Two chords ab and cd of a circle intersect each other at e such that ae=2.4cm , be=3.2 , ce = 1.6 find the length of de ,
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Answer:
DE=4.8 cm
Step-by-step explanation:
Since, we are given that two chords AB and CD of the circle intersect each other at E and AE= 2.4 cm , BE = 3.2cm and CE =1.6cm.
The length of DE can be found out as:
When the two chords of the circle intersect each other at any point ,then the product of two lengths of one chord is equal to the product of the two lengths of the other chord.
AE{\times}EB=DE{\times}EC
2.4{\times}3.2=DE{\times}1.6
\frac{7.68}{1.6}=DE
DE=4.8
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