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Svet_ta [14]
2 years ago
11

Given: Two concentric circles with AB tangent to smaller circle at R Prove: AR=RB

Mathematics
2 answers:
disa [49]2 years ago
5 0

Answer:

See explanation

Step-by-step explanation:

If segment AB is tangent to the smaller circle, than AB⊥OR. Consider two right triangles AOR and BOR. In these triangles:

  • OR is common leg;
  • AO=OB as radii of larger circle;
  • ∠ARO=∠BRO, because AB⊥OR.

By HL theorem, ttriangles AOR and BOR are congruent. This gives you that AR=RB.

Alexeev081 [22]2 years ago
5 0

Answer:

R is the mid point of AB so  AR = RB

Step-by-step explanation:

Points to remember

The diameter of a circle and a chord is mutually  perpendicular then the diameter divide the chord in two equal parts.

<u>To prove AR = RB</u>

From the figure we get, AB is the tangent at R of small circle.

Therefore OR ⊥ AB

O is the center of both circles.

AB is the chord of large circle.

So The diameter of large circle passing through R is perpendicular  to AB

Therefore AR = RB

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Step-by-step explanation:

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3 years ago
Circle 1, Circle 2, and Circle 3 have the same center and have radii, respectively, of r₁ cm, r₂ cm, r₃ cm, where r₁ &lt; r₂ &lt
ira [324]

Answer:

a) A₁ /A₂  =  r₁² / (r₂²   - r₁²)

b) A₂ /A₃  = (r₂²   - r₁²) / (r₃²   - r₂²)

Step-by-step explanation:

We have Circle 1  and area  A₁  

Area of circle 2 outside circle 1     =  A₂

Area of circle 3 outside circle 2    = A₃

On the other hand we have

A₁  = π*r₁²   area of circle 1

A₂´ =  π*r₂² area of circle 2

A₃´ = π*r₃²  area of circle 3

All areas in cm²

a) A₁ /A₂        

A₁  =  π*r₁²

According to problem statement  A₂  = π*r₂²  -   A₁

A₂  =  π*r₂²  -   π*r₁²      ⇒         A₂  =  π* (r₂²   - r₁²)

Then  A₁ /A₂  =   π*r₁² / [π* (r₂²   - r₁²)]

A₁ /A₂  =  r₁² / (r₂²   - r₁²)

b)  A₂ /A₃        

A₂  =   π* (r₂²   - r₁²)

And

A₃   =  π* (r₃²   - r₂²)

Therefore

A₂ /A₃  =   π* (r₂²   - r₁²) / π* (r₃²   - r₂²)  ⇒  A₂ /A₃  = (r₂²   - r₁²) / (r₃²   - r₂²)

A₂ /A₃  = (r₂²   - r₁²) / (r₃²   - r₂²)

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