Question-37220

Question Number 37220 by tawa tawa last updated on 10/Jun/18

Answered by ajfour last updated on 01/Jul/18

l e t x a x i s b e D E F

s u m o f p e r p e n d i c u l a r s o n t a n g e n t

(x a x i s n o w) = y_{A} + y_{B} + y_{C}

= 3 y_{G}

y_{G} b e i n g y c o o r d i n a t e o f c e n t r o i d

o f △ A B C (s a m e a s y c o o r d i n a t e

o f c i r c u m c e n t r e) = R

L e n g t h o f m e d i a n = R + \frac{R}{2} = \frac{3 R}{2}

h e n c e y_{A} + y_{B} + y_{C} = 3 y_{G}

________________________

W e c a n s e e t h a t c e n t r o i d a n d

c i r c u m c e n t r e a r e t h e s a m e f o r

a n e q u i l a t e r a l t r i a n g l e; y_{G} = R

a s c i r c l e t o u c h e s x - a x i s .

H e n c e y_{A} + y_{B} + y_{C} = 3 R

w h i c h i s t w i c e t h e l e n g t h o f a

m e d i a n (= R + R \sin 30 °) = \frac{3 R}{2}

________________________

= 2 (l e n g t h o f m e d i a n) .

Commented by tawa tawa last updated on 11/Jun/18

God bless you sir

Commented by tawa tawa last updated on 28/Jun/18

please sir, can you produce more explanation on the solution for me

to understand better . Thanks you sir and God bless you .

Commented by tawa tawa last updated on 01/Jul/18

?

Commented by tawa tawa last updated on 01/Jul/18

wow, i really appreciate sir . God bless you sir