From a sector to a Cone!

Considering the circular sector:

The arc length is $c=\dfrac{45}{360}(2\pi )(10)=2.5\pi$

Considering the right circular cone:

The circumference of the base is equal to the arc length of the circular sector. So we have

$2.5\pi = 2\pi r$

$r=1.25$

By pythagorean theorem, $h=\sqrt{10^2-1.25^2} \approx 9.92157$

Finally, the volume is

$V=\dfrac{1}{3}A_b h = \dfrac{1}{3}(\pi)(1.25^2)(9.92157) \approx \boxed{16.234~cm^3}$

Area of the sector=45°/360°×22/7×10×10=22/7×1.25 now equate it with curved surface area of cone i.e. 22/7×r×l (here l=radius of sector) u will get r=1.25 now use pythegoras theorem to find h of cone
u will get h=9.92(approx) now find volume of cone i.e.
1/3×22/7×1.25×1.25×9.92

In order to find the volume of the cone, we have to get its base radius and it’s height. If a sector is formed into a right circular cone, then its arc length is equal to the circumference of the base of the cone.

First, get the arc length of the sector.

s=θr s=(π/4)(10) s=5/2 π cm If s=circumference of the base of the cone, then we can could get the base radius by equating these equations. s=2πr 5/2 π cm=2πr

r= 5/4 cm

second, through pythagorian theorem, we can solve for the height.

h=sqrt[(10 cm)^2-(5/4cm)^2)} h=9.92157 cm

Lastly, solve for the cone’s volume. V=(πr^2 h)/3
V=(π(5/4 cm)^2 (9.921567 cm))/3

V=16.234126 (cm)^3 ≈ 16.234 (cm)^3