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Solution Sheet (Activity 2 - #1)

This document provides solutions to 10 geometry problems using various geometry theorems. It lists each problem lettered a-j, states the relevant theorem, shows the calculation, and provides the answer. The problems involve angles, radii, areas and use theorems about inscribed angles, supplementary angles, radii-tangent, center angle, tangent-secant exterior, triangle sum, radii being equal in a circle, and tangent-secant power.

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Adrian Barberan
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49 views2 pages

Solution Sheet (Activity 2 - #1)

This document provides solutions to 10 geometry problems using various geometry theorems. It lists each problem lettered a-j, states the relevant theorem, shows the calculation, and provides the answer. The problems involve angles, radii, areas and use theorems about inscribed angles, supplementary angles, radii-tangent, center angle, tangent-secant exterior, triangle sum, radii being equal in a circle, and tangent-secant power.

Uploaded by

Adrian Barberan
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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Solution Sheet (Activity 2 - #1)

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a.) m∠A • 2 | 36° • 2 = 72° [Inscribed Angle Theorem]


Answer: mArc BC = 72°

b.)180° - m∠COB | 180° - 72° = 108° [Supplementary Angle Theorem]


Answer: mArc AC = 108°

c.) m∠COE = 90° | 90° [Radius-Tangent Theorem]


Answer: m∠COE = 90°

d.) m∠AOC = Arc AC | m∠AOC = 108° [Center Angle Theorem]


Answer: m∠AOC = 108°

e.) m∠CDB = ½ ( mArc AC - mArc BC ) | ½ ( 108° - 72° ) = 18°


[Tangent Secant Exterior Theorem]
Answer: m∠CDB = 18°

f.) 180° - m∠CBA | 180° - 54° = 126° [Supplementary Angle Theorem]


Answer: m∠CBD = 126°
g.) 180° - ( m∠CBD + m∠CDB ) | 180° - ( 126° + 18° ) = 36°
[Triangle Sum Theorem]
Answer: m∠DCB = 36°

h.) OC = AC | OC = 12 [Radii in a circle are ≅]


Answer: OC = 12 units

i.) (CD)^2 = BD • AD | 81 = x • ( 24 + x ) -> x^2 + 24x - 81 = 0 ->


( x + 27 ) ( x - 3 ) = 0 -> x = -27 ; x = 3 [Tangent-Secant Power Theorem]
Answers: BD = 3 units

j.) πr^2 -> π(AO)^2 | π (12)^2 = 144π [Area of a circle]


Answer: Area of OC = 144π units2

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