Answer: In the given figure EC= 2 S AE= 2 3 S AS c ′ is the centroid of △ABC ⇒C ′ E= 3 1 AE= 2 3 S CC ′ = 3 S Now, In △DC ′ C Using Pythagoras theorem (DC ′ ) 2 =(CC ′ ) 2 +(CD) 2 h 2 + 3 S 2 =S 2 ⇒h 2 = 3 2S 2 ⇒S 2 = 2 3h 2 Reply

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Answer:In the given figure

EC=

2

S

AE=

2

3

S

AS c

′

is the centroid of △ABC

⇒C

′

E=

3

1

AE=

2

3

S

CC

′

=

3

S

Now, In △DC

′

C

Using Pythagoras theorem

(DC

′

)

2

=(CC

′

)

2

+(CD)

2

h

2

+

3

S

2

=S

2

⇒h

2

=

3

2S

2

⇒S

2

=

2

3h

2