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Vivek

· started a discussion

· 1 Months ago

request (Question: 113099.0 )

sir ji solution to teek se kar diya karo, kahin = nahi h aur kahin > nahi h.

Question:
If AD, BE and CF are the medians of \(\triangle\)ABC, then which one of the following statements is correct?
Options:
A) (AD + BE + CF) = (AB + BC + CA)
B) (AD + BE + CF)>\(\cfrac{3}{4}\)(AB + BC + CA)
C) (AD + BE + CF) <\(\cfrac{3}{4}\)(AB + BC + CA)
D) (AD + BE + CF)=\(\cfrac{1}{2}\) (AB + BC + CA)
Solution:
Ans: (b) 


From \(\Delta\)BCG

BG + GC > BC

(sum of any two sides of a traiangle is greater than the third

side)\(\cfrac{2}{3}BE+\cfrac{2}{3}FC>BC\)

(medians of a triangle divides the median in the ratio 2 : 1)

\(BE+CF>\cfrac{3}{2}BC\) ......(i)

Similarly \(CF+AD\cfrac{3}{2}CA\) ......(ii)

and \(AD+BE.\cfrac{3}{2}AB\) .......(iii)

Adding (i), (ii) and (iii),


2 (AD + BE + CF) >\(\cfrac{3}{4}\) (AB + BC + CA)

or AD + BE + CF > \(\cfrac{3}{4}\) \((AB + BC + CA)\)

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