Question

two chords AB and CD of a circle intersect at E such that AE= 2.4 cm , BE = 3.2cm and CE =1.6cm .the length of DE is :

Answer 1

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.

$AE{\times}EB=DE{\times}EC$

$2.4{\times}3.2=DE{\times}1.6$

$\frac{7.68}{1.6}=DE$

$DE=4.8$

Answer 2

Answer:

Step-by-step explanation:

Apply the rule, AE EB = CE ED

2 4 3 2. .= 1 6.  ED

ED = 4 8. cm