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ACT Math : Other Polygons

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Example Questions

Example Question #1 : Other Polygons

All segments of the polygon meet at right angles (90 degrees). The length of segment $\overline{AB}$ $\displaystyle \overline{AB}$ is 10. The length of segment $\overline{BC}$ $\displaystyle \overline{BC}$ is 8. The length of segment $\overline{DE}$ $\displaystyle \overline{DE}$ is 3. The length of segment $\overline{GH}$ $\displaystyle \overline{GH}$ is 2.

Find the perimeter of the polygon.

Possible Answers:

$\dpi{100} \small 48$ $\displaystyle \dpi{100} \small 48$

$\dpi{100} \small 40$ $\displaystyle \dpi{100} \small 40$

$\dpi{100} \small 42$ $\displaystyle \dpi{100} \small 42$

$\dpi{100} \small 44$ $\displaystyle \dpi{100} \small 44$

$\dpi{100} \small 46$ $\displaystyle \dpi{100} \small 46$

Correct answer:

$\dpi{100} \small 46$ $\displaystyle \dpi{100} \small 46$

Explanation:

The perimeter of the polygon is 46. Think of this polygon as a rectangle with two of its corners "flipped" inwards. This "flipping" changes the area of the rectangle, but not its perimeter; therefore, the top and bottom sides of the original rectangle would be 12 units long $\dpi{100} \small (10+2=12)$ $\displaystyle \dpi{100} \small (10+2=12)$. The left and right sides would be 11 units long $\dpi{100} \small (8+3=11)$ $\displaystyle \dpi{100} \small (8+3=11)$. Adding all four sides, we find that the perimeter of the recangle (and therefore, of this polygon) is 46.

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Example Question #1 : Other Polygons

In the figure below, each pair of intersecting line segments forms a right angle, and all the lengths are given in feet. What is the perimeter, in feet, of the figure?

Quad

Possible Answers:

$\displaystyle 56$

$\displaystyle 48$

$\displaystyle 50$

$\displaystyle 44$

$\displaystyle 38$

Correct answer:

$\displaystyle 50$

Explanation:

Fill in the missing sides by thinking about the entire figure as a big rectangle. In the figure below, the large rectangle is outlined in blue and the missing numbers are supplied in red.

Quad

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Example Question #592 : Plane Geometry

What is the perimeter of a regular, $\textup{13-sided}$ $\displaystyle \textup{13-sided}$ polygon with a side length of $4\textup{cm}$ $\displaystyle 4\textup{cm}$?

Possible Answers:

$\textup{52cm}$ $\displaystyle \textup{52cm}$

$\textup{56cm}$ $\displaystyle \textup{56cm}$

$\textup{50cm}$ $\displaystyle \textup{50cm}$

$\textup{44cm}$ $\displaystyle \textup{44cm}$

$\textup{4cm}$ $\displaystyle \textup{4cm}$

Correct answer:

$\textup{52cm}$ $\displaystyle \textup{52cm}$

Explanation:

To find the perimeter of an $\displaystyle n$-sided polygon with a side length of $\displaystyle a$, simply multiply the side length by the number of sides:
$4\textup{cm}*13=52\textup{cm}$ $\displaystyle 4\textup{cm}*13=52\textup{cm}$

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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 $\overline{AD}$ $\displaystyle \overline{AD}$?

Possible Answers:

120 $\displaystyle 120$

150 $\displaystyle 150$

130 $\displaystyle 130$

160 $\displaystyle 160$

140 $\displaystyle 140$

Correct answer:

140 $\displaystyle 140$

Explanation:

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

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

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

The other two angles are supplementary to these:

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

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

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

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

Thingy

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

$\displaystyle AD = AM + MN + ND$

$\displaystyle AD = AM + BC+ AM$

$AD = 2 \cdot AM + 55.56$ $\displaystyle AD = 2 \cdot AM + 55.56$

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

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

$AD \approx 2 \cdot 42.56 + 55.56 \approx 140.68$ $\displaystyle AD \approx 2 \cdot 42.56 + 55.56 \approx 140.68$

This makes 140 the best choice.

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Example Question #2 : Other Polygons

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

Possible Answers:

150 $\displaystyle 150$

160 $\displaystyle 160$

170 $\displaystyle 170$

140 $\displaystyle 140$

180 $\displaystyle 180$

Correct answer:

160 $\displaystyle 160$

Explanation:

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

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

The other two angles are supplementary to these:

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

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

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

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

Thingy

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

$\displaystyle AD = AM + MN + ND$

$\displaystyle AD = AM + BC+ AM$

$AD = 2 \cdot AM + 71.42$ $\displaystyle AD = 2 \cdot AM + 71.42$

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

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

$AD \approx 2 \cdot 44.53 + 71.42 \approx 160.48$ $\displaystyle AD \approx 2 \cdot 44.53 + 71.42 \approx 160.48$

This makes 160 the best choice.

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

Possible Answers:

140 $\displaystyle 140$

110 $\displaystyle 110$

100 $\displaystyle 100$

120 $\displaystyle 120$

130 $\displaystyle 130$

Correct answer:

130 $\displaystyle 130$

Explanation:

Congruent sides $\overline{AB}$ $\displaystyle \overline{AB}$ and $\overline{BC}$ $\displaystyle \overline{BC}$ and the diagonal $\overline{AC}$ $\displaystyle \overline{AC}$ form an isosceles triangle.

$\angle ABC$ $\displaystyle \angle ABC$, being an angle of a seven-sided regular polygon, has measure

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

The other two are congruent, and each has measure

$m \angle CAB = m \angle ACB = \frac{180^{\circ }- 128.57^{\circ }}{2}\approx 25.71^{\circ}$ $\displaystyle m \angle CAB = m \angle ACB = \frac{180^{\circ }- 128.57^{\circ }}{2}\approx 25.71^{\circ}$

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

$AB = BC = \frac{500}{7} \approx 71.42$ $\displaystyle AB = BC = \frac{500}{7} \approx 71.42$

$\displaystyle AC$ can be found using the Law of Sines:

$\frac{AC}{\sin m \angle B} = \frac{AB}{\sin m \angle ACB}$ $\displaystyle \frac{AC}{\sin m \angle B} = \frac{AB}{\sin m \angle ACB}$

$\frac{AC}{\sin 128.57^{\circ} } = \frac{71.42}{\sin 25.71^{\circ}}$ $\displaystyle \frac{AC}{\sin 128.57^{\circ} } = \frac{71.42}{\sin 25.71^{\circ}}$

$AC \approx \frac{71.42}{\sin 25.71^{\circ}}\cdot \sin 128.57^{\circ}$ $\displaystyle AC \approx \frac{71.42}{\sin 25.71^{\circ}}\cdot \sin 128.57^{\circ}$

$AC \approx \frac{71.42}{0.4339}\cdot 0.7818$ $\displaystyle AC \approx \frac{71.42}{0.4339}\cdot 0.7818$

$AC \approx 128.7$ $\displaystyle AC \approx 128.7$

Of the given choices, 130 comes closest.

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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 $\overline{AC}$ $\displaystyle \overline{AC}$?

Possible Answers:

100 $\displaystyle 100$

110 $\displaystyle 110$

$\displaystyle 90$

$\displaystyle 95$

105 $\displaystyle 105$

Correct answer:

105 $\displaystyle 105$

Explanation:

$\angle ABC$ $\displaystyle \angle ABC$, being an angle of a nine-sided regular polygon, has measure

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

The other two are congruent, and each has measure

$m \angle CAB = m \angle ACB = \frac{180^{\circ }- 140^{\circ }}{2}\approx 20^{\circ}$ $\displaystyle m \angle CAB = m \angle ACB = \frac{180^{\circ }- 140^{\circ }}{2}\approx 20^{\circ}$

The length of one side is one-ninth of 500, or

$AB = BC = \frac{500}{9} \approx 55.56$ $\displaystyle AB = BC = \frac{500}{9} \approx 55.56$

$\displaystyle AC$ can be found using the Law of Sines:

$\frac{AC}{\sin m \angle B} = \frac{AB}{\sin m \angle ACB}$ $\displaystyle \frac{AC}{\sin m \angle B} = \frac{AB}{\sin m \angle ACB}$

$\frac{AC}{\sin 140^{\circ} } = \frac{55.56}{\sin 20^{\circ}}$ $\displaystyle \frac{AC}{\sin 140^{\circ} } = \frac{55.56}{\sin 20^{\circ}}$

$AC \approx \frac{55.56}{\sin 20^{\circ}}\cdot \sin 140^{\circ}$ $\displaystyle AC \approx \frac{55.56}{\sin 20^{\circ}}\cdot \sin 140^{\circ}$

$AC \approx \frac{55.56}{0.3420}\cdot 0.6428$ $\displaystyle AC \approx \frac{55.56}{0.3420}\cdot 0.6428$

$AC \approx 104.4$ $\displaystyle AC \approx 104.4$

Of the given choices, 105 comes closest.

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

What is the side length of a 12-sided polygon with a perimeter of 48 cm? $\displaystyle 48 cm?$

Possible Answers:

576cm $\displaystyle 576cm$

26cm $\displaystyle 26cm$

2cm $\displaystyle 2cm$

4cm $\displaystyle 4cm$

12cm $\displaystyle 12cm$

Correct answer:

4cm $\displaystyle 4cm$

Explanation:

To find the side length, $\displaystyle s$, of an $\displaystyle n$-sided polygon with a perimeter of $\displaystyle p$.

Use the formula:
$\\s = p/n \newline s= 48/12 \\s= 4cm$ $\displaystyle \\s = p/n \newline s= 48/12 \\s= 4cm$

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Example Question #2 : Other Polygons

In the diagram below, what is $\theta$ $\displaystyle \theta$ equal to?

Hexagon_sides

Possible Answers:

$\tan \left ( \frac{s}{r} \right )$ $\displaystyle \tan \left ( \frac{s}{r} \right )$

$\tan^{-1}\left ( \frac{r}{s} \right )$ $\displaystyle \tan^{-1}\left ( \frac{r}{s} \right )$

$\sin^{-1}\left ( \frac{r}{s} \right )$ $\displaystyle \sin^{-1}\left ( \frac{r}{s} \right )$

$\tan \left ( \frac{r}{s} \right )$ $\displaystyle \tan \left ( \frac{r}{s} \right )$

$\tan^{-1}\left ( \frac{s}{r} \right )$ $\displaystyle \tan^{-1}\left ( \frac{s}{r} \right )$

Correct answer:

$\tan^{-1}\left ( \frac{s}{r} \right )$ $\displaystyle \tan^{-1}\left ( \frac{s}{r} \right )$

Explanation:

The figure given is a hexagon with an embedded triangle. The fact that it is embedded in a triangle is mainly to throw you off, as it has little to no consequence on the correct answer. Of the available answer choices, you must choose a relationship that would give the value of $\theta$ $\displaystyle \theta$. Tangent describes the relationship between an angle and the opposite and adjacent sides of that angle. Or in other words, tan $\theta$ $\displaystyle \theta$ = opposite side/adjacent side. However, when solving for an angle, we must use the inverse function. Therefore, if we know the opposite and adjacent sides are, we can use the inverse of the tangent, or arctangent (tan^-1), of $\frac{s}{r}$ $\displaystyle \frac{s}{r}$ to find $\theta$ $\displaystyle \theta$.

Thus,

$\theta =\tan^{-1}\left ( \frac{s}{r} \right )$ $\displaystyle \theta =\tan^{-1}\left ( \frac{s}{r} \right )$

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Example Question #601 : Plane Geometry

Shape_1

What is the value of angle $\displaystyle x$ in the figure above?

Possible Answers:

$50^{\circ}$ $\displaystyle 50^{\circ}$

$100^{\circ}$ $\displaystyle 100^{\circ}$

$75^{\circ}$ $\displaystyle 75^{\circ}$

$35^{\circ}$ $\displaystyle 35^{\circ}$

$110^{\circ}$ $\displaystyle 110^{\circ}$

Correct answer:

$100^{\circ}$ $\displaystyle 100^{\circ}$

Explanation:

Begin by noticing that the upper-right angle of this figure is supplementary to $70^{\circ}$ $\displaystyle 70^{\circ}$. This means that it is $180^{\circ}-70^{\circ}=110^{\circ}$ $\displaystyle 180^{\circ}-70^{\circ}=110^{\circ}$:

Shape_1

Now, a quadrilateral has a total of $360^{\circ}$ $\displaystyle 360^{\circ}$. This is computed by the formula $\displaystyle degrees = 180 * (s-2)$, where $\displaystyle s$ represents the number of sides. Thus, we know:

$110^{\circ}+70^{\circ}+80^{\circ}+x=360^{\circ}$ $\displaystyle 110^{\circ}+70^{\circ}+80^{\circ}+x=360^{\circ}$

This is the same as $260^{\circ} + x = 360^{\circ}$ $\displaystyle 260^{\circ} + x = 360^{\circ}$

Solving for $\displaystyle x$, we get:

$x=100^{\circ}$ $\displaystyle x=100^{\circ}$

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ACT Math : Other Polygons

Study concepts, example questions & explanations for ACT Math

All ACT Math Resources

Example Questions

Example Question #1 : Other Polygons

Example Question #1 : Other Polygons

Example Question #592 : Plane Geometry

Example Question #1 : Other Polygons

Example Question #2 : Other Polygons

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

Example Question #1 : Other Polygons

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

Example Question #2 : Other Polygons

Example Question #601 : Plane Geometry

All ACT Math Resources

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