Question : In a $\triangle ABC$, the median AD, BE, and CF meet at G, then which of the following is true?
Option 1: 4(AD + BE + CF) > 3(AB + BC + AC)
Option 2: 2(AD + BE + CF) > (AB + BC + AC)
Option 3: 3(AD + BE + CF) > 4(AB + BC + AC)
Option 4: AB + BC + AC > AD + BE + CF
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
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
Correct Answer: AB + BC + AC > AD + BE + CF
Solution : AB + AC > 2AD -------------(i) AB + BC > 2BE -------------(ii) AC + BC > 2CF -------------(iii) Adding (i), (ii), and (iii) ⇒ 2(AB + BC + AC) > 2 (AD + BE + CF) ⇒ AB + BC + AC > AD + BE + CF Hence, the correct answer is AB + BC + AC > AD + BE + CF.
Candidates can download this e-book to give a boost to thier preparation.
Result | Eligibility | Application | Admit Card | Answer Key | Preparation Tips | Cutoff
Question : If $AD, BE$ and $CF$ are medians of $\triangle ABC$, then which of the following statement is correct?
Question : E is the midpoint of the median AD of $\triangle ABC. BE$ is joined and produced to meet AC at F. F divides AC in ratio:
Question : In $\triangle ABC$, AB = BC = $k$, AC =$\sqrt2k$, then $\triangle ABC$ is a:
Question : In triangle ABC, AD BE and CF are the medians intersecting at point G and the area of triangle ABC is 156 cm2. What is the area (in cm2) of triangle FGE?
Question : Simplify the following expression. $(a+b+c)^2-(a-b+c)^2+4ac$
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile