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Section 10-5 Tangents Monday, May 21, 2012

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Page 1: Geometry Section 10-5 1112

Section 10-5Tangents

Monday, May 21, 2012

Page 2: Geometry Section 10-5 1112

Essential Questions

• How do you use properties of tangents?

• How do you solve problems involving circumscribed polygons?

Monday, May 21, 2012

Page 3: Geometry Section 10-5 1112

Vocabulary1. Tangent:

2. Point of Tangency:

3. Common Tangent:

Monday, May 21, 2012

Page 4: Geometry Section 10-5 1112

Vocabulary1. Tangent: A line in the same plane as a circle and

intersects the circle in exactly one point

2. Point of Tangency:

3. Common Tangent:

Monday, May 21, 2012

Page 5: Geometry Section 10-5 1112

Vocabulary1. Tangent: A line in the same plane as a circle and

intersects the circle in exactly one point

2. Point of Tangency: The point that a tangent intersects a circle

3. Common Tangent:

Monday, May 21, 2012

Page 6: Geometry Section 10-5 1112

Vocabulary1. Tangent: A line in the same plane as a circle and

intersects the circle in exactly one point

2. Point of Tangency: The point that a tangent intersects a circle

3. Common Tangent: A line, ray, or segment that is tangent to two different circles in the same plane

Monday, May 21, 2012

Page 7: Geometry Section 10-5 1112

TheoremsTheorem 10.10 - Tangents:

Theorem 11.11 - Two Segments:

Monday, May 21, 2012

Page 8: Geometry Section 10-5 1112

TheoremsTheorem 10.10 - Tangents: A line is tangent to a

circle IFF it is perpendicular to a radius drawn to the point of tangency

Theorem 11.11 - Two Segments:

Monday, May 21, 2012

Page 9: Geometry Section 10-5 1112

TheoremsTheorem 10.10 - Tangents: A line is tangent to a

circle IFF it is perpendicular to a radius drawn to the point of tangency

Theorem 11.11 - Two Segments: If two segments from the same exterior point are tangent to a circle, then the segments are congruent

Monday, May 21, 2012

Page 10: Geometry Section 10-5 1112

Example 1Draw in the common tangents for each figure. If no tangent

exists, state no common tangent.

a. b.

Monday, May 21, 2012

Page 11: Geometry Section 10-5 1112

Example 1Draw in the common tangents for each figure. If no tangent

exists, state no common tangent.

a. b.

No common tangent

Monday, May 21, 2012

Page 12: Geometry Section 10-5 1112

Example 1Draw in the common tangents for each figure. If no tangent

exists, state no common tangent.

a. b.

No common tangent

Monday, May 21, 2012

Page 13: Geometry Section 10-5 1112

Example 1Draw in the common tangents for each figure. If no tangent

exists, state no common tangent.

a. b.

No common tangent

Monday, May 21, 2012

Page 14: Geometry Section 10-5 1112

Example 2KL is a radius of ⊙K. Determine whether LM is tangent to

⊙K. Justify your answer.

Monday, May 21, 2012

Page 15: Geometry Section 10-5 1112

Example 2

a2 + b2 = c2

KL is a radius of ⊙K. Determine whether LM is tangent to ⊙K. Justify your answer.

Monday, May 21, 2012

Page 16: Geometry Section 10-5 1112

Example 2

a2 + b2 = c2

202 + 212 = 292

KL is a radius of ⊙K. Determine whether LM is tangent to ⊙K. Justify your answer.

Monday, May 21, 2012

Page 17: Geometry Section 10-5 1112

Example 2

a2 + b2 = c2

202 + 212 = 292

400+ 441 = 841

KL is a radius of ⊙K. Determine whether LM is tangent to ⊙K. Justify your answer.

Monday, May 21, 2012

Page 18: Geometry Section 10-5 1112

Example 2

a2 + b2 = c2

202 + 212 = 292

400+ 441 = 841

KL is a radius of ⊙K. Determine whether LM is tangent to ⊙K. Justify your answer.

Since the values hold for a right triangle, we know that LM is perpendicular to

KL, making it a tangent.

Monday, May 21, 2012

Page 19: Geometry Section 10-5 1112

Example 3In the figure, WE is tangent to ⊙D at W. Find the value of x.

Monday, May 21, 2012

Page 20: Geometry Section 10-5 1112

Example 3In the figure, WE is tangent to ⊙D at W. Find the value of x.

a2 + b2 = c2

Monday, May 21, 2012

Page 21: Geometry Section 10-5 1112

Example 3In the figure, WE is tangent to ⊙D at W. Find the value of x.

a2 + b2 = c2

x2 + 242 = ( x + 16)2

Monday, May 21, 2012

Page 22: Geometry Section 10-5 1112

Example 3In the figure, WE is tangent to ⊙D at W. Find the value of x.

a2 + b2 = c2

x2 + 242 = ( x + 16)2

x2 + 576 = x2 + 32x + 256

Monday, May 21, 2012

Page 23: Geometry Section 10-5 1112

Example 3In the figure, WE is tangent to ⊙D at W. Find the value of x.

a2 + b2 = c2

x2 + 242 = ( x + 16)2

x2 + 576 = x2 + 32x + 256

−x2 −x2

Monday, May 21, 2012

Page 24: Geometry Section 10-5 1112

Example 3In the figure, WE is tangent to ⊙D at W. Find the value of x.

a2 + b2 = c2

x2 + 242 = ( x + 16)2

x2 + 576 = x2 + 32x + 256

−x2 −x2

−256 −256

Monday, May 21, 2012

Page 25: Geometry Section 10-5 1112

Example 3In the figure, WE is tangent to ⊙D at W. Find the value of x.

a2 + b2 = c2

x2 + 242 = ( x + 16)2

x2 + 576 = x2 + 32x + 256

−x2 −x2

−256 −256

320 = 32x

Monday, May 21, 2012

Page 26: Geometry Section 10-5 1112

Example 3In the figure, WE is tangent to ⊙D at W. Find the value of x.

a2 + b2 = c2

x2 + 242 = ( x + 16)2

x2 + 576 = x2 + 32x + 256

−x2 −x2

−256 −256

320 = 32x x = 10

Monday, May 21, 2012

Page 27: Geometry Section 10-5 1112

Example 4AC and BC are tangent to ⊙Z. Find the value of x.

Monday, May 21, 2012

Page 28: Geometry Section 10-5 1112

Example 4AC and BC are tangent to ⊙Z. Find the value of x.

3x + 2 = 4x − 3

Monday, May 21, 2012

Page 29: Geometry Section 10-5 1112

Example 4AC and BC are tangent to ⊙Z. Find the value of x.

3x + 2 = 4x − 3

x = 5

Monday, May 21, 2012

Page 30: Geometry Section 10-5 1112

Example 5Some round cookies are marketed in a triangular package

to pique the consumer’s interest. If ∆QRS is circumscribed with ⊙T with NQ = 2 cm, QR = 8 cm, and

SM = 10 cm, find the perimeter of ∆QRS.

Monday, May 21, 2012

Page 31: Geometry Section 10-5 1112

Example 5Some round cookies are marketed in a triangular package

to pique the consumer’s interest. If ∆QRS is circumscribed with ⊙T with NQ = 2 cm, QR = 8 cm, and

SM = 10 cm, find the perimeter of ∆QRS.

2

810

Monday, May 21, 2012

Page 32: Geometry Section 10-5 1112

Example 5Some round cookies are marketed in a triangular package

to pique the consumer’s interest. If ∆QRS is circumscribed with ⊙T with NQ = 2 cm, QR = 8 cm, and

SM = 10 cm, find the perimeter of ∆QRS.

2

810

10

Monday, May 21, 2012

Page 33: Geometry Section 10-5 1112

Example 5Some round cookies are marketed in a triangular package

to pique the consumer’s interest. If ∆QRS is circumscribed with ⊙T with NQ = 2 cm, QR = 8 cm, and

SM = 10 cm, find the perimeter of ∆QRS.

2

810

10 2

Monday, May 21, 2012

Page 34: Geometry Section 10-5 1112

Example 5Some round cookies are marketed in a triangular package

to pique the consumer’s interest. If ∆QRS is circumscribed with ⊙T with NQ = 2 cm, QR = 8 cm, and

SM = 10 cm, find the perimeter of ∆QRS.

2

810

10 2

6

Monday, May 21, 2012

Page 35: Geometry Section 10-5 1112

Example 5Some round cookies are marketed in a triangular package

to pique the consumer’s interest. If ∆QRS is circumscribed with ⊙T with NQ = 2 cm, QR = 8 cm, and

SM = 10 cm, find the perimeter of ∆QRS.

2

810

10 2

66

Monday, May 21, 2012

Page 36: Geometry Section 10-5 1112

Example 5Some round cookies are marketed in a triangular package

to pique the consumer’s interest. If ∆QRS is circumscribed with ⊙T with NQ = 2 cm, QR = 8 cm, and

SM = 10 cm, find the perimeter of ∆QRS.

2

810

10 2

66

P = 10 + 10 + 2 + 2 + 6 + 6

Monday, May 21, 2012

Page 37: Geometry Section 10-5 1112

Example 5Some round cookies are marketed in a triangular package

to pique the consumer’s interest. If ∆QRS is circumscribed with ⊙T with NQ = 2 cm, QR = 8 cm, and

SM = 10 cm, find the perimeter of ∆QRS.

2

810

10 2

66

P = 10 + 10 + 2 + 2 + 6 + 6

P = 36 cm

Monday, May 21, 2012

Page 38: Geometry Section 10-5 1112

Check Your Understanding

p. 721 #1-8

Monday, May 21, 2012

Page 39: Geometry Section 10-5 1112

Problem Set

Monday, May 21, 2012

Page 40: Geometry Section 10-5 1112

Problem Set

p. 722 #9-31 odd, 45, 49, 55

“Stop thinking what's good for you, and start thinking what's good for everyone else. It changes the game.”

- Stephen Basilone & Annie MebaneMonday, May 21, 2012