Question

AD, BE and CF are the medians in ΔABC. If AD + BE + CF = a and AB + BC + CA = b, then which of the following is always correct?

• A. 2a = b
• B. $a<\frac{1}{2}b$
  
• C. $a<\frac{1}{4}b$
  
• D. $a<\frac{3}{4}b$
  

Answer 1

The correct option is D $a<\frac{3}{4}b$


Let G be the point of intersection of medians of ΔABC.
Sum of any two sides of a triangle is greater than the third side.
∴ BG + GA > AB …..(i)
and, CG + GA > AC …..(ii)
and, BG + CG > BC …..(iii)
Adding (i), (ii) and (iii), we get
BG + GA + CG + GA + BG + CG > AB + AC + BC
⇒ 2(BG + GA + CG) > AB + AC + BC …..(iv)
The point of intersection of medians (centroid) divides each median in the ratio 2 : 1.
$\therefore BG=\frac{2}{3}BE,CG=\frac{2}{3}CFandAG=\frac{2}{3}AD$
$|RightarrowBG+GA+CG=\frac{2}{3}(BE+CF+AD)$ ....(v)
From (iv) and (v), we get
$2\left(\frac{2}{3}a\right)>b$ $(\because BE+CF+AD=\text{a and AB + BC + CA = b)}$
$\Rightarrow \frac{4}{3}a>b$
$\Rightarrow a>\frac{3}{4}b$
Hence, the correct answer is option (d).