Other Polygons - ACT Math

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Question

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

Answer

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 $\theta$ . Tangent describes the relationship between an angle and the opposite and adjacent sides of that angle. Or in other words, tan $\theta$ = 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 $\frac{s}{r}$ to find $\theta$ .

Thus,

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

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Question

What is the value of angle in the figure above?

Answer

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

Now, a quadrilateral has a total of $360^{\circ}$ . This is computed by the formula , where represents the number of sides. Thus, we know:

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

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

Solving for , we get:

$x=100^{\circ}$

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Question

What is the angle measure for the largest unknown angle in the figure above? Round to the nearest hundredth.

Answer

The total degree measure of a given figure is given by the equation , where represents the number of sides in the figure. For this figure, it is: $180(6-2)=1804 = 720^{\circ}$

Therefore, we know that the sum of the angles must equal $720^{\circ}$ . This gives us the equation:

$90^{\circ}+120^{\circ}+100^{\circ}+x+2.5x+2x=720^{\circ}$

Simplifying, this is:

$310^{\circ}+5.5x=720^{\circ}$

Now, just solve for :

$x=\frac{410}{5.5}$

The largest of the unknown angles is or $2.5 * \frac{410}{5.5}=186.36363636363636...^{\circ}$

Rounding, this is $186.36^{\circ}$ .

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Question

What is the interior angle of a $\textup{9-sided}$ polygon (a nonagon)? Round answer to the nearest hundredth if necessary.

Answer

To find the interior angle of an sided polygon, first find the total number of degrees in the polygon by the formula:
$(n-2)180^{\circ}$ . For us that yields:
$(9-2)180^{\circ} = 1260^{\circ}$ . Next we divide the total number of degrees by the number of sides:

$1260^{\circ}/9 = 140^{\circ}$

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Question

What is the total number of degrees in a $\textup{12-sided}$ polygon?

Answer

To find the total number of degrees in an -sided polygon, use the formula:

thus we see that $1800^{\circ}$

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Question

What is the interior angle of a $\textup{16-sided}$ polygon?

Answer

To find the interior angle of a regular, -sided polygon, use the formula:

:

Thus we see that $(16-2)*180^{\circ} = 2520^{\circ}$ and

$2520^{\circ}/16 = 157.5^{\circ}$

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Question

What is the total number of degrees in a $\textup{7-sided}$ polygon?

Answer

To find the total number of degrees in an -sided polygon, use the formula:

$(n-2)180^{\circ}$ . Thus we see that:

$(7-2)180 = 900^{\circ}$

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Question

How many degrees are in the interior of an octagon (an 8-sided polygon)?

Answer

To find the total number of degrees in an -sided polygon, use the formula:

$(n-2)180^{\circ}$ . Thus, for an octagon we have:

$(8-2)180^{\circ} = 1080^{\circ}$

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Question

What is the interior angle of a ten-sided polygon?

Answer

To find the interior angle of an -sided polygon, use the equation:

$((n-2)180^{\circ})/n$

Plugging in 10 for yields:

$((10-2)180^{\circ}/10)=144^{\circ}$

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Question

What is the interior angle of a pentagon?

Answer

To find the interior angle of an -sided polygon, use the formula:

$((n-2)*180^{\circ})/n$
we see for our polygon this yields:

$((5-2)180^{\circ})/5=108^{\circ}$

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Question

Find the total number of degrees inside a hexagon.

Answer

To solve, simply use the following formula where is the number of sides. Thus,

$\textup{degrees}=(n-2)(180)=(6-2)(180)=4*180=720$

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Question

Find the total number of degrees in a heptagon.

Answer

To solve, simply use the formula for finding the degrees in a closed polygon, given n being the number of sides.

In this particular case, a heptagon has seven sides.

Thus,

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Question

A certain rectangle is 4 times as long as it is wide. Suppose the length and the width of this rectangle are both halved. The area of the first rectangle is how many times as large than the area of the second rectangle?

Answer

This question actually has a lot of extraneous information that you won't need to solve the problem. With any rectangle, when the dimensions are halved, the area of the resulting rectangle will always be $\frac {1}{4}$ of the original.

Another way to solve this is by calculating the area of the first rectangle. In this case we let be the width and be the length.

Therefore the area becomes:

$A=w\cdot4w=4w^2$

For the second rectangle our width becomes $\frac{1}{2}w$ and length becomes $\frac{1}{2}\cdot4w=2w$

Thus the area of this rectangle is:

$A=\frac{1}{2}w\cdot2w=w^2$

Therefore rectangle one has an area 4 times larger than that of rectangle two.

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Question

$Area=\frac{1}{2}a\cdot P$

A gardener is tending to a garden in the shape of a regular octagon. Its apothem is ft. What is the area of the garden?

Answer

In this type of problem, the first step is to determine what information are you missing that prevents you from solving the problem as it is? Considering that the formula to find the area of a regular polygon is $Area=\frac{1}{2}a\cdot P$ , where a is apothem and P is perimeter, it can be deduced that no information is provided about the length of each side of the octagon. However, enough information is provided for that piece of information to be calculated through the use of right triangles.

In order to proceed, the internal angles of a regular octagon must be caluclated. The arrow points to an internal angle. This can be solved using the equation: $\frac{180(n-2)}{n}$ , where n is the number of sides. The internal angle is calculated to be $\frac{180^\circ(8-2)}{8}=135^\circ$ . When creating the right triangle, as shown in the previous figure, the internal angle is bisected. As a result, the triangle may be seen as:

. Now, using trigonometric functions, the base of the triangle can be calculated to be $x=\frac{12}{\tan(67.5^{\circ})}=4.97$ . Because the base of the triangle is only half the total length of one of the sides of the octagon, this number has to be multiplied by 2. $4.97\cdot2=9.94$

9.94 will be used to calculate the octagon's perimeter: $9.94\cdot8=79.52 = P$

Now all that's left is to substitute in the numbers obtained for apothem and perimeter to determine the area: $Area=\frac{1}{}2\cdot12\cdot79.52=477.12ft^{2}$

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Question

Two circles are inscribed on the inside of a rectangle and are side-by-side. What is the area of the shaded region?

Answer

When attempting to solve the area for the space between two shapes, find the area of both shapes and subtract the area of the smaller shape (in this case, the circles) from the larger shape (the rectangle).

To do this, we must first figure out the area of the circles given.

When a circle is inscribed on the inside of a square or rectangle (in other words, it will touch all sides of the shape, the diameter of the circle will be the same as the square. In this case, since the shape is a rectangle, the width is equal to the diameter of one circle and the length is equal to the diameter of both circles.

Use this to solve for the area of the cirlces:

$4 \times 4 = 16 \times 3.14 = 50.24$ Each circle has an area of 50.24 square units, so the total area of the circles is 100.48

$50.24 \times 2 = 100.48$

Then, solve for the area of the rectangle:

$8 \times 16 = 128$

Then subtract the two in order to find the area of the shaded space.

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Question

Third grader Sammy drew a house by drawing a square with an equilateral triangle on top. The bottom of the square is long. Sammy would like to know the area of the house in order to procure the appropriately sized crayon to color it in.

Which of the following is nearest to the area of the house in square centimeters?

Answer

The area of the house is comprised of the area of the square base and the triangle roof. The square has side length , so the area of the square is $s^{2}=4^{2}=16$ and the area of the equilateral triangle roof is given by $\frac{1}{2}bh$ , where is the base of the triangle, and is the height.

The base of the triangle sits on top of the square, and therefore is the same length, .

The height of an equilateral triangle is given by $\frac{b}{2}\sqrt{3}$ .

This can be found by dividing the equilateral triangle into two right triangles ( right triangles to be more specific).

All together, the area of the square and triangle is

$16+\frac{1}{2}(4)\left(\frac{4}{2}\sqrt{3}\right)=16+4\sqrt{3}$

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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.

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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 $\overline{AD}$ ?

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.

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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.

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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.

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