In this figure the value of AE is
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4
In the given figure,
∆ABC and ∆ADE has,
- <ACB = <AED ; ( each 90°)
- <BAC = <DAE ; ( common angle)
So, ∆ABC ~ ∆ADE by AA criteria.
Therefore,
=
=
AE =
AE =
AE = 2.5 cm
Answered by
0
Step-by-step explanation:
AB=5 cm
OB=Radius=13 cm
∴OA=√(OB²)−(AB²)
=√13²−5²=√144
∴OA=12 cm
Hence, the shortest distance of the chord from the centre is 12 cm
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