3 years ago

BC is tangent to circle A at B and to circle D at C. AB= 10 BC= 21 DC= 8 find AD to the nearest tenth

1 answer:

4 0

The attached diagram is an interpretation of the question.
If there are other details, please post.

The distance AD is found by Pythagoras theorem, with
AD=sqrt(BC^2+(AB-DC)^2)=sqrt(21^2+2^2)=sqrt(441+4)=sqrt(445)
=21.095 (to the third decimal place)

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