Question : Let G be the centroid of the equilateral triangle ABC of perimeter 24 cm. Then the length of AG is:
Option 1: $2 \sqrt 3$
Option 2: $\frac{8}{\sqrt3}$
Option 3: $8 \sqrt 3$
Option 4: $4 \sqrt 3$
Latest: SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL complete guide
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
Correct Answer: $\frac{8}{\sqrt3}$
Solution : The perimeter of the equilateral triangle, P = 24 cm Side, a = $\frac{P}{3}$ = $\frac{24}{3}$ = 8 cm Height of the equilateral triangle = $\frac{\sqrt{3}×a}{2}$ = $\frac{\sqrt{3}×8}{2}$ = $4\sqrt3$ As the median is divided in the ratio 2 : 1 at the centroid, AG = $\frac{2×4\sqrt3}{3}$ = $\frac{8}{\sqrt3}$cm Hence, the correct answer is $\frac{8}{\sqrt3}$.
Candidates can download this ebook to know all about SSC CGL.
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Question : G is the centroid of the equilateral triangle ABC. If AB = 10 cm, then the length of AG (in cm) is:
Question : $\triangle ABC$ is an equilateral triangle with a side of 12 cm and AD is the median. Find the length of GD if G is the centroid of $\triangle ABC$.
Question : $G$ is the centroid of the equilateral triangle $ABC$. If $AB$ is $9\text{ cm}$, then $AG$ is equal to:
Question : PQR is an equilateral triangle and the centroid of triangle PQR is point A. If the side of the triangle is 12 cm, then what is the length of PA?
Question : In a triangle ${ABC}, {AB}={AC}$ and the perimeter of $\triangle {ABC}$ is $8(2+\sqrt{2}) $ cm. If the length of ${BC}$ is $\sqrt{2}$ times the length of ${AB}$, then find the area of $\triangle {ABC}$.
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile