How to find the length of a side of a polygon - ACT Math

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Question

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

Answer

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

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

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

The other two are congruent, and each has measure

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

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

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

can be found using the Law of Sines:

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

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

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

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

$AC \approx 104.4$

Of the given choices, 105 comes closest.

Compare your answer with the correct one above

Question

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

Answer

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

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

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

The other two are congruent, and each has measure

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

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

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

can be found using the Law of Sines:

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

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

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

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

$AC \approx 128.7$

Of the given choices, 130 comes closest.

Compare your answer with the correct one above

Question

What is the side length of a 12-sided polygon with a perimeter of

Answer

To find the side length, , of an -sided polygon with a perimeter of .

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

Compare your answer with the correct one above

Question

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

Answer

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

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

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

The other two are congruent, and each has measure

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

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

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

can be found using the Law of Sines:

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

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

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

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

$AC \approx 104.4$

Of the given choices, 105 comes closest.

Compare your answer with the correct one above

Question

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

Answer

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

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

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

The other two are congruent, and each has measure

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

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

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

can be found using the Law of Sines:

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

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

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

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

$AC \approx 128.7$

Of the given choices, 130 comes closest.

Compare your answer with the correct one above

Question

What is the side length of a 12-sided polygon with a perimeter of

Answer

To find the side length, , of an -sided polygon with a perimeter of .

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

Compare your answer with the correct one above

Question

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

Answer

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

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

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

The other two are congruent, and each has measure

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

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

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

can be found using the Law of Sines:

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

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

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

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

$AC \approx 104.4$

Of the given choices, 105 comes closest.

Compare your answer with the correct one above

Question

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

Answer

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

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

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

The other two are congruent, and each has measure

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

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

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

can be found using the Law of Sines:

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

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

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

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

$AC \approx 128.7$

Of the given choices, 130 comes closest.

Compare your answer with the correct one above

Question

What is the side length of a 12-sided polygon with a perimeter of

Answer

To find the side length, , of an -sided polygon with a perimeter of .

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

Compare your answer with the correct one above

Question

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

Answer

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

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

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

The other two are congruent, and each has measure

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

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

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

can be found using the Law of Sines:

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

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

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

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

$AC \approx 104.4$

Of the given choices, 105 comes closest.

Compare your answer with the correct one above

Question

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

Answer

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

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

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

The other two are congruent, and each has measure

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

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

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

can be found using the Law of Sines:

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

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

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

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

$AC \approx 128.7$

Of the given choices, 130 comes closest.

Compare your answer with the correct one above

Question

What is the side length of a 12-sided polygon with a perimeter of

Answer

To find the side length, , of an -sided polygon with a perimeter of .

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

Compare your answer with the correct one above

Question

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

Answer

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

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

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

The other two are congruent, and each has measure

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

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

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

can be found using the Law of Sines:

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

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

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

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

$AC \approx 104.4$

Of the given choices, 105 comes closest.

Compare your answer with the correct one above

Question

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

Answer

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

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

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

The other two are congruent, and each has measure

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

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

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

can be found using the Law of Sines:

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

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

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

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

$AC \approx 128.7$

Of the given choices, 130 comes closest.

Compare your answer with the correct one above

Question

What is the side length of a 12-sided polygon with a perimeter of

Answer

To find the side length, , of an -sided polygon with a perimeter of .

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

Compare your answer with the correct one above

Question

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

Answer

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

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

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

The other two are congruent, and each has measure

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

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

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

can be found using the Law of Sines:

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

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

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

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

$AC \approx 104.4$

Of the given choices, 105 comes closest.

Compare your answer with the correct one above

Question

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

Answer

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

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

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

The other two are congruent, and each has measure

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

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

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

can be found using the Law of Sines:

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

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

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

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

$AC \approx 128.7$

Of the given choices, 130 comes closest.

Compare your answer with the correct one above

Question

What is the side length of a 12-sided polygon with a perimeter of

Answer

To find the side length, , of an -sided polygon with a perimeter of .

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

Compare your answer with the correct one above

Question

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

Answer

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

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

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

The other two are congruent, and each has measure

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

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

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

can be found using the Law of Sines:

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

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

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

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

$AC \approx 104.4$

Of the given choices, 105 comes closest.

Compare your answer with the correct one above

Question

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

Answer

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

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

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

The other two are congruent, and each has measure

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

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

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

can be found using the Law of Sines:

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

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

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

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

$AC \approx 128.7$

Of the given choices, 130 comes closest.

Compare your answer with the correct one above

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