A right angled sector of radius $$r$$ cm is rolled up into a cone in such a way that the two binding radii are joined together. Then the curved surface area of the cone is
A. $$\pi {r^2}{\text{ c}}{{\text{m}}^2}$$
B. $$\frac{{\pi {r^2}}}{4}{\text{ c}}{{\text{m}}^2}$$
C. $$\frac{{\pi {r^2}}}{2}{\text{ c}}{{\text{m}}^2}$$
D. $$2\pi {r^2}{\text{ c}}{{\text{m}}^2}$$
Answer: Option B
Solution (By Examveda Team)
$$\eqalign{ & \Rightarrow {\text{Circumference of sectors}} = \frac{{\pi r}}{2} \cr & \Rightarrow {\text{Circumference of base of cone of radius}} = 2\pi {r_1} \cr & \frac{{\pi r}}{2} = 2\pi {r_1} \Rightarrow {r_1} = \frac{r}{4} \cr & \therefore {\text{Radius of cone}} = \frac{r}{4} \cr & {\text{Curved surface area of cone}} = \pi {r_1}l \cr & l = {\text{slant height}} \cr & l = r \cr & \therefore {\text{Surface area of cone}} = \pi \times \frac{r}{4} \times r \Rightarrow \frac{{\pi {r^2}}}{4} \cr} $$
This Question Belongs to Arithmetic Ability >> Mensuration 3D
Join The Discussion