Class 9 Exam  >  Class 9 Notes  >  RD Sharma Solutions for Class 9 Mathematics  >  RD Sharma Solutions -Ex-20.1(Part - 1), Surface Area And Volume Of Right Circular Cone, Class 9, Mat

Ex-20.1(Part - 1), Surface Area And Volume Of Right Circular Cone, Class 9, Mat RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics PDF Download

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

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

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

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

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

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

Area of the circular base = πr2

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

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

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

= 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

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

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

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

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

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

⇒ 88x2 = 792

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

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

C.S.A of 1 conical cap = πrl

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

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

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

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

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

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

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

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

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

Therefore ratio of rand r2 is 4:1.

The document Ex-20.1(Part - 1), Surface Area And Volume Of Right Circular Cone, Class 9, Mat RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.
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FAQs on Ex-20.1(Part - 1), Surface Area And Volume Of Right Circular Cone, Class 9, Mat RD Sharma Solutions - RD Sharma Solutions for Class 9 Mathematics

1. What is the formula for finding the surface area of a right circular cone?
Ans. The formula for finding the surface area of a right circular cone is given by S = πrl + πr², where S is the surface area, r is the radius of the base, and l is the slant height of the cone.
2. How is the volume of a right circular cone calculated?
Ans. The volume of a right circular cone is calculated using the formula V = (1/3)πr²h, where V is the volume, r is the radius of the base, and h is the height of the cone.
3. Can you explain the concept of slant height in a right circular cone?
Ans. The slant height of a right circular cone is the distance between the apex (top point) and any point on the circumference of the base. It is calculated using the Pythagorean theorem, where the slant height (l) is the hypotenuse, the radius (r) is one side, and the height (h) is the other side. The formula is l = √(r² + h²).
4. How do you find the height of a right circular cone if only the slant height and radius are known?
Ans. To find the height of a right circular cone when only the slant height (l) and radius (r) are known, you can use the Pythagorean theorem. Rearranging the formula for slant height, we have h = √(l² - r²).
5. Is the radius of the base the same as the slant height in a right circular cone?
Ans. No, the radius of the base and the slant height are different. The radius (r) is the distance from the center of the base to any point on the circumference, while the slant height (l) is the distance from the apex (top point) to any point on the circumference. The slant height is always longer than the radius.
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