SEBA Class 9 Maths Chapter 12 Surface Areas and Volumes

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SEBA Class 9 Maths Chapter 12 Surface Areas and Volumes

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Surface Areas and Volumes

Chapter: 12

Exercise 12.1

Assume π = 22/ 7, unless stated otherwise.

1. Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its can surface area.

Ans: Here, Diameter of the base of the cone = 10.5 cm

∴ Radius of the base of the cone, r =  5-25 cm /2 

Slant height of the cone, I = 10 cm 

Curved surface area of the cone = πrl

= 22/7 x 5-25 × 10cm²

= 165 cm²

2. Find the total surface area of a cone, if its slant height is 21 m and diameter of its bay is 24 m.

Ans:  Here, Slant height of the cone, 1 = 21 m 

Diameter of the base of the cone = 24 m

∴ Radius of the base of the cone, r = 24/2m = 12 m 

Total surface area of the cone = πr (l + r) 

= 22/7 ×12 x (21+12) m² 

= 22/7 ×12 × 33m²

= 1244.57 m²

3. Curved surface area of a cone is 308 cm² and its slant height is 14 cm. Find (i) radius of the base and (ii) total surface area of the cone.

Ans: (i) Slant height of the cone, l = 14 cm 

Let the radius of the base be r cm 

Curved surface area of the cone = 308 cm² 

⇒ πrl = 308 

⇒ 22/7 × rx14 = 308 

r= 308 × 7/ 22 × 14

r = 7 cm 

Therefore, radius of the base of the cone is 7 cm.

(ii) Total surface area of the cone = πr (l+r) 

= 22/7× (14+7)cm² 

= 22 × 21 cm². 

= 462 cm²

4. A conical tent is 10 m high and the radius of its base is 24 m. Find

(i) slant height of the tent.

(ii) cost of the canvas required to make the tent, if the cost of 1 m² canvas is Rs 70.

Ans: (i) Here, Height of the conical tent, h = 10 m 

Radius of the base, r = 24 m 

Slant height, l = ?

We know,

(ii) Curved surface area of the conical tent = πrl

= 22/7 × 24 × 26 m²

= 13728/7 m²

∴ Quantity of canvas required = 13728/m²

  Cost of 1 m² canvas Rs 70

∴ Cost of 13728/7 m² canvas = Rs 70 × 13728/7 = Rs 137280

5. What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6 m ? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm (Use π = 3.14).

Ans:  Here, Height of the conical tent, h = 8 m 

Base radius of the conical tent, r = 6 m

It is given that the width of the tarpaulin is 3 m

∴ Lenght of the tarpaulin = 188.4/3 m = 62.8 m

But the extra length of the material required for stitching margins and wastage in cutting is 20 cm = 0.2 m

∴ Actual length of the tarpaulin required = (62.8 + 0.2) m = 63 m  

6. The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white- washing its curved surface at the rate of Rs 210 per m^2..

Ans: Slant height of the conical tomb, l = 25 m 

Base diameter of the conical tomb, = 14 m 

∴ Base radius of the conical tomb, r = 14/ 2m =7m 

Curved surface area of the conical tomb = πrl 

                                                                 = 22/ 7 × 7 × 25m²  

                                                                 = 550 m²

Cost of white-washing per 100 m² = Rs 210                                  

∴ Cost of white-washing per m² = Rs 210/100 

∴ Cost of white-washing per 500 m² = Rs 210/100 × 550 = Rs 1155

7. 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.

Ans: Here, Base radius of the conical cap, r = 7 cm

Height of the cap, h = 24 cm

∴  Slant height of the conical cap, 

Curved surface area of 1 conical cap = πrl

                                                           = 22/7 × 7 × 25 cm²

                                                           = 550 cm²

∴ Sheet required to make 1 conical cap = 550 cm²

∴ Sheet required to make 10 such conical caps = 550 × 10 cm² = 5500 cm²

8. A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. 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? 

(Use π = 3.14 and take √1.04=1.02)

Ans: Here, Base diameter of each cone = 40 cm

∴ Base radius of each cone, r = 40/2 cm = 20 cm = 0.20 m

Height of each cone, h = 1 m

∴  Slant height of each cone, 

Curved surface area of 1 conical cap = πrl

                                                            = 3.14 × 0.2 × 1.02 m²

                                                            = 0.64056 m²

∴ Curved surface area of 50 such cones = 50 × 0.64056 m² = 32.028 m²

Cost of painting per m² of the curved surface = Rs 12

∴ Cost of painting 32.028 m² of the curved surface = Rs 12 × 32. 028 = rs 384.34

Hence, the cost of painting all the 50 cones is Rs 384.34 (approx.)

9. This question consists of two statements namely Assertion (A) and Reason (R). Assertion (A): The curved surface area of a cone in 330 cm² and its diameter is 14 cm. Then its slant height is 15 cm.

Reason (R): If radius and slant height of a right circular cone are r and I respectively, then its curved surface area is πrl.

Select the correct answer from the following options.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).

(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).

(c) Assertion (A) is true but Reason (R) is false.

(d) Assertion (A) is false but Reason (R) is true.

Ans: Assertion:

Curved surface Area = 330 cm²

⇒ πrl = 330 cm²

⇒ 22/7 × 7 × l = 330 cm²

⇒ l = 330/22 cm

⇒ l = 15cm 

Assertion is True.

Reason: Curved Surface Area of a cone = πrl

Reason is also True.

And Reason is correct explanation of Assertion.

The correct option is (a)

10. The total surface area of a cone of radius r/2 and slant height 2l is 

(a) 2πrl

(b) 2π r(l + r)

(d) πr (l+r)

Ans: T.S.A of a cone 

The correct option is (c) 

10. The total surface area of a cone of radius r/2  and slant height 2l is 

(a) 2πrl

(b) 2π r(1 + r)

(d) πr(l + r)

Ans:

Exercise 12.2

Q.1. Find the surface area of a sphere of radius:

(i) 10.5 cm

(ii) 5.6 cm

(iii) 14 cm

Ans: (i) r = 10.5cm

∴ Surface area = 4πr²

(ii) r = 5.6cm

(iii) r = 14cm

Q.2. Find the surface area of a sphere of radius:

(i) 14 cm

(ii) 21 cm

(iii) 3.5 cm

Ans: (i) Diameter = 14cm

(ii) Diameter = 21 cm

(iii) Diameter = 3.5 cm

Q.3. Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14

Ans: r = 10cm 

∴ Total surface area of the hemisphere

= 3πr² = 3 × 3.14 × (10)² = 942cm²

Q.4. The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.

Ans: Case I. r = 7cm

∴ Case II. r = 14cm

∴ Ratio of surface areas of the balloon

Q.5. A Hemisphere bowl made of brass has an inner diameter 10.5 cm. Find the cost of tin-plating it on the incident the rate of Rs. 16 per 100 cm²

Ans: Inner diameter = 10.5cm

∴ Inner surface area

∴ Cost of tin-plating at the rate of Rs. 16 per 100 cm²

Q.6. Find the radius of a sphere whose surface area is 154 cm². 

Ans: Let the radius of the sphere be r cm. 

Surface area = 154 cm²

Hence the radius of the sphere is 3.5 cm.

Q.7. The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.

Ans: Let the diameter of the earth be 2r.

∴ Surface area of the earth = 4πr²

∴ Ratio of their surface area

Q.8. A Hemisphere bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.

Ans: Inner radius of the bowl = 5 cm

Thickness of steel = 0.25 cm

∴ Outer radius of the bowl = 5 + 0.25 = 5.25cm 

∴ Outer curved surface of the bowl

Q.9. A right circular cylinder just encloses a sphere of radius r. Find

(i) surface area of the sphere.

(ii) curved surface area of the cylinder.

(iii) ratio of the areas obtained in (i) and (ii).

Ans:  (i) Surface area of the sphere = 4πr²

(ii) For cylinder 

Radius of the base = r 

Height = 2r

∴ Curved surface area of the cylinder = 2π(r)(2r) = 4πr²

(iii) Ratio of the areas obtained in (i) and (ii)

10. Two statements are given below – 

Statement-1 : Surface area of a sphere is πr² if the radius is r/2

Statement-2: If the radius is r/2 2, then the total surface area of a hemisphere is ¾ πr²

Choose the correct alternative.

(a) Both Statement-1 and Statement-2 are true

(b) Both Statement-1 and Statement-2 are false

(c) Statement-1 is true but Statement-2 is false

(d) Statement-1 is false but Statement-2 is true

Ans. Statement I

Surface Area of a Sphere = 4πr² [where, r = radius]

If radius is r/2, surface area 

Statement – 1 is true.

Statement – 2

We have, Total surface area of a hemisphere = 3πr²

When radius is r/2, Total Surface area of hemisphere

Statement – 2 is True.

The correct option is (a) 

11. The following question contains Assertion (A) and Reason (R) along with four choices (a), (b), (c) and (d), only one of which is the correct answer. Mark the correct choice.

Assertion (A): The radius of spherical ballon increases from 8cm to 12 cm. The ratio of the surface areas of the ballon is 4:9.

Reason (R): If the radius of two spheres are in the ratio : √S₁ : √S₂ then the ratio of their surface areas is S₁ : S₂

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A)

(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A)

(c) Assertion (A) is true but Reason (R) is false

(d) Assertion (A) is false but Reason (R) is true

Ans. Assertion (A):

Surface area of a sphere = 4πr² Ratio of surface area of spherical ballons 

= 4π. 8² : 4π.12² 

= 8² : 12² 

= 8 × 8 =12 × 12

= 4 : 9 

Assertion is True.

Reason (R):

If r₁ & r₂ are the radius of two sphere, then ratio of their surface area is r₁ ²: r₂²

Hence, If √S: √S are the ratio of radius of two sphere. 

Then, Ratio of Surface area

 = (√s₁)²  : (√s₂)²   = S₁ : S₂ 

Reason is True.

Reason (R) is correct explanation of Assertion (A).

The correct option is (a).