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

Card 0 of 18

Question

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

Answer

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

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

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

The other two angles are supplementary to these:

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

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

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

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

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

$AD = 2 \cdot AM + 71.42$

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

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

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

This makes 160 the best choice.

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 $\overline{AD}$ ?

Answer

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

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

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

The other two angles are supplementary to these:

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

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

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

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

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

$AD = 2 \cdot AM + 55.56$

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

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

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

This makes 140 the best choice.

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 $\overline{AD}$ ?

Answer

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

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

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

The other two angles are supplementary to these:

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

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

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

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

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

$AD = 2 \cdot AM + 71.42$

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

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

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

This makes 160 the best choice.

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 $\overline{AD}$ ?

Answer

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

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

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

The other two angles are supplementary to these:

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

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

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

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

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

$AD = 2 \cdot AM + 55.56$

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

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

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

This makes 140 the best choice.

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 $\overline{AD}$ ?

Answer

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

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

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

The other two angles are supplementary to these:

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

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

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

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

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

$AD = 2 \cdot AM + 71.42$

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

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

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

This makes 160 the best choice.

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 $\overline{AD}$ ?

Answer

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

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

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

The other two angles are supplementary to these:

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

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

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

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

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

$AD = 2 \cdot AM + 55.56$

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

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

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

This makes 140 the best choice.

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 $\overline{AD}$ ?

Answer

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

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

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

The other two angles are supplementary to these:

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

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

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

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

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

$AD = 2 \cdot AM + 71.42$

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

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

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

This makes 160 the best choice.

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 $\overline{AD}$ ?

Answer

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

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

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

The other two angles are supplementary to these:

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

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

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

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

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

$AD = 2 \cdot AM + 55.56$

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

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

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

This makes 140 the best choice.

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 $\overline{AD}$ ?

Answer

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

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

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

The other two angles are supplementary to these:

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

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

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

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

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

$AD = 2 \cdot AM + 71.42$

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

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

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

This makes 160 the best choice.

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 $\overline{AD}$ ?

Answer

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

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

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

The other two angles are supplementary to these:

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

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

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

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

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

$AD = 2 \cdot AM + 55.56$

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

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

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

This makes 140 the best choice.

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 $\overline{AD}$ ?

Answer

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

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

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

The other two angles are supplementary to these:

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

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

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

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

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

$AD = 2 \cdot AM + 71.42$

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

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

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

This makes 160 the best choice.

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 $\overline{AD}$ ?

Answer

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

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

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

The other two angles are supplementary to these:

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

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

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

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

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

$AD = 2 \cdot AM + 55.56$

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

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

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

This makes 140 the best choice.

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 $\overline{AD}$ ?

Answer

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

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

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

The other two angles are supplementary to these:

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

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

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

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

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

$AD = 2 \cdot AM + 71.42$

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

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

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

This makes 160 the best choice.

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 $\overline{AD}$ ?

Answer

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

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

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

The other two angles are supplementary to these:

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

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

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

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

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

$AD = 2 \cdot AM + 55.56$

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

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

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

This makes 140 the best choice.

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 $\overline{AD}$ ?

Answer

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

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

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

The other two angles are supplementary to these:

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

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

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

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

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

$AD = 2 \cdot AM + 71.42$

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

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

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

This makes 160 the best choice.

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 $\overline{AD}$ ?

Answer

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

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

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

The other two angles are supplementary to these:

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

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

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

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

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

$AD = 2 \cdot AM + 55.56$

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

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

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

This makes 140 the best choice.

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 $\overline{AD}$ ?

Answer

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

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

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

The other two angles are supplementary to these:

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

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

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

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

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

$AD = 2 \cdot AM + 71.42$

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

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

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

This makes 160 the best choice.

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 $\overline{AD}$ ?

Answer

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

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

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

The other two angles are supplementary to these:

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

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

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

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

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

$AD = 2 \cdot AM + 55.56$

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

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

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

This makes 140 the best choice.

Compare your answer with the correct one above

Tap the card to reveal the answer