High students so far, we have discussed about inscribing a circle in a square and doing that repeatedly
right in this video.
Let's talk about inscribing a circle, a hexagon and a triangle and all of these regular shapes, OK,
a regular hexagon and a regular triangle inside a square.
All right.
Over here you have a square.
Let it be of sight.
OK.
Now, the area of this square is equal to a square.
We have seen that.
All right.
Now let me inscribe a circle inside the square.
Now the radius of this circle will be able to write and therefore the area of the circle will be by
by four into a square.
And we have seen why that is.
If you draw the diameter over here, you will see that two are is equal to it.
And that's why we are able to.
So we have seen this before.
All right.
Now, let's extend this concept to a hexagon and an equilateral triangle.
All right.
Now, first we inscribe a hexagon inside this circle that would look like this.
OK, now let me make a copy of this shape over here.
And you can see that this regular hexagon can be divided into six equilateral triangles.
Right.
And the side of each equilateral triangle will be equal to about two.
Right.
Because over here you can see that this plus this is equal to the diameter of the circle, which is
equal to eight.
Right.
This is it.
Therefore, this match over here will be equal to a by two.
And that is the side of the hexagon.
Right.
So this is also equal to EBITA.
Now, we have found that the side of the hexagon is equal to about two and hence the area of the regular
hexagon will be six times Route three by four in the side square side is able to side squares is square
by four.
So six in the row, three by four in the square by four is the area of this hexagon.
And that is equal to three room, three by eight square.
All right.
Now let's proceed and create a regular triangle inside the hexagon.
And remember, a regular triangle is an equilateral triangle.
All right.
So that would look like this now over here.
Let's see how to find the length of the side of the triangle.
Over here, I have this equilateral triangle.
Right, which is part of the regular hexagon.
Now, if I drop a perpendicular from this vertex to this site, that would look like this right now,
what would be the length of this height?
We know that the height of an equilateral triangle is Route three by two into the side.
Right now we have found the side of this equilateral triangle as EBITA two, because that's all we found.
The side of the hexagon is able to rule three by two in the side is equal to room three by two into
a by two.
Right.
All right.
Now, this is the height of the equilateral triangle.
Now this is a three by two in the air by two.
That gives you room three by four into it.
So that is half of the side of the equilateral triangle.
Right.
Therefore, the side of the triangle will be two times Route three by four into a that's equal to room
three by two into it.
So that is the site of the equilateral triangle.
Now we know how to find the area of triangle.
That would be Route three by four sides square.
Right.
And that is equal to room three by four into room three by two.
Eight that is this the whole square.
Right, that simplifies to three through three by 16 is square.
Now, you don't need to buy any of these formulas.
All you need to know is this approach, because if you needed for a question, if you know this approach,
you can easily derive whatever you need and find the answer and get out of the question.
All right, now.
Let's also analyze the parameters of these shapes.
OK, so what is the perimeter of the square over here?
It's offsite, hence the perimeter is for it right now.
What about the circle?
We have found that the radius of the circle is about to therefore the circumference of the circle will
be to pay into a bitou that's equal to buy into it.
What about the hexagon?
We have found the site is able to therefore the perimeter of the hexagon will be six a by two.
That's equal to three, right?
This is three eight.
And what about the triangle?
We know that the site is Route three by two.
Therefore, the perimeter of the triangle will be three in the three by two a.