Geometry Section 10-5 1112
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Geometry Section 10-5 1112

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Tangents

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    Geometry Section 10-5 1112 Geometry Section 10-5 1112 Presentation Transcript

    • Section 10-5 TangentsMonday, May 21, 2012
    • Essential Questions • How do you use properties of tangents? • How do you solve problems involving circumscribed polygons?Monday, May 21, 2012
    • Vocabulary 1. Tangent: 2. Point of Tangency: 3. Common Tangent:Monday, May 21, 2012
    • Vocabulary 1. 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
    • Vocabulary 1. 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
    • Vocabulary 1. 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 planeMonday, May 21, 2012
    • Theorems Theorem 10.10 - Tangents: Theorem 11.11 - Two Segments:Monday, May 21, 2012
    • Theorems Theorem 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
    • Theorems Theorem 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 congruentMonday, May 21, 2012
    • Example 1 Draw in the common tangents for each figure. If no tangent exists, state no common tangent. a. b.Monday, May 21, 2012
    • Example 1 Draw in the common tangents for each figure. If no tangent exists, state no common tangent. a. b. No common tangentMonday, May 21, 2012
    • Example 1 Draw in the common tangents for each figure. If no tangent exists, state no common tangent. a. b. No common tangentMonday, May 21, 2012
    • Example 1 Draw in the common tangents for each figure. If no tangent exists, state no common tangent. a. b. No common tangentMonday, May 21, 2012
    • Example 2 KL is a radius of ⊙K. Determine whether LM is tangent to ⊙K. Justify your answer.Monday, May 21, 2012
    • Example 2 KL is a radius of ⊙K. Determine whether LM is tangent to ⊙K. Justify your answer. a +b =c 2 2 2Monday, May 21, 2012
    • Example 2 KL is a radius of ⊙K. Determine whether LM is tangent to ⊙K. Justify your answer. a +b =c 2 2 2 20 + 21 = 29 2 2 2Monday, May 21, 2012
    • Example 2 KL is a radius of ⊙K. Determine whether LM is tangent to ⊙K. Justify your answer. a +b =c 2 2 2 20 + 21 = 29 2 2 2 400 + 441 = 841Monday, May 21, 2012
    • Example 2 KL is a radius of ⊙K. Determine whether LM is tangent to ⊙K. Justify your answer. a +b =c 2 2 2 20 + 21 = 29 2 2 2 400 + 441 = 841 Since the values hold for a right triangle, we know that LM is perpendicular to KL, making it a tangent.Monday, May 21, 2012
    • Example 3 In the figure, WE is tangent to ⊙D at W. Find the value of x.Monday, May 21, 2012
    • Example 3 In the figure, WE is tangent to ⊙D at W. Find the value of x. a +b =c 2 2 2Monday, May 21, 2012
    • Example 3 In the figure, WE is tangent to ⊙D at W. Find the value of x. a +b =c 2 2 2 x + 24 = ( x + 16) 2 2 2Monday, May 21, 2012
    • Example 3 In the figure, WE is tangent to ⊙D at W. Find the value of x. a +b =c 2 2 2 x + 24 = ( x + 16) 2 2 2 x + 576 = x + 32x + 256 2 2Monday, May 21, 2012
    • Example 3 In the figure, WE is tangent to ⊙D at W. Find the value of x. a +b =c 2 2 2 x + 24 = ( x + 16) 2 2 2 x2 + 576 = x 2 + 32x + 256 2 2 −x −xMonday, May 21, 2012
    • Example 3 In the figure, WE is tangent to ⊙D at W. Find the value of x. a +b =c 2 2 2 x + 24 = ( x + 16) 2 2 2 x2 + 576 = x 2 + 32x + 256 2 2 −x −256 −x −256Monday, May 21, 2012
    • Example 3 In the figure, WE is tangent to ⊙D at W. Find the value of x. a +b =c 2 2 2 x + 24 = ( x + 16) 2 2 2 x2 + 576 = x 2 + 32x + 256 2 2 −x −256 −x −256 320 = 32xMonday, May 21, 2012
    • Example 3 In the figure, WE is tangent to ⊙D at W. Find the value of x. a +b =c 2 2 2 x + 24 = ( x + 16) 2 2 2 x2 + 576 = x 2 + 32x + 256 2 2 −x −256 −x −256 320 = 32x x = 10Monday, May 21, 2012
    • Example 4 AC and BC are tangent to ⊙Z. Find the value of x.Monday, May 21, 2012
    • Example 4 AC and BC are tangent to ⊙Z. Find the value of x. 3x + 2 = 4x − 3Monday, May 21, 2012
    • Example 4 AC and BC are tangent to ⊙Z. Find the value of x. 3x + 2 = 4x − 3 x=5Monday, May 21, 2012
    • Example 5 Some 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
    • Example 5 Some 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 8 10Monday, May 21, 2012
    • Example 5 Some 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 10 8 10Monday, May 21, 2012
    • Example 5 Some 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 10 2 8 10Monday, May 21, 2012
    • Example 5 Some 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 10 2 8 10 6Monday, May 21, 2012
    • Example 5 Some 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 10 2 8 10 6 6Monday, May 21, 2012
    • Example 5 Some 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 P = 10 + 10 + 2 + 2 + 6 + 6 10 2 8 10 6 6Monday, May 21, 2012
    • Example 5 Some 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 P = 10 + 10 + 2 + 2 + 6 + 6 10 2 P = 36 cm 8 10 6 6Monday, May 21, 2012
    • Check Your Understanding p. 721 #1-8Monday, May 21, 2012
    • Problem SetMonday, May 21, 2012
    • Problem Set p. 722 #9-31 odd, 45, 49, 55 “Stop thinking whats good for you, and start thinking whats good for everyone else. It changes the game.” - Stephen Basilone & Annie MebaneMonday, May 21, 2012