radius (r) of a cicrlce is divided into 10 equal sections. The area of the circle = sum of the areas of the pie-shaped sections.
We can estimate the area of the circle by thinking of each pie-shaped piece as a triangle with base B and height H.
The area of the circle will be approximately (bh/2)*8
The more sections you divide the circle into (n), the closer the approximation of the area: (bhn/2)*8
As n approaches to infinity, find the limit of h and bn
Very interesting question
First of all, if you divide the circle into 10 equal segments, then the
area can be approximated by (bh/2)*10 , not (bh/2)*8 as you stated.
as n ----> ∞ , h will approach r, the radius
b ----> 0 , b as the base of each triangle
your (bhn/2)*8 is a meaningless expression.
sorry statement written wrong
radius (r) of a circle is divided into 10 equal sections. The area of the circle = sum of the areas of the pie-shaped sections.
We can estimate the area of the circle by thinking of each pie-shaped piece as a triangle with base B and height H.
The area of the circle will be approximately (bh/2)*10
The more sections you divide the circle into (n), the closer the approximation of the area: (bhn/2)*10
As n approaches to infinity, find the limit of h and bn
actually, you mean the circumference is divided into n equal parts.
The triangles are isosceles, with the equal sides being r, and the vertex angle 2π/n
Thus, b = 2r*sin(π/n) and h = r*cos(π/n)
The area of each triangle is thus r^2/2 sin(2π/n)
So, the approximate area of the circle is nr^2/2 sin(2π/n)
See what you can do with the limit. Recall that
lim(x->0) sinx/x = 1
can u explain further
sigh
you want the limit of nr^2/2 sin(2π/n)
Now, you know that sinx/x -> 1, so you need to massage your expression so it looks like that
nr^2 sin(2π/n) / 2 = r^2 sin(2π/n) / (2/n) = πr^2 sin(2π/n)/(2π/n)
Now, if you let x = 2π/n, you have πr^2 sinx/x
as n->∞, x->0
so, the limit is πr^2 * 1 = πr^2, as desired
To find the limit as n approaches infinity for the values of h (the height of each triangle) and b (the base of each triangle), we need to consider how the division of the circle into triangular sections evolves as n increases.
First, let's analyze how the height, h, changes as n increases. The height of each triangle is equal to the radius of the circle, as given in the question. Therefore, h is a constant value that does not depend on the number of sections, n. So, as n approaches infinity, the height, h, remains constant.
Next, let's examine how the base, b, changes as n increases. The base of each triangle is determined by the angle formed by two adjacent radii of the circle. As n increases, the number of sections becomes larger, resulting in smaller angles between the radii. Therefore, the base of each triangle, b, becomes smaller as n increases.
As n approaches infinity, the angles between the radii become infinitesimally small, and the number of sections effectively becomes infinite. In this case, the base, b, approaches zero.
So, when considering the limit as n approaches infinity:
- The height, h, remains constant.
- The base, b, approaches zero.
Therefore, the limit of h is equal to the radius of the circle, and the limit of b is zero.
This means that the estimate of the circle's area using the formula (bhn/2)*8 becomes more accurate as n increases and approaches infinity, with the height, h, and the base, b, taking on their respective limits.