Class 9 Exam  >  Class 9 Notes  >  RD Sharma Solutions for Class 9 Mathematics  >  RD Sharma Solutions -Ex-20.1 (Part -2), Surface Area And Volume Of Right Circular Cone, Class 9, Mat

Ex-20.1 (Part -2), Surface Area And Volume Of Right Circular Cone, Class 9, Mat RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics PDF Download

Q 14: The diameters of two cones are equal. If their slant height are in the ratio 5:4,find the ratio of their curved surfaces.

Solution:

It is given that

Diameters of two cones are equal

Therefore their radius are also equal i.e

r1 = r2

Let the ratio of slant height be x

Therefore l= 5x

l= 4x

Therefore Ratio of curved surface area =Ex-20.1 (Part -2), 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 -2), 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 -2), 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 -2), Surface Area And Volume Of Right Circular Cone, Class 9, Mat RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

∴ Ratio of curved surface area is 5:4.

 

Q 15: Curved surface area of a cone is 308 cm2 and its slant height is 14cm. Find the radius of the base and total surface area of the cone.

Solution:

(1) It is given that

Slant height of cone = 14cm

Let radius of circular end of cone = r

Curved surface area of cone = πrl

⇒ 308 cm2  = Ex-20.1 (Part -2), Surface Area And Volume Of Right Circular Cone, Class 9, Mat RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics∗r∗14

⇒ r = Ex-20.1 (Part -2), Surface Area And Volume Of Right Circular Cone, Class 9, Mat RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics = 7cm

Thus radius of circular end of cone = 7 cm.

(ii) It is given that C.S.A = 308 cm2

We know that total surface area of a cone

= curved surface area of a cone + Area of base

=  πrl + πr2

= [308 + (Ex-20.1 (Part -2), Surface Area And Volume Of Right Circular Cone, Class 9, Mat RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics∗72)]

= 308 + 154

= 462cm2

Thus total surface area of the cone is 462cm2

 

Q 16: The slant height and base diameter of a conical tomb are 25m and 14m respectively. Find the cost of white washing its curved surface area at the rate of Rs 210 per 100m2.

Solution:

It is given that

Slant height of conical tomb(l) = 25m

Base radius (r) of tomb = Ex-20.1 (Part -2), Surface Area And Volume Of Right Circular Cone, Class 9, Mat RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics = 7m

Curved surface area of conical length tomb = πrl

= Ex-20.1 (Part -2), Surface Area And Volume Of Right Circular Cone, Class 9, Mat RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics∗7∗25

= 550 m2

Cost of white washing 100 marea = Rs 210

Cost of white washing 550m2 area = Ex-20.1 (Part -2), Surface Area And Volume Of Right Circular Cone, Class 9, Mat RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics = Rs 1155

Therefore the cost of white washing the whole tomb is Rs 1155.

 

Q 17: A conical tent is 10 m high and the radius of it base is 24 m. Find the slant height of the tent. If the cost of 1m2 canvas is Rs 70, find the cost of canvas required for the tent.

Solution:

It is given that

Height of the conical tent(h) = 10m

Radius of conical tent(r) = 24m

Let slant height of conical tent be l

l2 = h2 + r2

= 10+ 242 = 100 + 576

= 676 m2

⇒ l = 26m

Thus,the slant height of the conical tent is 26m.

 

(ii) It is given that:

Radius(r) = 24 m

Slant height (l) = 26 m

C.S.A of tent = πrl

= Ex-20.1 (Part -2), Surface Area And Volume Of Right Circular Cone, Class 9, Mat RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics∗24∗26

Ex-20.1 (Part -2), Surface Area And Volume Of Right Circular Cone, Class 9, Mat RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Cost of 1m2 canvas = Rs 70

Cost of Ex-20.1 (Part -2), Surface Area And Volume Of Right Circular Cone, Class 9, Mat RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematicsm2 canvas = Rs Ex-20.1 (Part -2), Surface Area And Volume Of Right Circular Cone, Class 9, Mat RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics*70 = Rs 1,37,280

Thus the cost of canvas required to make the tent is Rs 1,37,280.

 

 

Q 18: A tent is in the form of a right circular cylinder surmounted by a cone. The diameter of cylinder is 24 m. The height of the cylindrical portion is 11m while the vertex of the cone is 16 m above the ground. Find the area of the canvas required for the tent.

Solution:

It is given that

Diameter of cylinder = 24 m

 therefore radius = Ex-20.1 (Part -2), 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 -2), Surface Area And Volume Of Right Circular Cone, Class 9, Mat RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics  = 12 cm

Also radius of cone = 12 m

Height of cylinder = 11 m

Height of cone = 16−11 = 5m

Slant height of cone

Ex-20.1 (Part -2), 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 -2), Surface Area And Volume Of Right Circular Cone, Class 9, Mat RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

= 13m

Therefore area of canvas required for the tent = πrl + 2πrh

= Ex-20.1 (Part -2), Surface Area And Volume Of Right Circular Cone, Class 9, Mat RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics[(12∗13) + (2∗12∗11)]  = 490.286 + 829.714

= 1320m2

 

Q19. A circus tent is cylindrical to a height of 3 m and conical above it. If its diameter is 105 m and the slant height of the conical portion is 53 m, calculate the length of the canvas 5 m wide to make the required tent.

Solution:

Given diameter = 105 m,

Radius = Ex-20.1 (Part -2), Surface Area And Volume Of Right Circular Cone, Class 9, Mat RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematicsm = 52.5 m

Therefore curved surface area of circus tent = πrl + 2πrh

= (Ex-20.1 (Part -2), Surface Area And Volume Of Right Circular Cone, Class 9, Mat RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics∗52.5∗53) + (2∗Ex-20.1 (Part -2), Surface Area And Volume Of Right Circular Cone, Class 9, Mat RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics∗52.5∗3)

= 8745 + 990 = 9735 m2

Therefore length of the canvas required for tent

 Ex-20.1 (Part -2), 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 -2), Surface Area And Volume Of Right Circular Cone, Class 9, Mat RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics = 1947 m

 

Q20 The circumference of the base of a 10m height conical tent is 44m, calculate the length of canvas used in making the tent if width of canvas is 2m.

Solution:

We know that

C.S.A of cone = πrl

Given circumference = 2πr

⇒ 2∗Ex-20.1 (Part -2), Surface Area And Volume Of Right Circular Cone, Class 9, Mat RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics∗r = 44

Ex-20.1 (Part -2), Surface Area And Volume Of Right Circular Cone, Class 9, Mat RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

⇒ r = 7m

Therefore l = Ex-20.1 (Part -2), 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 -2), 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 -2), Surface Area And Volume Of Right Circular Cone, Class 9, Mat RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Therefore C.S.A of tent = πrl

Ex-20.1 (Part -2), 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 -2), Surface Area And Volume Of Right Circular Cone, Class 9, Mat RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Therefore the length of canvas used in making the tent

Ex-20.1 (Part -2), 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 -2), 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 -2), Surface Area And Volume Of Right Circular Cone, Class 9, Mat RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

= 134.2 m

 

Q 21: What length of tarpaulin 4 m wide will be required to make a conical tent of height 8 m and base radius 6 m?Assume that the extra length of material will be required for stitching margins and wastage in cutting is approximately 20 cm.

Solution:

Given that,

Height of conical tent(h) = 8m

Radius of base of tent(r) = 6m

Slant height(l) =  Ex-20.1 (Part -2), 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 -2), 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 -2), Surface Area And Volume Of Right Circular Cone, Class 9, Mat RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

C.S.A of conical tent = πrl

= (3.14*6*10) m2  = 188.4 m2

Let the length of tarpaulin sheet required be l

As 20 cm will wasted ,so effective

Length will be (l – 0.2 m)

Breadth of tarpaulin = 3m

Area of sheet = C.S.A of sheet

[l*0.2*3]m = 188.4 m2  = l – 0.2 m = 62.8 m

Accounting extra for wastage:

⇒ l = 63 m

Thus the length of the tarpaulin sheet will be = 63 m

 

Q 22: A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled card board. Each cone has a base diameter of 40 cm and height 1m. If the outer side of each of the cones is to be painted and the cost of painting is Rs 12 per m2,what will be the cost of painting all these cones?

Solution:

The area to be painted is the curved surface area of each cone.

The formula of the curved surface area of a cone with base radius and slant height 7 is given as

Curved Surface Area =  πrl

For each cone, we’re given that the base diameter is 0.40 m.

Hence the base radius r = 0.20 m.

The vertical height l 1 = 1 m.

To find the slant height ‘l’ to be used in the formula for Curved Surface Area we use the following relation

Slant height ,

Ex-20.1 (Part -2), 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 -2), 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 -2), 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 -2), Surface Area And Volume Of Right Circular Cone, Class 9, Mat RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

l = 1.02 m

Now substituting the values of r = 0.2 m and slant height 1 = 1.02 m and using pi = 3.14 in the formula of C.S.A.

We get Curved Surface Area = (3.14)(0.2)(1.02) = 0.64056 m2

This is the curved surface area of a single cone.

Since we need to paint 50 such cones the total area to be painted is,

Total area to be painted = (0.64056) (50) = 32.028 m2

The cost of painting is given as Rs. 12 per m2

Hence the total cost of painting = (12) (32.028) = 384.336

Hence, the total cost that would be incurred in the painting is Rs. 384.336

 

Q23. A cylinder and a cone have equal radii of their base and equal heights. If their curved surface area are in the ratio 8:5, show that the radius of each is to the height of each as 3:4.

Solution:

It is given that the base radius and the height of the cone and the cylinder are the same.

So let the base radius of each is ‘ r ‘ and the vertical height of each is ‘h’.

Let the slant height of the cone be ‘ l ’.

The curved surface area of the cone =  πrl

The curved surface area of the cylinder =  Extra close brace or missing open brace

It is said that the ratio of the curved surface areas of the cylinder to that of the cone is 8:5

So,

Ex-20.1 (Part -2), 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 -2), 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 -2), Surface Area And Volume Of Right Circular Cone, Class 9, Mat RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

But we know that l =Ex-20.1 (Part -2), 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 -2), Surface Area And Volume Of Right Circular Cone, Class 9, Mat RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Squaring on both sides we get:

Ex-20.1 (Part -2), 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 -2), 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 -2), 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 -2), 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 -2), Surface Area And Volume Of Right Circular Cone, Class 9, Mat RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Hence it is shown that the ratio of the radius to the height of the cone as well as the cylinder is: 3 : 4

The document Ex-20.1 (Part -2), 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 -2), 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 to find the surface area of a right circular cone?
Ans. The formula to find the surface area of a right circular cone is given by: Surface Area = πr(r + l), where r is the radius of the base and l is the slant height of the cone.
2. How do you calculate the volume of a right circular cone?
Ans. The volume of a right circular cone can be calculated using the formula: Volume = (1/3)πr^2h, where 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 from the vertex of the cone to any point on the circumference of its base. It is represented by the letter 'l' in the formulas for surface area and volume of a cone. The slant height can be calculated using the Pythagorean Theorem, where the slant height, radius, and height of the cone form a right-angled triangle.
4. How is a right circular cone different from a cylinder?
Ans. A right circular cone and a cylinder are both three-dimensional geometric shapes, but they have some key differences. - A cone has a circular base and tapers to a point called the vertex, while a cylinder has two parallel circular bases. - The height of a cone is the distance from the vertex to the base, while the height of a cylinder is the distance between the two bases. - The lateral surface area of a cone is curved, while the lateral surface area of a cylinder is flat. - The volume of a cone is one-third of the volume of a cylinder with the same base area and height.
5. How can the surface area and volume of a cone be useful in real-life applications?
Ans. The surface area and volume of a cone have various real-life applications: - In construction, the surface area of a cone is used to calculate the amount of paint or wallpaper needed to cover the surface of a cone-shaped roof or structure. - In manufacturing, the volume of a cone is used to determine the capacity of containers such as ice cream cones, party hats, or traffic cones. - In engineering, the surface area and volume of a cone are used in designing and analyzing structures such as traffic barriers, silos, and cooling towers. - In geometry, the surface area and volume of a cone are studied to understand the properties and relationships of different three-dimensional shapes.
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