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

Study concepts, example questions & explanations for ACT Math

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Example Questions

Example Question #1 : Other Polygons

Polygon \(\displaystyle ABCDEFGHI\) is a regular nine-sided polygon, or nonagon, with perimeter 500. Which choice comes closest to the length of diagonal \(\displaystyle \overline{AD}\)?

Possible Answers:

\(\displaystyle 120\)

\(\displaystyle 150\)

\(\displaystyle 130\)

\(\displaystyle 160\)

\(\displaystyle 140\)

Correct answer:

\(\displaystyle 140\)

Explanation:

Congruent sides \(\displaystyle \overline{AB}\)\(\displaystyle \overline{BC}\), and \(\displaystyle \overline{CD}\), and the diagonal \(\displaystyle \overline{AD}\) form an isosceles trapezoid. 

\(\displaystyle \angle ABC\) and \(\displaystyle \angle BCD\). being angles of a nine-sided regular polygon, have measure

\(\displaystyle m \angle ABC = m\angle BCDC = \frac{180 ^{\circ } (9-2)}{9} =140^{\circ }\)

The other two angles are supplementary to these:

\(\displaystyle m \angle DAB = m \angle ADC = 180 ^{\circ }- 140^{\circ } = 40 ^{\circ }\)

The length of one side of the nonagon is one-ninth of 500, so

\(\displaystyle AB = BC =CD= \frac{500}{9} \approx 55.56\)

The trapezoid formed is below (figure NOT drawn to scale):

Thingy

Altitudes \(\displaystyle \overline{BM}\) and \(\displaystyle \overline{CN}\) to the base have been drawn, so

\(\displaystyle AD = AM + MN + ND\)

\(\displaystyle AD = AM + BC+ AM\)

\(\displaystyle AD = 2 \cdot AM + 55.56\)

\(\displaystyle \frac{AM}{AB} = \cos 40^{ \circ}\)

\(\displaystyle AM=AB \cos 40^{ \circ} \approx 55.56 \cdot 0.7660 \approx 42.56\)

\(\displaystyle AD \approx 2 \cdot 42.56 + 55.56 \approx 140.68\)

This makes 140 the best choice.

Example Question #2 : Other Polygons

Polygon \(\displaystyle ABCDEFG\) is a regular seven-sided polygon, or heptagon, with perimeter 500. Which choice comes closest to the length of diagonal \(\displaystyle \overline{AD}\)?

Possible Answers:

\(\displaystyle 150\)

\(\displaystyle 160\)

\(\displaystyle 170\)

\(\displaystyle 140\)

\(\displaystyle 180\)

Correct answer:

\(\displaystyle 160\)

Explanation:

Congruent sides \(\displaystyle \overline{AB}\)\(\displaystyle \overline{BC}\), and \(\displaystyle \overline{CD}\), and the diagonal \(\displaystyle \overline{AD}\) form an isosceles trapezoid. 

\(\displaystyle \angle ABC\) and \(\displaystyle \angle BCD\). being angles of a seven-sided regular polygon, have measure

\(\displaystyle m \angle ABC = m \angle BCD = \frac{180 ^{\circ } (7-2)}{7} \approx 128.57^{\circ }\)

The other two angles are supplementary to these:

\(\displaystyle m \angle DAB = m \angle ADC \approx 180 ^{\circ }- 128.57^{\circ }\approx 51.43 ^{\circ }\)

The length of one side is one-seventh of 500, so

\(\displaystyle AB = BC =CD= \frac{500}{7} \approx 71.42\)

The trapezoid formed is below (figure NOT drawn to scale):

Thingy

Altitudes \(\displaystyle \overline{BM}\) and \(\displaystyle \overline{CN}\) to the base have been drawn, so

\(\displaystyle AD = AM + MN + ND\)

\(\displaystyle AD = AM + BC+ AM\)

\(\displaystyle AD = 2 \cdot AM + 71.42\)

\(\displaystyle \frac{AM}{AB} = \cos 51.43^{ \circ}\)

\(\displaystyle AM=AB \cos 51.43^{ \circ} \approx 71.42 \cdot 0.6235 \approx 44.53\)

\(\displaystyle AD \approx 2 \cdot 44.53 + 71.42 \approx 160.48\)

This makes 160 the best choice.

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