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ravindra

· started a discussion

· 1 Months ago

solution (Question: 364661.0 )

in ADE,
tanA=DE/AD
=>tan60=DE/2
=>DE=2 sqrt(3)
now EFC is an equilateral triangle so,
EF=2(you can find AE using pythagoras theorem)
now use pythagoras in DEF which gives
DF=4

Question:

ABCD is an equilateral triangle of side 6 units. Point D on AB is at a distance of 2 units from point A. Angles ADE and DEF are right angles. Find the length of segment DF.

  

Options:
A)

5

B)

3

C)

4

D)

6

Solution:

Ans:(c)


Deepak

· commented

· 1 Months ago

Yet to be approved!

mayank davaggar

· commented

· 1 Months ago

nice one
Thanks
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