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.
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
5 answers
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
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
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
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
3 answers