RD Sharma Solutions -Ex-20.1(Part - 1), Surface Area And Volume Of Right Circular Cone, Class 9, Mat Class 9 Notes | EduRev

RD Sharma Solutions for Class 9 Mathematics

Created by: Abhishek Kapoor

Class 9 : RD Sharma Solutions -Ex-20.1(Part - 1), Surface Area And Volume Of Right Circular Cone, Class 9, Mat Class 9 Notes | EduRev

The document RD Sharma Solutions -Ex-20.1(Part - 1), Surface Area And Volume Of Right Circular Cone, Class 9, Mat Class 9 Notes | EduRev is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.
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Q 1: Find the curved surface area of a cone, if its slant height is 60 cm and the radius of its base is 21 cm.

Solution:

It is given that :

Slant height of cone(l) = 60 cm

Radius of the base of the cone(r) = 21 cm

Curved surface area(C.S.A) = ?

C.S.A =  πrl  = RD Sharma Solutions -Ex-20.1(Part - 1), Surface Area And Volume Of Right Circular Cone, Class 9, Mat Class 9 Notes | EduRev*21*60 = 3960cm2

Therefore the curved surface area of the right circular cone is 3960cm2

 

Q 2 :The radius of a cone is 5cm and vertical height is 12cm. Find the area of the curved surface.

Solution:

It is given that:

Radius of cone (r) = 5 cm

Height of the tent (h) = 12 cm

Slant Height of tent (l) = ?

Curved surface area (C.S.A) = ?

Now we know we that,

RD Sharma Solutions -Ex-20.1(Part - 1), Surface Area And Volume Of Right Circular Cone, Class 9, Mat Class 9 Notes | EduRev

l2 = r2 + h2

l2 = 52 + 122

l2 = = 25 + 144

⇒ l = 13 cm

Now,

C.S.A =  πrl  = 3.14*5*12 = 204.28cm2

Therefore the curved surface area of the cone is 204.28cm2

 

Q3 : The radius of a cone is 7 cm and area of curved surface is 176cm2 .Find the slant height.

Solution:

It is given that:

Radius of cone(r) = 7 cm

Curved surface area(C.S.A) = 176cm2

Slant Height of tent (l) = ?

Now we know,

C.S.A = πrl

⇒ πrl  = 176

RD Sharma Solutions -Ex-20.1(Part - 1), Surface Area And Volume Of Right Circular Cone, Class 9, Mat Class 9 Notes | EduRev∗7∗l = 176

 RD Sharma Solutions -Ex-20.1(Part - 1), Surface Area And Volume Of Right Circular Cone, Class 9, Mat Class 9 Notes | EduRev = 8 cm

Therefore the slant height of the cone is 8 cm.

 

Q 4: The height of a cone 21 cm. Find the area of the base if the slant height is 28 cm.

Solution:

It is given that :

Height of the traffic cone(h) = 21 cm

Slant height of the traffic cone(l) = 28 cm

Now we know that,

l2 = r2 + h2

282 = r2 + 212

r2 = 282 − 212

RD Sharma Solutions -Ex-20.1(Part - 1), Surface Area And Volume Of Right Circular Cone, Class 9, Mat Class 9 Notes | EduRev

Area of the circular base = πr2

 RD Sharma Solutions -Ex-20.1(Part - 1), Surface Area And Volume Of Right Circular Cone, Class 9, Mat Class 9 Notes | EduRev= 1078 cm2

Therefore the area of the base is 1078cm2.

 

Q 5: Find the total surface area of a right circular cone with radius 6 cm and height 8 cm.

Solution:

It is given that :

Radius of the cone(r) = 6 cm

Height of the cone(h) = 8 cm

Total Surface area of the cone (T.S.A) = ?

l2 = r2 + h2

= 62 + 82

= 36 + 64 = 100

⇒ l = 10 cm

T.S.A = Curved surface area of cone + Area of circular base

=  π∗r∗l +  π∗r2

=  (RD Sharma Solutions -Ex-20.1(Part - 1), Surface Area And Volume Of Right Circular Cone, Class 9, Mat Class 9 Notes | EduRev∗6∗10) + (RD Sharma Solutions -Ex-20.1(Part - 1), Surface Area And Volume Of Right Circular Cone, Class 9, Mat Class 9 Notes | EduRev∗6∗6)

RD Sharma Solutions -Ex-20.1(Part - 1), Surface Area And Volume Of Right Circular Cone, Class 9, Mat Class 9 Notes | EduRev

= 301.171cm2.

Therefore the area of the base is 301.71cm2.

 

Q 6: Find the curved surface area of a cone with base radius 5.25 cm and slant height 10 cm.

Solution:

It is given that :

Base radius of the cone(r) = 5.25 cm

Slant height of the cone(l) = 10 cm

Curved surface area (C.S.A) =  πrl

= RD Sharma Solutions -Ex-20.1(Part - 1), Surface Area And Volume Of Right Circular Cone, Class 9, Mat Class 9 Notes | EduRev*5.25*10 = 165cm2.

Therefore the curved surface area of the cone is 165cm2

 

Q 7: Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.

Solution:

It is given that :

Diameter of the cone(d) = 24 m

Radius of the cone(r) = RD Sharma Solutions -Ex-20.1(Part - 1), Surface Area And Volume Of Right Circular Cone, Class 9, Mat Class 9 Notes | EduRev = 12m

Slant height of the cone(l) = 21 m

T.S.A = Curved surface area of cone + Area of circular base

= π∗r∗l +  π∗r2

=  (RD Sharma Solutions -Ex-20.1(Part - 1), Surface Area And Volume Of Right Circular Cone, Class 9, Mat Class 9 Notes | EduRev∗12∗21) + (RD Sharma Solutions -Ex-20.1(Part - 1), Surface Area And Volume Of Right Circular Cone, Class 9, Mat Class 9 Notes | EduRev∗12∗12)  = RD Sharma Solutions -Ex-20.1(Part - 1), Surface Area And Volume Of Right Circular Cone, Class 9, Mat Class 9 Notes | EduRev∗12(12 + 21)  = 1244.57m2.

Therefore the total surface area of the cone is 1244.57m2.

 

Q 8: The area of the curved surface of a cone is 60 π cm2 . If the slant height of the cone be 8 cm, find the radius of the base.

Solution:

It is given that :

Curved surface area(C.S.A) = 60 π cm2

Slant height of the cone(l) = 8 cm

Radius of the cone(r) = ?

Now we know,

Curved surface area(C.S.A ) = πrl

⇒  πrl  = 60 π cm2

⇒ r*8 = 60

⇒ r = 7.5 cm

Therefore the radius of the base of the cone is 7.5 cm.

 

Q 9: The curved surface area of a cone is 4070 cm2 and diameter is 70 cm .What is its slant height?

Solution:

Diameter of the cone(d) = 70 cm

Radius of the cone(r) =  RD Sharma Solutions -Ex-20.1(Part - 1), Surface Area And Volume Of Right Circular Cone, Class 9, Mat Class 9 Notes | EduRev  = 35cm

Slant height of the cone(l) = ?

Now,

Curved surface area = 4070cm2

⇒ πrl  = 4070

Where, r = radius of the cone

l = slant height of the cone

Therefore πrl  = 4070

RD Sharma Solutions -Ex-20.1(Part - 1), Surface Area And Volume Of Right Circular Cone, Class 9, Mat Class 9 Notes | EduRev∗35∗l = 4070

RD Sharma Solutions -Ex-20.1(Part - 1), Surface Area And Volume Of Right Circular Cone, Class 9, Mat Class 9 Notes | EduRev  = 37 cm

Therefore slant height of the cone is 37 cm.

 

Q 10: The radius and slant height of a cone are in the ratio 4:7.If its curved surface area is 792 cm2, find its radius.

Solution:

It is given that

Curved surface area = πrl  = 792

Let the radius(r) = 4x

Height(h) = 7x

Now,

C.S.A = 792

RD Sharma Solutions -Ex-20.1(Part - 1), Surface Area And Volume Of Right Circular Cone, Class 9, Mat Class 9 Notes | EduRev∗4x∗7x = 792

⇒ 88x2 = 792

⇒  x2 =  RD Sharma Solutions -Ex-20.1(Part - 1), Surface Area And Volume Of Right Circular Cone, Class 9, Mat Class 9 Notes | EduRev = 9

⇒ x = 3

Therefore Radius = 4x = 4*3 = 12 cm

 

Q 11: A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.

Solution:

Given that,

Radius of conical cap(r) = 7 cm

Height of the conical cap(h) = 24 cm

Slant height(l) of conical cap =RD Sharma Solutions -Ex-20.1(Part - 1), Surface Area And Volume Of Right Circular Cone, Class 9, Mat Class 9 Notes | EduRev

RD Sharma Solutions -Ex-20.1(Part - 1), Surface Area And Volume Of Right Circular Cone, Class 9, Mat Class 9 Notes | EduRevcm = 25 cm

C.S.A of 1 conical cap = πrl

=  RD Sharma Solutions -Ex-20.1(Part - 1), Surface Area And Volume Of Right Circular Cone, Class 9, Mat Class 9 Notes | EduRev∗7∗25

= 550cm2

Curved surface area of 0 such conical caps = 5500 cm2

Thus, 5500 cm2 sheet will be required for making 10 caps.

 

Q 12: Find the ratio of the curved surface area of two cones if their diameter of the bases are equal and slant heights are in the ratio 4:3.

Solution:

Given that,

Diameter of two coins are equal.

Therefore their radius are equal.

Let r1  = r2 = r

Let ratio be x

Therefore slant height l1 of 1st cone = 4x

Similarly slant height l2 of 2nd cone = 3x

Therefore RD Sharma Solutions -Ex-20.1(Part - 1), Surface Area And Volume Of Right Circular Cone, Class 9, Mat Class 9 Notes | EduRev

RD Sharma Solutions -Ex-20.1(Part - 1), Surface Area And Volume Of Right Circular Cone, Class 9, Mat Class 9 Notes | EduRev

RD Sharma Solutions -Ex-20.1(Part - 1), Surface Area And Volume Of Right Circular Cone, Class 9, Mat Class 9 Notes | EduRev

= RD Sharma Solutions -Ex-20.1(Part - 1), Surface Area And Volume Of Right Circular Cone, Class 9, Mat Class 9 Notes | EduRev

Hence ratio is 4:3

 

Q 13: There are two cones .The curved surface area of one is twice that of the other. The slant height of the later is twice that of the former. Find the ratio of their radii.

Solution:

Let the curved surface area of 1st cone = 2x

C.S.A of 2nd cone = x

Slant height of 1st cone = h

Slant height of 2nd cone = 2h

Therefore RD Sharma Solutions -Ex-20.1(Part - 1), Surface Area And Volume Of Right Circular Cone, Class 9, Mat Class 9 Notes | EduRev

RD Sharma Solutions -Ex-20.1(Part - 1), Surface Area And Volume Of Right Circular Cone, Class 9, Mat Class 9 Notes | EduRev

RD Sharma Solutions -Ex-20.1(Part - 1), Surface Area And Volume Of Right Circular Cone, Class 9, Mat Class 9 Notes | EduRev

RD Sharma Solutions -Ex-20.1(Part - 1), Surface Area And Volume Of Right Circular Cone, Class 9, Mat Class 9 Notes | EduRev

RD Sharma Solutions -Ex-20.1(Part - 1), Surface Area And Volume Of Right Circular Cone, Class 9, Mat Class 9 Notes | EduRev

Therefore ratio of rand r2 is 4:1.

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