✏️POWER THEOREMS
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[tex]\underline{\mathbb{ANSWERS:}}[/tex]
[tex]\qquad\LARGE \rm » \: \: 1. \: \green{x ≈ 6.67}[/tex]
[tex]\qquad\LARGE \rm » \: \: 2. \: \green{x = 8}[/tex]
[tex]\qquad\LARGE \rm » \: \: 3. \: \green{x ≈ 11.25}[/tex]
[tex]\qquad\LARGE \rm » \: \: 4. \: \green{x ≈ 12.83}[/tex]
[tex]\qquad\LARGE \rm » \: \: 5. \: \green{x = 2}[/tex]
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[tex]\underline{\mathbb{SOLUTIONS:}}[/tex]
#1. Solve the value of x by applying the Chord-Chord Power Theorem.
- [tex](LS)(GS) = (AS)(FS)[/tex]
- [tex](3)(x) = (4)(5)[/tex]
- [tex] \frac{ \cancel3x}{ \cancel3} = \frac{20}{3} \\ [/tex]
[tex]\therefore[/tex] The length of the segment x is 6.67 units
[tex]\rm[/tex]
#2. Solve for x by applying the Chord-Chord Power Theorem again.
- [tex](GE)(IE) = (UE)(DE)[/tex]
- [tex](x)(6) = (4)(12)[/tex]
- [tex] \frac{ \cancel6x}{ \cancel6} = \frac{48}{6} \\ [/tex]
[tex]\therefore[/tex] The length of the segment x is 8 units.
[tex]\rm[/tex]
#3. Solve for x by applying the Secant-Secant Power Theorem.
- [tex](IS)(IH) = (IF + FT)(FT)[/tex]
- [tex](16)(x) = (8 + 10)(10)[/tex]
- [tex]16x = (18)(10)[/tex]
- [tex]\frac{\cancel{16}x}{\cancel{16}}= \frac{180}{16}\\[/tex]
[tex]\therefore[/tex] The length of the segment x is 11.25 units.
[tex]\rm[/tex]
#4. Solve for x by applying the Secant-Secant Power Theorem again.
- [tex](AS)(AN) = (AE + JE)(JE) [/tex]
- [tex](12)(x) = (4 + 11)(11)[/tex]
- [tex]12x = (14)(11)[/tex]
- [tex]\frac{\cancel{12}x}{\cancel{12}}= \frac{154}{12}\\[/tex]
[tex]\therefore[/tex] The length of the segment x is 12.83 units.
[tex]\rm[/tex]
#5. Solve for x by applying the Secant-Secant Power Theorem again.
- [tex](GA +MA)(MA) = (GI)(IC)[/tex]
- [tex](x + 6)(6) = (6)(8)[/tex]
- [tex]\frac{\cancel{6}x}{\cancel6}= \frac{12}{6}\\[/tex]
[tex]\therefore[/tex] The length of the segment x is 2 units.
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