How to find the area of a hexagon - PSAT Math

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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.

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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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