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Pa Help Sa Math Tyty class=

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KAntoineDoixgo

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

  • 1. If AG = 24 cm, what is AC?

  • 2. If OA = 5 cm and OG = 3 cm, what is CG?

  • 3. If OG = 6 cm and AC = 16 cm, what is BG?

  • 4. If BG = 2 cm and OC = 10 cm, what is AG?

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

[tex] \qquad \Large \: \rm{1) \: AC = 48cm} [/tex]

[tex] \qquad \Large \: \rm{2) \: CG = 4cm} [/tex]

[tex] \qquad \Large \: \rm{3) \: BG = 4cm} [/tex]

[tex] \qquad \Large \: \rm{4) \: AG = 6cm} [/tex]

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

First, we need to look in some of its features:

  • Point O is the center of the circle. OC, OA, OB are radii, making them equidistant to each other.

  • Drawing chord AC creating an isosceles ∆AOC.

  • Radius OB is perpendicular to chord AC intersects at point G. Thus, it bisects AC which makes AG ≅ CG.

  • Since OC ≅ OA, OG ≅ OG, AG ≅ CG, then we have created two congruent right triangles ∆CGO and ∆AGO

[tex] \: [/tex]

Solve for the measures:

#1: If AG = 24 cm, what is AC?

  • AG is half of AC. Thus, twice of AG is equal to AC. We can say that twice of 24cm is 48cm. Therefore, AC is 48cm

[tex] \: [/tex]

#2: If OA = 5 cm and OG = 3 cm, what is CG?

» Focus on ∆AGO. Find AG using the Pythagorean Theorem.

  • [tex] (OG)^2 + (AG)^2 = (OA)^2 [/tex]

  • [tex] (3cm)^2 + (AG)^2 = (5cm)^2 [/tex]

  • [tex] 9cm^2 + (AG)^2 = 25cm^2 [/tex]

  • [tex] (AG)^2 = 25cm^2 - 9cm^2 [/tex]

  • [tex] (AG)^2 = 16cm^2 [/tex]

  • [tex] \sqrt{(AG)^2} = \sqrt{16cm^2} [/tex]

  • [tex] AG = 4cm [/tex]

» Thus, AG measures 4cm. We can say that AG is congruent to CG. Therefore, CG is 4cm as well.

[tex] \: [/tex]

#3: If OG = 6 cm and AC = 16 cm, what is BG?

» Focus on any right triangles, like ∆AGO, because we will be finding for the radius. Solve for OA using the Pythagorean Theorem.

  • [tex] (OG)^2 + (AG)^2 = (OA)^2 [/tex]

  • [tex] (6cm)^2 + (AG)^2 = (OA)^2 [/tex]

» We know that AG is half of AC. Thus, half of 16cm is 8cm, which is AG.

  • [tex] (6cm)^2 + (8cm)^2 = (OA)^2 [/tex]

  • [tex] 36cm^2 + 64cm^2 = (OA)^2 [/tex]

  • [tex] 100cm^2 = (OA)^2 [/tex]

  • [tex] \sqrt{100cm^2} = \sqrt{(OA)^2} [/tex]

  • [tex] 10cm = OA [/tex]

» OA measures 10cm. Thus, OB is 10cm as well since it is also a radius. Find BG that is the difference of OB and OG since the sum of the measures of BG and OG is OB.

  • [tex] BG = OB - OG [/tex]

  • [tex] BG = 10cm - 6cm [/tex]

  • [tex] BG = 4cm [/tex]

[tex] \therefore [/tex] BG measures 4cm

[tex] \: [/tex]

#:. If BG = 2 cm and OC = 10 cm, what is AG?

» Focus on ∆AGO. OC is equidistant to OA. Thus, OA is 10cm as well. Find AG using the Pythagorean Theorem.

  • [tex] (OG)^2 + (AG)^2 = (OA)^2 [/tex]

  • [tex] (OG)^2 + (AG)^2 = (10cm)^2 [/tex]

» OB is also a radius, measuring 10cm. The difference of OB and BG is OG.

  • [tex] OG = OB - BG [/tex]

  • [tex] OG = 10cm - 2cm [/tex]

  • [tex] OG = 8cm [/tex]

» OG is 8cm. Substitute it to our solution to find AG.

  • [tex] (8cm)^2 + (AG)^2 = (10cm)^2 [/tex]

  • [tex] 64cm^2 + (AG)^2 = 100cm^2 [/tex]

  • [tex] (AG)^2 = 100cm^2 - 64cm^2 [/tex]

  • [tex] (AG)^2 = 36cm^2 [/tex]

  • [tex] \sqrt{(AG)^2} = \sqrt{36cm^2} [/tex]

  • [tex] AG = 6cm [/tex]

[tex] \therefore [/tex] AG measures 6cm

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