Surface Areas and Volumes: 9 Math

NCERT Exercise 13.2 part-2: 9th math

Surface Areas And Volumes Class nine Math NCERT Exercise 13.2 Question (6) Curved surface area of a right circular cylinder is 4.4 m². If the radius of the base of the cylinder is 0.7 m, find its height.

Solution

Given,

Curved surface area of the cylinder = 4.4 m²

And, Radius (r) = 0.7 m

Therefore, Height of the given cylinder = ?

We know that, Curved surface area of a cylinder = 2 ℼ r h

⇒ 4.4 m²= 2 × `22/7` × 0.7 m × h

⇒ 4.4 m² = 2 × 22 × 0.1 m × h

⇒ 4.4 m² = 44 × 0.1 m × h

⇒ 4.4 m² = 4.4 m × h

`=>h = (4.4m^2)/(4.4m)`

⇒ h = 1 m

Therefore, height of the given cylinder = 1 m Answer

Surface Areas And Volumes Class nine Math NCERT Exercise 13.2 Question(7) The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find

(i) Its inner curved surface area.

(ii) The cost of plastering this curved surface at the rate of Rs 40 per m²

Solution:

Given, Inner diameter of circular well = 3.5 m

Therefore, inner radius (r) = 3.5/2 m = 1.75 m

Height (depth) of the well = 10 m

(i) Its inner curved surface area =?

We know that inner curved surface area of a circular cylinder = 2 ℿ r h

Therefore, inner curved surface area of given circular well

= 2 × `22/7` × 1.75 m × 10 m

= 2 × 22 × 0.25 m × 10 m

= 44 × 2.5 m

=110 m2

Thus inner curved surface area of given well = 110 m² Answer

(ii) The cost of plastering this curved surface at the rate of Rs 40 per m² =?

As given the cost of plastering for 1 m² = Rs 40

Therefore, cost of plastering for 110 m²

= 40 × 110 = Rs 4400

Thus, cost of plastering = Rs 4400 Answer

Surface Areas And Volumes Class nine Math NCERT Exercise 13.2Question(8) In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.

Solution

Given, Length of the pipe = 28 m

And, Diameter of the pipe = 5 cm

= 5/100 m = 0.05 m

Therefore, Radius (r) of the pipe = 0.05/2 m

⇒ Radius (r) = 0.025 m

Thus, total radiating surface in the system = ?

Here, since outer surface of the pipe of a heating system radiates heat, thus curved surface area of the pipe will be equal to the area of radiating surface.

Now, we know that, Curved Surface Area of a Cylinder = 2 π r h

Therefore, curved surface area of the given pipe

= 2 × `22/7` × 0.025 m × 28 m

= 2 × 22 × 0.025 m × 4 m

= 44 × 0.1 m²

= 4.4 m²

Thus, radiating surface area of given pipe = 4.4 m² Answer

Surface Areas And Volumes Class nine Math NCERT Exercise 13.2Question(9) Find

(i) The lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5 m height.

(ii) How much steel was actually used, if `1/12` of the steel actually used was wasted in making the tank.

Solution:

Given,

Diameter of petrol storage tank = 4.2 m

Therefore, Radius of petrol storage tank (r) = 4.2/2 m

⇒ Radius (r) = 2.1 m

And, Height of petrol storage tank = 4.5 m

Steel wasted in making of tank = 1/2 of steel actually used.

Thus, curved surface area of closed given cylindrical tank = ?

And, Steel actually used in making of tank =?

(i) The lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5 m height.

We know that, curved surface area of a cylinder = 2 π r h

Therefore, curved surface area of given cylindrical petrol storage tank

= 2 × `22/7` × 2.1 m × 4.5 m

= `44/7` × 9.45 m²

= 59.4 m²

Thus, curved surface area of given cylindrical petrol tank storage = 59.4 m² Answer

(ii) How much steel was actually used, if `1/12` of the steel actually used was wasted in making the tank.

Calculation of Total Surface Area of the given Cylindrical Petrol Tank

Steel used in making of tank will be equal to the total surface area of the tank.

We know that, Total Surface Area of a Closed Cylinder = Curved Surface Area + 2 × Area of base

Here, curved surface area = 59.4 m²

[As calculated in the section (i) of this question]

Here, radius of the tank = 2.1 m

Thus, total surface area of the given cylindrical petrol tank

= 59.4 + (2 × π r²)

= 59.4 + (2 × `22/7` × 2.1 × 2.1)

= 59.4 + (2 × 22 × 0.3 × 2.1)

= 59.4 + 27.72 = 87.12 m²

Thus, total surface area of the given cylindrical petrol tank = 87.12 m²

Calculation of Steel used in making of given cylindrical petrol tank

Let steel used in making of given petrol tank `=x`

And , as given in question, steel wasted = 1/12 of total steel used

⇒ Steel wasted `=(1x)/12`

Therefore, actually steel used `= 11/12\xx\x`

Now, area of steel used `= 11/12\xm^2`

`=> 11/12\x` = Total surface area of cylinder

`=>11/12\x = 87.126`

`=>x = 87.12 xx12/11`

⇒ x = 95.04 m²

Thus, steel actually used in making of given petrol tank = 95.04 m² Answer

Surface Areas And Volumes Class nine Math NCERT Exercise 13.2Question(10) In figure you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade.

Solution

Given, Base of diameter of the lampshade = 20 cm

Therefore, radius = 20/2= 10 cm

And, height of the lampshade = 30 cm

And, margin to cover with a decorative cloth for folding it over the top and bottom = 2.5 cm

Therefore, effective height of the lampshade to cover with cloth = height + margin

= 30 cm + 2.5 cm + 2.5 cm

Thus, effective height of lampshade = 35 cm

Thus, cloth required to cover the lampshade = ?

Now, Area of sheet required for covering the lamp shade = Curved surface area of the lamp shade with effective height

= 2 π r h

= 2 × `22/7` × 10 cm × 35 cm

= 2 × 22 × 10 cm × 5 cm

= 440 cm × 5 cm

= 2200 cm ²

Therefore area of sheet required for covering the lamp shade = 2200 cm² Answer

Surface Areas And Volumes Class nine Math NCERT Exercise 13.2Question(11) The students of a vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard to be bought for the competition?

Solution

Given, Radius of pen holder = 3 cm

And, Height of pen holder = 10.5 cm

Number of competitors in the vidyalaya = 35

Therefore, area of cardboard to be bought for making of penholders = ?

Here, the area of cardboard required for making one pen holder = Curved surface area of a box

= curved surface area of pen holder + Area of base of pen holder

Thus, total area of cardboard to make 35 penholders

= 35(curved surface area of one pen holder + Area of base of one pen holder)

= 35(2 π r h + π r²)

= 35 × π r (2 h + r)

`=35xx22/7xx3(2xx10.5+3)`

= 5 × 22 × 3 (21 + 3)

= 110 × 3 × 24

= 110 × 72

= 7920 cm²

Thus, cardboard is to be purchased for all the students = 7920 cm² Answer

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