10-6: Find Segment Lengths in Circles Geometry Chapter 10 10-6: Find Segment Lengths in Circles

Warm-Up Solve the following equations for x. 1.) 𝑥 𝑥+4 = 1 2 ( 2𝑥 2 +24) 3.) 5𝑥−6= 𝑥 2 2.) 𝑥 2 +4𝑥+5= 2𝑥 2 +5 4.) 2𝑥 2 +3𝑥= 𝑥 2 +4

Warm-Up Solve the following equations for x. 1.) 𝑥 𝑥+4 = 1 2 ( 2𝑥 2 +24) 𝑥 2 +4𝑥= 𝑥 2 +12 4𝑥=12 𝒙=𝟑

Warm-Up Solve the following equations for x. 2.) 𝑥 2 +4𝑥+5= 2𝑥 2 +5 2.) 𝑥 2 +4𝑥+5= 2𝑥 2 +5 4𝑥= 𝑥 2 𝒙=𝟒

Warm-Up Solve the following equations for x. 3.) 5𝑥−6= 𝑥 2 𝑥 2 −5𝑥+6=0

Warm-Up Solve the following equations for x. 3.) 5𝑥−6= 𝑥 2 𝑥 2 −5𝑥+6=0 𝑥 2 −2𝑥−3𝑥+6=0 𝑥 𝑥−2 −3 𝑥−2 =0 𝑥−2 𝑥−3 =0

Warm-Up Solve the following equations for x. 3.) 5𝑥−6= 𝑥 2 𝑥 2 −5𝑥+6=0 𝑥 2 −2𝑥−3𝑥+6=0 𝑥 𝑥−2 −3 𝑥−2 =0 𝑥−2 𝑥−3 =0 𝑥−2=0 𝑥−3=0 𝒙=𝟐 𝒙=𝟑

Warm-Up Solve the following equations for x. 4.) 2𝑥 2 +3𝑥= 𝑥 2 +4 𝑥 2 +3𝑥−4=0

Warm-Up Solve the following equations for x. 4.) 2𝑥 2 +3𝑥= 𝑥 2 +4 𝑥 2 +3𝑥−4=0 𝑥 2 −𝑥+4𝑥−4=0 𝑥 𝑥−1 +4 𝑥−1 =0 𝑥+4 𝑥−1 =0

Warm-Up Solve the following equations for x. 4.) 2𝑥 2 +3𝑥= 𝑥 2 +4 𝑥 2 +3𝑥−4=0 𝑥 2 −𝑥+4𝑥−4=0 𝑥 𝑥−1 +4 𝑥−1 =0 𝑥+4 𝑥−1 =0 𝑥+4=0 𝑥−1=0 𝒙=−𝟒 𝒙=𝟏

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

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 𝑬𝑨 and 𝑬𝑩 are the segments of chord 𝐴𝐵 𝑬𝑪 and 𝑬𝑫 are the segments of chord 𝐶𝐷

Theorem 10.14 Theorem 10.14 – 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 𝑬𝑨∙𝑬𝑩=𝑬𝑪∙𝑬𝑫

Example 1 Find 𝑴𝑳 and 𝑱𝑲. 𝑴 𝑳 𝑵 𝑲 𝑱 𝒙+𝟐 𝒙+𝟏 𝒙 𝒙+𝟒

Example 1 Find 𝑴𝑳 and 𝑱𝑲. By Thm. 10.14, 𝑴 𝑀𝑁∙𝑁𝐿=𝐾𝑁∙𝑁𝐽 𝒙+𝟐 𝑱 𝑵 𝒙+𝟒 𝑲 𝒙 𝒙+𝟏 𝒙 𝒙+𝟒

Example 1 Find 𝑴𝑳 and 𝑱𝑲. By Thm. 10.14, 𝑴 𝑀𝑁∙𝑁𝐿=𝐾𝑁∙𝑁𝐽 𝑥+2 𝑥+1 =𝑥(𝑥+4) 𝑥+2 𝑥+1 =𝑥(𝑥+4) 𝑥 2 +3𝑥+2= 𝑥 2 +4𝑥 𝒙=𝟐 𝑴 𝑳 𝑵 𝑲 𝑱 𝒙+𝟐 𝒙+𝟏 𝒙 𝒙+𝟒

Example 1 Find 𝑴𝑳 and 𝑱𝑲. For ML 𝑀𝐿=𝑥+2+𝑥+1 𝑀𝐿= 2 +2+ 2 +1 𝑴𝑳=𝟕 By Thm. 10.14, 𝑀𝑁∙𝑁𝐿=𝐾𝑁∙𝑁𝐽 𝑥+2 𝑥+1 =𝑥(𝑥+4) 𝑥 2 +3𝑥+2= 𝑥 2 +4𝑥 𝒙=𝟐 𝑴 𝑳 𝑵 𝑲 𝑱 𝒙+𝟐 𝒙+𝟏 𝒙 𝒙+𝟒

Example 1 Find 𝑴𝑳 and 𝑱𝑲. For ML 𝑀𝐿=𝑥+2+𝑥+1 𝑀𝐿= 2 +2+ 2 +1 𝑴𝑳=𝟕 By Thm. 10.14, 𝑀𝑁∙𝑁𝐿=𝐾𝑁∙𝑁𝐽 𝑥+2 𝑥+1 =𝑥(𝑥+4) 𝑥 2 +3𝑥+2= 𝑥 2 +4𝑥 𝒙=𝟐 𝑴 𝑳 𝑵 𝑲 𝑱 𝒙+𝟐 𝒙+𝟏 𝒙 𝒙+𝟒 For KJ 𝐾𝐽=𝑥+𝑥+4 𝐾𝐽= 2 + 2 +4 𝑲𝑱=𝟖

Example 2 Find 𝑹𝑻 and 𝑺𝑼. 𝑹 𝑽 𝑻 𝑼 𝑺 𝒙+𝟏 𝟒𝒙 𝒙 𝟑𝒙

Example 2 Find 𝑹𝑻 and 𝑺𝑼. 𝑺 By Thm. 10.14, 𝑹 𝑅𝑉∙𝑉𝑇=𝑈𝑉∙𝑉𝑆 𝒙 𝑽 𝑻 𝑼 𝑺 𝒙+𝟏 𝟒𝒙 𝒙 𝟑𝒙 By Thm. 10.14, 𝑅𝑉∙𝑉𝑇=𝑈𝑉∙𝑉𝑆 3𝑥 𝑥+1 =𝑥∙4𝑥 3𝑥 2 +3𝑥=4 𝑥 2 3𝑥= 𝑥 2 𝒙=𝟑

Example 2 Find 𝑹𝑻 and 𝑺𝑼. 𝑹 𝑽 𝑻 𝑼 𝑺 𝒙+𝟏 𝟒𝒙 𝒙 𝟑𝒙 For RT 𝑅𝑇=3𝑥+𝑥+1 𝑅𝑇=3 3 + 3 +1 𝑅𝑇=9+3+1 𝑹𝑻=𝟏𝟑 By Thm. 10.14, 𝑅𝑉∙𝑉𝑇=𝑈𝑉∙𝑉𝑆 3𝑥 𝑥+1 =𝑥∙4𝑥 3𝑥 2 +3𝑥=4 𝑥 2 3𝑥= 𝑥 2 𝒙=𝟑

Example 2 Find 𝑹𝑻 and 𝑺𝑼. 𝑹 𝑽 𝑻 𝑼 𝑺 𝒙+𝟏 𝟒𝒙 𝒙 𝟑𝒙 For RT 𝑅𝑇=3𝑥+𝑥+1 𝑅𝑇=3 3 + 3 +1 𝑅𝑇=9+3+1 𝑹𝑻=𝟏𝟑 By Thm. 10.14, 𝑅𝑉∙𝑉𝑇=𝑈𝑉∙𝑉𝑆 3𝑥 𝑥+1 =𝑥∙4𝑥 3𝑥 2 +3𝑥=4 𝑥 2 3𝑥= 𝑥 2 𝒙=𝟑 For SU 𝑆𝑈=𝑥+4𝑥 𝑆𝑈= 3 +4 3 𝑆𝑈=3+12 𝑺𝑼=𝟏𝟓

Tangents and Secants 𝑨𝑪 is the external segment of secant 𝑨𝑩 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 𝑨𝑪 is the external segment of secant 𝑨𝑩 𝑨𝑫 is a tangent segment

Theorem 10.15 Theorem 10.15 – 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 𝑬𝑨∙𝑬𝑩=𝑬𝑪∙𝑬𝑫

Example 3a Find the value(s)of x. 𝟔 𝟗 𝒙 𝟓

Example 3a Find the value(s)of x. 𝟔 𝟗 𝒙 𝟓 𝟏𝟓 𝒙+𝟓

Example 3a Find the value(s)of x. 𝟗 𝟔 𝒙 𝟓 By Thm. 10.15, 15∙6=5(𝑥+5) 𝟏𝟓 𝒙+𝟓

Example 3a Find the value(s)of x. 𝟗 𝟔 𝒙 𝟓 By Thm. 10.15, 15∙6=5(𝑥+5) 90=5𝑥+25 65=5𝑥 𝒙=𝟏𝟑 𝟔 𝟗 𝒙 𝟓 𝟏𝟓 𝒙+𝟓

Example 3b Find the value(s)of x. 𝟑 𝒙 𝟒 𝟐

Example 3b Find the value(s)of x. 𝒙 𝟑 𝟒 𝟐 By Thm. 10.15, 4∙6=3(𝑥+3) 𝟔 𝒙+𝟑

Example 3b Find the value(s)of x. 𝒙 𝟑 𝟒 𝟐 By Thm. 10.15, 4∙6=3(𝑥+3) 24=3𝑥+9 15=3𝑥 𝒙=𝟓 𝟑 𝒙 𝟒 𝟐 𝟔 𝒙+𝟑

Example 3c Find the value(s)of x. 𝟑 𝒙+𝟐 𝒙+𝟏 𝒙−𝟏

Example 3c Find the value(s)of x. 𝟑 𝒙+𝟐 𝒙+𝟏 𝒙−𝟏 𝟐𝒙 𝒙+𝟓

Example 3c Find the value(s)of x. 𝒙+𝟓 𝟐𝒙 By Thm. 10.15, 3(𝑥+5)=2𝑥(𝑥+1) 3𝑥+15= 2𝑥 2 +2𝑥 2𝑥 2 −𝑥−15=0 𝟑 𝒙+𝟐 𝒙+𝟏 𝒙−𝟏 𝟐𝒙 𝒙+𝟓

Example 3c Find the value(s)of x. 𝒙+𝟓 𝟐𝒙 By Thm. 10.15, 3(𝑥+5)=2𝑥(𝑥+1) 3𝑥+15= 2𝑥 2 +2𝑥 2𝑥 2 −𝑥−15=0 2𝑥+5 𝑥−3 =0 𝟑 𝒙+𝟐 𝒙+𝟏 𝒙−𝟏 𝟐𝒙 𝒙+𝟓 2𝑥+5=0 𝒙=− 𝟓 𝟐 𝑥−3=0 𝒙=𝟑

Example 3c Find the value(s)of x. 𝒙+𝟓 𝟐𝒙 By Thm. 10.15, 3(𝑥+5)=2𝑥(𝑥+1) 3𝑥+15= 2𝑥 2 +2𝑥 2𝑥 2 −𝑥−15=0 2𝑥+5 𝑥−3 =0 𝟑 𝒙+𝟐 𝒙+𝟏 𝒙−𝟏 𝟐𝒙 𝒙+𝟓 𝑥−3=0 𝒙=𝟑 2𝑥+5=0 𝒙=− 𝟓 𝟐 Why?

Theorem 10.16 Theorem 10.16 – 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 𝑬𝑨 𝟐 =𝑬𝑪∙𝑬𝑫

Example 4 Find the value of x. 𝟏𝟐 𝟐𝒙 𝒙

Example 4 Find the value of x. 𝟏𝟐 𝟐𝒙 𝒙 𝒙+𝟏𝟐

Example 4 Find the value of x. 𝒙 𝟐𝒙 𝟏𝟐 𝒙+𝟏𝟐 By Thm. 10.16, (2𝑥) 2 =𝑥(𝑥+12) 𝟏𝟐 𝟐𝒙 𝒙 𝒙+𝟏𝟐

Example 4 Find the value of x. 𝒙 𝟐𝒙 𝟏𝟐 𝒙+𝟏𝟐 By Thm. 10.16, (2𝑥) 2 =𝑥(𝑥+12) 4𝑥 2 = 𝑥 2 +12𝑥 3𝑥 2 =12𝑥 3𝑥=12 𝒙=𝟒 𝟏𝟐 𝟐𝒙 𝒙 𝒙+𝟏𝟐

Example 5 Find the value of x. 𝟏𝟔 𝟐𝟒 𝒙

Example 5 Find the value of x. 𝟏𝟔 𝟐𝟒 𝒙 𝒙+𝟏𝟔

Example 5 Find the value of x. 𝒙 𝟐𝟒 𝟏𝟔 𝒙+𝟏𝟔 By Thm. 10.16, (24) 2 =16(𝑥+16) 576=16𝑥+256 320=16𝑥 𝒙=𝟐𝟎 𝟏𝟔 𝟐𝟒 𝒙 𝒙+𝟏𝟔

Extra Practice Find the value of x. State which theorem you used. 1.) 𝟕 𝟓 𝒙 2.) 𝟏𝟖 𝟗 𝒙 𝟏𝟔

Extra Practice Find the value of x. State which theorem you used. 3.) 4.) 𝟏𝟐 𝟏𝟎 𝒙 𝟖 𝟐𝟕 𝒙 𝟏𝟐

Extra Practice 1.) Find the value of x. State which theorem you used. 𝟕 𝟓 𝒙

Extra Practice 1.) Find the value of x. State which theorem you used. By Thm. 10.16, 7 2 =5(𝑥+5) 49=5𝑥+25 24=5𝑥 𝒙=𝟒.𝟖 𝟕 𝟓 𝒙 𝒙+𝟓

Extra Practice 2.) Find the value of x. State which theorem you used. 𝟏𝟖 𝟗 𝒙 𝟏𝟔

Extra Practice 2.) Find the value of x. State which theorem you used. By Thm. 10.14, 𝑥∙18=9∙16 18𝑥=144 𝒙=𝟖 𝟏𝟖 𝟗 𝒙 𝟏𝟔

Extra Practice 3.) Find the value of x. State which theorem you used. 𝟏𝟐 𝟏𝟎 𝒙

Extra Practice 3.) Find the value of x. State which theorem you used. By Thm. 10.16, 12 2 =𝑥(𝑥+10) 144= 𝑥 2 +10𝑥 𝑥 2 +10𝑥−144=0 (𝑥+18)(𝑥−8) 𝒙+𝟏𝟖=𝟎 𝒙−𝟖=𝟎 𝒙=−𝟏𝟖 𝒙=𝟖 𝟏𝟐 𝟏𝟎 𝒙 𝒙+𝟏𝟎

Extra Practice 4.) Find the value of x. State which theorem you used. 𝟖 𝟐𝟕 𝒙 𝟏𝟐

Extra Practice 4.) Find the value of x. State which theorem you used. By Thm. 10.15, 8∙20=𝑥(𝑥+27) 160= 𝑥 2 +27𝑥 𝑥 2 +27𝑥−160=0 𝑥+32 𝑥−5 =0 𝒙+𝟑𝟐=𝟎 𝒙−𝟓=𝟎 𝒙=−𝟑𝟐 𝒙=𝟓 𝟖 𝟐𝟕 𝒙 𝟏𝟐 𝟐𝟎 𝒙+𝟐𝟕