Properties of polygons and circles - HiSET

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A quadrilateral is shown, and the angle measures of 3 interior angles are given. Find x, the missing angle measure.

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The sum of the measures of the interior angles of a quadrilateral is 360 degrees. The sum of the measures of the interior angles of any polygon can be determined using the following formula:

$Sum \ of\ Interior \ Angles = $180^{\circ}$(n-2)$ , where is the number of sides.

For example, with a quadrilateral, which has 4 sides, you obtain the following calculation:

$Sum \ of\ Interior \ Angles = $180^{\circ}$$(4-2)=180^{\circ}$$\cdot2=360^{\circ}$$

Solving for requires setting up an algebraic equation, adding all 4 angles to equal 360 degrees:

Solving for is straightforward: subtract the values of the 3 known angles from both sides:

is the center of the above circle. Calculate $m \angle AOB$ .

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$\angle AOB$ is the central angle that intercepts $\overarc{AB}$ , so

$m \angle AOB = m \overarc{AB}$ .

Therefore, we need to find $m \overarc{AB}$ to obtain our answer.

If the sides of an angle with vertex outside the circle are both tangent to the circle, the angle formed is half the difference of the measures of the arcs. Therefore,

$$\frac{m \overarc{ACB}$ - m \overarc{AB} }{2} = m \angle AVB$

Letting $x= m \overarc{AB}$ , since the total arc measure of a circle is 360 degrees,

$m \overarc{ACB} = 360 - x$

We are also given that

$m \angle AVB = 45$

Making substitutions, and solving for :

$$\frac{(360- x )- x }{2}$ =45$

$$\frac{360- 2 x }{2}$ =45$

Multiply both sides by 2:

$$\frac{360- 2 x }{2}$ \cdot 2 =45 \cdot 2$

Subtract 360 from both sides:

Divide both sides by :

$-2x \div (-2) = -270 \div (-2)$

,

the measure of $\overarc{AB}$ and, consequently, that of $\angle AOB$ .

The above figure shows a regular seven-sided polygon, or heptagon, inscribed inside a circle. is the common center of the figures.

Give the measure of $\angle DBO$ .

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Consider the figure below, which adds some radii of the heptagon (and circle):

$\overline{OB}$ , as a radius of a regular polygon, bisects $\angle EBD$ . The measure of this angle can be calculated using the formula

$m = $\frac{180 (n-2)}{n}$$ ,

where :

$m \angle EBD = $$\frac{180^{\circ }$$ (7-2)}{7} = $$\frac{180^{\circ }$$ (5)}{7} = $$\frac{900}{7}$^{\circ }$ = 1 28 $$\frac{4}{7}$^{\circ }$$

Consequently,

$\angle DBO = $\frac{1}{2}$ \cdot m \angle EBD = $\frac{1}{2}$ \cdot 1 28 $$\frac{4}{7}$^{\circ }$ = 64 $$\frac{2}{7}$^{\circ }$$ ,

the correct response.

The above figure shows a regular seven-sided polygon, or heptagon, inscribed inside a circle. is the common center of the figures.

Give the measure of $\angle AOB$ .

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Examine the diagram below, which divides $\angle AOB$ into three congruent angles, one of which is $\angle 1$ :

The measure of a central angle of a regular -sided polygon which intercepts one side of the polygon is $$\frac{360 ^{\circ }$}{n}$ ; setting , the measure of $\angle 1$ is

$m \angle 1 = $\frac{360 ^{\circ }$}{7} = 51 $$\frac{3}{7}$^{\circ }$$ . $\angle AOC$ has measure three times this; that is,

$m \angle AOC = 3 \times51 $$\frac{3}{7}$^{\circ }$ = 154 $$\frac{2}{7}$^{\circ }$$

The above figure shows a regular seven-sided polygon, or heptagon, inscribed inside a circle. is the common center of the figures.

Give the measure of $\angle AOC$ .

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Examine the diagram below, which divides $\angle AOC$ into two congruent angles, one of which is $\angle 1$ :

The measure of a central angle of a regular -sided polygon which intercepts one side of the polygon is $$\frac{360 ^{\circ }$}{n}$ ; setting , the measure of $\angle 1$ is

$m \angle 1 = $\frac{360 ^{\circ }$}{7} = 51 $$\frac{3}{7}$^{\circ }$$ . $\angle AOC$ has measure twice this; that is,

$m \angle AOC = 2 \times51 $$\frac{3}{7}$^{\circ }$ = 102 $$\frac{6}{7}$^{\circ }$$

The above figure shows a regular ten-sided polygon, or decagon, inscribed inside a circle. is the common center of the figures.

Give the measure of $\angle AOB$ .

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Examine the diagram below, which divides $\angle AOB$ into two congruent angles, one of which is $\angle 1$ :

The measure of a central angle of a regular -sided polygon which intercepts one side of the polygon is $$\frac{360 ^{\circ }$}{n}$ ; setting , the measure of $\angle 1$ is

$m \angle 1 = $\frac{360 ^{\circ }$}{10} $=36^{\circ }$$ . $\angle AOB$ has measure twice this; that is,

$m \angle AOB = 2 \times 36 ^{\circ } = 72 ^{\circ }$ .

The above figure shows a regular ten-sided polygon, or decagon, inscribed inside a circle. is the common center of the figures.

Give the measure of $\angle OBC$ .

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Consider the triangle $\bigtriangleup OBC$ . Since $\overline{OB}$ and $\overline{OC}$ are radii, they are congruent, and by the Isosceles Triangle Theorem, $m \angle OBC = m \angle OCB$ .

Now, examine the figure below, which divides $\angle BOC$ into three congruent angles, one of which is $\angle 1$ :

The measure of a central angle of a regular -sided polygon which intercepts one side of the polygon is $$\frac{360 ^{\circ }$}{n}$ ; setting , the measure of $\angle 1$ is

$m \angle 1 = $\frac{360 ^{\circ }$}{10} $=36^{\circ }$$ . $\angle BOC$ has measure three times this; that is,

$m \angle BOC = 3 \times 36 ^{\circ } = 108 ^{\circ }$ .

The measures of the interior angles of a triangle total , so

$m \angle OCB + m \angle OBC+m \angle BOC = 180$

Substituting 108 for $\angle BOC$ and $m \angle OBC$ for $m \angle OCB$ :

$m \angle OBC+ m \angle OBC+108 = 180$

$2 m \angle OBC+108 = 180$

$2 m \angle OBC+108- 108 = 180- 108$

$2 m \angle OBC=72$

$2 m \angle OBC \div 2 =72 \div 2$

$m \angle OBC= 36$

The above figure shows a regular ten-sided polygon, or decagon, inscribed inside a circle. is the common center of the figures.

Give the measure of $\angle CBE$ .

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Through symmetry, it can be seen that Quadrilateral is a trapezoid, such that $\overline{BC}$ || $\overline{DE }$ . By the Same-Side Interior Angle Theorem, $\angle CBE$ and $\angle DEB$ are supplementary - that is,

$m \angle CBE +\angle DEB = 180$ .

The measure of $\angle DEB$ can be calculated using the formula

$m = $\frac{180 (n-2)}{n}$$ ,

where :

$m \angle DEB= $$\frac{180^{\circ }$$ (10-2)}{10} = $$\frac{180^{\circ }$$ (8)}{10} = $$\frac{1,440}{10}$^{\circ }$ =144 ^{\circ }$

Substituting:

$m \angle CBE +144= 180$

$m \angle CBE +144 - 144 = 180 - 144$

$m \angle CBE = 36$

If two angles are supplementary and one angle measures , what is the measurement of the second angle?

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Step 1: Define supplementary angles. Supplementary angles are two angles whose sum is .

Step 2: Find the other angle by subtracting the given angle from the maximum sum of the two angles.

So,

The missing angle (or second angle) is

$\angle 1$ and $\angle 2$ are complementary angles.

$\angle 1$ and $\angle 3$ are supplementary angles.

$\angle 2 \cong \angle 4$

$m \angle 3= $112^{\circ }$$

Evaluate $m \angle 4$ .

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$\angle 1$ and $\angle 3$ are supplementary angles, so, by definition,

$m \angle 1 + m \angle 3 = $180^{\circ }$$

$m \angle 3= $112^{\circ }$$ , so substitute and solve for $m \angle 1$ :

$m \angle 1 + $112^{\circ }$ = $180^{\circ }$$

$m \angle 1 + $112^{\circ }$ $-112^{\circ }$ = $180^{\circ }$$-112^{\circ }$$

$m \angle 1 = 68 ^{\circ }$

$\angle 1$ and $\angle 2$ are complementary angles, so, by definition,

$m \angle 1 + m \angle 2 = $90^{\circ }$$

Substitute and solve for $m \angle 2$ :

$68 ^{\circ } + m \angle 2 = $90^{\circ }$$

$68 ^{\circ } + m \angle 2 - 68 ^{\circ } = $90^{\circ }$-68 ^{\circ }$

$m \angle 2 $=22^{\circ }$$

$\angle 2 \cong \angle 4$ - that is, the angles have the same measure. Therefore,

$m \angle 4 $=22^{\circ }$$ .

$\angle 1$ and $\angle 2$ are a pair of vertical angles.

$\angle 3$ and $\angle 4$ are a linear pair.

$\angle 2$ and $\angle 3$ are the two acute angles of a right triangle.

Which of the following must be true?

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$\angle 1$ and $\angle 2$ are a pair of vertical angles; it follows that

$m \angle 1 = m \angle 2$

$\angle 3$ and $\angle 4$ are a linear pair; it follows that they are supplementary - that is,

$m \angle 3 + m \angle 4 = $180^{\circ }$$ .

$\angle 2$ and $\angle 3$ are the two acute angles of a right triangle; it follows that they are complementary - that is,

$m \angle 2 + m \angle 3 = $90^{\circ }$$ .

Therefore, we have the three statements

$m \angle 1 = m \angle 2$

$m \angle 2 + m \angle 3 = $90^{\circ }$$

$m \angle 3 + m \angle 4 = $180^{\circ }$$

From the second statement, we can subtract $m \angle 2$ from both sides to get

$m \angle 2 + m \angle 3 - m \angle 2 = $90^{\circ }$ - m \angle 2$

$m \angle 3 = $90^{\circ }$ - m \angle 2$

Substitute this expression for $m \angle 3$ in the third expression to get

Substitute $m \angle 1$ for $m \angle 2$ :

Add $m \angle 1 - $90^{\circ }$$ to both sides:

$m \angle 4 = $90^{\circ }$+m \angle 1$ ,

or, rearranged,

$m \angle 1 + $90^{\circ }$= m \angle 4$ .

The depicted circle has a radius of 3 cm. The arc length between the two points shown on the circle is $\pi$ cm. Find the area of the enclosed sector (highlighted in green).

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First, find the area and circumference of the circle using the radius and the following formulae for circles:

$Circumference = 2\pi r$

$Area = \pi $r^2$$

Substituting in 3 for yields:

$Circumference = 6\pi \ cm$

$Area = 9\pi \ $cm^2$$

Next, find what fraction of the total circumference is between the two points on the circle (the arc length).

$\frac{Arc \ Length }{Circumference }$ = $\frac{\pi}{6\pi}$ = $\frac{1}{6}$

Finally, use this fraction to calculate the area of the enclosed sector. Note that this area is proportional to the above fraction. In other words:

$\frac{Arc \ Length }{Circumference }$ = $\frac{Sector \ Area}{Total \ Area}$

So the Sector Area is one sixth of the total area.

$\frac{1 }{6}$ = $\frac{Sector \ Area}{9\pi}$

Cross multiply:

$Sector \ Area\times6 = 9\pi$

Divide both sides by 6, then simplify to get the final answer:

$Sector \ Area = $\frac{9\pi}{6}$ = $\frac{3\pi}{2}$; $cm^2$$

is the center of the above circle, and . Evaluate the length of $\overarc{AB}$ .

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The radius of the circle is given to be . The total circumference of the circle is $2 \pi$ times this, or

$C = 2 \pi r = 2 \pi (6)= 12\pi$ .

The length of $\overarc{AB}$ is equal to

$l = $\frac{m \overarc {AB}$}{360} \cdot C$

Thus, it is first necessary to find the degree measure of $\overarc{AB}$ . If the sides of an angle with vertex outside the circle are both tangent to the circle, the angle formed is half the difference of the measures of the arcs. Therefore,

$$\frac{m \overarc{ACB}$ - m \overarc{AB} }{2} = m \angle AVB$

Letting $x= m \overarc{AB}$ , since the total arc measure of a circle is 360 degrees,

$m \overarc{ACB} = 360 - x$

We are also given that

$m \angle AVB = 45$

Making substitutions, and solving for :

$$\frac{(360- x )- x }{2}$ =45$

$$\frac{360- 2 x }{2}$ =45$

Multiply both sides by 2:

$$\frac{360- 2 x }{2}$ \cdot 2 =45 \cdot 2$

Subtract 360 from both sides:

Divide both sides by :

$-2x \div (-2) = -270 \div (-2)$

,

the degree measure of $\overarc{AB}$ .

Thus, the length of $\overarc{AB}$ is

$l = $\frac{m \overarc {AB}$}{360} \cdot C$

$= $\frac{225}{360}$ \cdot (12 \pi)$

$= $\frac{3}{8}$ (12\pi)$

$= $\frac{36}{8}$ \pi$

$= $\frac{9}{2}$ \pi$

is the center of the above circle, and . Evaluate the area of the shaded sector.

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The radius of the circle is given to be . The total area of the circle can be found using the area formula:

$A = \pi $r^{2}$$

$A = \pi $(8^{2}$)$

$A =64 \pi$

The area of the sector is equal to

$a = $\frac{m \overarc {AB}$}{360} \cdot A$

Thus, it is first necessary to find the degree measure of $\overarc{AB}$ . If the sides of an angle with vertex outside the circle are both tangent to the circle, the angle formed is half the difference of the measures of the arcs. Therefore,

$$\frac{m \overarc{ACB}$ - m \overarc{AB} }{2} = m \angle AVB$

Letting $x= m \overarc{AB}$ , since the total arc measure of a circle is 360 degrees,

$m \overarc{ACB} = 360 - x$

We are also given that

$m \angle AVB = 45$

Making substitutions, and solving for :

$$\frac{(360- x )- x }{2}$ =45$

$$\frac{360- 2 x }{2}$ =45$

Multiply both sides by 2:

$$\frac{360- 2 x }{2}$ \cdot 2 =45 \cdot 2$

Subtract 360 from both sides:

Divide both sides by :

$-2x \div (-2) = -270 \div (-2)$

,

the degree measure of $\overarc{AB}$ .

Thus, the area of the shaded sector is

$a = $\frac{m \overarc {AB}$}{360} \cdot A$

$= $\frac{135}{360}$ \cdot (64 \pi)$

$= $\frac{3}{8}$ \cdot 64 \pi$

$= $\frac{192}{8}$ \pi$

$=24 \pi$

is the center of the above circle, and . Evaluate the area of the shaded sector.

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The radius of the circle is given to be . The total area of the circle can be found using the area formula:

$A = \pi $r^{2}$$

$A = \pi $(6^{2}$)$

$A =36\pi$

The area of the sector is equal to

$a = $\frac{m \overarc {ACB}$}{360} \cdot A$

Thus, it is first necessary to find the degree measure of $\overarc{ACB}$ . If the sides of an angle with vertex outside the circle are both tangent to the circle, the angle formed is half the difference of the measures of the arcs. Therefore,

$$\frac{m \overarc{ACB}$ - m \overarc{AB} }{2} = m \angle AVB$

Letting $x= m \overarc{ACB}$ , since the total arc measure of a circle is 360 degrees,

$m \overarc{AB} = 360 - x$

We are also given that

$m \angle AVB = 45$

Making substitutions, and solving for :

$$\frac{x- (360- x ) }{2}$ =45$

$$\frac{ 2 x -360 }{2}$ =45$

Multiply both sides by 2:

$$\frac{2 x -360}{2}$ \cdot 2 =45 \cdot 2$

Add 360 to both sides:

Divide both sides by 2:

$2 x \div 2 = 450 \div 2$

,

the degree measure of $\overarc{ACB}$ .

Therefore, the area of the shaded sector is

$a = $\frac{m \overarc {ACB}$}{360} \cdot A$

$= $\frac{225}{360}$ \cdot 36 \pi$

$= $\frac{5}{8 }$ \cdot 36 \pi$

$= $\frac{180}{8 }$ \pi$

$= $\frac{45}{2}$ \pi$

is the center of the above circle, and . Evaluate the length of $\overarc{ACB}$ .

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The radius of the circle is given to be .

The total circumference of the circle is $2 \pi$ times this, or

$C = 2 \pi r = 2 \pi (8)= 16\pi$ .

The length of $\overarc{ACB}$ is equal to

$l= $\frac{m \overarc {ACB}$}{360} \cdot C$

Thus, it is first necessary to find the degree measure of $\overarc{ACB}$ . If the sides of an angle with vertex outside the circle are both tangent to the circle, the angle formed is half the difference of the measures of the arcs. Therefore,

$$\frac{m \overarc{ACB}$ - m \overarc{AB} }{2} = m \angle AVB$

Letting $x= m \overarc{ACB}$ , since the total arc measure of a circle is 360 degrees,

$m \overarc{AB} = 360 - x$

We are also given that

$m \angle AVB = 45$

Making substitutions, and solving for :

$$\frac{x- (360- x ) }{2}$ =45$

$$\frac{ 2 x -360 }{2}$ =45$

Multiply both sides by 2:

$$\frac{2 x -360}{2}$ \cdot 2 =45 \cdot 2$

Add 360 to both sides:

Divide both sides by 2:

$2 x \div 2 = 450 \div 2$

,

the degree measure of $\overarc{ACB}$ .

Thus, the length of $\overarc{ACB}$ is

$l = $\frac{m \overarc {ACB}$}{360} \cdot C$

$= $\frac{225}{360}$ \cdot 16 \pi$

$= $\frac{5}{8}$ \cdot 16 \pi$

$= $\frac{80}{8}$ \pi$

$= 10 \pi$

Refer to the above figure. Give the ratio of the area of Sector 2 to that of Sector 1.

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The ratio of the area of the larger Sector 2 to that of smaller Sector 1 is equal to the ratio of their respective arc measures - that is,

$m \overarc{ACB} : m \overarc{AD}$ .

Therefore, it is sufficient to find these arc measures.

If the sides of an angle with vertex outside the circle are both tangent to the circle, the angle formed is half the difference of the measures of the arcs. Therefore,

$$\frac{m \overarc{ACB}$ - m \overarc{AB} }{2} = m \angle AVB$

Letting $x= m \overarc{ACB}$ , since the total arc measure of a circle is 360 degrees,

$m \overarc{AB} = 360 - x$

We are also given that

$m \angle AVB = 45$

Making substitutions, and solving for :

$$\frac{x- (360- x ) }{2}$ =45$

$$\frac{ 2 x -360 }{2}$ =45$

Multiply both sides by 2:

$$\frac{2 x -360}{2}$ \cdot 2 =45 \cdot 2$

Add 360 to both sides:

Divide both sides by 2:

$2 x \div 2 = 450 \div 2$

,

the degree measure of $\overarc{ACB}$ .

It follows that

$m \overarc{AB} = 360 - 225 = 135$

By the Arc Addition Principle,

$m \overarc{AB} = m \overarc{AD} + m \overarc{DB}$

Since $\angle DOB$ , the central angle which intercepts $\overarc{DB}$ , is a right angle, $m \overarc{DB} = 90$ . By substitution,

$135 = m \overarc{AD} + 90$ ,

and

$135 - 90 = m \overarc{AD} + 90 - 90$

$45= m \overarc{AD}$

The ratio $m \overarc{ACB} : m \overarc{AD}$ is equal to

$\frac{225}{45}$ = $\frac{225 \div 45}{45 \div 45 }$ = $\frac{5}{1}$ ,

a 5 to 1 ratio.

Find the area of a square with the following side length:

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We can find the area of a circle using the following formula:

In this equation the variable, , represents the length of a single side.

Substitute and solve.

The perimeter of a square is . In terms of , give the area of the square.

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Since a square comprises four segments of the same length, the length of one side is equal to one fourth of the perimeter of the square, which is $$\frac{1}{4}$P$ . The area of the square is equal to the square of this sidelength, or

$\left ($\frac{1}{4}$P \right $)^{2}$ = \left ($\frac{1}{4}$ \right $)^{2}$ $P^{2}$ = $$\frac{1}{16}$P^{2}$$ .

The volume of a sphere is equal to $108 \pi$ . Give the surface area of the sphere.

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The volume of a sphere can be calculated using the formula

$V= $\frac{4}{3}$ \pi $r^{3}$$

Solving for :

Set $V = 108 \pi$ . Multiply both sides by $\frac{3}{4}$ :

$$\frac{3}{4}$ \cdot $\frac{4}{3}$ \pi $r^{3}$ = $\frac{3}{4}$ \cdot 108 \pi$

$\pi $r^{3}$ = 81\pi$

Divide by $\pi$ :

$\frac{\pi $r^{3}$$ }{\pi}= $\frac{81\pi}{\pi}$

Take the cube root of both sides:

$r = \sqrt[3]{81} = \sqrt[3]{27(3)} = \sqrt[3]{27} \cdot \sqrt[3]{3} = 3 \sqrt[3]{3}$

Now substitute for in the surface area formula:

$A = 4 \pi $r^{2}$$

$A = 4 \pi (3 $\sqrt[3]{3})^{2}$= 4 \pi (3 $)^{2}$( $\sqrt[3]{3})^{2}$ = 4 \pi (9) ( $\sqrt[3]{3^{2}$}) = 36 \pi \sqrt[3]{9}$ ,

the correct response.