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

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asked Jun 12, 2019 103k views

1 Answer

Sventevit · Answer 1 · 2019-06-16T02:45:28+0000

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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)

BC is tangent to circle A at B and to circle D at C. AB= 10 BC= 21 DC= 8 find AD to-example-1

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

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