# A sector of a circle has radius 5cm and an angle of subtended at the centre is cut out of cardboard then curved around to form a cone. find SA and V.SA stands for surface area and V stands for volume.

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We are given a circle of radius 5. A sector with central angle `alpha` has been cut from the circle and formed into a cone. We are asked to find the surface area and volume of the resulting cone.

(1) The length of the arc of the sector is the circumference of the base of the cone. The length of the arc is found by `s=alpha/360 pi d` where s is the arc length, alpha the central angle, and d the diameter of the circle.

In this case we have d=10, so `s=alpha/360 10pi=(alpha pi)/36`

The circumference of the base of the cone is the same as the arc length so `C=2pir=(alpha pi)/36` where C is the circumference and r the radius of the base of the cone. This implies that `r=alpha/72` .

(2) Assuming we are forming a right circular cone, we can form a right triangle whose legs are the radius of the base of the cone and the height of the cone, and whose hypotenuse is the slant height `l` of the cone. Note that the slant height of the cone is the radius of the circle, so `l=5` .

Using `r=alpha/72` and `l=5` we find `r^2+h^2=l^2=>h=sqrt(25-(alpha/72)^2)`

(3) The surface area of a cone is given by `SA=LA+B=pi r l + pi r^2` where SA is the surface area, LA the lateral area, B the area of the base, r the radius of the base of the cone and l the slant height of the cone. (Even though the cardboard will not form the base, you asked for the surface area. If you just want the area of the cardboard, you would find the lateral area)

**So `SA=pi(alpha/72)(5)+pi(alpha/72)^2` **

(4) The volume of the cone is given by `V=1/3 pi r^2 h` where V is the volume, r is the radius of the base of the cone, and h is the height of the cone.

**So `V=1/3 pi (alpha/72)^2(sqrt(25-(alpha/72)^2))` **

** If you know the central angle, just replace `alpha` with that angle (in degrees) and evaluate. For example, if `alpha=36^@` then the volume is `pi/3 (1/2)^2sqrt(25-(1/2)^2)=pi/12 (3sqrt(11))/2=(pi sqrt(11))/8` cubic cm and the surface area is `pi(1/2)(5)+pi(1/2)^2=(5pi)/2+pi/4=(11pi)/4` square cm **

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