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sweet [91]
3 years ago
11

Given: ∠AOB is a central angle and ∠ACB is a circumscribed angle. Prove: △ACO ≅ △BCO Circle O is shown. Line segments A O and B

O are radii. Tangents C B and C B intersect at point C outside of the circle. A line is drawn to connect points C and O. We are given that angle AOB is a central angle of circle O and that angle ACB is a circumscribed angle of circle O. We see that AO ≅ BO because . We also know that AC ≅ BC since . Using the reflexive property, we see that . Therefore, we conclude that △ACO is congruent to △BCO by the .
Mathematics
2 answers:
charle [14.2K]3 years ago
6 0

Answer:

all radii of the same circle are congruent

tangents to a circle that intersect are congruent

side CO is congruent to side CO

SSS congruency theorem

Step-by-step explanation:

m_a_m_a [10]3 years ago
6 0

Answer:

1,3,3,1

Step-by-step explanation:

Trust the process

edge 2021

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Answer:

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Step-by-step explanation:

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enot [183]

Answer:

Composite Area = 1105.94

Step-by-step explanation:

Remark

The two semi circles at each end make an entire circle with radius 11. The width of the central rectangle = the diameter of the circle. The area should be able to be found.

Width

2*radius = width

width = 2 * 11

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Area = pi * r^2

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Total Area = Area of Rectangle + Area of the Circle

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Digiron [165]

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