How to find the length of the diagonal of a hexagon - PSAT Math

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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

Regular Hexagon has a diagonal $\overline{AE}$ with length 1.

Give the length of diagonal $\overline{A D}$ .

Answer

The key is to examine $\Delta AED$ in thie figure below:

Each interior angle of a regular hexagon, including $\angle F$ , measures $120^{\circ }$ , so it can be easily deduced by way of the Isosceles Triangle Theorem that $m \angle FEA = 30^{\circ }$ . $m \angle FED = 120^{\circ }$ , so by angle addition,

$m \angle AED = 90^{\circ }$ .

Also, by symmetry,

$m \angle EDA = \frac{1}{2} \cdot 120^{\circ } = 60^{\circ }$ ,

so $m \angle EAD= 30^{\circ }$ ,

and $\Delta AED$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle whose long leg $\overline{AE}$ has length .

By the $30^{\circ}-60^{\circ}-90^{\circ}$ Theorem, $\overline{AD}$ , which is the hypotenuse of $\Delta AED$ , has length $\frac{2\sqrt{3}}{3}$ times that of the long leg, so $AD =\frac{2\sqrt{3}}{3}$ .

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Question

Regular Hexagon has a diagonal $\overline{AD}$ with length 1.

Give the length of diagonal $\overline{A E}$ .

Answer

The key is to examine $\Delta AED$ in thie figure below:

Each interior angle of a regular hexagon, including $\angle F$ , measures $120^{\circ }$ , so it can be easily deduced by way of the Isosceles Triangle Theorem that $m \angle FEA = 30^{\circ }$ . $m \angle FED = 120^{\circ }$ , so by angle addition,

$m \angle AED = 90^{\circ }$ .

Also, by symmetry,

$m \angle EDA = \frac{1}{2} \cdot 120^{\circ } = 60^{\circ }$ ,

so $m \angle EAD= 30^{\circ }$ ,

and $\Delta AED$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle whose hypotenuse $\overline{AD}$ has length .

By the $30^{\circ}-60^{\circ}-90^{\circ}$ Theorem, the long leg $\overline{AE}$ of $\Delta AED$ has length $\frac{\sqrt{3}}{2}$ times that of hypotenuse $\overline{AD}$ , so $AE = \frac{\sqrt{3}}{2}$ .

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Question

Regular hexagon has side length 1.

Give the length of diagonal $\overline{A E}$ .

Answer

The key is to examine $\Delta AED$ in thie figure below:

Each interior angle of a regular hexagon, including $\angle F$ , measures $120^{\circ }$ , so it can be easily deduced by way of the Isosceles Triangle Theorem that $m \angle FEA = 30^{\circ }$ . To find $m \angle AED$ we subtract $m \angle FEA = 30^{\circ }$ from $m \angle FED = 120^{\circ }$ . Thus resullting in $m \angle AED=90^{\circ}$

Also, by symmetry,

$m \angle EDA = \frac{1}{2} \cdot 120^{\circ } = 60^{\circ }$ ,

so $m \angle EAD= 30^{\circ }$ ,

and $\Delta AED$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle whose short leg $\overline{ED}$ has length .

The long leg $\overline{AE}$ of this $30^{\circ}-60^{\circ}-90^{\circ}$ triangle measures $\sqrt{3}$ times the length of short leg $\overline{ED}$ , so $AE= \sqrt{3}$ .

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Question

Regular hexagon has side length of 1.

Give the length of diagonal $\overline{A D}$ .

Answer

The key is to examine $\Delta AED$ in thie figure below:

Each interior angle of a regular hexagon, including $\angle F$ , measures $120^{\circ }$ , so it can be easily deduced by way of the Isosceles Triangle Theorem that $m \angle FEA = 30^{\circ }$ . To find $m \angle AED$ we can subtract $m \angle FEA = 30^{\circ }$ from $m \angle FED = 120^{\circ }$ . Thus resulting in:

$m \angle AED = 90^{\circ }$

Also, by symmetry,

$m \angle EDA = \frac{1}{2} \cdot 120^{\circ } = 60^{\circ }$ ,

so $m \angle EAD= 30^{\circ }$ .

Therefore, $\Delta AED$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle whose short leg $\overline{ED}$ has length .

The hypotenuse $\overline{AD}$ of this $30^{\circ}-60^{\circ}-90^{\circ}$ triangle measures twice the length of short leg $\overline{ED}$ , so .

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