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

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Question

Regular Octagon has sidelength 1.

Give the length of diagonal $\overline{AD}$ .

Answer

The trick is to construct segments perpendicular to $\overline{AD}$ from and , calling the points of intersection and respectively.

Each interior angle of a regular octagon measures

$\frac{\left ( 8 - 2\right ) 180^{\circ }}{8} = 135^{\circ }$ ,

and by symmetry, $m \angle HAD = m \angle ADE = 90^{\circ }$ ,

so $m \angle BAX = m \angle CDY= 45^{\circ }$ .

This makes $\Delta BAX$ and $\Delta CDY$ $45^{\circ } -45^{\circ } -90^{\circ }$ triangles.

Since their hypotenuses are sides of the octagon with length 1, then their legs - in particular, $\overline{AX}$ and $\overline{YD}$ - have length $\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$ .

Also, since a rectangle was formed when the perpendiculars were drawn, .

The length of diagonal $\overline{AD}$ is

$AD = AX + XY + YD = \frac{\sqrt{2}}{2}+ 1 + \frac{\sqrt{2}}{2} = 1 + \sqrt{2}$ .

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Question

Regular Polygon (a twelve-sided polygon, or dodecagon) has sidelength 1.

Give the length of diagonal $\overline{AD}$ to the nearest tenth.

Answer

The trick is to construct segments perpendicular to $\overline{AD}$ from and , calling the points of intersection and respectively.

Each interior angle of a regular dodecagon measures

$\frac{\left ( 12 - 2\right ) 180^{\circ }}{12} = 150^{\circ }$ .

Since $\overline{BX}$ and $\overline{CY}$ are perpendicular to $\overline{AD}$ , it can be shown via symmetry that they are also perpendicular to $\overline{BC}$ . Therefore,

$\angle ABX$ and $\angle DCY$ both measure $60^{\circ }$

and $\Delta ABX$ and $\Delta DCY$ are $30^{\circ}-60^{\circ}-90^{\circ}$ triangles with long legs $\overline{AX}$ and $\overline{YD}$ . Since their hypotenuses are sides of the dodecagon and therefore have length 1,

$AX = YD = \frac{\sqrt{3}}{2}$ .

Also, since Quadrilateral is a rectangle, .

The length of diagonal $\overline{AD}$ is $AD = AX + XY + YD = \frac{\sqrt{3}}{2}+ 1 + \frac{\sqrt{3}}{2} = 1 + \sqrt{3}$ .

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