Question

A cone is made by rolling up a sector of central angle 90° and radius 12 cm.
(i)Find the slant height of the cone.
(ii)Find the radius of the cone.​

Answer 1

Answer:

Sland height =12

Radius=l/r=x/360

12/r=90/360

12*4=48

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

• A sector of radius 12 cm and central angle 90° is rolled to form a cone.

Let 'r' represents the radius of sector and central angle be 'θ'.

So,

we have

• r = 12 cm

• θ = 90°.

And

Let radius of cone be represented by 'R' and slant height be represented by 'L'

Now,

When sector is rolled to form a cone,

• i. Radius of sector become slant height.

• ii. the length of the arc of sector becomes the circumference of cone.

So,

it implies,

• Slant height of cone, L = r = 12 cm

Now,

We know that,

Length of arc of sector is

$\rm :\longmapsto\:Length_{(arc)}, \: l \: = \: 2\pi \: r \: \dfrac{ \theta}{360}$

and

Circumference of base of cone

$\rm :\longmapsto\:Circumference_{(base cone)} = 2\pi \: R$

According to condition ii. ,

$\rm :\longmapsto\:Length{(arc)} = Circumference_{(base cone)}$

$\rm :\longmapsto\: \: \cancel{2\pi }\: r \: \dfrac{ \theta}{360} = \cancel{ 2\pi }\: R$

$\rm :\longmapsto\:12 \times \dfrac{90}{360} = R$

$\bf\implies \:R = 3 \: cm$

Additional Information :-

Volume of cylinder = πr²h

T.S.A of cylinder = 2πrh + 2πr²

Volume of cone = ⅓ πr²h

C.S.A of cone = πrl

T.S.A of cone = πrl + πr²

Volume of cuboid = l × b × h

C.S.A of cuboid = 2(l + b)h

T.S.A of cuboid = 2(lb + bh + lh)

C.S.A of cube = 4a²

T.S.A of cube = 6a²

Volume of cube = a³

Volume of sphere = 4/3πr³

Surface area of sphere = 4πr²

Volume of hemisphere = ⅔ πr³

C.S.A of hemisphere = 2πr²

T.S.A of hemisphere = 3πr²