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Sagot :

UK2019idn

✏️POWER THEOREMS

[tex]\red{••••••••••••••••••••••••••••••••••••••••••••••••••}[/tex]

[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]

[tex]\red{••••••••••••••••••••••••••••••••••••••••••••••••••}[/tex]

[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]3x = 20[/tex]

  • [tex] \frac{ \cancel3x}{ \cancel3} = \frac{20}{3} \\ [/tex]

  • [tex]x ≈ 6.67[/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]6x = 48[/tex]

  • [tex] \frac{ \cancel6x}{ \cancel6} = \frac{48}{6} \\ [/tex]

  • [tex]x = 8[/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]16x = 180[/tex]

  • [tex]\frac{\cancel{16}x}{\cancel{16}}= \frac{180}{16}\\[/tex]

  • [tex]x ≈ 11.25[/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]12x = 154[/tex]

  • [tex]\frac{\cancel{12}x}{\cancel{12}}= \frac{154}{12}\\[/tex]

  • [tex]x ≈ 12.83[/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]6x + 36 = 48[/tex]

  • [tex]6x = 48 - 36[/tex]

  • [tex]6x = 12[/tex]

  • [tex]\frac{\cancel{6}x}{\cancel6}= \frac{12}{6}\\[/tex]

  • [tex]x = 2[/tex]

[tex]\therefore[/tex] The length of the segment x is 2 units.

[tex]\red{••••••••••••••••••••••••••••••••••••••••••••••••••}[/tex]

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