{3}}{3})=50\sqrt{3}:units^2$

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Question

What is the area of a regular hexagon with a side length of miles? Simplify all fractions and square roots in your answer.

Answer

For a hexagon with side length , the formula for the area is
$\frac{3\sqrt{3}}{2}a^2$ .
We have a side length of 4 miles, so we plug that into the equation and simplify the fraction.
$\ \frac{3\sqrt{3}}{2}(4miles)^2 = \frac{3\sqrt{3}}{2}16miles^2 \ \= \frac{48\sqrt{3}}{2}miles^2 \ \= 24\sqrt{3}miles^2$

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Question

What is the area of a hexagon with a side of length two? Simplify all fractions and square roots.

Answer

To find the area of a hexagon with a given side length, , use the formula:

$\frac{3\sqrt{3}}{2}a^2$

Plugging in 2 for and reducing we get:

$\\frac{3\sqrt{3}}{2}2^2\ \=\frac{3\sqrt{3}}{2}\cdot 4\ \=\frac{12\sqrt3}{2}$

$=6\sqrt{3}$ . (remember order of operations, square first!)

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