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The centre of gravity of an equilateral triangle with each side a, is __________ from any of the three sides.
- a)3a/2
- b)23a
- c)a/23
- d)32a
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
The centre of gravity of an equilateral triangle with each side a, is ...
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).
This question is part of UPSC exam. View all Mechanical Engineering courses
Most Upvoted Answer
The centre of gravity of an equilateral triangle with each side a, is ...
Finding the centre of gravity of an equilateral triangle
- The centre of gravity of an equilateral triangle is the point where the medians intersect.
- The medians of an equilateral triangle are the line segments drawn from each vertex to the midpoint of the opposite side.
- Since all sides of an equilateral triangle are equal, the medians are also equal in length and intersect at a point that is two-thirds of the distance from each vertex to the midpoint of the opposite side.
- Let's label the vertices of the equilateral triangle as A, B, and C, and let's draw the medians from each vertex to the midpoint of the opposite side.
- Let's label the point where the medians intersect as G, which is the centre of gravity of the equilateral triangle.
Finding the distance from the centre of gravity to any side of the equilateral triangle
- Let's draw a perpendicular line from the centre of gravity G to side AB of the equilateral triangle.
- Let's label the point where this line intersects side AB as D.
- We know that AD = BD = a/2, where a is the length of each side of the equilateral triangle.
- We also know that AG = (2/3)AD = (2/3)(a/2) = a/3, since G is two-thirds of the distance from vertex A to the midpoint of side BC.
- Therefore, GD = AD - AG = a/2 - a/3 = a/6.
- Similarly, we can find that the distance from the centre of gravity G to sides BC and AC of the equilateral triangle is also a/6.
- Therefore, the centre of gravity of an equilateral triangle with each side a is a/2√3 or a/√23 from any of the three sides.
- The centre of gravity of an equilateral triangle is the point where the medians intersect.
- The medians of an equilateral triangle are the line segments drawn from each vertex to the midpoint of the opposite side.
- Since all sides of an equilateral triangle are equal, the medians are also equal in length and intersect at a point that is two-thirds of the distance from each vertex to the midpoint of the opposite side.
- Let's label the vertices of the equilateral triangle as A, B, and C, and let's draw the medians from each vertex to the midpoint of the opposite side.
- Let's label the point where the medians intersect as G, which is the centre of gravity of the equilateral triangle.
Finding the distance from the centre of gravity to any side of the equilateral triangle
- Let's draw a perpendicular line from the centre of gravity G to side AB of the equilateral triangle.
- Let's label the point where this line intersects side AB as D.
- We know that AD = BD = a/2, where a is the length of each side of the equilateral triangle.
- We also know that AG = (2/3)AD = (2/3)(a/2) = a/3, since G is two-thirds of the distance from vertex A to the midpoint of side BC.
- Therefore, GD = AD - AG = a/2 - a/3 = a/6.
- Similarly, we can find that the distance from the centre of gravity G to sides BC and AC of the equilateral triangle is also a/6.
- Therefore, the centre of gravity of an equilateral triangle with each side a is a/2√3 or a/√23 from any of the three sides.
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The centre of gravity of an equilateral triangle with each side a, is __________ from any of the three sides.a)3a/2b)23ac)a/23d)32aCorrect answer is option 'C'. Can you explain this answer?
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