Title: Chapter 10
Course: quiz duplicate error

Q: Use the diagram to find the measure of the given angle.
[Image]

[Image]
- 95° ([Image] and [Image] are vertical angles and are therefore congruent. So set the two expressions equal and solve for x. 3x + 35 = 4x + 15 and x = 20. Substituting 20 in place of x in the expression for [Image] gives 4(20) + 15 = 80 + 15 = 95°.)
- 20° ([Image] and [Image] are vertical angles and are therefore congruent. So set the two expressions equal and solve for x. 3x + 35 = 4x + 15 and x = 20. Substituting 20 in place of x in the expression for [Image] gives 4(20) + 15 = 80 + 15 = 95°.)
- 50° ([Image] and [Image] are vertical angles and are therefore congruent. So set the two expressions equal and solve for x. 3x + 35 = 4x + 15 and x = 20. Substituting 20 in place of x in the expression for [Image] gives 4(20) + 15 = 80 + 15 = 95°.)
- 85° ([Image] and [Image] are vertical angles and are therefore congruent. So set the two expressions equal and solve for x. 3x + 35 = 4x + 15 and x = 20. Substituting 20 in place of x in the expression for [Image] gives 4(20) + 15 = 80 + 15 = 95°.)
A: 95°

Q: In [Image], [Image] and [Image] are diameters, and [Image].
[Image]
Find m arc BC.
- 270 (Since the diameters [Image] and [Image] are perpendicular, they form right angles which measure 90. Additionally, since [Image], all of those angles are [Image] which is 45°. The measure of an arc is equal to the measure of its central angle, so add any angles that are necessary to find the measure of the given angle and its intercepted arc.)
- 90 (Since the diameters [Image] and [Image] are perpendicular, they form right angles which measure 90. Additionally, since [Image], all of those angles are [Image] which is 45°. The measure of an arc is equal to the measure of its central angle, so add any angles that are necessary to find the measure of the given angle and its intercepted arc.)
- 225 (Since the diameters [Image] and [Image] are perpendicular, they form right angles which measure 90. Additionally, since [Image], all of those angles are [Image] which is 45°. The measure of an arc is equal to the measure of its central angle, so add any angles that are necessary to find the measure of the given angle and its intercepted arc.)
- 315 (Since the diameters [Image] and [Image] are perpendicular, they form right angles which measure 90. Additionally, since [Image], all of those angles are [Image] which is 45°. The measure of an arc is equal to the measure of its central angle, so add any angles that are necessary to find the measure of the given angle and its intercepted arc.)
A: 90

Q: In [Image], TS = 15, UQ = US. Find m [Image].
[Image]
- 40 (Since [Image] and [Image] are congruent and perpendicular to separate chords, the chords [Image] and [Image] must also be congruent. Additionally, [Image] and [Image] bisect these chords, so TS = RS. So take the given measure of [Image] and double it. That will be the measure of [Image] and also [Image].)
- 31 (Since [Image] and [Image] are congruent and perpendicular to separate chords, the chords [Image] and [Image] must also be congruent. Additionally, [Image] and [Image] bisect these chords, so TS = RS. So take the given measure of [Image] and double it. That will be the measure of [Image] and also [Image].)
- 15 (Since [Image] and [Image] are congruent and perpendicular to separate chords, the chords [Image] and [Image] must also be congruent. Additionally, [Image] and [Image] bisect these chords, so TS = RS. So take the given measure of [Image] and double it. That will be the measure of [Image] and also [Image].)
- 30 (Since [Image] and [Image] are congruent and perpendicular to separate chords, the chords [Image] and [Image] must also be congruent. Additionally, [Image] and [Image] bisect these chords, so TS = RS. So take the given measure of [Image] and double it. That will be the measure of [Image] and also [Image].)
A: 30

Q: In [Image], [Image], [Image] = 7x, [Image] = 5x + 12, and [Image] and [Image] are diameters.
[Image]
Find m arc [Image].
- 56 (Since [Image] is a diameter, [Image]. Solve the equation, then substitute the value of x to find [Image]. Since [Image] and [Image] are vertical angles, they are congruent. Additionally, use the fact that m arc DC = [Image].)
- 46 (Since [Image] is a diameter, [Image]. Solve the equation, then substitute the value of x to find [Image]. Since [Image] and [Image] are vertical angles, they are congruent. Additionally, use the fact that m arc DC = [Image].)
- 50 (Since [Image] is a diameter, [Image]. Solve the equation, then substitute the value of x to find [Image]. Since [Image] and [Image] are vertical angles, they are congruent. Additionally, use the fact that m arc DC = [Image].)
- 49 (Since [Image] is a diameter, [Image]. Solve the equation, then substitute the value of x to find [Image]. Since [Image] and [Image] are vertical angles, they are congruent. Additionally, use the fact that m arc DC = [Image].)
A: 49

Q: The radius, diameter, or circumference of a circle is given. Find the missing measures. Round to the nearest hundredth if necessary.

d = 22.3 km, r = ?   , C = ?
- r = 44.6 km, C = 35.03 km (Radius = Diameter ¸ 2 Circumference = (2 ´ Radius ´ p) or (Diameter ´ p))
- r = 11.15 km, C = 35.03 km (Radius = Diameter ¸ 2 Circumference = (2 ´ Radius ´ p) or (Diameter ´ p))
- r = 11.15 km, C = 70.06 km (Radius = Diameter ¸ 2 Circumference = (2 ´ Radius ´ p) or (Diameter ´ p))
- r = 44.6 km, C = 70.06 km (Radius = Diameter ¸ 2 Circumference = (2 ´ Radius ´ p) or (Diameter ´ p))
A: r = 11.15 km, C = 70.06 km

Q: In [Image], [Image] and m arc [Image]. Find [Image].
[Image]
- 85 ([Image], so begin by finding the measures of [Image] and [Image]. Since [Image] and [Image] are both inscribed in semicircles, they are right triangles ( [Image] and [Image] are right angles). Additionally, since [Image], [Image] is isosceles. That makes [Image]. Now for [Image]. Since [Image] is an inscribed angle, its measure is one-half its intercepted arc (arc CE). That makes [Image]. Finally, add the measures of the two angles to get the final answer.)
- 108 ([Image], so begin by finding the measures of [Image] and [Image]. Since [Image] and [Image] are both inscribed in semicircles, they are right triangles ( [Image] and [Image] are right angles). Additionally, since [Image], [Image] is isosceles. That makes [Image]. Now for [Image]. Since [Image] is an inscribed angle, its measure is one-half its intercepted arc (arc CE). That makes [Image]. Finally, add the measures of the two angles to get the final answer.)
- 117 ([Image], so begin by finding the measures of [Image] and [Image]. Since [Image] and [Image] are both inscribed in semicircles, they are right triangles ( [Image] and [Image] are right angles). Additionally, since [Image], [Image] is isosceles. That makes [Image]. Now for [Image]. Since [Image] is an inscribed angle, its measure is one-half its intercepted arc (arc CE). That makes [Image]. Finally, add the measures of the two angles to get the final answer.)
- 110 ([Image], so begin by finding the measures of [Image] and [Image]. Since [Image] and [Image] are both inscribed in semicircles, they are right triangles ( [Image] and [Image] are right angles). Additionally, since [Image], [Image] is isosceles. That makes [Image]. Now for [Image]. Since [Image] is an inscribed angle, its measure is one-half its intercepted arc (arc CE). That makes [Image]. Finally, add the measures of the two angles to get the final answer.)
A: 110

Q: In [Image], [Image] and AE = 10.
[Image]
Find m [Image].
- 14 (Since [Image], [Image] and [Image] form a right triangle, you can use the Pythagorean Theorem to find [Image]. Since [Image] is a segment that passes through the center of the circle and is perpendicular to chord [Image], it also bisects [Image]. That means [Image].)
- 12 (Since [Image], [Image] and [Image] form a right triangle, you can use the Pythagorean Theorem to find [Image]. Since [Image] is a segment that passes through the center of the circle and is perpendicular to chord [Image], it also bisects [Image]. That means [Image].)
- 10 (Since [Image], [Image] and [Image] form a right triangle, you can use the Pythagorean Theorem to find [Image]. Since [Image] is a segment that passes through the center of the circle and is perpendicular to chord [Image], it also bisects [Image]. That means [Image].)
- 16 (Since [Image], [Image] and [Image] form a right triangle, you can use the Pythagorean Theorem to find [Image]. Since [Image] is a segment that passes through the center of the circle and is perpendicular to chord [Image], it also bisects [Image]. That means [Image].)
A: 12

Q: Use the diagram to find the measure of the given angle.
[Image]

[Image]
- 140 ([Image] forms a linear pair with [Image], so their sum is 180. Therefore, [Image] is 180 – 50 = 130.)
- 120 ([Image] forms a linear pair with [Image], so their sum is 180. Therefore, [Image] is 180 – 50 = 130.)
- 130 ([Image] forms a linear pair with [Image], so their sum is 180. Therefore, [Image] is 180 – 50 = 130.)
- 150 ([Image] forms a linear pair with [Image], so their sum is 180. Therefore, [Image] is 180 – 50 = 130.)
A: 130

Q: Use the diagram to find the measure of the given angle.
[Image]

[Image]
- 180 ([Image] is a right angle, so its measure is 90.)
- 90 ([Image] is a right angle, so its measure is 90.)
- 60 ([Image] is a right angle, so its measure is 90.)
- 30 ([Image] is a right angle, so its measure is 90.)
A: 90

Q: Use the diagram to find the measure of the given angle.
[Image]

[Image]
- 95° ([Image] and [Image] are vertical angles and are therefore congruent. So set the two expressions equal and solve for x. 3x + 35 = 4x + 15 and x = 20. Substituting 20 in place of x in the expression for [Image] gives 3(20) + 35 = 60 + 35 = 95°.)
- 20° ([Image] and [Image] are vertical angles and are therefore congruent. So set the two expressions equal and solve for x. 3x + 35 = 4x + 15 and x = 20. Substituting 20 in place of x in the expression for [Image] gives 3(20) + 35 = 60 + 35 = 95°.)
- 50° ([Image] and [Image] are vertical angles and are therefore congruent. So set the two expressions equal and solve for x. 3x + 35 = 4x + 15 and x = 20. Substituting 20 in place of x in the expression for [Image] gives 3(20) + 35 = 60 + 35 = 95°.)
- 85° ([Image] and [Image] are vertical angles and are therefore congruent. So set the two expressions equal and solve for x. 3x + 35 = 4x + 15 and x = 20. Substituting 20 in place of x in the expression for [Image] gives 3(20) + 35 = 60 + 35 = 95°.)
A: 95°

Q: Quadrilateral ABCD is inscribed in [Image] such that [Image] and [Image]. Find [Image].
[Image]
- 48 (First note that m arc BC = [Image] because a central angle of a circle is always congruent to its intercepted arc. Secondly, [Image] is one-half m arc BC, as the measure of an inscribed angle is half the measure of its intercepted arc. Since [Image], [Image] because alternate interior angles are congruent. So [Image].)
- 44 (First note that m arc BC = [Image] because a central angle of a circle is always congruent to its intercepted arc. Secondly, [Image] is one-half m arc BC, as the measure of an inscribed angle is half the measure of its intercepted arc. Since [Image], [Image] because alternate interior angles are congruent. So [Image].)
- 46 (First note that m arc BC = [Image] because a central angle of a circle is always congruent to its intercepted arc. Secondly, [Image] is one-half m arc BC, as the measure of an inscribed angle is half the measure of its intercepted arc. Since [Image], [Image] because alternate interior angles are congruent. So [Image].)
- 42 (First note that m arc BC = [Image] because a central angle of a circle is always congruent to its intercepted arc. Secondly, [Image] is one-half m arc BC, as the measure of an inscribed angle is half the measure of its intercepted arc. Since [Image], [Image] because alternate interior angles are congruent. So [Image].)
A: 42

Q: Use the diagram to find the measure of the given angle.
[Image]

[Image]
- 110 ([Image] is equal to the sum of [Image] and [Image], so its measure is 90 + 40 = 130.)
- 120 ([Image] is equal to the sum of [Image] and [Image], so its measure is 90 + 40 = 130.)
- 130 ([Image] is equal to the sum of [Image] and [Image], so its measure is 90 + 40 = 130.)
- 140 ([Image] is equal to the sum of [Image] and [Image], so its measure is 90 + 40 = 130.)
A: 130

Q: The radius, diameter, or circumference of a circle is given. Find the missing measures. Round to the nearest hundredth if necessary.

r = 13.1 km, d = ?   , C = ?
- d = 26.2 km, C = 82.31 km (Diameter = 2 ´ Radius Circumference = (2 ´ Radius ´ p) or (Diameter ´ p))
- d = 26.2 km, C = 41.15 km (Diameter = 2 ´ Radius Circumference = (2 ´ Radius ´ p) or (Diameter ´ p))
- d = 6.55 km, C = 82.31 km (Diameter = 2 ´ Radius Circumference = (2 ´ Radius ´ p) or (Diameter ´ p))
- d = 6.55 km, C = 41.15 km (Diameter = 2 ´ Radius Circumference = (2 ´ Radius ´ p) or (Diameter ´ p))
A: d = 26.2 km, C = 82.31 km

Q: Use the diagram to find the measure of the given angle.
[Image]

[Image]
- 170 ([Image] is equal to the sum of [Image] and [Image], so its measure is 50 + 90 = 140.)
- 160 ([Image] is equal to the sum of [Image] and [Image], so its measure is 50 + 90 = 140.)
- 150 ([Image] is equal to the sum of [Image] and [Image], so its measure is 50 + 90 = 140.)
- 140 ([Image] is equal to the sum of [Image] and [Image], so its measure is 50 + 90 = 140.)
A: 140

Q: Find the exact circumference of the circle.
[Image]
- 12p [Image] mm (The circumference formula is Diameter ´ p. The diameter shown also happens to be the diagonal of a square, so it can be found by multiplying the side of the square by [Image]. So the diameter would be side ´ [Image] and the circumference would be side ´ p ´ [Image].)
- 24p [Image] mm (The circumference formula is Diameter ´ p. The diameter shown also happens to be the diagonal of a square, so it can be found by multiplying the side of the square by [Image]. So the diameter would be side ´ [Image] and the circumference would be side ´ p ´ [Image].)
- 12p mm (The circumference formula is Diameter ´ p. The diameter shown also happens to be the diagonal of a square, so it can be found by multiplying the side of the square by [Image]. So the diameter would be side ´ [Image] and the circumference would be side ´ p ´ [Image].)
- 6p [Image] mm (The circumference formula is Diameter ´ p. The diameter shown also happens to be the diagonal of a square, so it can be found by multiplying the side of the square by [Image]. So the diameter would be side ´ [Image] and the circumference would be side ´ p ´ [Image].)
A: 12p [Image] mm

Q: Use the diagram to find the measure of the given angle.
[Image]

[Image]
- 50 ([Image] is complementary with [Image], so its measure is 90 – 50 = 40.)
- 40 ([Image] is complementary with [Image], so its measure is 90 – 50 = 40.)
- 60 ([Image] is complementary with [Image], so its measure is 90 – 50 = 40.)
- 30 ([Image] is complementary with [Image], so its measure is 90 – 50 = 40.)
A: 40

Q: [Image]
If [Image] = 2x + 2, [Image] = 9x, find [Image].
- 72 (m [Image] = 90 since it is inscribed in a semicircle. Since the sum of the angles in any triangle is 180°, [Image] + [Image] = 90. Substitute the given values for [Image] and [Image] into that equation. Then substitute the value found for x into the expression for [Image].)
- 19 (m [Image] = 90 since it is inscribed in a semicircle. Since the sum of the angles in any triangle is 180°, [Image] + [Image] = 90. Substitute the given values for [Image] and [Image] into that equation. Then substitute the value found for x into the expression for [Image].)
- 75 (m [Image] = 90 since it is inscribed in a semicircle. Since the sum of the angles in any triangle is 180°, [Image] + [Image] = 90. Substitute the given values for [Image] and [Image] into that equation. Then substitute the value found for x into the expression for [Image].)
- 18 (m [Image] = 90 since it is inscribed in a semicircle. Since the sum of the angles in any triangle is 180°, [Image] + [Image] = 90. Substitute the given values for [Image] and [Image] into that equation. Then substitute the value found for x into the expression for [Image].)
A: 18

Q: Use the diagram to find the measure of the given angle.
[Image]

[Image]
- 85° ([Image] and [Image] are a linear pair and are therefore supplementary. First find [Image]. [Image] and [Image] are vertical angles and are therefore congruent. So set the two expressions equal and solve for x. 3x + 35 = 4x + 15 and x = 20. Substituting 20 in place of x in the expression for [Image] gives 3(20) + 35 = 60 + 35 = 95. So [Image] = 180 – [Image] which gives 180 – 95 = 85.)
- 20° ([Image] and [Image] are a linear pair and are therefore supplementary. First find [Image]. [Image] and [Image] are vertical angles and are therefore congruent. So set the two expressions equal and solve for x. 3x + 35 = 4x + 15 and x = 20. Substituting 20 in place of x in the expression for [Image] gives 3(20) + 35 = 60 + 35 = 95. So [Image] = 180 – [Image] which gives 180 – 95 = 85.)
- 50° ([Image] and [Image] are a linear pair and are therefore supplementary. First find [Image]. [Image] and [Image] are vertical angles and are therefore congruent. So set the two expressions equal and solve for x. 3x + 35 = 4x + 15 and x = 20. Substituting 20 in place of x in the expression for [Image] gives 3(20) + 35 = 60 + 35 = 95. So [Image] = 180 – [Image] which gives 180 – 95 = 85.)
- 95° ([Image] and [Image] are a linear pair and are therefore supplementary. First find [Image]. [Image] and [Image] are vertical angles and are therefore congruent. So set the two expressions equal and solve for x. 3x + 35 = 4x + 15 and x = 20. Substituting 20 in place of x in the expression for [Image] gives 3(20) + 35 = 60 + 35 = 95. So [Image] = 180 – [Image] which gives 180 – 95 = 85.)
A: 85°

Q: [Image]
If [Image] = 40, [Image] = 120, and [Image] = 110, find [Image].
- 40 (The measure of arc BC is 2 ´ [Image]. Since the full circle measures 360, then m arc AD is 360 – ( [Image] + [Image] + [Image]). Finally, since [Image] is an inscribed angle for arc [Image], its measure is half m arc AD.)
- 25 (The measure of arc BC is 2 ´ [Image]. Since the full circle measures 360, then m arc AD is 360 – ( [Image] + [Image] + [Image]). Finally, since [Image] is an inscribed angle for arc [Image], its measure is half m arc AD.)
- 50 (The measure of arc BC is 2 ´ [Image]. Since the full circle measures 360, then m arc AD is 360 – ( [Image] + [Image] + [Image]). Finally, since [Image] is an inscribed angle for arc [Image], its measure is half m arc AD.)
- 80 (The measure of arc BC is 2 ´ [Image]. Since the full circle measures 360, then m arc AD is 360 – ( [Image] + [Image] + [Image]). Finally, since [Image] is an inscribed angle for arc [Image], its measure is half m arc AD.)
A: 25

Q: Find the exact circumference of the circle.
[Image]
- 7p cm (The circumference formula is Diameter ´ p. The diameter shown also happens to be the hypotenuse of the right triangle inscribed in the circle, so it can be found by using the Pythagorean Theorem. Solve the equation [Image], which results in [Image].)
- 5p cm (The circumference formula is Diameter ´ p. The diameter shown also happens to be the hypotenuse of the right triangle inscribed in the circle, so it can be found by using the Pythagorean Theorem. Solve the equation [Image], which results in [Image].)
- 10p cm (The circumference formula is Diameter ´ p. The diameter shown also happens to be the hypotenuse of the right triangle inscribed in the circle, so it can be found by using the Pythagorean Theorem. Solve the equation [Image], which results in [Image].)
- 4p cm (The circumference formula is Diameter ´ p. The diameter shown also happens to be the hypotenuse of the right triangle inscribed in the circle, so it can be found by using the Pythagorean Theorem. Solve the equation [Image], which results in [Image].)
A: 5p cm

Q: Use the diagram to find the measure of the given angle.
[Image]

[Image]
- 50 ([Image] is a vertical angle with [Image], so they are congruent. Therefore, [Image] is 50.)
- 60 ([Image] is a vertical angle with [Image], so they are congruent. Therefore, [Image] is 50.)
- 130 ([Image] is a vertical angle with [Image], so they are congruent. Therefore, [Image] is 50.)
- 40 ([Image] is a vertical angle with [Image], so they are congruent. Therefore, [Image] is 50.)
A: 50

Q: Use the diagram to find the measure of the given angle.
[Image]

[Image]
- 85° ([Image] and [Image] are a linear pair and are therefore supplementary. First find [Image]. [Image] and [Image] are vertical angles and are therefore congruent. So set the two expressions equal and solve for x. 3x + 35 = 4x + 15 and x = 20. Substituting 20 in place of x in the expression for [Image] gives 4(20) + 15 = 80 + 15 = 95. So [Image] = 180 – [Image] which gives 180 – 95 = 85.)
- 20° ([Image] and [Image] are a linear pair and are therefore supplementary. First find [Image]. [Image] and [Image] are vertical angles and are therefore congruent. So set the two expressions equal and solve for x. 3x + 35 = 4x + 15 and x = 20. Substituting 20 in place of x in the expression for [Image] gives 4(20) + 15 = 80 + 15 = 95. So [Image] = 180 – [Image] which gives 180 – 95 = 85.)
- 50° ([Image] and [Image] are a linear pair and are therefore supplementary. First find [Image]. [Image] and [Image] are vertical angles and are therefore congruent. So set the two expressions equal and solve for x. 3x + 35 = 4x + 15 and x = 20. Substituting 20 in place of x in the expression for [Image] gives 4(20) + 15 = 80 + 15 = 95. So [Image] = 180 – [Image] which gives 180 – 95 = 85.)
- 95° ([Image] and [Image] are a linear pair and are therefore supplementary. First find [Image]. [Image] and [Image] are vertical angles and are therefore congruent. So set the two expressions equal and solve for x. 3x + 35 = 4x + 15 and x = 20. Substituting 20 in place of x in the expression for [Image] gives 4(20) + 15 = 80 + 15 = 95. So [Image] = 180 – [Image] which gives 180 – 95 = 85.)
A: 85°

Q: Find the measure of the numbered angle.

[Image]
- 92 (When two secants intersect in the interior of a circle, then the measure of an angle formed by this intersection is equal to one-half the sum of the measures of the arcs intercepted by the angle and its vertical angle. In this diagram the measures of the intercepted arcs for [Image] are not given. However, the full circle measures 360°, so 2x° + 2x° +4x° +4x° = 360°. Solving this equation, x = 30°. So the intercepted arcs for [Image] are 60° and 120°. That means [Image].)
- 95 (When two secants intersect in the interior of a circle, then the measure of an angle formed by this intersection is equal to one-half the sum of the measures of the arcs intercepted by the angle and its vertical angle. In this diagram the measures of the intercepted arcs for [Image] are not given. However, the full circle measures 360°, so 2x° + 2x° +4x° +4x° = 360°. Solving this equation, x = 30°. So the intercepted arcs for [Image] are 60° and 120°. That means [Image].)
- 94 (When two secants intersect in the interior of a circle, then the measure of an angle formed by this intersection is equal to one-half the sum of the measures of the arcs intercepted by the angle and its vertical angle. In this diagram the measures of the intercepted arcs for [Image] are not given. However, the full circle measures 360°, so 2x° + 2x° +4x° +4x° = 360°. Solving this equation, x = 30°. So the intercepted arcs for [Image] are 60° and 120°. That means [Image].)
- 90 (When two secants intersect in the interior of a circle, then the measure of an angle formed by this intersection is equal to one-half the sum of the measures of the arcs intercepted by the angle and its vertical angle. In this diagram the measures of the intercepted arcs for [Image] are not given. However, the full circle measures 360°, so 2x° + 2x° +4x° +4x° = 360°. Solving this equation, x = 30°. So the intercepted arcs for [Image] are 60° and 120°. That means [Image].)
A: 90

Q: Find x. Assume that any segment that appears to be tangent is tangent.

[Image]
- 35 (When a secant and tangent intersect in the exterior of a circle, then the measure of the angle formed is equal to one-half the positive difference of the measures of the intercepted arcs. Here that gives the equation [Image]. Solving this equation results in an answer of x = 25. Note that the intercepted arc nearest the angle is 40 since 360 – 230 – 90 = 40.)
- 20 (When a secant and tangent intersect in the exterior of a circle, then the measure of the angle formed is equal to one-half the positive difference of the measures of the intercepted arcs. Here that gives the equation [Image]. Solving this equation results in an answer of x = 25. Note that the intercepted arc nearest the angle is 40 since 360 – 230 – 90 = 40.)
- 25 (When a secant and tangent intersect in the exterior of a circle, then the measure of the angle formed is equal to one-half the positive difference of the measures of the intercepted arcs. Here that gives the equation [Image]. Solving this equation results in an answer of x = 25. Note that the intercepted arc nearest the angle is 40 since 360 – 230 – 90 = 40.)
- 30 (When a secant and tangent intersect in the exterior of a circle, then the measure of the angle formed is equal to one-half the positive difference of the measures of the intercepted arcs. Here that gives the equation [Image]. Solving this equation results in an answer of x = 25. Note that the intercepted arc nearest the angle is 40 since 360 – 230 – 90 = 40.)
A: 25

Q: Find x. Assume that segments that appear tangent are tangent.
[Image]
- 7 (The triangle shown is a right triangle since the tangent segment, FE, intersects a radius, DE, which always results in a right angle. So to solve for x, use the Pythagorean Theorem. Note that [Image] since they are both radii of the same circle. The equation from the Pythagorean Theorem is [Image].)
- 6 (The triangle shown is a right triangle since the tangent segment, FE, intersects a radius, DE, which always results in a right angle. So to solve for x, use the Pythagorean Theorem. Note that [Image] since they are both radii of the same circle. The equation from the Pythagorean Theorem is [Image].)
- 14 (The triangle shown is a right triangle since the tangent segment, FE, intersects a radius, DE, which always results in a right angle. So to solve for x, use the Pythagorean Theorem. Note that [Image] since they are both radii of the same circle. The equation from the Pythagorean Theorem is [Image].)
- 5 (The triangle shown is a right triangle since the tangent segment, FE, intersects a radius, DE, which always results in a right angle. So to solve for x, use the Pythagorean Theorem. Note that [Image] since they are both radii of the same circle. The equation from the Pythagorean Theorem is [Image].)
A: 6

Q: Find x. Round to the nearest tenth if necessary.

[Image]
- 2 (The products of the segments for each intersecting chord are equal, so 10 · x = 4 · 8. Solving this equation, x = 3.2.)
- 2.4 (The products of the segments for each intersecting chord are equal, so 10 · x = 4 · 8. Solving this equation, x = 3.2.)
- 3.2 (The products of the segments for each intersecting chord are equal, so 10 · x = 4 · 8. Solving this equation, x = 3.2.)
- 3.5 (The products of the segments for each intersecting chord are equal, so 10 · x = 4 · 8. Solving this equation, x = 3.2.)
A: 3.2

Q: Find the measure of the numbered angle.

[Image]
- 62.5 (When two secants intersect in the interior of a circle, then the measure of an angle formed by this intersection is equal to one-half the sum of the measures of the arcs intercepted by the angle and its vertical angle. In this diagram, the measure of one of the intercepted arcs for [Image] is not given, but it can be found since the sum of all of the arcs must be 360. Subtracting the sum of the other 3 arcs from 360, leaves 110. That means [Image].)
- 105 (When two secants intersect in the interior of a circle, then the measure of an angle formed by this intersection is equal to one-half the sum of the measures of the arcs intercepted by the angle and its vertical angle. In this diagram, the measure of one of the intercepted arcs for [Image] is not given, but it can be found since the sum of all of the arcs must be 360. Subtracting the sum of the other 3 arcs from 360, leaves 110. That means [Image].)
- 112.5 (When two secants intersect in the interior of a circle, then the measure of an angle formed by this intersection is equal to one-half the sum of the measures of the arcs intercepted by the angle and its vertical angle. In this diagram, the measure of one of the intercepted arcs for [Image] is not given, but it can be found since the sum of all of the arcs must be 360. Subtracting the sum of the other 3 arcs from 360, leaves 110. That means [Image].)
- 115 (When two secants intersect in the interior of a circle, then the measure of an angle formed by this intersection is equal to one-half the sum of the measures of the arcs intercepted by the angle and its vertical angle. In this diagram, the measure of one of the intercepted arcs for [Image] is not given, but it can be found since the sum of all of the arcs must be 360. Subtracting the sum of the other 3 arcs from 360, leaves 110. That means [Image].)
A: 112.5

Q: Find x. Assume that any segment that appears to be tangent is tangent.

[Image]
- 10 (When two secants intersect in the exterior of a circle, then the measure of the angle formed is equal to one-half the positive difference of the measures of the intercepted arcs. Here that gives the equation [Image]. Note that the measure of the intercepted arc nearest the angle is 10 because 180 – 130 – 40 = 10.)
- 5 (When two secants intersect in the exterior of a circle, then the measure of the angle formed is equal to one-half the positive difference of the measures of the intercepted arcs. Here that gives the equation [Image]. Note that the measure of the intercepted arc nearest the angle is 10 because 180 – 130 – 40 = 10.)
- 12 (When two secants intersect in the exterior of a circle, then the measure of the angle formed is equal to one-half the positive difference of the measures of the intercepted arcs. Here that gives the equation [Image]. Note that the measure of the intercepted arc nearest the angle is 10 because 180 – 130 – 40 = 10.)
- 15 (When two secants intersect in the exterior of a circle, then the measure of the angle formed is equal to one-half the positive difference of the measures of the intercepted arcs. Here that gives the equation [Image]. Note that the measure of the intercepted arc nearest the angle is 10 because 180 – 130 – 40 = 10.)
A: 15

Q: Find x. Assume that any segment that appears to be tangent is tangent.

[Image]
- 65 (When two secants intersect in the exterior of a circle, then the measure of the angle formed is equal to one-half the positive difference of the measures of the intercepted arcs. Here that gives the equation [Image]. Note that the measure of the intercepted arc farthest from the angle is 168 because 360 – 92 – 38 – 62 = 168.)
- 66 (When two secants intersect in the exterior of a circle, then the measure of the angle formed is equal to one-half the positive difference of the measures of the intercepted arcs. Here that gives the equation [Image]. Note that the measure of the intercepted arc farthest from the angle is 168 because 360 – 92 – 38 – 62 = 168.)
- 68 (When two secants intersect in the exterior of a circle, then the measure of the angle formed is equal to one-half the positive difference of the measures of the intercepted arcs. Here that gives the equation [Image]. Note that the measure of the intercepted arc farthest from the angle is 168 because 360 – 92 – 38 – 62 = 168.)
- 62 (When two secants intersect in the exterior of a circle, then the measure of the angle formed is equal to one-half the positive difference of the measures of the intercepted arcs. Here that gives the equation [Image]. Note that the measure of the intercepted arc farthest from the angle is 168 because 360 – 92 – 38 – 62 = 168.)
A: 65

Q: Find the measure of the numbered angle.

[Image]
- 60 (When two secants intersect in the interior of a circle, then the measure of an angle formed by this intersection is equal to one-half the sum of the measures of the arcs intercepted by the angle and its vertical angle. In this diagram, the measures of the intercepted arcs for [Image] are not given, but they must have a sum of 120° since the arcs shown have a sum of 240 (360 – 240 = 120). That means [Image].)
- 70 (When two secants intersect in the interior of a circle, then the measure of an angle formed by this intersection is equal to one-half the sum of the measures of the arcs intercepted by the angle and its vertical angle. In this diagram, the measures of the intercepted arcs for [Image] are not given, but they must have a sum of 120° since the arcs shown have a sum of 240 (360 – 240 = 120). That means [Image].)
- 80 (When two secants intersect in the interior of a circle, then the measure of an angle formed by this intersection is equal to one-half the sum of the measures of the arcs intercepted by the angle and its vertical angle. In this diagram, the measures of the intercepted arcs for [Image] are not given, but they must have a sum of 120° since the arcs shown have a sum of 240 (360 – 240 = 120). That means [Image].)
- 65 (When two secants intersect in the interior of a circle, then the measure of an angle formed by this intersection is equal to one-half the sum of the measures of the arcs intercepted by the angle and its vertical angle. In this diagram, the measures of the intercepted arcs for [Image] are not given, but they must have a sum of 120° since the arcs shown have a sum of 240 (360 – 240 = 120). That means [Image].)
A: 60

Q: Find x. Round to the nearest tenth if necessary.

[Image]
- 7 (The products of the segments for each intersecting chord are equal, so 2 · x = 3 · 6. Solving this equation, x = 9.)
- 8 (The products of the segments for each intersecting chord are equal, so 2 · x = 3 · 6. Solving this equation, x = 9.)
- 9 (The products of the segments for each intersecting chord are equal, so 2 · x = 3 · 6. Solving this equation, x = 9.)
- 10 (The products of the segments for each intersecting chord are equal, so 2 · x = 3 · 6. Solving this equation, x = 9.)
A: 9

Q: Find x. Assume that any segment that appears to be tangent is tangent.

[Image]
- 15 (When two secants intersect in the exterior of a circle, then the measure of the angle formed is equal to one-half the positive difference of the measures of the intercepted arcs. Here that gives the equation [Image].)
- 35 (When two secants intersect in the exterior of a circle, then the measure of the angle formed is equal to one-half the positive difference of the measures of the intercepted arcs. Here that gives the equation [Image].)
- 25 (When two secants intersect in the exterior of a circle, then the measure of the angle formed is equal to one-half the positive difference of the measures of the intercepted arcs. Here that gives the equation [Image].)
- 45 (When two secants intersect in the exterior of a circle, then the measure of the angle formed is equal to one-half the positive difference of the measures of the intercepted arcs. Here that gives the equation [Image].)
A: 25

Q: Find the measure of the numbered angle.

[Image]
- 70 (When a secant intersects a tangent at the point of tangency, then the measure of the angle formed is one-half the measure of the intercepted arc. In this diagram, the measure of the intercepted arc for [Image] is 140 since 360 – 220 = 140. That means [Image].)
- 75 (When a secant intersects a tangent at the point of tangency, then the measure of the angle formed is one-half the measure of the intercepted arc. In this diagram, the measure of the intercepted arc for [Image] is 140 since 360 – 220 = 140. That means [Image].)
- 80 (When a secant intersects a tangent at the point of tangency, then the measure of the angle formed is one-half the measure of the intercepted arc. In this diagram, the measure of the intercepted arc for [Image] is 140 since 360 – 220 = 140. That means [Image].)
- 85 (When a secant intersects a tangent at the point of tangency, then the measure of the angle formed is one-half the measure of the intercepted arc. In this diagram, the measure of the intercepted arc for [Image] is 140 since 360 – 220 = 140. That means [Image].)
A: 70

Q: Find x. Assume that any segment that appears to be tangent is tangent.

[Image]
- 22 (When a secant and tangent intersect in the exterior of a circle, then the measure of the angle formed is equal to one-half the positive difference of the measures of the intercepted arcs. Here that gives the equation [Image]. Solving this equation results in an answer of x = 22.)
- 9 (When a secant and tangent intersect in the exterior of a circle, then the measure of the angle formed is equal to one-half the positive difference of the measures of the intercepted arcs. Here that gives the equation [Image]. Solving this equation results in an answer of x = 22.)
- 18 (When a secant and tangent intersect in the exterior of a circle, then the measure of the angle formed is equal to one-half the positive difference of the measures of the intercepted arcs. Here that gives the equation [Image]. Solving this equation results in an answer of x = 22.)
- 11 (When a secant and tangent intersect in the exterior of a circle, then the measure of the angle formed is equal to one-half the positive difference of the measures of the intercepted arcs. Here that gives the equation [Image]. Solving this equation results in an answer of x = 22.)
A: 22

Q: Find x. Assume that any segment that appears to be tangent is tangent.

[Image]
- 50 (When two secants intersect in the exterior of a circle, then the measure of the angle formed is equal to one-half the positive difference of the measures of the intercepted arcs. Here that gives the equation [Image].)
- 40 (When two secants intersect in the exterior of a circle, then the measure of the angle formed is equal to one-half the positive difference of the measures of the intercepted arcs. Here that gives the equation [Image].)
- 60 (When two secants intersect in the exterior of a circle, then the measure of the angle formed is equal to one-half the positive difference of the measures of the intercepted arcs. Here that gives the equation [Image].)
- 70 (When two secants intersect in the exterior of a circle, then the measure of the angle formed is equal to one-half the positive difference of the measures of the intercepted arcs. Here that gives the equation [Image].)
A: 40

Q: Find x. Assume that any segment that appears to be tangent is tangent.

[Image]
- 15 (When two secants intersect in the exterior of a circle, then the measure of the angle formed is equal to one-half the positive difference of the measures of the intercepted arcs. Here that gives the equation [Image]. Solving this equation results in the answer x = 15.)
- 30 (When two secants intersect in the exterior of a circle, then the measure of the angle formed is equal to one-half the positive difference of the measures of the intercepted arcs. Here that gives the equation [Image]. Solving this equation results in the answer x = 15.)
- 20 (When two secants intersect in the exterior of a circle, then the measure of the angle formed is equal to one-half the positive difference of the measures of the intercepted arcs. Here that gives the equation [Image]. Solving this equation results in the answer x = 15.)
- 10 (When two secants intersect in the exterior of a circle, then the measure of the angle formed is equal to one-half the positive difference of the measures of the intercepted arcs. Here that gives the equation [Image]. Solving this equation results in the answer x = 15.)
A: 15

Q: Find x. Assume that segments that appear tangent are tangent.
[Image]
- 9 (The triangle shown is a right triangle since the tangent segment, CB, intersects a radius, AB, which always results in a right angle. So to solve for x, use the Pythagorean Theorem. The equation is [Image] which results in [Image].)
- 7 (The triangle shown is a right triangle since the tangent segment, CB, intersects a radius, AB, which always results in a right angle. So to solve for x, use the Pythagorean Theorem. The equation is [Image] which results in [Image].)
- 12 (The triangle shown is a right triangle since the tangent segment, CB, intersects a radius, AB, which always results in a right angle. So to solve for x, use the Pythagorean Theorem. The equation is [Image] which results in [Image].)
- 17 (The triangle shown is a right triangle since the tangent segment, CB, intersects a radius, AB, which always results in a right angle. So to solve for x, use the Pythagorean Theorem. The equation is [Image] which results in [Image].)
A: 12

Q: Find x. Round to the nearest tenth if necessary.

[Image]
- 2 (The products of the segments for each intersecting chord are equal, so 8 · x = 4 · 6. Solving this equation, x = 3.)
- 3 (The products of the segments for each intersecting chord are equal, so 8 · x = 4 · 6. Solving this equation, x = 3.)
- 2.5 (The products of the segments for each intersecting chord are equal, so 8 · x = 4 · 6. Solving this equation, x = 3.)
- 3.5 (The products of the segments for each intersecting chord are equal, so 8 · x = 4 · 6. Solving this equation, x = 3.)
A: 3

Q: Find x. Round to the nearest tenth if necessary.

[Image]
- 10.5 (The products of the segments for each intersecting chord are equal, so 2 · x = 3 · 7. Solving this equation, x = 10.5.)
- 2.4 (The products of the segments for each intersecting chord are equal, so 2 · x = 3 · 7. Solving this equation, x = 10.5.)
- 3.2 (The products of the segments for each intersecting chord are equal, so 2 · x = 3 · 7. Solving this equation, x = 10.5.)
- 3.5 (The products of the segments for each intersecting chord are equal, so 2 · x = 3 · 7. Solving this equation, x = 10.5.)
A: 10.5

Q: Find the measure of the numbered angle.

[Image]
- 90 (When a secant intersects a tangent at the point of tangency, then the measure of the angle formed is one-half the measure of the intercepted arc. In this diagram, the measure of the intercepted arc for [Image] is 220. That means [Image].)
- 180 (When a secant intersects a tangent at the point of tangency, then the measure of the angle formed is one-half the measure of the intercepted arc. In this diagram, the measure of the intercepted arc for [Image] is 220. That means [Image].)
- 220 (When a secant intersects a tangent at the point of tangency, then the measure of the angle formed is one-half the measure of the intercepted arc. In this diagram, the measure of the intercepted arc for [Image] is 220. That means [Image].)
- 110 (When a secant intersects a tangent at the point of tangency, then the measure of the angle formed is one-half the measure of the intercepted arc. In this diagram, the measure of the intercepted arc for [Image] is 220. That means [Image].)
A: 110

Q: Find x. Assume that segments that appear tangent are tangent.

[Image]
[Image]
- 11 (Recall that two tangents from the same external point are congruent. So, for example, [Image] and [Image]. Those two equalities, plus the fact that [Image], lead us to the equation [Image]. Substitute the appropriate values into this equation and solve for x.)
- 29 (Recall that two tangents from the same external point are congruent. So, for example, [Image] and [Image]. Those two equalities, plus the fact that [Image], lead us to the equation [Image]. Substitute the appropriate values into this equation and solve for x.)
- 13 (Recall that two tangents from the same external point are congruent. So, for example, [Image] and [Image]. Those two equalities, plus the fact that [Image], lead us to the equation [Image]. Substitute the appropriate values into this equation and solve for x.)
- 16 (Recall that two tangents from the same external point are congruent. So, for example, [Image] and [Image]. Those two equalities, plus the fact that [Image], lead us to the equation [Image]. Substitute the appropriate values into this equation and solve for x.)
A: 11

Q: Find the measure of the numbered angle.

[Image]
- 55 (When two secants intersect in the interior of a circle, then the measure of an angle formed by this intersection is equal to one-half the sum of the measures of the arcs intercepted by the angle and its vertical angle. In this diagram, that means [Image].)
- 65 (When two secants intersect in the interior of a circle, then the measure of an angle formed by this intersection is equal to one-half the sum of the measures of the arcs intercepted by the angle and its vertical angle. In this diagram, that means [Image].)
- 75 (When two secants intersect in the interior of a circle, then the measure of an angle formed by this intersection is equal to one-half the sum of the measures of the arcs intercepted by the angle and its vertical angle. In this diagram, that means [Image].)
- 60 (When two secants intersect in the interior of a circle, then the measure of an angle formed by this intersection is equal to one-half the sum of the measures of the arcs intercepted by the angle and its vertical angle. In this diagram, that means [Image].)
A: 65

Q: Find the measure of the numbered angle.

[Image]
- 115 (When two secants intersect in the interior of a circle, then the measure of an angle formed by this intersection is equal to one-half the sum of the measures of the arcs intercepted by the angle and its vertical angle. In this diagram, the measures of the intercepted arcs for [Image] are not given, but they have a sum of 250 since the arcs shown have a sum of 110 (360 – 110 = 250). That means [Image].)
- 125 (When two secants intersect in the interior of a circle, then the measure of an angle formed by this intersection is equal to one-half the sum of the measures of the arcs intercepted by the angle and its vertical angle. In this diagram, the measures of the intercepted arcs for [Image] are not given, but they have a sum of 250 since the arcs shown have a sum of 110 (360 – 110 = 250). That means [Image].)
- 120 (When two secants intersect in the interior of a circle, then the measure of an angle formed by this intersection is equal to one-half the sum of the measures of the arcs intercepted by the angle and its vertical angle. In this diagram, the measures of the intercepted arcs for [Image] are not given, but they have a sum of 250 since the arcs shown have a sum of 110 (360 – 110 = 250). That means [Image].)
- 130 (When two secants intersect in the interior of a circle, then the measure of an angle formed by this intersection is equal to one-half the sum of the measures of the arcs intercepted by the angle and its vertical angle. In this diagram, the measures of the intercepted arcs for [Image] are not given, but they have a sum of 250 since the arcs shown have a sum of 110 (360 – 110 = 250). That means [Image].)
A: 125

Q: Find x. Assume that any segment that appears to be tangent is tangent.

[Image]
- 15 (When a secant and tangent intersect in the exterior of a circle, then the measure of the angle formed is equal to one-half the positive difference of the measures of the intercepted arcs. Here that gives the equation [Image]. Solving this equation results in an answer of x = 25.)
- 20 (When a secant and tangent intersect in the exterior of a circle, then the measure of the angle formed is equal to one-half the positive difference of the measures of the intercepted arcs. Here that gives the equation [Image]. Solving this equation results in an answer of x = 25.)
- 25 (When a secant and tangent intersect in the exterior of a circle, then the measure of the angle formed is equal to one-half the positive difference of the measures of the intercepted arcs. Here that gives the equation [Image]. Solving this equation results in an answer of x = 25.)
- 30 (When a secant and tangent intersect in the exterior of a circle, then the measure of the angle formed is equal to one-half the positive difference of the measures of the intercepted arcs. Here that gives the equation [Image]. Solving this equation results in an answer of x = 25.)
A: 25

Q: Find x. Assume that segments that appear tangent are tangent.
[Image]
- 7 (Recall that two tangents from the same external point are congruent. So, for example, [Image] and [Image]. Those two equalities, plus the fact that [Image] allows us to make the equality [Image]. Substitute the appropriate values into that equation and solve for x.)
- 5 (Recall that two tangents from the same external point are congruent. So, for example, [Image] and [Image]. Those two equalities, plus the fact that [Image] allows us to make the equality [Image]. Substitute the appropriate values into that equation and solve for x.)
- 9 (Recall that two tangents from the same external point are congruent. So, for example, [Image] and [Image]. Those two equalities, plus the fact that [Image] allows us to make the equality [Image]. Substitute the appropriate values into that equation and solve for x.)
- 3 (Recall that two tangents from the same external point are congruent. So, for example, [Image] and [Image]. Those two equalities, plus the fact that [Image] allows us to make the equality [Image]. Substitute the appropriate values into that equation and solve for x.)
A: 5

Q: Find x. Assume that any segment that appears to be tangent is tangent.

[Image]
- 47 (When a secant and tangent intersect in the exterior of a circle, then the measure of the angle formed is equal to one-half the positive difference of the measures of the intercepted arcs. Here that gives the equation [Image]. Solving this equation results in an answer of x = 43.)
- 48 (When a secant and tangent intersect in the exterior of a circle, then the measure of the angle formed is equal to one-half the positive difference of the measures of the intercepted arcs. Here that gives the equation [Image]. Solving this equation results in an answer of x = 43.)
- 44 (When a secant and tangent intersect in the exterior of a circle, then the measure of the angle formed is equal to one-half the positive difference of the measures of the intercepted arcs. Here that gives the equation [Image]. Solving this equation results in an answer of x = 43.)
- 43 (When a secant and tangent intersect in the exterior of a circle, then the measure of the angle formed is equal to one-half the positive difference of the measures of the intercepted arcs. Here that gives the equation [Image]. Solving this equation results in an answer of x = 43.)
A: 43

Q: Find the measure of the numbered angle.

[Image]
- 230 (When two secants intersect in the interior of a circle, then the measure of an angle formed by this intersection is equal to one-half the sum of the measures of the arcs intercepted by the angle and its vertical angle. In this diagram, that means [Image].)
- 115 (When two secants intersect in the interior of a circle, then the measure of an angle formed by this intersection is equal to one-half the sum of the measures of the arcs intercepted by the angle and its vertical angle. In this diagram, that means [Image].)
- 130 (When two secants intersect in the interior of a circle, then the measure of an angle formed by this intersection is equal to one-half the sum of the measures of the arcs intercepted by the angle and its vertical angle. In this diagram, that means [Image].)
- 125 (When two secants intersect in the interior of a circle, then the measure of an angle formed by this intersection is equal to one-half the sum of the measures of the arcs intercepted by the angle and its vertical angle. In this diagram, that means [Image].)
A: 115

Q: Find the measure of the numbered angle.

[Image]
- 100 (When a secant intersects a tangent at the point of tangency, then the measure of the angle formed is one-half the measure of the intercepted arc. In this diagram, the measure of the intercepted arc for [Image] is 180 because the chord is also a diameter of the circle. That means [Image].)
- 180 (When a secant intersects a tangent at the point of tangency, then the measure of the angle formed is one-half the measure of the intercepted arc. In this diagram, the measure of the intercepted arc for [Image] is 180 because the chord is also a diameter of the circle. That means [Image].)
- 90 (When a secant intersects a tangent at the point of tangency, then the measure of the angle formed is one-half the measure of the intercepted arc. In this diagram, the measure of the intercepted arc for [Image] is 180 because the chord is also a diameter of the circle. That means [Image].)
- 95 (When a secant intersects a tangent at the point of tangency, then the measure of the angle formed is one-half the measure of the intercepted arc. In this diagram, the measure of the intercepted arc for [Image] is 180 because the chord is also a diameter of the circle. That means [Image].)
A: 90

Q: Find x. Round to the nearest tenth if necessary.

[Image]
- 2 (The products of the segments for each intersecting chord are equal, so 8 · x = 4 · 4. Solving this equation, x = 2.)
- 1 (The products of the segments for each intersecting chord are equal, so 8 · x = 4 · 4. Solving this equation, x = 2.)
- 3 (The products of the segments for each intersecting chord are equal, so 8 · x = 4 · 4. Solving this equation, x = 2.)
- 4 (The products of the segments for each intersecting chord are equal, so 8 · x = 4 · 4. Solving this equation, x = 2.)
A: 2

Q: Find x. Round to the nearest tenth if necessary.

[Image]
- 5 (The products of the segments for each intersecting chord are equal, so 3 · x = 6 · 2. Solving this equation, x = 4.)
- 6 (The products of the segments for each intersecting chord are equal, so 3 · x = 6 · 2. Solving this equation, x = 4.)
- 3 (The products of the segments for each intersecting chord are equal, so 3 · x = 6 · 2. Solving this equation, x = 4.)
- 4 (The products of the segments for each intersecting chord are equal, so 3 · x = 6 · 2. Solving this equation, x = 4.)
A: 4

Q: Find x. Assume that any segment that appears to be tangent is tangent.

[Image]
- 12 (When two secants intersect in the exterior of a circle, then the measure of the angle formed is equal to one-half the positive difference of the measures of the intercepted arcs. Here that gives the equation [Image]. Solving this equation results in the answer x = 10.)
- 14 (When two secants intersect in the exterior of a circle, then the measure of the angle formed is equal to one-half the positive difference of the measures of the intercepted arcs. Here that gives the equation [Image]. Solving this equation results in the answer x = 10.)
- 8 (When two secants intersect in the exterior of a circle, then the measure of the angle formed is equal to one-half the positive difference of the measures of the intercepted arcs. Here that gives the equation [Image]. Solving this equation results in the answer x = 10.)
- 10 (When two secants intersect in the exterior of a circle, then the measure of the angle formed is equal to one-half the positive difference of the measures of the intercepted arcs. Here that gives the equation [Image]. Solving this equation results in the answer x = 10.)
A: 10

Q: Find the measure of the numbered angle.

[Image]
- 81 (When two secants intersect in the interior of a circle, then the measure of an angle formed by this intersection is equal to one-half the sum of the measures of the arcs intercepted by the angle and its vertical angle. In this diagram the measures of the intercepted arcs for [Image] are not given. However, the full circle measures 360°, so 3a° + 4a° +6a° +7a° = 360°. Solving this equation, a = 18. So the intercepted arcs for [Image] are 54° and 108°. That means [Image].)
- 90 (When two secants intersect in the interior of a circle, then the measure of an angle formed by this intersection is equal to one-half the sum of the measures of the arcs intercepted by the angle and its vertical angle. In this diagram the measures of the intercepted arcs for [Image] are not given. However, the full circle measures 360°, so 3a° + 4a° +6a° +7a° = 360°. Solving this equation, a = 18. So the intercepted arcs for [Image] are 54° and 108°. That means [Image].)
- 94 (When two secants intersect in the interior of a circle, then the measure of an angle formed by this intersection is equal to one-half the sum of the measures of the arcs intercepted by the angle and its vertical angle. In this diagram the measures of the intercepted arcs for [Image] are not given. However, the full circle measures 360°, so 3a° + 4a° +6a° +7a° = 360°. Solving this equation, a = 18. So the intercepted arcs for [Image] are 54° and 108°. That means [Image].)
- 102 (When two secants intersect in the interior of a circle, then the measure of an angle formed by this intersection is equal to one-half the sum of the measures of the arcs intercepted by the angle and its vertical angle. In this diagram the measures of the intercepted arcs for [Image] are not given. However, the full circle measures 360°, so 3a° + 4a° +6a° +7a° = 360°. Solving this equation, a = 18. So the intercepted arcs for [Image] are 54° and 108°. That means [Image].)
A: 81

Q: Write an equation for a circle with center at (–6, 10) and diameter = 6.
- [Image] (The equation of a circle is [Image] where (h, k) is the center and r is the radius. For example, if the center is [Image] and the radius is 4, the equation of the circle would be [Image]. In this problem, the center is given, but not the radius. The radius is one-half the diameter, so first divide the given diameter by 2 to get the radius.)
- [Image] (The equation of a circle is [Image] where (h, k) is the center and r is the radius. For example, if the center is [Image] and the radius is 4, the equation of the circle would be [Image]. In this problem, the center is given, but not the radius. The radius is one-half the diameter, so first divide the given diameter by 2 to get the radius.)
- [Image] (The equation of a circle is [Image] where (h, k) is the center and r is the radius. For example, if the center is [Image] and the radius is 4, the equation of the circle would be [Image]. In this problem, the center is given, but not the radius. The radius is one-half the diameter, so first divide the given diameter by 2 to get the radius.)
- [Image] (The equation of a circle is [Image] where (h, k) is the center and r is the radius. For example, if the center is [Image] and the radius is 4, the equation of the circle would be [Image]. In this problem, the center is given, but not the radius. The radius is one-half the diameter, so first divide the given diameter by 2 to get the radius.)
A: [Image]

A drawing of the area between four buildings at Bay City Recreation Center is shown below. The diagonals shown represent concrete walkways. Points have been added for reference.
[Image]
a.      Which diagonal segments would have to be congruent for ABCD to be a rectangle?
b.      Suppose the length of rectangle ABCD is 3 more than 2 times the width and the perimeter is 150 meters. What are the dimensions of the rectangle?

Q: Find x. Round to the nearest tenth if necessary. Assume that segments that appear to be tangent are tangent.

[Image]
- 5 (When a secant segment and a tangent segment intersect in the exterior of a circle, set the product of each external part of the secant segment and the entire secant segment equal to the square of the tangent segment. Here that gives the equation x · x = 2 · (2 + 6). Solving this equation results in x = 4.)
- 2.8 (When a secant segment and a tangent segment intersect in the exterior of a circle, set the product of each external part of the secant segment and the entire secant segment equal to the square of the tangent segment. Here that gives the equation x · x = 2 · (2 + 6). Solving this equation results in x = 4.)
- 3.5 (When a secant segment and a tangent segment intersect in the exterior of a circle, set the product of each external part of the secant segment and the entire secant segment equal to the square of the tangent segment. Here that gives the equation x · x = 2 · (2 + 6). Solving this equation results in x = 4.)
- 4 (When a secant segment and a tangent segment intersect in the exterior of a circle, set the product of each external part of the secant segment and the entire secant segment equal to the square of the tangent segment. Here that gives the equation x · x = 2 · (2 + 6). Solving this equation results in x = 4.)
A: 4

Q: Find x. Round to the nearest tenth if necessary. Assume that segments that appear to be tangent are tangent.

[Image]
- 9 (When a secant segment and a tangent segment intersect in the exterior of a circle, set the product of each external part of the secant segment and the entire secant segment equal to the square of the tangent segment. Here that gives the equation 6 · 6 = 3 · (x + 3). Solving this equation results in x = 9.)
- 2 (When a secant segment and a tangent segment intersect in the exterior of a circle, set the product of each external part of the secant segment and the entire secant segment equal to the square of the tangent segment. Here that gives the equation 6 · 6 = 3 · (x + 3). Solving this equation results in x = 9.)
- 3 (When a secant segment and a tangent segment intersect in the exterior of a circle, set the product of each external part of the secant segment and the entire secant segment equal to the square of the tangent segment. Here that gives the equation 6 · 6 = 3 · (x + 3). Solving this equation results in x = 9.)
- 8 (When a secant segment and a tangent segment intersect in the exterior of a circle, set the product of each external part of the secant segment and the entire secant segment equal to the square of the tangent segment. Here that gives the equation 6 · 6 = 3 · (x + 3). Solving this equation results in x = 9.)
A: 9

Q: Find x. Round to the nearest tenth if necessary. Assume that segments that appear to be tangent are tangent.

[Image]
- 8.5 (When two secant segments intersect in the exterior of a circle, set an equality between the product of each external segment and the entire segment. Here that gives the equation 3.5 · (8.2 + 3.5) = 3.2 · (x + 3.2). Solving this equation results in x = 9.6.)
- 9.0 (When two secant segments intersect in the exterior of a circle, set an equality between the product of each external segment and the entire segment. Here that gives the equation 3.5 · (8.2 + 3.5) = 3.2 · (x + 3.2). Solving this equation results in x = 9.6.)
- 9.3 (When two secant segments intersect in the exterior of a circle, set an equality between the product of each external segment and the entire segment. Here that gives the equation 3.5 · (8.2 + 3.5) = 3.2 · (x + 3.2). Solving this equation results in x = 9.6.)
- 9.6 (When two secant segments intersect in the exterior of a circle, set an equality between the product of each external segment and the entire segment. Here that gives the equation 3.5 · (8.2 + 3.5) = 3.2 · (x + 3.2). Solving this equation results in x = 9.6.)
A: 9.6

Q: Find x. Round to the nearest tenth if necessary.

[Image]
- 4.2 (The products of the segments for each intersecting chord are equal, so 4 · x = 5 · 3. Solving this equation, x = 3.75, which rounds to 3.8.)
- 3.8 (The products of the segments for each intersecting chord are equal, so 4 · x = 5 · 3. Solving this equation, x = 3.75, which rounds to 3.8.)
- 3.2 (The products of the segments for each intersecting chord are equal, so 4 · x = 5 · 3. Solving this equation, x = 3.75, which rounds to 3.8.)
- 3.7 (The products of the segments for each intersecting chord are equal, so 4 · x = 5 · 3. Solving this equation, x = 3.75, which rounds to 3.8.)
A: 3.8

Q: Find x. Round to the nearest tenth if necessary.

[Image]
- 6 (The products of the segments for each intersecting chord are equal, so x · (x + 3) = 2 · 9. To solve this equation, first rewrite it as [Image]. This equation factors to [Image]. The solutions are then –6 and 3. Since x represents a measurement of length, it cannot be a negative number. So the answer is x = 3.)
- 5 (The products of the segments for each intersecting chord are equal, so x · (x + 3) = 2 · 9. To solve this equation, first rewrite it as [Image]. This equation factors to [Image]. The solutions are then –6 and 3. Since x represents a measurement of length, it cannot be a negative number. So the answer is x = 3.)
- 4 (The products of the segments for each intersecting chord are equal, so x · (x + 3) = 2 · 9. To solve this equation, first rewrite it as [Image]. This equation factors to [Image]. The solutions are then –6 and 3. Since x represents a measurement of length, it cannot be a negative number. So the answer is x = 3.)
- 3 (The products of the segments for each intersecting chord are equal, so x · (x + 3) = 2 · 9. To solve this equation, first rewrite it as [Image]. This equation factors to [Image]. The solutions are then –6 and 3. Since x represents a measurement of length, it cannot be a negative number. So the answer is x = 3.)
A: 3

Q: Graph the equation.

[Image]
- [Image] (The graph of an equation of the form [Image] will be a circle centered at (0, 0) and with radius r. For example, if the equation is [Image], the graph will be a circle centered at the origin with a radius of 9.)
- [Image] (The graph of an equation of the form [Image] will be a circle centered at (0, 0) and with radius r. For example, if the equation is [Image], the graph will be a circle centered at the origin with a radius of 9.)
- [Image] (The graph of an equation of the form [Image] will be a circle centered at (0, 0) and with radius r. For example, if the equation is [Image], the graph will be a circle centered at the origin with a radius of 9.)
- [Image] (The graph of an equation of the form [Image] will be a circle centered at (0, 0) and with radius r. For example, if the equation is [Image], the graph will be a circle centered at the origin with a radius of 9.)
A: [Image]

Q: Find x. Round to the nearest tenth if necessary. Assume that segments that appear to be tangent are tangent.

[Image]
- 4 (When two secant segments intersect in the exterior of a circle, set an equality between the product of each external segment and the entire segment. Here that gives the equation 6 · (6 + 10 + x) = x · (x + 10 + x). Solving this equation results in x = 6 or x = –8. Since x is a measurement and cannot be negative, the answer is x = 6.)
- 5 (When two secant segments intersect in the exterior of a circle, set an equality between the product of each external segment and the entire segment. Here that gives the equation 6 · (6 + 10 + x) = x · (x + 10 + x). Solving this equation results in x = 6 or x = –8. Since x is a measurement and cannot be negative, the answer is x = 6.)
- 6 (When two secant segments intersect in the exterior of a circle, set an equality between the product of each external segment and the entire segment. Here that gives the equation 6 · (6 + 10 + x) = x · (x + 10 + x). Solving this equation results in x = 6 or x = –8. Since x is a measurement and cannot be negative, the answer is x = 6.)
- 7 (When two secant segments intersect in the exterior of a circle, set an equality between the product of each external segment and the entire segment. Here that gives the equation 6 · (6 + 10 + x) = x · (x + 10 + x). Solving this equation results in x = 6 or x = –8. Since x is a measurement and cannot be negative, the answer is x = 6.)
A: 6

Q: Graph the equation.

[Image]
- [Image] (The graph of an equation of the form [Image] will be a circle centered at [Image] and with radius r. For example, if the equation is [Image], the correct graph would have a center at [Image] and a radius of 3.)
- [Image] (The graph of an equation of the form [Image] will be a circle centered at [Image] and with radius r. For example, if the equation is [Image], the correct graph would have a center at [Image] and a radius of 3.)
- [Image] (The graph of an equation of the form [Image] will be a circle centered at [Image] and with radius r. For example, if the equation is [Image], the correct graph would have a center at [Image] and a radius of 3.)
- [Image] (The graph of an equation of the form [Image] will be a circle centered at [Image] and with radius r. For example, if the equation is [Image], the correct graph would have a center at [Image] and a radius of 3.)
A: [Image]

Q: Write an equation for a circle with a diameter that has endpoints at (–10, 1) and (–8, 5). Round to the nearest tenth if necessary.
- [Image] (The equation of a circle is [Image] where (h, k) is the center and r is the radius. The center is the midpoint of the endpoints of the diameter. Use the distance formula to find the diameter, which is the distance between the endpoints. The radius is one-half the diameter, so the divide the diameter by 2 to get the radius.)
- [Image] (The equation of a circle is [Image] where (h, k) is the center and r is the radius. The center is the midpoint of the endpoints of the diameter. Use the distance formula to find the diameter, which is the distance between the endpoints. The radius is one-half the diameter, so the divide the diameter by 2 to get the radius.)
- [Image] (The equation of a circle is [Image] where (h, k) is the center and r is the radius. The center is the midpoint of the endpoints of the diameter. Use the distance formula to find the diameter, which is the distance between the endpoints. The radius is one-half the diameter, so the divide the diameter by 2 to get the radius.)
- [Image] (The equation of a circle is [Image] where (h, k) is the center and r is the radius. The center is the midpoint of the endpoints of the diameter. Use the distance formula to find the diameter, which is the distance between the endpoints. The radius is one-half the diameter, so the divide the diameter by 2 to get the radius.)
A: [Image]

A segment with endpoints C(-3, 3)and D(7, 3) is the diameter of a circle.
a.      Graph the points and draw the circle.
b.      What is the center of the circle?
c.      What is the length of the radius?
d.      What is the circumference of the circle?
e.      What is the equation of the circle?

Q: Find x. Round to the nearest tenth if necessary. Assume that segments that appear to be tangent are tangent.

[Image]
- 7.2 (When a secant segment and a tangent segment intersect in the exterior of a circle, set the product of each external part of the secant segment and the entire secant segment equal to the square of the tangent segment. Here that gives the equation 8 · 8 = x · (x + 12). To solve this equation, first rewrite it as [Image]. This equation factors to [Image]. The solutions are then –16 and 4. Since x represents a measurement of length, it cannot be a negative number. So the answer is x = 4.)
- 4 (When a secant segment and a tangent segment intersect in the exterior of a circle, set the product of each external part of the secant segment and the entire secant segment equal to the square of the tangent segment. Here that gives the equation 8 · 8 = x · (x + 12). To solve this equation, first rewrite it as [Image]. This equation factors to [Image]. The solutions are then –16 and 4. Since x represents a measurement of length, it cannot be a negative number. So the answer is x = 4.)
- 1.7 (When a secant segment and a tangent segment intersect in the exterior of a circle, set the product of each external part of the secant segment and the entire secant segment equal to the square of the tangent segment. Here that gives the equation 8 · 8 = x · (x + 12). To solve this equation, first rewrite it as [Image]. This equation factors to [Image]. The solutions are then –16 and 4. Since x represents a measurement of length, it cannot be a negative number. So the answer is x = 4.)
- 3 (When a secant segment and a tangent segment intersect in the exterior of a circle, set the product of each external part of the secant segment and the entire secant segment equal to the square of the tangent segment. Here that gives the equation 8 · 8 = x · (x + 12). To solve this equation, first rewrite it as [Image]. This equation factors to [Image]. The solutions are then –16 and 4. Since x represents a measurement of length, it cannot be a negative number. So the answer is x = 4.)
A: 4

Q: Find x. Round to the nearest tenth if necessary. Assume that segments that appear to be tangent are tangent.

[Image]
- 3 (When two secant segments intersect in the exterior of a circle, set an equality between the product of each external segment and the entire segment. Here that gives the equation 2 · 12 = x · (x + 5). Solving this equation results in x = –8, 3. Since x is a measurement and cannot be negative, the answer is x = 3.)
- 7 (When two secant segments intersect in the exterior of a circle, set an equality between the product of each external segment and the entire segment. Here that gives the equation 2 · 12 = x · (x + 5). Solving this equation results in x = –8, 3. Since x is a measurement and cannot be negative, the answer is x = 3.)
- 2 (When two secant segments intersect in the exterior of a circle, set an equality between the product of each external segment and the entire segment. Here that gives the equation 2 · 12 = x · (x + 5). Solving this equation results in x = –8, 3. Since x is a measurement and cannot be negative, the answer is x = 3.)
- 4 (When two secant segments intersect in the exterior of a circle, set an equality between the product of each external segment and the entire segment. Here that gives the equation 2 · 12 = x · (x + 5). Solving this equation results in x = –8, 3. Since x is a measurement and cannot be negative, the answer is x = 3.)
A: 3

Q: Find x. Round to the nearest tenth if necessary. Assume that segments that appear to be tangent are tangent.

[Image]
- 8.7 (When a secant segment and a tangent segment intersect in the exterior of a circle, set the product of each external part of the secant segment and the entire secant segment equal to the square of the tangent segment. Here that gives the equation x · x = 5 · (5 + 15). Solving this equation results in x = 10.)
- 7 (When a secant segment and a tangent segment intersect in the exterior of a circle, set the product of each external part of the secant segment and the entire secant segment equal to the square of the tangent segment. Here that gives the equation x · x = 5 · (5 + 15). Solving this equation results in x = 10.)
- 5 (When a secant segment and a tangent segment intersect in the exterior of a circle, set the product of each external part of the secant segment and the entire secant segment equal to the square of the tangent segment. Here that gives the equation x · x = 5 · (5 + 15). Solving this equation results in x = 10.)
- 10 (When a secant segment and a tangent segment intersect in the exterior of a circle, set the product of each external part of the secant segment and the entire secant segment equal to the square of the tangent segment. Here that gives the equation x · x = 5 · (5 + 15). Solving this equation results in x = 10.)
A: 10

Q: Find x. Round to the nearest tenth if necessary.

[Image]
- 6 (The products of the segments for each intersecting chord are equal, so x · x = 4 · 9. To solve this equation, rewrite it as [Image]. The solutions are then –6 and 6. Since x represents a measurement of length, it cannot be a negative number. So the answer is x = 6.)
- 6.5 (The products of the segments for each intersecting chord are equal, so x · x = 4 · 9. To solve this equation, rewrite it as [Image]. The solutions are then –6 and 6. Since x represents a measurement of length, it cannot be a negative number. So the answer is x = 6.)
- 7 (The products of the segments for each intersecting chord are equal, so x · x = 4 · 9. To solve this equation, rewrite it as [Image]. The solutions are then –6 and 6. Since x represents a measurement of length, it cannot be a negative number. So the answer is x = 6.)
- 7.2 (The products of the segments for each intersecting chord are equal, so x · x = 4 · 9. To solve this equation, rewrite it as [Image]. The solutions are then –6 and 6. Since x represents a measurement of length, it cannot be a negative number. So the answer is x = 6.)
A: 6

Q: Find x. Round to the nearest tenth if necessary. Assume that segments that appear to be tangent are tangent.

[Image]
- 2.3 (When a secant segment and a tangent segment intersect in the exterior of a circle, set the product of each external part of the secant segment and the entire secant segment equal to the square of the tangent segment. Here that gives the equation 9.2 · 9.2 = 6.4 · (3x + 6.4). Solving this equation results in x = 2.3.)
- 0.9 (When a secant segment and a tangent segment intersect in the exterior of a circle, set the product of each external part of the secant segment and the entire secant segment equal to the square of the tangent segment. Here that gives the equation 9.2 · 9.2 = 6.4 · (3x + 6.4). Solving this equation results in x = 2.3.)
- 2.8 (When a secant segment and a tangent segment intersect in the exterior of a circle, set the product of each external part of the secant segment and the entire secant segment equal to the square of the tangent segment. Here that gives the equation 9.2 · 9.2 = 6.4 · (3x + 6.4). Solving this equation results in x = 2.3.)
- 3.4 (When a secant segment and a tangent segment intersect in the exterior of a circle, set the product of each external part of the secant segment and the entire secant segment equal to the square of the tangent segment. Here that gives the equation 9.2 · 9.2 = 6.4 · (3x + 6.4). Solving this equation results in x = 2.3.)
A: 2.3

Q: Find x. Round to the nearest tenth if necessary.

[Image]
- 4 (The products of the segments for each intersecting chord are equal, so x · (x + 1) = 3 · 10. To solve this equation, first rewrite it as [Image]. This equation factors to [Image]. The solutions are then –6 and 5. Since x represents a measurement of length, it cannot be a negative number. So the answer is x = 5.)
- 5.5 (The products of the segments for each intersecting chord are equal, so x · (x + 1) = 3 · 10. To solve this equation, first rewrite it as [Image]. This equation factors to [Image]. The solutions are then –6 and 5. Since x represents a measurement of length, it cannot be a negative number. So the answer is x = 5.)
- 6 (The products of the segments for each intersecting chord are equal, so x · (x + 1) = 3 · 10. To solve this equation, first rewrite it as [Image]. This equation factors to [Image]. The solutions are then –6 and 5. Since x represents a measurement of length, it cannot be a negative number. So the answer is x = 5.)
- 5 (The products of the segments for each intersecting chord are equal, so x · (x + 1) = 3 · 10. To solve this equation, first rewrite it as [Image]. This equation factors to [Image]. The solutions are then –6 and 5. Since x represents a measurement of length, it cannot be a negative number. So the answer is x = 5.)
A: 5

Q: Find x. Round to the nearest tenth if necessary. Assume that segments that appear to be tangent are tangent.

[Image]
- 8 (When two secant segments intersect in the exterior of a circle, set an equality between the product of each external segment and the entire segment. Here that gives the equation 6 · 8 = 4 · (x + 4). Solving this equation results in x = 8.)
- 10 (When two secant segments intersect in the exterior of a circle, set an equality between the product of each external segment and the entire segment. Here that gives the equation 6 · 8 = 4 · (x + 4). Solving this equation results in x = 8.)
- 3 (When two secant segments intersect in the exterior of a circle, set an equality between the product of each external segment and the entire segment. Here that gives the equation 6 · 8 = 4 · (x + 4). Solving this equation results in x = 8.)
- 4 (When two secant segments intersect in the exterior of a circle, set an equality between the product of each external segment and the entire segment. Here that gives the equation 6 · 8 = 4 · (x + 4). Solving this equation results in x = 8.)
A: 8

Q: Find x. Round to the nearest tenth if necessary. Assume that segments that appear to be tangent are tangent.

[Image]
- 3.3 (When two secant segments intersect in the exterior of a circle, set an equality between the product of each external segment and the entire segment. Here that gives the equation 7 · (x + 10) = x · (5x). Solving this equation, and rounding to the nearest tenth, results in x = 4.5 or x = -3.1. Since x is a measurement, x cannot be negative. So the answer is x = 4.5.)
- 2.5 (When two secant segments intersect in the exterior of a circle, set an equality between the product of each external segment and the entire segment. Here that gives the equation 7 · (x + 10) = x · (5x). Solving this equation, and rounding to the nearest tenth, results in x = 4.5 or x = -3.1. Since x is a measurement, x cannot be negative. So the answer is x = 4.5.)
- 3.1 (When two secant segments intersect in the exterior of a circle, set an equality between the product of each external segment and the entire segment. Here that gives the equation 7 · (x + 10) = x · (5x). Solving this equation, and rounding to the nearest tenth, results in x = 4.5 or x = -3.1. Since x is a measurement, x cannot be negative. So the answer is x = 4.5.)
- 4.5 (When two secant segments intersect in the exterior of a circle, set an equality between the product of each external segment and the entire segment. Here that gives the equation 7 · (x + 10) = x · (5x). Solving this equation, and rounding to the nearest tenth, results in x = 4.5 or x = -3.1. Since x is a measurement, x cannot be negative. So the answer is x = 4.5.)
A: 4.5