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10-6: Find Segment Lengths in Circles
Geometry Chapter 10 10-6: Find Segment Lengths in Circles
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Warm-Up Solve the following equations for x.
1.) π₯ π₯+4 = 1 2 ( 2π₯ 2 +24) ) 5π₯β6= π₯ 2 2.) π₯ 2 +4π₯+5= 2π₯ ) 2π₯ 2 +3π₯= π₯ 2 +4
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Warm-Up Solve the following equations for x.
1.) π₯ π₯+4 = 1 2 ( 2π₯ 2 +24) π₯ 2 +4π₯= π₯ 2 +12 4π₯=12 π=π
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Warm-Up Solve the following equations for x. 2.) π₯ 2 +4π₯+5= 2π₯ 2 +5
2.) π₯ 2 +4π₯+5= 2π₯ 2 +5 4π₯= π₯ 2 π=π
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Warm-Up Solve the following equations for x. 3.) 5π₯β6= π₯ 2 π₯ 2 β5π₯+6=0
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Warm-Up Solve the following equations for x. 3.) 5π₯β6= π₯ 2 π₯ 2 β5π₯+6=0
π₯ 2 β2π₯β3π₯+6=0 π₯ π₯β2 β3 π₯β2 =0 π₯β2 π₯β3 =0
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Warm-Up Solve the following equations for x. 3.) 5π₯β6= π₯ 2 π₯ 2 β5π₯+6=0
π₯ 2 β2π₯β3π₯+6=0 π₯ π₯β2 β3 π₯β2 =0 π₯β2 π₯β3 =0 π₯β2= π₯β3=0 π=π π=π
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Warm-Up Solve the following equations for x. 4.) 2π₯ 2 +3π₯= π₯ 2 +4
π₯ 2 +3π₯β4=0
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Warm-Up Solve the following equations for x. 4.) 2π₯ 2 +3π₯= π₯ 2 +4
π₯ 2 +3π₯β4=0 π₯ 2 βπ₯+4π₯β4=0 π₯ π₯β1 +4 π₯β1 =0 π₯+4 π₯β1 =0
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Warm-Up Solve the following equations for x. 4.) 2π₯ 2 +3π₯= π₯ 2 +4
π₯ 2 +3π₯β4=0 π₯ 2 βπ₯+4π₯β4=0 π₯ π₯β1 +4 π₯β1 =0 π₯+4 π₯β1 =0 π₯+4= π₯β1=0 π=βπ π=π
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Find Segment Lengths in Circles
Objective: Students will be able to find segment lengths of chords, secants, and tangents of circles. Agenda Chord Lengths Secant Lengths Tangent Lengths
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Segments of Chords When two chords intersect in the interior of a circle, each chord is divided into two segments that are called segments of the chord. A B E C D π¬π¨ and π¬π© are the segments of chord π΄π΅ π¬πͺ and π¬π« are the segments of chord πΆπ·
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Theorem 10.14 Theorem β Segments of Chords Theorem: If two chords intersect in the interior of a circle, then the product of the lengths of the segments of one chord is equal to the product of the lengths of the segments of the other chord. A B E C D π¬π¨βπ¬π©=π¬πͺβπ¬π«
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Example 1 Find π΄π³ and π±π². π΄ π³ π΅ π² π± π+π π+π π π+π
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Example 1 Find π΄π³ and π±π². By Thm. 10.14, π΄ ππβππΏ=πΎπβππ½ π+π π± π΅ π+π π² π
π+π π π+π
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Example 1 Find π΄π³ and π±π². By Thm. 10.14, π΄ ππβππΏ=πΎπβππ½ π₯+2 π₯+1 =π₯(π₯+4)
π₯+2 π₯+1 =π₯(π₯+4) π₯ 2 +3π₯+2= π₯ 2 +4π₯ π=π π΄ π³ π΅ π² π± π+π π+π π π+π
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Example 1 Find π΄π³ and π±π². For ML ππΏ=π₯+2+π₯+1 ππΏ= 2 +2+ 2 +1
π΄π³=π By Thm , ππβππΏ=πΎπβππ½ π₯+2 π₯+1 =π₯(π₯+4) π₯ 2 +3π₯+2= π₯ 2 +4π₯ π=π π΄ π³ π΅ π² π± π+π π+π π π+π
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Example 1 Find π΄π³ and π±π². For ML ππΏ=π₯+2+π₯+1 ππΏ= 2 +2+ 2 +1
π΄π³=π By Thm , ππβππΏ=πΎπβππ½ π₯+2 π₯+1 =π₯(π₯+4) π₯ 2 +3π₯+2= π₯ 2 +4π₯ π=π π΄ π³ π΅ π² π± π+π π+π π π+π For KJ πΎπ½=π₯+π₯+4 πΎπ½= π²π±=π
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Example 2 Find πΉπ» and πΊπΌ. πΉ π½ π» πΌ πΊ π+π ππ π ππ
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Example 2 Find πΉπ» and πΊπΌ. πΊ By Thm. 10.14, πΉ π
πβππ=ππβππ π
π½ π» πΌ πΊ π+π ππ π ππ By Thm , π
πβππ=ππβππ 3π₯ π₯+1 =π₯β4π₯ 3π₯ 2 +3π₯=4 π₯ 2 3π₯= π₯ 2 π=π
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Example 2 Find πΉπ» and πΊπΌ. πΉ π½ π» πΌ πΊ π+π ππ π ππ For RT π
π=3π₯+π₯+1
π
π= π
π=9+3+1 πΉπ»=ππ By Thm , π
πβππ=ππβππ 3π₯ π₯+1 =π₯β4π₯ 3π₯ 2 +3π₯=4 π₯ 2 3π₯= π₯ 2 π=π
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Example 2 Find πΉπ» and πΊπΌ. πΉ π½ π» πΌ πΊ π+π ππ π ππ For RT π
π=3π₯+π₯+1
π
π= π
π=9+3+1 πΉπ»=ππ By Thm , π
πβππ=ππβππ 3π₯ π₯+1 =π₯β4π₯ 3π₯ 2 +3π₯=4 π₯ 2 3π₯= π₯ 2 π=π For SU ππ=π₯+4π₯ ππ= ππ=3+12 πΊπΌ=ππ
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Tangents and Secants π¨πͺ is the external segment of secant π¨π©
A secant segment is a segment that contains a chord of the circle, and has exactly one endpoint outside the circle. The part of the secant segment that is outside the circle is called the external segment. A B C D π¨πͺ is the external segment of secant π¨π© π¨π« is a tangent segment
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Theorem 10.15 Theorem β Segments of Secants Theorem: If two secant segments share the same endpoint, then the product of the lengths of each secant segment, along with its external segment, are equal. A B E C D π¬π¨βπ¬π©=π¬πͺβπ¬π«
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Example 3a Find the value(s)of x. π π π π
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Example 3a Find the value(s)of x. π π π π ππ π+π
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Example 3a Find the value(s)of x. π π π π By Thm. 10.15, 15β6=5(π₯+5)
ππ π+π
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Example 3a Find the value(s)of x. π π π π By Thm. 10.15, 15β6=5(π₯+5)
90=5π₯+25 65=5π₯ π=ππ π π π π ππ π+π
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Example 3b Find the value(s)of x. π π π π
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Example 3b Find the value(s)of x. π π π π By Thm. 10.15, 4β6=3(π₯+3)
π π+π
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Example 3b Find the value(s)of x. π π π π By Thm. 10.15, 4β6=3(π₯+3)
24=3π₯+9 15=3π₯ π=π π π π π π π+π
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Example 3c Find the value(s)of x. π π+π π+π πβπ
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Example 3c Find the value(s)of x. π π+π π+π πβπ ππ π+π
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Example 3c Find the value(s)of x. π+π ππ By Thm. 10.15, 3(π₯+5)=2π₯(π₯+1)
3π₯+15= 2π₯ 2 +2π₯ 2π₯ 2 βπ₯β15=0 π π+π π+π πβπ ππ π+π
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Example 3c Find the value(s)of x. π+π ππ By Thm. 10.15, 3(π₯+5)=2π₯(π₯+1)
3π₯+15= 2π₯ 2 +2π₯ 2π₯ 2 βπ₯β15=0 2π₯+5 π₯β3 =0 π π+π π+π πβπ ππ π+π 2π₯+5=0 π=β π π π₯β3=0 π=π
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Example 3c Find the value(s)of x. π+π ππ By Thm. 10.15, 3(π₯+5)=2π₯(π₯+1)
3π₯+15= 2π₯ 2 +2π₯ 2π₯ 2 βπ₯β15=0 2π₯+5 π₯β3 =0 π π+π π+π πβπ ππ π+π π₯β3=0 π=π 2π₯+5=0 π=β π π Why?
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Theorem 10.16 Theorem β Segments of Secants and Tangents Theorem: If a secant and a tangent segment share the same endpoint, then the product of the lengths of the secant segment and its external segment equal the square of the length of the tangent segment. A E C D π¬π¨ π =π¬πͺβπ¬π«
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Example 4 Find the value of x. ππ ππ π
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Example 4 Find the value of x. ππ ππ π π+ππ
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Example 4 Find the value of x. π ππ ππ π+ππ By Thm. 10.16,
(2π₯) 2 =π₯(π₯+12) ππ ππ π π+ππ
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Example 4 Find the value of x. π ππ ππ π+ππ By Thm. 10.16,
(2π₯) 2 =π₯(π₯+12) 4π₯ 2 = π₯ 2 +12π₯ 3π₯ 2 =12π₯ 3π₯=12 π=π ππ ππ π π+ππ
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Example 5 Find the value of x. ππ ππ π
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Example 5 Find the value of x. ππ ππ π π+ππ
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Example 5 Find the value of x. π ππ ππ π+ππ By Thm. 10.16,
(24) 2 =16(π₯+16) 576=16π₯+256 320=16π₯ π=ππ ππ ππ π π+ππ
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Extra Practice Find the value of x. State which theorem you used. 1.)
π π π 2.) ππ π π ππ
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Extra Practice Find the value of x. State which theorem you used. 3.)
4.) ππ ππ π π ππ π ππ
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Extra Practice 1.) Find the value of x. State which theorem you used.
π π π
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Extra Practice 1.) Find the value of x. State which theorem you used.
By Thm , 7 2 =5(π₯+5) 49=5π₯+25 24=5π₯ π=π.π π π π π+π
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Extra Practice 2.) Find the value of x. State which theorem you used.
ππ π π ππ
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Extra Practice 2.) Find the value of x. State which theorem you used.
By Thm , π₯β18=9β16 18π₯=144 π=π ππ π π ππ
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Extra Practice 3.) Find the value of x. State which theorem you used.
ππ ππ π
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Extra Practice 3.) Find the value of x. State which theorem you used.
By Thm , 12 2 =π₯(π₯+10) 144= π₯ 2 +10π₯ π₯ 2 +10π₯β144=0 (π₯+18)(π₯β8) π+ππ=π πβπ=π π=βππ π=π ππ ππ π π+ππ
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Extra Practice 4.) Find the value of x. State which theorem you used.
π ππ π ππ
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Extra Practice 4.) Find the value of x. State which theorem you used.
By Thm , 8β20=π₯(π₯+27) 160= π₯ 2 +27π₯ π₯ 2 +27π₯β160=0 π₯+32 π₯β5 =0 π+ππ=π πβπ=π π=βππ π=π π ππ π ππ ππ π+ππ
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