Mechanical Engineering - Engineering Mechanics - Discussion

Discussion Forum : Engineering Mechanics - Section 1 (Q.No. 32)
32.
The centre of gravity of an equilateral triangle with each side a, is __________ from any of the three sides.
3a/2
23a
a/23
32a
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
11 comments Page 1 of 2.

Junaid Shaikh said:   1 decade ago
According to Pythagoras theorem:

a^2 = h^2+(a^2/4).

Therefore, h = (roor3/2)*a.

And we know that,

C.G of triangle from its base is h/3.

Hence h = a/(2*root3).

Jay said:   1 decade ago
Isn't the answer h = root(3)*a/6?

Mounika said:   1 decade ago
How would we be known to take Pythagoras theorem?

Ram Kadam said:   10 years ago
Answer is ( root(3)*a)/6).

Felix said:   10 years ago
If you rationalize [a/2*root (3)], it will give [a*root (3)/6].

Check it out.

J.Jayakumar said:   10 years ago
How is it come?

Clement said:   9 years ago
They just converted the same answer.

(root(3) * a/6)= a/(2 * root3).

Raju said:   8 years ago
(a*root3)/6 = (a*root3)/2*3=(a*root3)/2*root3*root3 = a/2*root3.

MD SALAHUDDIN said:   7 years ago
h2 = a2-(a/2)2,
h=root3*a/2,
cg=h/3, =a/2√3.

Hriday said:   7 years ago
Draw a triangle take its height h & side a.

Then apply Pythagoras theorem for the half of the equilateral triangle.
Wll get a*2=h*2 +(a/2) *2.
Find the value of h.
Put in cg formula = h/3.


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