Question:
What is m $\overset{\frown}{FG}$ ?

m $\overset{\frown}{FG}$ = $\Box$ °

Answer: m \$\overset{\frown}{FG}\$ =90°

1. Given points E, F, G on the circumference of a circle, let the center of the circle be O, and these three points are arranged in a clockwise direction. Therefore, the sum of the central angles corresponding to the circumferential angles is 360°.

2. According to the angle properties of the circle, arc EF, arc FG, and arc EG cover the entire circumference, so arc EF + arc FG + arc EG = 360°.

3. The problem states that arc EF = 140° and arc EG = 130°.

4. Substitute the known arc measures into the formula: 140° + arc FG + 130° = 360°.

5. Combine the known angles of arc EF and arc EG, yielding 140° + 130° = 270°.

6. Substitute 270° into the equation to get 270° + arc FG = 360°.

7. Rearrange the equation to find arc FG: arc FG = 360° - 270° = 90°.

8. Through the above reasoning, the final answer is 90°.

Definition of Central Angle

In the diagram of this problem, in the circle <note>O</note>, point <note>E</note> and point <note>G</note> are two points on the circle, and the center of the circle is <note>point O</note>. The angle <note>∠EOG</note> formed by the lines <note>OE</note> and <note>OG</note> is called the central angle. Point <note>E</note> and point <note>F</note> are two points on the circle, and the center of the circle is <note>point O</note>. The angle <note>∠EOF</note> formed by the lines <note>OE</note> and <note>OF</note> is called the central angle. Point <note>F</note> and point <note>G</note> are two points on the circle, and the center of the circle is <note>point O</note>. The angle <note>∠FOG</note> formed by the lines <note>OF</note> and <note>OG</note> is called the central angle.

Definition of Arc

<note>There are three points E, F, and G on the circumference</note>, arc EF is <note>a segment of a curve connecting point E and point F</note>, arc FG is <note>a segment of a curve connecting point F and point G</note>, arc EG is <note>a segment of a curve connecting point E and point G</note>.

Angle Properties of a Circle

Arc EF, arc FG, arc GE make a complete circle, the sum of their arc lengths is 360 degrees, and the sum of their corresponding central angles is also 360 degrees, that is, angle EOF + angle FOG + angle GOE = 360 degrees.


Question:
What is m $\overset{\frown}{RS}$ ?

m $\overset{\frown}{RS}$ = $\Box$ °

Answer: m \$\overset{\frown}{RS}\$ =120°

1. From the figure, it can be seen that the center of the circle is connected to points R, Q, and S, dividing the circumference into three arcs, namely \\overset{\\frown}{RS}, \\overset{\\frown}{QR}, and \\overset{\\frown}{QS}.

2. According to the angle properties of the circle (i.e., in a circle, the sum of the angles of the three sectors formed from the center is 360°), the following equation can be obtained: m \\overset{\\frown}{RS} + m \\overset{\\frown}{QR} + m \\overset{\\frown}{QS} = 360°.

3. Given m \\overset{\\frown}{QR} = 110°, m \\overset{\\frown}{QS} = 130°, substituting them into the above equation, we get: m \\overset{\\frown}{RS} + 110° + 130° = 360°.

4. By calculation, we get: m \\overset{\\frown}{RS} + 240° = 360°.

5. By subtracting 240° from both sides of the equation, we get: m \\overset{\\frown}{RS} = 120°.

6. Through the above reasoning, the final answer is 120°.

The total measure of the angles in a complete circular arc is 360 degrees, and the sum of the central angles subtended by the entire circumference of a circle is 360 degrees.

<note>In the figure of this problem, let the center of the circle be O</note>, the arcs RQ, QS, and RS make a full circle, with a total radian measure of 360 degrees, and their corresponding <note>central angles also sum to 360 degrees</note>, that is, <note>angle ROS + angle SOQ + angle ROQ = 360 degrees</note>, thus <note>angle ROQ = 110 degrees</note>.


Question:
What is the area of the trapezoid?

 $\Box$ square miles

Answer: 30square miles

1. Given a trapezoid with two bases, the upper base b1=3 miles and the lower base b2=9 miles. The height of the trapezoid h=5 miles. We need to calculate the area of the trapezoid.

2. The formula for the area of a trapezoid is A=1/2 * (b1+b2) * h, where b1 and b2 are the two parallel sides of the trapezoid, and h is the distance between them.

3. Substitute the given b1, b2, and h into the area formula to get: A=1/2 * (3+9) * 5.

4. Calculate the sum inside the parentheses: 3+9=12.

5. Continue calculating the area: A=1/2 * 12 * 5.

6. Multiply 12 by 5: 12 * 5=60.

7. Finally, multiply 60 by 1/2 to get: A=1/2 * 60=30.

8. Through the above reasoning, the final answer is 30 square miles.

Definition of Trapezoid

A quadrilateral is a trapezoid if and only if it has exactly one pair of parallel sides.

<note>The sides 3 mi and 9 mi are parallel</note>, while the other two sides are not parallel. Therefore, according to the definition of a trapezoid, this quadrilateral is a trapezoid because it <note>has exactly one pair of parallel sides</note>.

Trapezoid Area Formula

Given a trapezoid with the lengths of its two parallel sides as \(a\) and \(b\), and the height between these sides as \(h\), the area of the trapezoid is given by \(\frac{(a + b) \cdot h}{2}\).

<note>Side 3 mi and side 9 mi are two parallel sides</note>, with lengths a and b respectively, and the height between them is h. Therefore, <note>the area of the trapezoid is (a + b) * h / 2</note>.


Question:
Find the area of △UVW.

Write your answer as an integer or as a decimal rounded to the nearest tenth.
 $\Box$  m ^ 2

Answer: 8.0 m ^ 2

1. Given in triangle △UVW, ∠U = 136°, ∠W = 23°, UV = 5 m.

2. According to the triangle angle sum theorem, the sum of the interior angles of a triangle is 180°, i.e., ∠U + ∠V + ∠W = 180°. Substituting the given angles, we get 136° + ∠V + 23° = 180°.

3. Simplifying the equation, we get ∠V = 180° - 159°, i.e., ∠V = 21°.

4. According to the sine rule, in any triangle, the ratio of the length of a side to the sine of its opposite angle is equal, i.e., a/sin(A) = b/sin(B) = c/sin(C).

5. Applying the sine rule, using the known ∠W and the length of its opposite side UV, we find the length of the side VW opposite ∠U. Let VW = u, then 5/sin(23°) = u/sin(136°).

8. Now, knowing VW = 8.8892 m, UV = 5 m, and the included angle ∠V = 21°, we can use the triangle area formula (using the sine function).

9. The area formula for triangle △UVW is Area = 1/2 * UV * VW * sin(∠V). Substituting the values, we get Area = 1/2 * 8.8892 * 5 * sin(21°).

11. The final calculated result should be rounded to one decimal place, i.e., the area of the triangle is 8.0 m².

12. Through the above reasoning, the final answer is 8.0 m².

Definition of Triangle

A geometric figure is called a triangle if it is formed by three non-collinear points and the line segments connecting them.

<note>The triangle UVW is a geometric figure composed of three non-collinear points U, V, W and their connecting line segments UV, VW, WU</note>.<note>Points U, V, W are the three vertices of the triangle</note>,<note>Line segments UV, VW, WU are the three sides of the triangle</note>.

Definition of an Angle

An angle is a geometric figure formed by two rays sharing a common endpoint. This common endpoint is called the vertex of the angle, and the two rays are referred to as the sides of the angle.

In the figure of this problem, <note>angle ∠UVW is a geometric figure formed by two rays UV and VW</note>, these two rays <note>share a common endpoint V</note>. This <note>common endpoint V is called the vertex of angle ∠UVW</note>, and <note>rays UV and VW are called the sides of angle ∠UVW</note>.

Triangle Angle Sum Theorem

The sum of the interior angles of any triangle is 180°.

<note>In triangle UVW, angles ∠U, ∠V, and ∠W are the three interior angles of triangle UVW</note>, according to the Triangle Angle Sum Theorem, <note>∠U + ∠V + ∠W = 180°</note>.

Sine Theorem

In any triangle, the ratio of the length of each side to the sine of its opposite angle is constant and equal to the diameter of the circumcircle, i.e., \( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2r = D \) (where \( r \) is the radius of the circumcircle, and \( D \) is its diameter).

In the <note>triangle △UVW</note>, the <note>sides UV, VW, and UW correspond to angles ∠W, ∠U, and ∠V</note> respectively. According to the Sine Theorem, the ratio of the lengths of each side to the sine of its opposite angle is equal and also equal to the diameter of the circumscribed circle, i.e., <note>UV/sin(∠W) = VW/sin(∠U) = UW/sin(∠V) = 2r = D</note> (where r is the radius of the circumscribed circle and D is the diameter). In this problem, to calculate the <note>length of side VW in triangle △UVW</note>, let <note>VW = u</note>. Given <note>angle ∠W = 23°</note> and <note>corresponding side UV = 5 m</note>, using the Sine Theorem, we have: <note>5/sin(23°) = u/sin(136°)</note>.

Triangle Area Formula (Using Sine Function)

The area \( S \) of any triangle can be expressed as \( S = \frac{1}{2} \cdot a \cdot b \cdot \sin(C) \), where \( a \) and \( b \) are the lengths of two sides, and \( C \) is the angle between these two sides.

In the diagram of this problem, in triangle UVW, side <note>UV</note> and side <note>VW</note> are <note>a and b</note> respectively, and <note>angle UVW</note> is the <note>included angle C</note> between these two sides. According to the triangle area formula, the area <note>S</note> of triangle UVW can be expressed as <note>S = (1/2) * a * b * sin(C)</note>, that is, <note>S = (1/2) * UV * VW * sin(UVW)</note>.


Question:
Triangle PQR is inscribed in circle C.

 Select all of the statements that must be true.

- The bisectors of  $\angle $ P,  $\angle $ Q, and  $\angle $ R intersect at C.
- Triangle CMR is a right triangle.
- Triangle QCR is an isosceles triangle.
- The perpendicular bisectors of $\overline{PQ}$ , $\overline{QR}$ , and $\overline{RP}$ intersect at C.

Answer:
Triangle QCR is an isosceles triangle.
 The perpendicular bisectors of \$\overline{PQ}\$ , \$\overline{QR}\$ , and \$\overline{RP}\$ intersect at C.

1. Based on the given information, triangle PQR is inscribed in circle C, and point C is the center of the circle.

2. Since point C is the center of the circle, segments QC and RC are both radii of the circle and have equal lengths.

3. From QC = RC, it follows that triangle QCR is an isosceles triangle.

4. Regarding the perpendicular bisectors of triangle PQR: according to the 'circumcenter description in the perpendicular bisector theorem,' the circumcenter of a triangle is the intersection point of the perpendicular bisectors of its sides.

5. Because the problem states that circle C is the circumcircle of triangle PQR, the center C is the intersection point of the perpendicular bisectors of the three sides of triangle PQR, i.e., the circumcenter of triangle PQR.

6. Based on the above reasoning, the statements 'triangle QCR is an isosceles triangle' and 'the perpendicular bisectors of segments PQ, QR, RP intersect at C' must be true.

7. The statement 'the angle bisectors of angle P, angle Q, and angle R intersect at C' represents that these angle bisectors intersect at the incenter of the triangle. Since no additional information is provided to indicate that the center C is also the incenter of triangle PQR, this statement may not necessarily be true.

8. The statement 'triangle CMR is a right triangle' requires proving that angle M is a right angle. Due to the lack of further information about the specific location of point M or the characteristics of triangle CMR, it is impossible to determine whether this statement is true.

9. In summary, the statements that can be confirmed as true are 'triangle QCR is an isosceles triangle' and 'the perpendicular bisectors of PQ, QR, RP intersect at C.'

Definition of Isosceles Triangle

A triangle is an isosceles triangle if and only if it has at least two congruent (≅) sides.

<note>Side QC and side RC are equal</note>, therefore <note>triangle QCR is an isosceles triangle</note>.


Question:
Find the area of the shaded region.

Round your answer to the nearest tenth.
 $\Box$  square meters

Answer: 10.3 square meters

1. Given that the center of the circle is Y, the radius YS = YT = 6 meters, and the central angle ∠SYT = 90°.

2. According to the properties of the central angle, the degree of the arc of sector SYT is 90°, and the area of the entire circle is π multiplied by the square of the radius, so the area of the entire circle is π×6^2 = 36π square meters.

3. By calculating the area of the sector portion, we can use the formula for the area of a sector: Sector area = (Central angle degree / 360°)×Entire circle area, resulting in the area of sector SYT being (90°/360°)×36π = 9π square meters.

4. Since ∠SYT is a right triangle, to calculate the area of △SYT, we need to use the formula for the area of a triangle: Area = 1/2×base×height, resulting in the area of △SYT being 1/2×YS×YT = 1/2×6×6 = 18 square meters.

5. To find the area of the shaded portion, we need to subtract the area of △SYT from the area of the sector, so the area of the shaded portion = Area of sector SYT - Area of △SYT = 9π - 18 square meters.

6. Calculating the result 9π - 18 as a decimal approximation, the area of the shaded portion is approximately 10.27433.

7. Rounding the result to one decimal place, the final answer is 10.3 square meters.

Formula for the Area of a Sector

The area \( X \) of a sector can be calculated using the formula \( X = \frac{\theta}{360} \times \pi \times r^2 \), where \( \theta \) is the measure of the central angle in degrees, and \( r \) is the radius length.

In the diagram of this problem, <note>sector SYT</note>, <note>the central angle of sector SYT is 90°</note>, <note>the length of radius SY is 6</note>. According to the <note>formula for the area of a sector</note>, the area A of the sector can be calculated using the formula A = (θ/360) * π * r², where θ is the central angle in degrees and r is the radius length. Therefore, <note>the area of sector SYT A = (90/360) * π * 6²</note>.

Area Formula of a Circle

The area of a circle is given by the formula A = πr², where A represents the area, π (pi) is the mathematical constant approximately equal to 3.14159, and r is the radius of the circle.

<note>In circle Y, the radius SY represents the radius of the circle</note>, and the area A represents the area of the circle. According to the area formula of a circle, the area of the circle A = πr^2, where π is the ratio of the circumference to the diameter. Therefore, <note>the area of circle Y can be calculated using the radius SY, that is, A = π(SY)^2</note>.

Area of Right Triangle

The area of a right triangle is equal to half the product of the two legs that form the right angle, i.e., Area = 1/2 * base * height.

In the figure of this problem, in the right triangle <note>SYT</note>, <note>angle SYT is a right angle (90 degrees)</note>, sides <note>SY</note> and <note>TY</note> are the legs, one leg as the base and the other leg as the height, so the area of the right triangle is equal to half the product of these two legs, i.e., Area = 1/2 * side <note>SY</note> * side <note>TY</note>.

Property of Central Angle

The degree measure of an arc is equal to the degree measure of the central angle that subtends the arc.

The central angle corresponding to <note>arc ST</note> is <note>angle SYT</note>, and <note>arc ST degrees = angle SYT degrees</note>.


Question:
Quadrilateral ABCD is inscribed in circle E. $\overline{AC}$ and $\overline{BD}$ are diameters of the circle.

 Select all of the statements that must be true.

-  $\angle $ C and  $\angle $ A are supplementary angles.
-  $\angle $ C and  $\angle $ D are supplementary angles.
- △ABD is a right triangle.
- △AED is a right triangle.

Answer:
-  \$\angle \$ C and  \$\angle \$ A are supplementary angles.
-  \$\angle \$ C and  \$\angle \$ D are supplementary angles.
- △ABD is a right triangle.
- △AED is a right triangle.

1. Given that quadrilateral ABCD is inscribed in circle E, and AC and BD are the diameters of the circle. In a cyclic quadrilateral, the sum of opposite angles is 180° {(Inscribed Angle Theorem Corollary 3) Cyclic Quadrilateral Opposite Angles Theorem}, so angles CDA and ABC are supplementary angles, and angles DAB and BCD are supplementary angles.

2. Since BD is a diameter, the measure of arc DAB is 180°, according to the Inscribed Angle Theorem, the corresponding inscribed angle BCD is 90°, i.e., angle C is 90°. Similarly, since AC is a diameter, the measure of arc ABC is also 180°, the corresponding inscribed angle D is 90°, thus angle ADC is 90°.

3. Because angles C and D are both 90°, their sum is 180°, therefore angles C and D are supplementary angles.

4. In △ABD, BD is a diameter, according to (Inscribed Angle Theorem Corollary 2) the inscribed angle opposite the diameter is a right angle, angle DAB is a right angle. Therefore △ABD is a right triangle.

5. In △AED, angle BEC is a central angle corresponding to arc BC, and the angle DEC opposite arc DC is vertical, therefore according to the definition of vertical angles, angle AED is a right angle, thus △AED is a right triangle.

6. In summary, all options are true: angles C and A are supplementary angles; angles C and D are supplementary angles; △ABD is a right triangle; △AED is a right triangle.

Cyclic Quadrilateral

A quadrilateral is called a cyclic quadrilateral if and only if all four vertices lie on the circumference of a single circle. This circle is referred to as the circumcircle of the quadrilateral.

In the figure of this problem, the quadrilateral <note>ABCD's four vertices A, B, C, and D are all on the same circle E</note>. This <note>circle is called the circumcircle of the quadrilateral ABCD</note>. Therefore, <note>quadrilateral ABCD is a cyclic quadrilateral</note>. According to the properties of cyclic quadrilaterals, it can be concluded that <note>the sum of opposite angles is equal to 180 degrees</note>, i.e., <note>angle A + angle C = 180 degrees</note>, <note>angle B + angle D = 180 degrees</note>.

Definition of Diameter

A diameter is a line segment that passes through the center of a circle and has its endpoints on the circle. It is the longest chord of the circle, with a length equal to twice the radius of the circle.

<note>Line segment AC</note> and <note>line segment BD</note> both pass through <note>the center E of the circle</note> and their endpoints are on <note>the circumference of circle E</note>, therefore <note>AC</note> and <note>BD</note> are <note>the diameters of circle E</note>.

Corollary 3 of the Inscribed Angle Theorem: Diagonal Supplementary Theorem for Cyclic Quadrilateral

In a cyclic quadrilateral (a quadrilateral inscribed in a circle), the sum of each pair of opposite angles is equal to 180°. Additionally, any exterior angle is equal to its interior opposite angle.

<note>In the figure of this problem, quadrilateral ABCD is a cyclic quadrilateral, with vertices A, B, C, and D all on the circle</note>. According to the Diagonal Supplementary Theorem for Cyclic Quadrilateral, <note>the sum of each pair of opposite angles of quadrilateral ABCD is 180°</note>. Specifically, <note>∠A + ∠C = 180°</note>; <note>∠B + ∠D = 180°</note>.

Corollary to the Inscribed Angle Theorem 2: The Angle Subtended by the Diameter

The angle subtended by the diameter of a semicircle is a right angle; conversely, if an inscribed angle is 90°, then the chord subtending that angle is a diameter.

In the figure of this problem, <note>the angle subtended by the diameter AC in circle E, ∠ADC, is a right angle (90 degrees)</note>, <note>the angle subtended by the diameter BD, ∠BCD, is a right angle (90 degrees)</note>.

Definition of Right Triangle

A triangle is classified as a right triangle if it has one interior angle measuring exactly 90 degrees.

<note>In the figure of this problem, in triangle ABD, angle DAB is a right angle (90 degrees)</note>, therefore triangle ABD is a right triangle. <note>Side AD and side AB are the legs, side DB is the hypotenuse</note>. <note>In triangle AED, angle AED is a right angle (90 degrees)</note>, therefore triangle AED is a right triangle. <note>Side AE and side ED are the legs, side AD is the hypotenuse</note>.


Question:
Allie draws the quadrilateral shown below.

 Find the value of m that would allow Allie to circumscribe a circle about the quadrilateral.
m= $\Box$

Answer: m=-23

1. The problem requires Allie to draw a quadrilateral that can be circumscribed by a circle; in such a case, the quadrilateral satisfies the definition of a cyclic quadrilateral, i.e., the sum of the opposite angles of the quadrilateral equals 180 degrees.

2. It is known that two opposite angles of the quadrilateral are (−6m−90)° and (−5m+17)°.

3. According to the definition of a cyclic quadrilateral: the sum of angles A and C should be equal to 180°, therefore let ∠A represent (−6m−90)° and ∠C represent (−5m+17)°, we have: ∠A + ∠C = 180°.

4. Substitute the angle expressions into the equation, we get: (−6m−90) + (−5m+17) = 180.

5. Simplify the equation: (−6m−90) + (−5m+17) = -11m−73 = 180.

6. By solving the equation step by step, we get: −11m = 180 + 73, thus -11m = 253.

7. Solve for m: m = −253/11;

8. Through the above reasoning, the final answer is m = -23.

Cyclic Quadrilateral

<note>The four vertices of the quadrilateral are on the same circle</note>. This circle is called the circumcircle of the quadrilateral. Therefore, the quadrilateral is a cyclic quadrilateral. According to the properties of cyclic quadrilaterals, it can be concluded that <note>the sum of opposite angles is equal to 180 degrees, that is, angle A + angle C = 180 degrees</note>.

<note>In the figure of this problem, the vertices of the quadrilateral are all on the circle</note>. According to the <note>Diagonal Supplementary Theorem for Cyclic Quadrilateral</note>, the sum of <note>each pair of opposite angles of the quadrilateral is equal to 180°</note>.


Question:
In circle G,  $\angle $ RGS measures t° and the length of radius $\overline{RG}$ is b.

If circle G were dilated about its center by a scale factor of 2.5, what would the new length of $\overset{\frown}{RS}$ be?

- 𝜋bt/144
- 𝜋b ^ 2t/72
- 𝜋b ^ 2t/144
- 𝜋bt/72

Answer: - 𝜋bt/72

1. First, we need to clarify the properties of circle G: the center is G, and the radius length is b.

2. The problem states that ∠RGS measures t°, which means that the arc RS corresponds to a central angle of t° in circle G.

3. According to the problem, circle G is enlarged with its center as the center of scaling, and the scaling factor is 2.5. Thus, the new circle, denoted as G', has a radius R'G' with a length of 2.5b.

4. Since the enlargement is centered at G, the central angle ∠R'G'S' remains unchanged, still t°.

5. Now let's calculate the circumference of the enlarged circle G'. According to the circumference formula: C = 2πr, where r is the radius. Therefore, the circumference of G', C' = 2 × π × 2.5b = 5πb.

6. According to the definition of the arc length, the length of arc R'S' can be calculated using the formula: 𝓁 = m/360 × C, where m is the degree measure of the central angle, and C is the circumference of the circle.

7. For circle G', the length of arc R'S' is: 𝓁 = t/360 × 5πb.

8. Simplifying the above expression: 𝓁 = (5πbt)/360.

9. Further simplification gives: 𝓁 = πbt/72.

10. Through the above reasoning, the final answer is πbt/72.

Definition of Circle

A circle is the locus of all points in a plane that are at a fixed distance from a given point.

<note>In circle G, point G is the center</note>, <note>the radius is b</note>. All points in the figure that are <note>at a distance of b from point G are on circle G</note>. In circle G', <note>point G' is the center</note>, <note>the radius is b</note>. All points in the figure that are <note>at a distance of b from point G' are on circle G'</note>.

Definition of Arc

An arc is a curve on a circle that is defined by two distinct points on the circle.

In the figure of this problem, there are two points <note>R</note> and <note>S</note> on the circle <note>G</note>, and the <note>arc RS</note> is a segment of the curve connecting these two points. According to the definition of an arc, the <note>arc RS</note> is a segment of the curve between the two points <note>R</note> and <note>S</note> on the circle. There are two points <note>R'</note> and <note>S'</note> on the circle <note>G'</note>, and the <note>arc R'S'</note> is a segment of the curve connecting these two points. According to the definition of an arc, the <note>arc R'S'</note> is a segment of the curve between the two points <note>R'</note> and <note>S'</note> on the circle.

Definition of Central Angle

An angle formed by two radii connecting two points on the circumference of a circle to the center of the circle is called a central angle.

In <note>circle G</note>, <note>point R and point S are two points on the circle</note>, <note>the center of the circle is point G</note>. The angle ∠RGS formed by the lines GR and GS is called the central angle, and its measure is <note>t°</note>.

Circumference Formula of Circle

The circumference of a circle is the length around the circular boundary. It can be calculated using the formula \( C = 2\pi r \), where \( C \) represents the circumference and \( r \) represents the radius.

<note>In circle G, point G is the center of the circle</note>, <note>line segment RG is the radius b</note>. According to the circumference formula of a circle, the circumference C of the circle is equal to <note>2π multiplied by the radius r</note>, that is, <note>C=2πr</note>. Therefore, <note>the radius of the enlarged circle G' is 2.5b</note>, <note>the circumference C' of circle G' is equal to 2π multiplied by the radius 2.5b</note>, that is, <note>C'=2π×2.5b=5πb</note>.

Arc Length Formula of a Circle

The arc length refers to the length of a segment of the circumference of a circle. It can be calculated using the formula 𝓁 = C × (m/360), where m is the measure of the central angle in degrees.

On <note>circle G</note>, the length of arc <note>R'S'</note> is <note>L = (t/360) × 5πb</note>, which simplifies to <note>L = πbt/72</note>.


Question:
Find the area of the shaded region.

 Round your answer to the nearest tenth.
 $\Box$  square centimeters

Answer: 4.6 square centimeters

1. Given that the radius of the circle is 4 cm and the central angle ∠ABC is 90°, calculating the area of sector ABC is the first step in finding the area of the shaded region.

2. The formula for the area of a sector is: Area = central angle in degrees/360° × area of the circle, where the area of the circle is πr².

3. Substitute the given data into the formula: Area of sector ABC = 90°/360° × π × 4² = 4π square centimeters.

4. Next, calculate the area of triangle ABC. Since triangle ABC is a right triangle and the two legs of the right angle are the radii of the circle, each with a length of 4 cm.

5. The formula for the area of a right triangle is: Area = 1/2 × base × height.

6. Substitute the given data into the formula: Area of triangle ABC = 1/2 × 4 × 4 = 8 square centimeters.

7. The area of the shaded region is equal to the area of the sector minus the area of the triangle.

8. Calculate the area of the shaded region: Area of the shaded region = 4π - 8 square centimeters.

9. For an approximate calculation, take π as 3.14159, then 4 × 3.14159 = 12.56636.

10. The area of the shaded region = 12.56636 - 8 = 4.56636 square centimeters.

11. Round the result to the nearest decimal place to get the area of the shaded region as approximately 4.6 square centimeters.

12. Through the above reasoning, the final answer is approximately 4.6 square centimeters.

Definition of Central Angle

<note>Point A and Point C are two points on the circle</note>, the center of the circle is <note>Point B</note>. The angle ∠ABC formed by the lines <note>BA and BC</note> is called the central angle, and <note>∠ABC = 90°</note>.

Definition of Sector

A sector is a figure formed by two radii of a circle and the arc enclosed between them.

<note>In sector ABC, radius AB and radius BC are two radii of the circle</note>, <note>arc AC is the arc enclosed by these two radii</note>, so according to the definition of a sector, the figure formed by these two radii and the enclosed <note>arc AC is a sector</note>.

Definition of Right Triangle

In the figure of this problem, in triangle ABC, <note>angle ∠ABC is a right angle (90 degrees)</note>, therefore <note>triangle ABC is a right triangle</note>. <note>Side AB and side BC are the legs</note>, <note>side AC is the hypotenuse</note>.

Area Formula of a Circle

<note>The radius of the circle is 4 centimeters</note>, according to the <note>area formula of a circle</note>, the area A of the circle is equal to pi multiplied by the <note>square of the radius 4</note>, that is, A = π4².

Formula for the Area of a Sector

In the figure of this problem, <note>in sector ABC, the central angle ABC measures θ degrees</note>, <note>the length of the radius AB is r</note>. According to the formula for the area of a sector, the area A of the sector can be calculated using the formula <note>A = (θ/360) * π * r²</note>, where <note>θ is the measure of the central angle, r is the length of the radius</note>. Therefore, <note>the area of sector ABC A = (θ/360) * π * r²</note>.

Area Formula of a Triangle

The area of any triangle is equal to its base multiplied by its height, divided by 2.

In the triangle ABC, <note>side AB is the base</note>, <note>line segment BC is the height on this base</note>, so the area of triangle ABC equals the base <note>AB multiplied by the height BC divided by 2</note>, that is, area = <note>(AB * BC) / 2</note>.


Question:
What is the area of the trapezoid?

 $\Box$ square centimeters

Answer: 54square centimeters

1. According to the problem statement, it is known that the upper base b1 of the trapezoid is 3 cm, the lower base b2 is 9 cm, and the height h is 9 cm.

2. The area formula of the trapezoid is: Area = 1/2 × (upper base + lower base) × height.

3. Substitute the known bases and height into the formula, we get Area = 1/2 × (3 + 9) × 9.

4. Calculate the sum inside the parentheses: 3 + 9 = 12.

5. Substitute the sum into the area formula, we get Area = 1/2 × 12 × 9.

6. Calculate 1/2 × 12 = 6.

7. Multiply the result from the previous step by the height, we get Area = 6 × 9.

8. Calculate the product: 6 × 9 = 54.

9. Based on the above steps, the final area of the trapezoid is 54 square cm.

Definition of Trapezoid

A quadrilateral is a trapezoid if and only if it has exactly one pair of parallel sides.

<note>In the figure of this problem, the upper base of the trapezoid is 3 cm</note>, <note>the lower base is 9 cm</note>, the height is 9 cm, and one side is 10 cm. According to the definition of a trapezoid, a trapezoid is a quadrilateral that has <note>one pair of parallel sides</note>.

Trapezoid Area Formula

Given a trapezoid with the lengths of its two parallel sides as \(a\) and \(b\), and the height between these sides as \(h\), the area of the trapezoid is given by \(\frac{(a + b) \cdot h}{2}\).

<note>The upper base of the trapezoid is 3 cm</note>, <note>the lower base is 9 cm</note>, <note>the height is 9 cm</note>, so the area of the trapezoid is (3 + 9) * 9 / 2 = 54 square cm.


Question:
The circumcenter of △ABC is partially constructed below.

Put the remaining steps of the construction in the correct order.
Mark the intersection of DE and FG with point H.Mark the intersections of circle A and the smaller circle B with points F and G.Draw a circle with radius AB centered at B and a circle with radius AB centered at A.Draw a line through F and G.

Answer: Draw a circle with radius AB centered at B and a circle with radius AB centered at A.Mark the intersections of circle A and the smaller circle B with points F and G.Draw a line through F and G.Mark the intersection of DE and FG with point H.

1. Given: ∠A, ∠B, and ∠C are the vertices of △ABC. It is required to construct the circumcenter H of △ABC. The circumcenter is the intersection point of the perpendicular bisectors of the sides of the triangle. The perpendicular bisector DE of side BC is already given in the figure.

2. Next, construct the perpendicular bisector of another side AB. First, according to the definition of a circle, draw a circle with center B and radius AB; then draw another circle with center A and the same radius AB.

3. According to the definition of the intersection points of circles, mark the intersection points of the above two circles as points F and G. This makes the segment FG the perpendicular bisector of AB.

4. According to the geometric principle that two points determine a unique straight line, draw the line passing through points F and G, i.e., the line FG.

5. According to geometric properties, the intersection point of the perpendicular bisectors DE and FG is the circumcenter H. Mark the intersection point of DE and FG as point H.

6. Through the above reasoning, the final answer is to mark the intersection point of DE and FG as point H.

Circumcenter of a Triangle

In a triangle, the circumcenter is the point where the three perpendicular bisectors of the sides intersect. This point is equidistant from the three vertices of the triangle.

<note>In the diagram of this problem, point H is the circumcenter of triangle ABC</note>. The perpendicular bisectors of the three sides AB, BC, and CA of triangle ABC intersect at <note>point H</note>. According to the definition of the circumcenter of a triangle, <note>the distances from point H to the three vertices A, B, and C of triangle ABC are equal</note>.

Definition of Circle

<note>In the figure of this problem, in circle B, point B is the center, and the radius is AB</note>; <note>in circle A, point A is the center, and the radius is AB</note>. <note>All points in the figure that are at a distance equal to AB from point B are on circle B</note>, <note>all points that are at a distance equal to AB from point A are on circle A</note>. <note>The intersection points of these two circles are points D and E</note>.

Properties of the Perpendicular Bisector

Any point on the perpendicular bisector of a line segment is equidistant from the endpoints of the segment.

In the figure of this problem, <note>the perpendicular bisector of segment BC is line DE</note>, <note>point H is on line DE</note>. According to the properties of the perpendicular bisector, <note>the distance from point H to the endpoints B and C of segment BC is equal, i.e., HB = HC</note>.


Question:
In circle G,  $\angle $ RGS measures t° and the length of radius $\overline{RG}$ is b.

If circle G were dilated about its center by a scale factor of 2.5, what would the new length of $\overset{\frown}{RS}$ be?

- 𝜋b ^ 2t/72
- 𝜋b ^ 2t/144
- 𝜋bt/144
- 𝜋bt/72

Answer: - 𝜋bt/72

1. First, according to the problem statement, the radius of circle G is \\( \\overline{RG} = b \\), and the central angle subtended by the arc is \\( \\angle RGS = t^\\circ \\).

2. Enlarge circle G by a factor of 2.5 centered at its center. According to the properties of similar figures, the radius \\( \\overline{RG'} = 2.5 \\times b \\).

3. Since dilation does not change the measure of angles, \\( \\angle R'G'S' = \\angle RGS = t^\\circ \\).

4. After enlargement, we obtain the new circumference \\( C' = 2 \\pi \\times 2.5b = 5\\pi b \\).

5. Using the formula for arc length on a circle, the formula for the arc length \\( \\overset{\\frown}{R'S'} \\) is: \\( \\frac{m}{360^\\circ} \\times C \\), where \\( m \\) is the measure of the central angle subtended by the arc.

6. Therefore, the arc length \\( \\overset{\\frown}{R'S'} = \\frac{t}{360} \\times 5\\pi b \\).

7. Further simplification gives \\( \\frac{t}{360} \\times 5\\pi b = \\frac{5t \\pi b}{360} = \\frac{\\pi bt}{72} \\).

8. Through the above reasoning, the final answer is \\( \\frac{\\pi bt}{72} \\).

In <note>circle G</note>, <note>point G is the center</note>, <note>the radius is b</note>. All points in the figure that are at a distance of b from <note>point G</note> are on <note>circle G</note>.

Circumference Formula of Circle

In the figure of this problem, <note>in circle G, point G is the center</note>, <note>line segment RG is the radius b</note>. According to the <note>circumference formula of circle</note>, the circumference C of the circle is equal to 2π multiplied by <note>radius b</note>, that is, C=2πb. After magnification, <note>the new circumference is C' = 2π × 2.5b = 5πb</note>.

Arc Length Formula of a Circle

The calculation formula for <note>arc length \( \overset{\frown}{RS} \)</note> is: <note>\( \frac{t}{360^\circ} \times 5\pi b \)</note>.

Definition of an Angle

An angle is a geometric figure formed by two rays sharing a common endpoint. This common endpoint is called the vertex of the angle, and the two rays are referred to as the sides of the angle.

In the diagram of this problem, <note>angle \( \angle RGS \)</note> is a geometric figure composed of two <note>rays \( \overline{RG} \) and \( \overline{GS} \)</note>, which share a common endpoint <note>G</note>. This common endpoint <note>G</note> is called the <note>vertex of angle \( \angle RGS \)</note>, and the <note>rays \( \overline{RG} \) and \( \overline{GS} \)</note> are called the <note>sides of angle \( \angle RGS \)</note>, with a measure of t°.

Definition of Central Angle

In the diagram of this problem, <note>circle G</note>, <note>points R and S are two points on the circle</note>, <note>the center of the circle is point G</note>. <note>The angle ∠RGS formed by the lines GR and GS is called the central angle</note>, and its measure is t°.

Similarity Theorem for Triangles (AA)

Two triangles are similar if two of their corresponding angles are congruent.

In the figure of this problem, the circle is enlarged by a factor of 2.5, and the radius of the new circle <note>\( \overline{RG'} \)</note> is proportional to the radius of the original circle <note>\( \overline{RG} \)</note>, with <note>\( \overline{RG'} = 2.5 \times b \)</note>. The corresponding angle <note>\( \angle RGS \)</note> remains unchanged at t°.

Arc Length Formula of a Circle

The arc length formula for <note>arc \( \overset{\frown}{RS} \)</note> is: <note>\( \frac{t}{360} \times 5\pi b \)</note>.


Question: AD is the bisector of  $\angle $ A, and CE is the bisector of  $\angle $ C. Construct the inscribed circle of △ABC.

1. Given that AD is the bisector of ∠BAC, and CE is the bisector of ∠ACB. According to the definition of angle bisectors, the intersection point of AD and CE is the incenter of triangle ABC. The intersection point F satisfies that the distances from AF, BF, and CF to the corresponding sides BC, AC, and AB are equal.

2. Since F is the intersection point of AD and CE, according to the definition of angle bisectors, F is the intersection point of the bisectors of ∠BAC and ∠ACB, thus it is the incenter of triangle ABC.

3. After determining point F, because the incenter F is equidistant to the three sides BC, AC, and AB of the triangle, a circle can be drawn that is tangent to all three sides, which is the incircle of triangle ABC.

4. To construct the incircle, a line passing through point F and perpendicular to one of the sides AC needs to be drawn. The intersection point of this line with AC is one of the tangency points, denoted as I.

5. After determining the tangency point I, draw a circle with center F and radius FI. This circle is the incircle of triangle ABC, as it passes through point I and is tangent to all three sides.

n. Through the above reasoning, it is concluded that this circle is the required incircle of triangle ABC.

Definition of Angle Bisector

An angle bisector is a line that originates from the vertex of an angle and divides the angle into two congruent angles.

In the figure of this problem, <note>the vertex of angle BAC is point A</note>, a line AD is drawn from <note>point A</note>, this line divides <note>angle BAC into two equal angles, that is, angle BAD and angle CAD are equal</note>. Therefore, <note>line AD is the angle bisector of angle BAC</note>. Similarly, <note>the vertex of angle ACB is point C</note>, a line CE is drawn from <note>point C</note>, this line divides <note>angle ACB into two equal angles, that is, angle BCE and angle ECA are equal</note>. Therefore, <note>line CE is the angle bisector of angle ACB</note>.

Incenter of a Triangle

The incenter of a triangle is the intersection point of the angle bisectors of the three interior angles of the triangle. It is also the center of the triangle's inscribed circle (incircle).

In the diagram of this problem, in triangle ABC, <note>point F is the incenter of the triangle</note>. The incenter of a triangle is formed by the <note>intersection of the three angle bisectors of the triangle</note>. Specifically, <note>the angle bisector of angle BAC is AD</note>, <note>the angle bisector of angle ACB is CE</note>, and <note>the angle bisector of angle ABC intersects at point F</note>. <note>Point F is the center of the triangle's inscribed circle</note>, and the distance to <note>the three sides BC, AC, and AB of the triangle is equal</note>.

<note>Point F is the incenter of triangle ABC</note>, draw a line through <note>point F perpendicular to one side AC of the triangle</note>, <note>the intersection point is I</note>, which is the tangent point on that side. <note>FI is the radius of the incircle</note>. Then, <note>with F as the center and FI as the radius, draw a circle</note>, this circle is the desired incircle.


Question:
$\overline{LK}$ is a diameter of circle Z.

 Find the area of the shaded region.
 Round your answer to the nearest tenth.
 $\Box$  square inches

Answer: 57.1 square inches

1. Determine the radius of the circle to be 10 meters. Therefore, the hypotenuse LK of triangle LJK is the sum of LJ and JK, both of which are the radius of the circle, with a length of 10 meters, making LK equal to 20.

2. The formula to calculate the area of circle Z is πr^2, where r is the radius of the circle. Since r = 10 meters, the area of the circle is 100π square meters.

3. Since LK is the diameter, the sector formed by arc LJK is a semicircle, with an area of 50π square meters.

4. The formula to calculate the area of triangle LJK is A = 1/2 · b · h, where b = LK = 20 meters, and h = ZJ = 10 meters. Therefore, the area of the triangle is 1/2 · 20 · 10 = 100 square meters.

5. The area of the shaded region is the area of the semicircular sector minus the area of triangle LJK, which is 50π - 100.

6. Calculating the above expression gives 50π - 100 ≈ 57.07963.

7. Rounding the result to one decimal place, the final area of the shaded region is 57.1 square meters.

Definition of Diameter

A diameter is a line segment that passes through the center of a circle and has its endpoints on the circle. It is the longest chord of the circle, with a length equal to twice the radius of the circle.

<note>LK is the diameter</note>, connecting <note>the center Z</note> and <note>points L and K on the circumference</note>, with a length of <note>2 times the radius</note>, that is, <note>LK = 20 meters</note>.

Definition of Radius

A radius is a line segment joining the center of a circle to any point on the circumference of the circle.

In the figure of this problem, in circle Z, <note>point Z is the center of the circle</note>, <note>points L, K, and J are any points on the circle</note>, <note>line segments LZ, KZ, and JZ are segments from the center Z to any point on the circle</note>, therefore <note>line segments LZ, KZ, and JZ are the radii of the circle</note>.

Definition of Sector

In the <note>sector LJK</note>, <note>radius LZ and radius KZ</note> are two radii of the circle, and <note>arc LJK</note> is the arc enclosed by these two radii. Therefore, according to the definition of a sector, the figure formed by these two radii and the enclosed <note>arc LJK</note> is a sector.

Area Formula of a Circle

The area of a circle is given by the formula A = πr², where A represents the area, π (pi) is the mathematical constant approximately equal to 3.14159, and r is the radius of the circle.

In the figure of this problem, <note>the radius of circle Z is 10 meters</note>, according to <note>the area formula of a circle</note>, <note>the area A of the circle is equal to pi multiplied by the square of the radius 10</note>, that is, A = π * 10² = 100π square meters.

Area Formula of a Triangle

In the triangle LJK, side LK is the base, and the line segment JZ is the height from this base, so the area of triangle LJK is equal to the base LK multiplied by the height JZ divided by 2, that is, area = (LK * JZ) / 2.


Question:
Quadrilateral ABCD is inscribed in circle E. $\overline{AC}$ and $\overline{BD}$ are diameters of the circle.

 Select all of the statements that must be true.

-  $\angle $ C and  $\angle $ A are supplementary angles.
- △ABD is a right triangle.
-  $\angle $ C and  $\angle $ D are supplementary angles.
- △AED is a right triangle.

Answer:
-  \$\angle \$ C and  \$\angle \$ A are supplementary angles.
- △ABD is a right triangle.
-  \$\angle \$ C and  \$\angle \$ D are supplementary angles.
- △AED is a right triangle.

1. Because quadrilateral ABCD is inscribed in circle E, it follows that angles ∠A and ∠C are opposite angles of the inscribed quadrilateral. According to the theorem of supplementary opposite angles in an inscribed quadrilateral, the sum of the opposite angles is 180°. Specifically, the formula is: ∠A + ∠C = 180°, therefore ∠A and ∠C are supplementary angles.

2. Because segments AC and BD are diameters, according to (corollary 2 of the inscribed angle theorem) the inscribed angle subtended by a diameter is a right angle. Each endpoint of the diameter forms a right angle at the circumference, thus the angles corresponding to the intersecting segments in the quadrilateral are right angles.

3. Because diameter BD passes through points A and D, in △ABD, angle ∠A is 90°. According to the property of complementary acute angles in a right triangle, it can be verified that ∠ADB and ∠ABD are acute angles, hence △ABD is a right triangle.

4. Since diameters AC and BD intersect, angles ∠ABC and ∠ADC are corresponding angles, which gives ∠B + ∠D = 180°, thus ∠D and ∠B are also supplementary angles.

5. For △AED, follow similar steps using the determination of right angles caused by the diameter. Segments AE and ED form right angles with diameter BD (according to corollary 2 of the inscribed angle theorem), thus ∠AED is 90°, hence triangle AED is a right triangle.

6. Through the above reasoning, the final answer is that all given options are valid.

In a cyclic quadrilateral (a quadrilateral inscribed in a circle), the sum of each pair of opposite angles is equal to 180°. Additionally, any exterior angle is equal to its interior opposite angle.

<note>In the figure of this problem, quadrilateral ABCD is inscribed in a circle, and the vertices A, B, C, D of the quadrilateral are all on the circle</note>. According to the Diagonal Supplementary Theorem for Cyclic Quadrilateral, <note>the sum of each pair of opposite angles of quadrilateral ABCD is equal to 180°</note>. Specifically, <note>∠A + ∠C = 180°</note>; <note>∠B + ∠D = 180°</note>.

Definition of Diameter

In the figure of this problem, <note>line segment AC and line segment BD are the diameters of circle E</note>, connecting <note>the center E</note> and <note>points A and C on the circumference</note> as well as <note>points B and D</note>, with a length of <note>twice the radius</note>, that is, <note>AC = BD</note>.

Complementary Acute Angles in a Right Triangle

In a right triangle, the sum of the two non-right angles is 90°.

<note>In right triangle ABD, angle DAB is a right angle (90 degrees)</note>, <note>angle ADB and angle ABD are the two acute angles other than the right angle</note>, according to the complementary acute angles property of right triangles, <note>the sum of angle ADB and angle ABD is 90 degrees, that is, angle ABD + angle ADB = 90°</note>.

The angle subtended by the diameter of a semicircle is a right angle; conversely, if an inscribed angle is 90°, then the chord subtending that angle is a diameter.

<note>In circle E, the angle subtended by the diameter BD at the circumference, angle DAB, is a right angle (90 degrees)</note>.<note>In circle E, the angle subtended by the diameter BD at the circumference, angle DCB, is a right angle (90 degrees)</note>.


Question:
Find the area of the shaded region.

Round your answer to the nearest tenth.
 $\Box$  square meters

Answer: 10.3 square meters

1. The figure provided in the problem is a circle with center at point Y. Points S and T are on the circle. It is known that the radius YS = YT = 6 meters, and in triangle SYT, angle SYT is a right angle (90°), thus triangle SYT is a right triangle.

2. The angle of arc ST used to define sector SYT is ∠SYT = 90°.

3. According to the formula for the area of a circle A = πr^2, the area of the entire circle is π * 6^2 = 36π square meters.

4. Using the formula for the area of a sector K = (m/360) * A, where m is the central angle. For sector SYT, m = 90°, so the area of sector SYT is (90/360) * 36π = 9π square meters.

5. Triangle SYT is a right triangle, and both legs (YS and YT) are 6 meters. Using the formula for the area of a triangle A = (YS * YT) / 2, the area of triangle SYT is (6 * 6) / 2 = 18 square meters.

6. The area of the shaded region is the area of sector SYT minus the area of triangle SYT: 9π - 18 square meters.

7. Substituting the approximate value of π (about 3.14159), 9π - 18 is approximately equal to 10.27433 square meters.

8. Rounding the result to the nearest tenth, the area of the shaded region is 10.3 square meters.

9. Through the above reasoning, the final answer is 10.3 square meters.

Definition of Circle

A circle is the locus of all points in a plane that are at a fixed distance from a given point.

In the figure of this problem, in the circle, <note>point Y is the center of the circle</note>, <note>the radius is 6 meters</note>. All <note>points that are 6 meters away from point Y are on the circle</note>.

Definition of Arc

An arc is a curve on a circle that is defined by two distinct points on the circle.

<note>There are two points S and T on the circle</note>, <note>arc ST is a segment of the curve connecting these two points</note>. According to the definition of arc, <note>arc ST is a segment of the curve between the two points S and T on the circle</note>.

Definition of Sector

A sector is a figure formed by two radii of a circle and the arc enclosed between them.

<note>In sector SYT, radius YS and radius YT are two radii of the circle, and arc ST is the arc enclosed by these two radii</note>, so according to the definition of a sector, the figure formed by these two radii and the enclosed <note>arc ST</note> is a sector.

Definition of Right Triangle

A triangle is classified as a right triangle if it has one interior angle measuring exactly 90 degrees.

In the figure of this problem, in triangle SYT, <note>angle SYT is a right angle (90 degrees)</note>, so triangle SYT is a right triangle. <note>Side YS and side YT are the legs</note>, <note>side ST is the hypotenuse</note>.

Area Formula of a Circle

In <note>Circle Y</note>, <note>the radius of the circle is 6 meters</note>, according to the <note>Area Formula of a Circle</note>, the area A of the circle is equal to pi π multiplied by <note>the square of radius 6</note>, that is, <note>A = π6²</note>.

Formula for the Area of a Sector

The area \( X \) of a sector can be calculated using the formula \( X = \frac{\theta}{360} \times \pi \times r^2 \), where \( \theta \) is the measure of the central angle in degrees, and \( r \) is the radius length.

<note>The central angle of sector SYT is m=90°</note>, the area of the entire circle A=36π square meters, therefore <note>the area of sector SYT is (90/360)* 36π=9π square meters</note>.

Area Formula of a Triangle

The area of any triangle is equal to its base multiplied by its height, divided by 2.

In the figure of this problem, in triangle SYT, <note>side YS is the base, and segment YT is the height on this base</note>, so <note>the area of triangle SYT is equal to the base YS multiplied by the height YT divided by 2</note>, that is, <note>area A = (YS * YT) / 2</note>.


Question:
Quadrilateral QUAD is inscribed in a circle.

 Select all of the statements that must be true.

- m $\overset{\frown}{QU}$ =m $\overset{\frown}{UA}$
-  $\angle $ Q and  $\angle $ D are complementary angles.
- m $\angle $ U=140°
- m $\overset{\frown}{QPD}$ =180°

Answer:
- m \$\overset{\frown}{QU}\$ =m \$\overset{\frown}{UA}\$
-  \$\angle \$ Q and  \$\angle \$ D are complementary angles.
- m \$\angle \$ U=140°

1. Given that quadrilateral QUAD is a cyclic quadrilateral, according to the corollary of the inscribed angle theorem (corollary 3), the opposite angles of a cyclic quadrilateral are supplementary, so ∠U and ∠D are supplementary.

2. According to the problem statement, check whether the measure of arc UA is equal to the measure of arc QU. Since ∠A is an inscribed angle on arc UQD, the measure of ∠A is 130°, then the measure of arc UQD is 2 × 130° = 260°.

3. Since arc UQ and arc QPD together make up arc UQD, the measure of arc QPD is 260° - 40° = 220°.

4. The total measure of a circle is 360°, and arcs QPD, AD, UA, and UQ make up the entire circle, so the measure of arc UA is 360° - 220° - 60° - 40° = 40°.

5. From the above calculations, the measures of arc UA and arc QU are both 40°, so arc QU = arc UA.

6. Analysis of the relationship between ∠Q and ∠D: ∠Q is an inscribed angle intercepting arc UD, and the total measure of arc UD is 40° + 60° = 100°, so the measure of ∠Q is 100° ÷ 2 = 50°.

7. ∠D is an inscribed angle intercepting arc QA, and the total measure of arc QA is 40° + 40° = 80°, so the measure of ∠D is 80° ÷ 2 = 40°.

8. Since ∠Q + ∠D = 50° + 40° = 90°, according to the definition of a right angle, ∠Q and ∠D are complementary angles, so the conclusion that they are supplementary angles is valid.

9. Check the measure of ∠U: ∠U is an inscribed angle intercepting arc QPA, and the measure of arc QPA is 220° + 60° = 280°, so the measure of ∠U is 280° ÷ 2 = 140°.

10. Check whether arc QPD in the problem is equal to 180°; from the previous calculations, the actual measure of arc QPD is 220°, not 180°.

11. Through the above reasoning, the final answer is: arc QU = arc UA, ∠Q and ∠D are supplementary angles, and ∠U = 140°. The above statements are correct, while arc QPD = 180° is incorrect.

Definition of Circle

<note>The circle is the blue circumference in the diagram</note>, the center is not marked, <note>the radius is the distance from the center to any point on the circumference</note>. <note>All points in the diagram that are equidistant from the center are on the circle</note>.

Definition of Arc

<note>In the figure of this problem, there are points Q, U, A, P, and D on the circle</note>. <note>Arc QU is a segment of the curve connecting points Q and U</note>, <note>Arc UA is a segment of the curve connecting points U and A</note>, <note>Arc UQD is a segment of the curve connecting points U and D</note>, <note>Arc QPD is a segment of the curve connecting points Q and D</note>.

Cyclic Quadrilateral

A quadrilateral is called a cyclic quadrilateral if and only if all four vertices lie on the circumference of a single circle. This circle is referred to as the circumcircle of the quadrilateral.

In the figure of this problem, <note>the four vertices Q, U, A, and D of the quadrilateral QUAD are all on the same circle</note>. This circle is called <note>the circumcircle of the quadrilateral QUAD</note>. Therefore, <note>the quadrilateral QUAD is a cyclic quadrilateral</note>. According to <note>the properties of cyclic quadrilaterals</note>, it can be concluded that <note>the sum of opposite angles is equal to 180 degrees</note>, i.e., <note>angle QUA + angle QDA = 180 degrees</note>, <note>angle UQD + angle UAD = 180 degrees</note>.

Definition of Inscribed Angle

An inscribed angle is an angle whose vertex lies on the circumference of a circle and whose sides intersect the circle at two points.

The vertex Q of <note>angle ∠Q</note> is on the circumference of the circle, and the two sides of <note>angle ∠Q</note> intersect the circle O at points U and D respectively. Therefore, <note>angle ∠Q is an inscribed angle</note>. Similarly, <note>angle ∠U</note>, <note>angle ∠A</note>, and <note>angle ∠D</note> are also inscribed angles.

<note>Quadrilateral QUAD is a cyclic quadrilateral</note>, therefore <note>∠QUD and ∠QAD sum to 50° + 130° = 180°</note>, thus they are supplementary.

Circumference Formula of Circle

The circumference of a circle is the length around the circular boundary. It can be calculated using the formula \( C = 2\pi r \), where \( C \) represents the circumference and \( r \) represents the radius.

<note>In circle O, point O is the center of the circle</note>, <note>line segments OA, OU, OQ, OD, OP are the radius r</note>. According to the circumference formula of the circle, the circumference C of the circle is equal to 2π multiplied by the radius r, i.e., C=2πr.


Question:
In the diagram below, there are two circles centered at L. The length of radius $\overline{LN}$ is 4.5 centimeters, and the length of diameter $\overline{OQ}$ is 12 centimeters. $\overset{\frown}{OQP}$ measures 240°.

 Find the area of the shaded region.
Round your answer to the nearest tenth.
 $\Box$  square centimeters

Answer: 16.5 square centimeters

1. Given the center L and radius LN as 4.5 cm, the radius of the smaller circle is 4.5 cm. Given the diameter OQ as 12 cm, the radius of the larger circle LO is 6 cm.

2. Given that the arc OQP measures 240°. Since the whole circle measures 360°, the arc OP measures 360° - 240° = 120°.

3. According to the formula for the area of a sector A = (m/360°) * π * r^2, where m is the degree measure of the arc and r is the radius, calculate the area of sector OLP.

4. The arc OP of sector OLP measures 120° and the radius is 6 cm. Therefore, the area of sector OLP is (120/360) * π * 6^2 = 12π square cm.

5. Since points M and N are on segments OL and PL respectively, and the central angle MLN is the same as OQP, which is 240°, the arc MN also measures 360° - 240° = 120°.

6. The arc MN of sector MLN also measures 120° and the radius is 4.5 cm. Therefore, the area of sector MLN is (120/360) * π * 4.5^2 = 6.75π square cm.

7. Find the difference in the areas of the sectors corresponding to the shaded region, which is the area of sector OLP minus the area of sector MLN.

8. Calculate the area of the shaded region: 12π - 6.75π = 5.25π square cm.

9. Calculate the numerical value, 5.25π is approximately 5.25 * 3.14159 ≈ 16.49336 square cm.

10. Round the result to one decimal place, obtaining the area of the shaded region as 16.5 square cm.

11. Through the above reasoning, the final answer is 16.5 square cm.

Definition of Circle

<note>Point L is the center of the circle</note>, with radii of <note>4.5 cm and 6 cm</note>. All points in the figure that are <note>4.5 cm away from point L</note> are on the <note>smaller circle</note>, and all points that are <note>6 cm away from point L</note> are on the <note>larger circle</note>.

Definition of Radius

In the diagram of this problem, in the circle, <note>point L is the center of the circle</note>, <note>point N is any point on the circle</note>, <note>segment LN is the segment from the center to any point on the circle</note>, therefore <note>segment LN is the radius of the circle</note>; similarly, in the circle, <note>point L is the center of the circle</note>, <note>point O is any point on the circle</note>, <note>segment LO is the segment from the center to any point on the circle</note>, therefore <note>segment LO is the radius of the circle</note>.

Definition of Diameter

<note>Line segment OQ is the diameter of the large circle</note>, connecting <note>the center L</note> and <note>points O and Q on the circumference</note>, with a length of twice the radius, that is <note>OQ = 2 * LO = 12 cm</note>.

Definition of Central Angle

An angle formed by two radii connecting two points on the circumference of a circle to the center of the circle is called a central angle.

In the figure of this problem, in the circle, <note>points O and P are two points on the circle</note>, <note>the center of the circle is point L</note>. <note>The angle ∠OLP formed by the lines LO and LP is called the central angle</note>.

Definition of Sector

In the diagram of this problem, <note>sector OLP, radius OL and radius PL are two radii of the circle, and the arc OP is the arc enclosed by these two radii</note>, so according to the definition of sector, the figure composed of these two radii and the arc OP they enclose is <note>sector OLP</note>. <note>Sector MLN, radius ML and radius NL are two radii of the circle, and the arc MN is the arc enclosed by these two radii</note>, so according to the definition of sector, the figure composed of these two radii and the arc MN they enclose is <note>sector MLN</note>.

Inscribed Angle Theorem

In a circle, an inscribed angle is equal to half of the central angle that subtends the same arc.

In the figure of this problem, <note>circle L</note>, <note>points O, Q, P are on the circle</note>, <note>the central angle corresponding to arc OQP is ∠OLP</note>, <note>the inscribed angle is ∠OQP</note>. According to the Inscribed Angle Theorem, <note>∠OQP is equal to half of the central angle ∠OLP corresponding to the arc OQP</note>, that is, <note>∠OQP = 1/2 ∠OLP</note>.

Formula for the Area of a Sector

<note>The area of the sector OLP of the larger circle</note> can be calculated using the formula A = (120/360) * π * 6^2, yielding 12π square centimeters; <note>The area of the sector MLN of the smaller circle</note> can be calculated using the formula A = (120/360) * π * 4.5^2, yielding 6.75π square centimeters.


Question:
In circle S,  $\angle $ OSP measures q° and the length of radius $\overline{OS}$ is h.

If circle S were dilated about its center by a scale factor of 1/2, what would the new area of sector OSP be?

- 𝜋h ^ 2q/360
- 𝜋hq/1,440
- 𝜋hq/360
- 𝜋h ^ 2q/1,440

Answer: - 𝜋h ^ 2q/1,440

1. According to the problem statement, we need to scale circle S by a factor of 1/2 about its center. The radius OS of the original circle has a length of h.

2. According to the scaling theorem of circles, the radius of the new circle S' after scaling becomes O'S' = 1/2 * h, thus the radius length is h/2.

3. According to the scaling theorem of circles, the central angle remains unchanged, so in the new circle S', the angle O'S'P' is still q°.

4. To calculate the area of the sector O'S'P' in the new circle. The area of the sector can be obtained using the sector area formula K = m/360 * A, where m is the degree measure of the central angle of the sector, and A is the total area of the circle.

5. According to the area formula of a circle, calculate the area of the new circle S': S' = π * (h/2)^2 = π * h^2 / 4.

6. The area of the sector O'S'P' is calculated as: K = q/360 * (π * h^2 / 4).

7. After simplification: K = π * h^2 * q / 1440.

8. Therefore, the area of the sector O'S'P' in the new circle S' is: π * h^2 * q / 1440.

Definition of Circle

In <note>circle S</note>, <note>point S is the center</note>, <note>with radius h</note>. All points in the figure that are <note>at a distance of h from point S</note> are on <note>circle S</note>.

Definition of Radius

In the diagram of this problem, in circle S, <note>point S is the center of the circle</note>, <note>point O is any point on the circle</note>, <note>line segment OS is the line segment from the center to any point on the circle</note>, therefore <note>line segment OS is the radius of the circle</note>, with a length of h.

Definition of Sector

In the figure of this problem, <note>sector OSP</note>, <note>radius OS</note> and <note>radius SP</note> are two radii of the circle, and <note>arc OP</note> is the arc enclosed by these two radii. Therefore, according to the <note>definition of sector</note>, the figure composed of these two <note>radii OS, SP</note> and the enclosed <note>arc OP</note> is a <note>sector</note>.

Formula for the Area of a Sector

In the problem diagram, the formula for calculating the area of sector <note>OSP</note> is <note>(q°/360) * πh^2</note>. The formula for calculating the area of sector <note>O'S'P'</note> of the new circle S' is <note>(q°/360) * (π * (h/2)^2/4)</note>.

Area Formula of a Circle

<note>In the figure of this problem, in circle S, the radius of the circle is h</note>, according to the area formula of a circle, the area A of the circle is equal to the circumference π multiplied by <note>the square of the radius h, that is A = πh²</note>. <note>The radius of the new circle S' is h/2</note>, therefore <note>the area of the new circle S' is π * (h/2)² = πh²/4</note>.

Circle Scaling Theorem

If a circle undergoes scaling transformation with its center as the reference point and the scaling factor is k, then the radius of the scaled circle is k times the radius of the original circle, and all angles remain equal before and after the scaling.

<note>Circle S with radius h</note> is centered at the center, and a <note>scaling transformation with a scaling factor of 1/2</note> is performed. <note>The radius of the scaled circle is 1/2*h</note>, and <note>the angle OSP does not change</note>.


Question:
The diagonals of this rhombus are 4 miles and 9 miles.

What is the area of the rhombus?
 $\Box$  square miles

Answer: 18 square miles

1. Given: The diagonals of the rhombus are 4 miles and 9 miles respectively. Let the intersection point of the diagonals be O, and the rhombus be ABCD, with AC as the horizontal diagonal and BD as the vertical diagonal.

2. According to the formula for the area of a rhombus, the area can be determined by its diagonals, i.e., Area = (d1 × d2) / 2, where d1 and d2 are the two diagonals of the rhombus.

3. Substitute d1 = 4 and d2 = 9 into the formula to calculate, yielding (4 × 9) / 2 = 18

4. Through the above reasoning, the final answer is 18.

Definition of Rhombus

A quadrilateral is a rhombus if and only if all four sides are of equal length, and its diagonals are perpendicular bisectors of each other.

<note>In the diagram of this problem, all sides AB, BC, CD, and DA of quadrilateral ABCD are equal</note>, therefore quadrilateral ABCD is a rhombus. Additionally, <note>the diagonals AC and BD of quadrilateral ABCD are perpendicular bisectors of each other</note>, that is, <note>diagonals AC and BD intersect at point O</note>, and <note>angle AOB is a right angle (90 degrees)</note>, and <note>AO=OC and BO=OD</note>.

Rhombus Area Formula

The area of a rhombus is equal to half the product of its diagonals.

<note>The diagonals of the rhombus are 4 miles and 9 miles respectively</note>. According to the rhombus area formula, the area of the rhombus is equal to half the product of the two diagonals, i.e., Area = (<note>4 miles</note> * <note>9 miles</note>) / 2 = 18 square miles.


Question:
Circle H is inscribed in △JKL.

 Select all of the statements that must be true.

- The bisector of  $\angle $ L passes through point H.
- The line that passes through point H and the midpoint of $\overline{JL}$ is perpendicular to $\overline{JL}$ .
- △LHM is a right triangle.
- △JHL is an isosceles triangle.

Answer:
- The bisector of  \$\angle \$ L passes through point H.
- △LHM is a right triangle.

1. From the problem, it is known that circle H is the inscribed circle of △JKL. According to the properties of the inscribed circle of a triangle, the center of the inscribed circle (incenter) is the intersection point of all the angle bisectors inside the triangle.

2. Therefore, it can be concluded that the angle bisectors of ∠JLK, ∠KJL, and ∠JKL must pass through the center H. Thus, the bisector of angle L must pass through point H.

3. Therefore, the first conclusion can be drawn: the bisector of ∠L passes through point H.

4. Analyzing △LHM, point H is the incenter, so segment LM is tangent to the inscribed circle of △JKL. According to the properties of the tangent line of a circle, ∠LMH must be a right angle because the tangent line is perpendicular to the radius.

5. Therefore, the second conclusion can be drawn: △LHM is a right triangle.

6. As for the statement 'the line passing through point H and the midpoint of segment JL is perpendicular to JL,' it cannot be derived from the existing information because there is not enough information about the specific position of H relative to the midpoint of JL to confirm the perpendicular relationship.

7. Regarding whether triangle JHL is an isosceles triangle, due to the lack of consistent data about the side lengths, it cannot be confirmed whether triangle JHL meets the characteristics of an isosceles triangle.

8. Based on the above analysis, the correct statements are:

   a) The bisector of ∠L passes through point H.

   b) △LHM is a right triangle.

Definition of Angle Bisector

An angle bisector is a line that originates from the vertex of an angle and divides the angle into two congruent angles.

In the figure of this problem, <note>the vertices of angles ∠KJL, ∠JKL, and ∠JLK are points J, K, and L respectively</note>, <note>the angle bisectors drawn from these vertices respectively pass through point H</note>, these angle bisectors divide their respective angles into <note>two equal angles</note>. Therefore, <note>the bisector of ∠L must pass through point H</note>.

Definition of Right Triangle

<note>Angle ∠LMH is a right angle (90 degrees)</note>, therefore triangle LHM is a right triangle. Side LM and side HM are the legs, and side LH is the hypotenuse.

Definition of Isosceles Triangle

A triangle is an isosceles triangle if and only if it has at least two congruent (≅) sides.

It cannot be confirmed whether <note>triangle JHL</note> is an isosceles triangle because there is no consistent data on <note>side length information</note>.

Property of the Tangent Line to a Circle

A tangent line to a circle is perpendicular to the radius that passes through the point of tangency.

<note>In circle H, point M is the point of tangency of line LM with the circle</note>, <note>segment HM is the radius of the circle</note>. According to the property of the tangent line to a circle, <note>the tangent line LM is perpendicular to the radius HM at the point of tangency M</note>, <note>i.e., ∠LMH = 90 degrees</note>.

Property of Incircle in Triangle

The point where the angle bisectors of all three angles of a triangle intersect is the center of the incircle.

In the figure of this problem, <note>circle H is the incircle of triangle KJL</note>, then <note>the bisector of angle J and the bisector of angle L intersect at the center H of circle H</note>.


Question:
In the diagram, $\overline{SU}$ is a diagonal of parallelogram STUV. Also, m $\angle $ VSU=43°, m $\angle $ UST=(6x–2)°, and m $\angle $ T=(23x–6)°.

What is m $\angle $ SUV?

 $\Box$ °

Answer: 28°

1. Given that quadrilateral STUV is a parallelogram, ∠VSU=43°, ∠UST=(6x–2)°, ∠T=(23x–6)°.

2. According to the supplementary angles theorem of parallelograms, ∠VSU + ∠UST + ∠T = 180°.

3. Substitute the given angles into the equation: 43 + (6x–2) + (23x–6) = 180.

4. Simplify the equation: 6x + 23x + 43 - 2 - 6 = 180.

5. Combine like terms to get: 29x + 35 = 180.

6. Solve the equation for x: 29x = 180 - 35.

7. Further simplify: 29x = 145.

8. Calculate to get x = 145 ÷ 29 = 5.

9. Substitute x to calculate the angle ∠UST, ∠UST = 6x - 2 = 6 × 5 - 2 = 30 - 2 = 28°.

10. According to the definition of parallelograms, ST ∥ VU, and SU is the diagonal, therefore ∠SUV = ∠UST (alternate interior angles, according to the parallel lines axiom 2).

11. Therefore, ∠SUV = 28°.

12. Through the above reasoning, the final answer is 28°.

Definition of Parallelogram

A quadrilateral is a parallelogram if and only if it has two pairs of opposite sides that are both parallel (∥) and congruent (≅).

<note>In the figure of this problem, quadrilateral STUV is a parallelogram</note>, <note>side ST is parallel and equal to side UV</note>, <note>side SV is parallel and equal to side TU</note>.

Definition of Alternate Interior Angles

When a straight line (referred to as a transversal) intersects two parallel lines, the angles that are located between the two parallel lines but on opposite sides of the transversal are called alternate interior angles.

<note>Two parallel lines ST and VU</note> are intersected by <note>a line SU</note>, where <note>angle ∠UST and angle ∠SUV are between the two parallel lines and on opposite sides of the intersecting line SU</note>, thus <note>angle ∠UST and angle ∠SUV are alternate interior angles</note>. Alternate interior angles are equal, that is, <note>angle ∠UST is equal to angle ∠SUV</note>.

Parallel Postulate 2 of Parallel Lines

If two parallel lines are cut by a transversal, then the corresponding angles are equal, the alternate interior angles are equal, and the consecutive interior angles on the same side of the transversal are supplementary.

In this problem diagram, <note>two parallel lines ST and VU</note> are intersected by a third <note>line SU</note>, forming the following geometric relationship: <note>alternate interior angles: ∠SUV and ∠UST are equal</note>. These relationships illustrate that when two parallel lines are intersected by a third line, the alternate interior angles are equal.

In a parallelogram, each pair of adjacent interior angles are supplementary, meaning their sum is 180°.

<note>In the figure of this problem, in parallelogram STUV, angle S and angle T are adjacent interior angles, and angle U and angle V are also adjacent interior angles</note>. According to the Adjacent Angles Supplementary Theorem of Parallelogram, <note>angle S + angle T = 180°, angle U + angle V = 180°</note>.


Question:
Meg is making props for her school play, including a large pencil made out of cardboard, shown below.

After making the pencil, she plans to paint its entire surface. What is the total surface area Meg plans to paint?
Round your answer to the nearest hundredth.

 $\Box$  square inches

Answer: 195.56 square inches

1. Divide the large pencil into two parts: the cone and the cylinder.

2. Calculate the lateral surface area of the cone. The base of the cone is embedded in the cylinder, so only its lateral surface area needs to be considered.

3. According to the formula for the lateral surface area of a cone S = πrℓ, where r is the radius and ℓ is the slant height. In this problem, the diameter of the cone is 3 inches, so its radius r = 1.5 inches, and the slant height ℓ = 4 inches.

4. Substitute into the formula to get S = π × 1.5 × 4 = 6π. The surface area of the cone part is 6π square inches.

5. Calculate the surface area of the cylinder part. The left base of the cylinder is covered by the cone, so only its lateral surface area and the right base area need to be considered.

6. According to the formula for the surface area of a cylinder S = πr^2 + 2πrh, where r is the radius and h is the height. In this problem, the diameter of the cylinder is 3 inches, so its radius r = 1.5 inches, and the height of the cylinder h = 18 inches.

7. Substitute into the formula to get S = π × 1.5^2 + 2 × π × 1.5 × 18 = 56.25π. The surface area of the cylinder part is 56.25π square inches.

8. Add the surface areas of the two parts to get the total surface area of the large pencil. The total surface area is 6π + 56.25π = 62.25π.

9. Calculate and round to two decimal places to get the total surface area Meg plans to paint as 195.56 square inches.

Definition of Cone

A cone in three-dimensional space is a surface generated by rotating a line segment (generatrix) around a fixed straight line (axis) belonging to a plane of a circle, which is not collinear with the endpoint (vertex). The surface consists of the conical surface and the base (circle).

<note>The base of the cone is a circle with a diameter of 3 inches</note>, its <note>vertex is at the tip of the pencil</note>. <note>The slant height of the cone ℓ = 4 inches</note>, <note>radius r = 1.5 inches (half of the base diameter)</note>, <note>lateral surface area formula S = πrℓ</note>.

Cylinder

A cylinder is a geometric solid consisting of two parallel and congruent circular bases and a curved lateral surface connecting the bases.

<note>A cylinder consists of two parallel and identical circular bases and a lateral surface</note>. The <note>diameter of the base is 3 inches</note>, so its <note>radius r = 1.5 inches</note>, <note>the height of the cylinder h = 18 inches</note>. The lateral surface area formula <note>S = 2πrh</note>, base area formula <note>S = πr²</note>.

Lateral Surface Area of a Cone

The lateral surface area \(A\) of a cone is calculated using the formula \(A = \pi r l\), where \(r\) is the radius of the base circle and \(l\) is the slant height of the cone.

<note>The diameter of the base of the cone is 3 inches (so the radius r = 1.5 inches)</note>, <note>the slant height ℓ = 4 inches</note>. Substituting into the formula, the lateral surface area of the cone is S = π × 1.5 × 4 = 6π square inches.

Surface Area Formula for a Cylinder

The surface area of a cylinder is equal to the sum of the areas of the two bases and the lateral surface area. The total surface area (SA) is given by the formula: SA = 2πr² + 2πrh, where r is the radius of the base circle, and h is the height of the cylinder.

Original: <note>The diameter of the base of the cylinder is 3 inches (so the radius r = 1.5 inches), height h = 18 inches</note>. Since one base is covered by a cone, only the lateral surface area and the area of the other base are considered, the formula is <note>S = 2πrh + πr²</note>, substituting the values we get <note>S = 2 × π × 1.5 × 18 + π × 1.5² = 56.25π square inches</note>.


Question:
In the diagram below, quadrilateral PQRS is inscribed in circle T.

 Select all of the statements that must be true.

- m $\angle $ R=70°
- m $\overset{\frown}{PQ}$ =50°
- m $\angle $ S+m $\angle $ Q=180°
- m $\angle $ Q=110°

Answer:
- m \$\angle \$ R=70°
- m \$\overset{\frown}{PQ}\$ =50°
- m \$\angle \$ S+m \$\angle \$ Q=180°

1. Quadrilateral PQRS is inscribed in circle T. According to the properties of a cyclic quadrilateral, ∠S and ∠Q are opposite angles, thus ∠S + ∠Q = 180°. Therefore, m∠S + m∠Q = 180° is true.

2. Given m∠P = 110°, then according to the properties of a cyclic quadrilateral: opposite angles are supplementary, thus m∠R + m∠P = 180°. Therefore, m∠R = 180° - m∠P = 70°. So, m∠R = 70° is true.

3. Given m∠P = 110°, the arc opposite to ∠P is arc QRS. According to the definition of a central angle, the angle subtended by arc QRS is ∠QTS. According to the definition of an inscribed angle, ∠P is an inscribed angle. According to the definition of an inscribed angle, the angle subtended by arc QRS is twice the measure of ∠P, thus arc QRS = 2 × 110° = 220°.

4. We know the total circumference of the circle is 360°. We can divide this 360° into arc PQ + arc QRS + arc PS.

5. According to step 3, arc QRS = 220°. Arc PS = 360° - arc QRS - arc PQ. Given that the opposite arc PS = 90°.

6. Therefore, according to steps 4 and 5, we can find arc PQ = 360° - 220° - 90° = 50°. Thus, arc PQ = 50° is true.

7. For the option m∠Q = 110°, we do not have enough information to determine the measure of m∠Q, thus it is not necessarily true.

8. In conclusion, the following statements must be correct: m∠R = 70°, arc PQ = 50°, and m∠S + m∠Q = 180°. However, m∠Q = 110° is not necessarily true.

n. After the above reasoning, the final answer is: m∠R = 70°, arc PQ = 50°, m∠S + m∠Q = 180°.

Definition of Inscribed Angle

An inscribed angle is an angle whose vertex lies on the circumference of a circle and whose sides intersect the circle at two points.

In the diagram of this problem, in circle T, the vertices of <note>angles ∠PQR, ∠QRS, ∠RSP, and ∠SPQ</note> are respectively on the <note>circumference</note>, and the sides of the angles intersect <note>circle T</note> at <note>points P, Q, R, and S</note>. Therefore, <note>angles ∠PQR, ∠QRS, ∠RSP, and ∠SPQ</note> are all <note>inscribed angles</note>.

Definition of Central Angle

An angle formed by two radii connecting two points on the circumference of a circle to the center of the circle is called a central angle.

In <note>circle T</note>, <note>point Q and point S are two points on the circle</note>, <note>the center of the circle is point T</note>. The angle ∠QTS formed by <note>line segments TQ and TS</note> is called the central angle.

Inscribed Angle Theorem

In a circle, an inscribed angle is equal to half of the central angle that subtends the same arc.

<note>In circle T, points P, Q, and S are on the circle</note>, <note>the central angle corresponding to arc QRS is ∠QTS</note>, <note>the inscribed angle is ∠P</note>. According to the Inscribed Angle Theorem, <note>∠QTS is equal to half of the central angle ∠QTS corresponding to the arc QRS</note>, that is, <note>∠P = 1/2 ∠QTS</note>.

Definition of Arc

An arc is a curve on a circle that is defined by two distinct points on the circle.

In the diagram of this problem, there are two points <note>P</note> and <note>Q</note> on the circle <note>T</note>, and <note>arc PQ</note> is a segment of the curve connecting these two points. According to the definition of an arc, <note>arc PQ</note> is a segment of the curve between the two points <note>P</note> and <note>Q</note> on the circle <note>T</note>.

In a cyclic quadrilateral (a quadrilateral inscribed in a circle), the sum of each pair of opposite angles is equal to 180°. Additionally, any exterior angle is equal to its interior opposite angle.

In the <note>cyclic quadrilateral PQRS</note>, the <note>vertices P, Q, R, S</note> are all on the <note>circle T</note>. According to the <note>Diagonal Supplementary Theorem for Cyclic Quadrilateral</note>, the sum of each pair of opposite angles in the <note>quadrilateral PQRS</note> is equal to 180°. Specifically, <note>∠PQR + ∠PSR = 180°</note>; <note>∠QPS + ∠QRS = 180°</note>.


Question:
$\overline{LK}$ is a diameter of circle Z.

 Find the area of the shaded region.
 Round your answer to the nearest tenth.
 $\Box$  square inches

Answer: 57.1 square inches

1. Given that \$\overline{LK}\$ is the diameter of circle \$Z\$, \$\overline{ZL} = \overline{ZK} = \overline{ZJ} = 10\$ meters (because they are radii), and the central angle \$\angle LZK = 180^\circ\$.

2. The sector area corresponding to the central angle \$\angle LZK\$ can be represented as half of the entire circle area because \$\angle LZK = 180^\circ\$, so it is a semicircle; the area of circle \$Z\$ is \$\pi \cdot r^2 = \pi \cdot 10^2 = 100\pi\$ square meters. Therefore, the sector area is \$\frac{1}{2} \times 100\pi = 50\pi\$ square meters.

3. Since \$\overline{LK}\$ is the diameter, and \$\angle LJK\$ is a right angle (according to the corollary of the inscribed angle theorem that the angle subtended by a diameter is a right angle), triangle \$LJK\$ is a right triangle with \$\overline{LK}\$ as the hypotenuse, measuring \$20\$ meters.

4. The area of triangle \$LJK\$ can be calculated using the formula for the area of a right triangle: area \$A = \frac{1}{2} \times \text{base} \times \text{height}\$. Let \$\overline{LK}\$ and \$\overline{JZ}\$ be the base and height of the right triangle, with the base being \$20\$ meters and the height being \$10\$ meters. Therefore: \$A = \frac{1}{2} \times 20 \times 10 = 100\$ square meters.

5. The shaded area is the difference between the sector area and the triangle area, so the area of the shaded region is \$50\pi - 100\$ square meters.

6. Calculating \$50\pi - 100\$, we get approximately \$57.07963\$ square meters.

7. Rounding this value to one decimal place, the area of the shaded region is approximately \$57.1\$ square meters.

Definition of Circle

A circle is the locus of all points in a plane that are at a fixed distance from a given point.

In <note>circle $Z$</note>, <note>point $Z$ is the center</note>, <note>with a radius of $10$ meters</note>. All points in the figure that are <note>at a distance of $10$ meters from point $Z$</note> are on <note>circle $Z$</note>, including <note>line segment $\overline{ZL}$</note>, <note>line segment $\overline{ZK}$</note>, and <note>line segment $\overline{ZJ}$</note>.

Definition of Diameter

A diameter is a line segment that passes through the center of a circle and has its endpoints on the circle. It is the longest chord of the circle, with a length equal to twice the radius of the circle.

<note>LK is the diameter</note>, connecting <note>the center Z and the points L and K on the circumference</note>, with a length of <note>2 times the radius</note>, that is, <note>LK = 20 meters</note>.

Definition of Sector

A sector is a figure formed by two radii of a circle and the arc enclosed between them.

In <note>sector LJK</note>, <note>radius LZ and radius KZ</note> are two radii of the circle, and <note>arc LJK</note> is the arc enclosed by these two radii. Therefore, according to the definition of a sector, the figure composed of these two radii and the enclosed <note>arc LJK</note> is a <note>sector</note>.

The angle subtended by the diameter of a semicircle is a right angle; conversely, if an inscribed angle is 90°, then the chord subtending that angle is a diameter.

<note>In circle Z, the inscribed angle LJK subtended by the diameter LK is a right angle (90 degrees)</note>. (Or <note>The inscribed angle LJK is 90 degrees, so the chord LK subtended by it is the diameter</note>.)

Area Formula of a Circle

The area of a circle is given by the formula A = πr², where A represents the area, π (pi) is the mathematical constant approximately equal to 3.14159, and r is the radius of the circle.

<note>The radius of circle Z is 10 meters</note>, according to the area formula of a circle, the area A of the circle is equal to pi π multiplied by the square of the radius 10, that is, A = π * 10² = 100π square meters.

Area Formula of a Triangle

The area of any triangle is equal to its base multiplied by its height, divided by 2.

In the triangle LJK, side LK is the base, and segment JZ is the height from that base, so the area of triangle LJK is equal to the base <note>LK multiplied by the height JZ divided by 2, i.e., Area = (LK * JZ) / 2</note>.


Question:
What is the surface area of this sphere?

Write an exact answer in terms of 𝜋.

 $\Box$ 𝜋/ square centimeters

Answer: 289𝜋 square centimeters

1. Given the diameter of the sphere is 17 cm.

2. According to the radius of the sphere, the radius is half of the diameter, so the radius of the sphere is 17 cm ÷ 2 = 8.5 cm.

3. According to the formula for the surface area of a sphere S = 4πr², where r is the radius.

4. Substitute the known radius r = 8.5 cm into the formula, resulting in S = 4 × π × (8.5)².

5. Calculate the square, (8.5)² = 72.25.

6. Substitute the square result into the surface area formula, S = 4 × π × 72.25.

7. Calculate the multiplication, 4 × 72.25 = 289.

8. Therefore, the surface area of the sphere is 289π square cm.

9. Through the above reasoning, the final answer is 289π square cm.

Diameter of a Sphere

The diameter of a sphere is a line segment that passes through the center of the sphere and connects two points on the surface of the sphere.

In the figure of this problem, the <note>Diameter of a Sphere</note> is a <note>17 cm line segment</note>. The <note>17 cm line segment passes through the center of the sphere</note> and <note>connects two points on the surface of the sphere</note>.

Radius of a Sphere

The radius R of a sphere is the distance from the center of the sphere to any point on its surface.

In the diagram of this problem, within the sphere, <note>a line segment from any point on the surface of the sphere to the center of the sphere is the radius of the sphere</note>. According to the definition of the radius of the sphere, <note>the radius R of the sphere is the distance from the center of the sphere to any point on the surface</note>, therefore <note>the length of the line segment is the radius R of the sphere</note>.

Sphere Surface Area Formula

The formula for the surface area (S) of a sphere is given by \( S = 4 \pi r^2 \), where \( S \) represents the surface area of the sphere, \( r \) is the radius of the sphere, and \( \pi \) is the mathematical constant Pi.

In this problem diagram, <note>in the sphere, radius r represents the radius of the sphere, and surface area S represents the surface area of the sphere</note>, according to the <note>sphere surface area formula, S = 4 * π * r^2, where π is the constant pi</note>. Therefore, <note>the surface area S of the sphere is equal to 4 times pi multiplied by the square of the radius r</note>.


Question:
A cylinder has a surface area of 384𝜋 square inches. Its base is shown below.

What is the height of the cylinder?

 $\Box$  inches

Answer: 16 inches

1. Given that the diameter of the cylinder is 16 inches, we can obtain the radius r as half of the diameter, i.e., r = 16 / 2 = 8 inches.

2. The surface area S of the cylinder is known to be 384𝜋 square inches. The formula for the surface area of a cylinder is S = 2πr² + 2πrh, where r is the radius of the base of the cylinder and h is the height of the cylinder.

3. Substitute the known radius r = 8 inches into the surface area formula S = 2πr² + 2πrh, we get 384𝜋 = 2π(8)² + 2π(8)h.

4. Calculate 2π(8)², we get 2π*64 = 128π, thus the formula becomes 384π = 128π + 16πh.

5. Move 128π from the left side of the equation to the right side, i.e., 384π - 128π = 16πh, we get 256π = 16πh.

6. Divide both sides by 16π, we get 256π / 16π = h, simplifying to h = 16.

7. Through the above reasoning, the final answer is 16 inches.

Cylinder

A cylinder is a geometric solid consisting of two parallel and congruent circular bases and a curved lateral surface connecting the bases.

In the diagram of this problem, the cylinder consists of <note>two parallel and identical circular bases</note> and <note>a lateral surface</note>. <note>The bases are two completely identical circles</note>, with <note>a radius of r</note>, <note>a diameter of 16 inches</note>, and their <note>centers are on the same straight line</note>. <note>The lateral surface is a rectangle, which, when unfolded, has a height equal to the height h of the cylinder and a width equal to the circumference of the base circle</note>. <note>Given that the diameter of the base circle is 16 inches</note>, it can be deduced that <note>the radius r is half of the diameter, i.e., r = 16 / 2 = 8 inches</note>.

Surface Area Formula for a Cylinder

<note>The radius of the base of the cylinder r is known to be 8 inches</note>, <note>the surface area S is known to be 384π square inches</note>, substituting into the formula <note>S = 2π(8)² + 2π(8)h</note>, we get <note>384π = 2π(8)² + 2π(8)h</note>, through calculation and algebraic transformation, we finally obtain the answer <note>h = 16 inches</note>.


Question:
What is the surface area of this sphere?

Round your answer to the nearest hundreth.

 $\Box$  square feet

Answer: 804.25 square feet

1. Given the radius of the sphere is 8 feet, according to the formula for the surface area of a sphere S=4𝜋r^2, where r represents the radius of the sphere, derive the formula for calculating the surface area S.

2. Substitute the radius of the sphere into the formula: S=4∙𝜋∙(8)^2.

3. Calculate the square of 8, obtaining 8^2=64.

4. Substitute the calculation result into the surface area formula: S=4∙𝜋∙64.

5. Calculate the product of 4 and 64, obtaining 4×64=256.

6. Substitute the result into the formula, obtaining: S=256𝜋.

7. Use 𝜋≈3.14159 for approximate calculation, obtaining S≈256∙3.14159.

8. The calculation result is S≈804.24704.

9. Round the result to the nearest hundredth as required by the problem.

10. Obtain the final result as S≈804.25 square feet.

11. Through the above reasoning, the final answer is 804.25 square feet.

Definition of Sphere

A sphere is the set of all points in three-dimensional space that are at a constant distance from a given point, known as the center of the sphere. This constant distance is called the radius of the sphere.

<note>The distance from any point on the sphere to the center of the sphere is the radius</note>. All points that are at a distance equal to the radius from the center of the sphere form the surface of the sphere. Therefore, <note>a sphere is the set of all points that are equidistant from the center of the sphere</note>.

Radius of a Sphere

In the figure of this problem, <note>the line segment from any point on the surface of the sphere to the center of the sphere is the radius of the sphere</note>. According to the definition of the radius of the sphere, <note>the radius R of the sphere is the distance from the center of the sphere to any point on the surface</note>, therefore <note>the length of the line segment is the radius R of the sphere</note>.

Sphere Surface Area Formula

In the figure of this problem, in the sphere, <note>radius r represents the radius of the sphere</note>, <note>surface area S represents the surface area of the sphere</note>, according to the sphere surface area formula, S = 4 * π * r^2, where π is the pi. Therefore, <note>the surface area S of the sphere is equal to 4 times π multiplied by the square of the radius r</note>.


Question:
In rhombus KLMN, diagonals $\overline{KM}$ and $\overline{NL}$ intersect at point O, m $\angle $ OKL=(9x)°, and m $\angle $ KNM=(7x+5)°.

What is m $\angle $ NML?

 $\Box$ °

Answer: 126°

1. In rhombus KLMN, it is known that angle OKL = (9x)°, angle KNM = (7x+5)°.

2. According to the properties of the diagonals of a rhombus, the diagonals of a rhombus are perpendicular bisectors of each other and they bisect the interior angles of the rhombus, so angle OKL and angle OKN are equal, i.e., angle OKN = (9x)°.

3. According to the supplementary angles property of a rhombus, the adjacent angles of a rhombus are supplementary, therefore, angle KNM + angle NKL = 180°.

4. Since angle OKN = (9x)° and angle OKL = (9x)°, angle NKL is the sum of these two angles, hence angle NKL = 2 * (9x)° = (18x)°.

5. Substituting the relationship between angle KNM and angle NKL into the supplementary angle relationship, we have: (7x+5)° + (18x)° = 180°.

6. Solving the equation:

   25x + 5 = 180

   25x = 175

   x = 7.

7. Now calculate the degree of angle KNM: angle KNM = (7x+5)° = (7*7+5)° = 54°.

8. According to the supplementary angles property of a rhombus, therefore angle NML = 180° - angle KNM = 180° - 54° = 126°.

9. Therefore, angle NML = 126°.

10. Through the above reasoning, the final answer is 126°.

Definition of Rhombus

A quadrilateral is a rhombus if and only if all four sides are of equal length, and its diagonals are perpendicular bisectors of each other.

In the figure of this problem, in quadrilateral KLMN, <note>all sides KL, LM, MN, and NK are equal</note>, therefore quadrilateral KLMN is a rhombus. Additionally, the diagonals <note>KM and NL are perpendicular bisectors of each other</note>, that is, <note>diagonals KM and NL intersect at point O</note>, and <note>angle KON is a right angle (90 degrees)</note>, and <note>KO=OM and NO=OL</note>.

Definition of Diagonal

A diagonal is a line segment connecting one vertex of a polygon to another vertex that is not adjacent to it.

In the figure of this problem, <note>the diagonals of rhombus KLMN are KM and NL, which connect the non-adjacent vertices K and M, as well as N and L</note>.

Properties of the Diagonals of a Rhombus

In a rhombus, the diagonals bisect each other and are perpendicular to each other.

In <note>rhombus KLMN</note>, <note>diagonals KM and NL bisect each other and are perpendicular</note>. Specifically, <note>point O is the intersection of diagonals KM and NL</note>, and <note>KO=OM</note> and <note>NO=OL</note>. Additionally, <note>angles KON and MOL are right angles (90 degrees)</note>, so <note>diagonals KM and NL are perpendicular</note>.

Adjacent Angles of Rhombus are Supplementary

Any pair of adjacent angles in a rhombus are supplementary.

<note>In rhombus KLMN, angle KNM and angle NKL are a pair of adjacent angles</note>, according to the properties of the rhombus, <note>any pair of adjacent angles in a rhombus are supplementary</note>, that is, <note>angle KNM + angle NKL = 180°</note>. Similarly, <note>angle KNM and angle NML are also a pair of adjacent angles</note>, satisfying <note>angle KNM + angle NML = 180 degrees</note>.