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given the figure below solve for the value x​

Given The Figure Below Solve For The Value X class=

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✒️CIRCLE

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[tex] \large\underline{\mathbb{DIRECTIONS}:} [/tex]

  • Given the figure below, solve for the value of x.

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[tex] \large\underline{\mathbb{ANSWERS}:} [/tex]

[tex] \qquad \Large \rm{1) \; 15 \: units\:} [/tex]

[tex] \qquad \Large \rm{2) \; 20 \: units\:} [/tex]

[tex] \qquad \Large \rm{3) \; 52\degree\:} [/tex]

[tex] \qquad \Large \rm{4) \; 5 \: units\:} [/tex]

[tex] \qquad \Large \rm{5) \; 124 \degree\:} [/tex]

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[tex] \large\underline{\mathbb{SOLUTIONS}:} [/tex]

Number 1:

» Radius BA is perpendicular to tangent segment AC. Thus, ∆BAC is a right triangle. Find x using the Pythagorean theorem.

  • [tex] (BC)^2 = (BA)^2 + (AC)^2 [/tex]

  • [tex] (x)^2 = (9)^2 + (12)^2 [/tex]

  • [tex] x^2 = 81 + 144 [/tex]

  • [tex] x^2 = 225 [/tex]

  • [tex] \sqrt{x^2} = \sqrt{225} [/tex]

  • [tex] x = 15 [/tex]

[tex] \therefore [/tex] The length of segment x is 15 units

[tex] \: [/tex]

Number 2:

» Radius CQ is perpendicular to tangent segment QM. Thus, ∆CQM is a right triangle. Find x using the Pythagorean Theorem.

  • [tex] (QM)^2 + (CQ)^2 = (MC)^2 [/tex]

  • [tex] (x)^2 + (15)^2 = (25)^2 [/tex]

  • [tex] x^2 + 225 = 625 [/tex]

  • [tex] x^2 = 625 - 225 [/tex]

  • [tex] x^2 = 400 [/tex]

  • [tex] \sqrt{x^2} = \sqrt{400} [/tex]

  • [tex] x = 20 [/tex]

[tex] \therefore [/tex] The length of segment x is 20 units

[tex] \: [/tex]

Number 3:

» Angles T and S are supplementary. Thus, angles R and O are also supplementary.

  • [tex] m\angle R + m\angle O = 180\degree [/tex]

  • [tex] x + 128\degree = 180\degree [/tex]

  • [tex] x = 180\degree - 128\degree [/tex]

  • [tex] x = 52\degree [/tex]

[tex] \therefore [/tex] The measure of angle x is 52°

[tex] \: [/tex]

Number 4:

» Tangent segments AB and AD both drawn on circle C. Thus, their lengths are equal.

  • [tex] AB = AD [/tex]

  • [tex] 2x^2 + 4 = 54 [/tex]

  • [tex] 2x^2 = 54 - 4 [/tex]

  • [tex] 2x^2 = 50 [/tex]

  • [tex] \frac{\,2x^2\,}{2} = \frac{\,50\,}{2} \\ [/tex]

  • [tex] x^2 = 25 [/tex]

  • [tex] \sqrt{x^2} = \sqrt{25} [/tex]

  • [tex] x = 5 [/tex]

[tex] \therefore [/tex] The length of segment x is 5 units

[tex] \: [/tex]

» Angles Q and S are supplementary, Thus, angles R and A are also supplementary.

  • [tex] m\angle A + m\angle R = 180\degree [/tex]

  • [tex] m\angle A + 56\degree = 180\degree [/tex]

  • [tex] m\angle A = 180\degree - 56\degree [/tex]

  • [tex] m\angle A = 124\degree [/tex]

» The measure of a central angle subtended by the two radii is equal to its intercepted arc.

  • [tex] m\overset{\frown}{QS} = m\angle A [/tex]

  • [tex] x = 124\degree [/tex]

[tex] \therefore [/tex] The measure of the arc x is 124°

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