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

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Example Question #1 : How To Find The Length Of A Diagonal Of A Polygon

Regular Octagon $\displaystyle ABCDEFGH$ has sidelength 1.

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

Possible Answers:

$\frac{2 + \sqrt{3}}{2}$ $\displaystyle \frac{2 + \sqrt{3}}{2}$

$\frac{2 + \sqrt{2}}{2}$ $\displaystyle \frac{2 + \sqrt{2}}{2}$

$1 + \sqrt{3}$ $\displaystyle 1 + \sqrt{3}$

$1 + \sqrt{2}$ $\displaystyle 1 + \sqrt{2}$

$\displaystyle 2$

Correct answer:

$1 + \sqrt{2}$ $\displaystyle 1 + \sqrt{2}$

Explanation:

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

Octagon_1

Each interior angle of a regular octagon measures

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

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

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

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

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

Also, since a rectangle was formed when the perpendiculars were drawn, $\displaystyle XY = BC = 1$.

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

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

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Example Question #2 : How To Find The Length Of A Diagonal Of A Polygon

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

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

Possible Answers:

$\frac{2 + \sqrt{2}}{2}$ $\displaystyle \frac{2 + \sqrt{2}}{2}$

$\displaystyle 2$

$1 + \sqrt{3}$ $\displaystyle 1 + \sqrt{3}$

$1 + \sqrt{2}$ $\displaystyle 1 + \sqrt{2}$

$\frac{2 + \sqrt{3}}{2}$ $\displaystyle \frac{2 + \sqrt{3}}{2}$

Correct answer:

$1 + \sqrt{3}$ $\displaystyle 1 + \sqrt{3}$

Explanation:

Dodecagon

Each interior angle of a regular dodecagon measures

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

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

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

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

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

Also, since Quadrilateral BCXY $\displaystyle BCXY$ is a rectangle, $\displaystyle XY = BC = 1$.

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

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

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Example Question #1 : How To Find The Length Of A Diagonal Of A Polygon

Example Question #2 : How To Find The Length Of A Diagonal Of A Polygon

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