Question : If $AD, BE$ and $CF$ are medians of $\triangle ABC$, then which of the following statement is correct?
Option 1: $(AD + BE + CF) > (AB + BC + CA)$
Option 2: $(AD + BE + CF) < (AB + BC + CA)$
Option 3: $(AD + BE + CF ) = (AB + BC + CA)$
Option 4: $(AD + BE + CF ) = \sqrt2(AB+BC+CA)$
New: SSC CHSL Tier 2 answer key released | SSC CHSL 2024 Notification PDF
Recommended: How to crack SSC CHSL | SSC CHSL exam guide
Don't Miss: Month-wise Current Affairs | Upcoming government exams
Correct Answer: $(AD + BE + CF) < (AB + BC + CA)$
Solution :
Here, $AB+AC>2AD$ ---------------(1)
$AB+BC>2BE$ -------------------(2)
$BC+AC>2CF$ -------------------(3)
By adding equation (1), (2), and (3), we get,
$2(AB+BC+AC)>2(AD+BE+CF)$
$⇒(AB+BC+AC)>(AD+BE+CF)$
Hence, the correct answer is $(AD + BE + CF) < (AB + BC + CA)$.
Related Questions
Know More about
Staff Selection Commission Combined High ...
Result | Eligibility | Application | Admit Card | Answer Key | Preparation Tips | Cutoff
Get Updates BrochureYour Staff Selection Commission Combined Higher Secondary Level Exam brochure has been successfully mailed to your registered email id “”.