Question

A sector of a circlr of radius 12cm has the angle 150 degree.It is rolled up so that two bounding radii are joined together to form a cone.The height of the cone is

Answer 1

I can get this answer but something error is there in question that we don't have to find height we have to find slant height

Answer 2

The slant height of the cone is 13 cm

Step-by-step explanation:

Given radius of circle = 12 cm

We know that,

$\text{Area of the sector}=\frac{\pi r^{2} \Theta}{360}$

$=\frac{\pi \times 12^{2} \times 150}{360}$

$\Rightarrow \text { Area of the sector }=60 \pi \ \mathrm{cm}$

Area of the sector is nothing but the curved surface area of the sector.

$60 \pi=\pi \times r \times l$

$60=r \times l$

$60=12 \times l$

$\Rightarrow l=\frac{60}{12}$

$\therefore l=5 \ \mathrm{cm}$

We know that,

$\text{Slant height of the cone} =\sqrt{l^{2}+r^{2}}$

$=\sqrt{5^{2}+12^{2}}$

$=\sqrt{25+144}$

$=\sqrt{169}$

$\therefore \text { Slant height of the cone }=13 \ \mathrm{cm}$

Note: We cannot find height for this problem.