9th Maths 13.3

Chapter 13

Surface Areas and Volumes

NCERT Class 9th solution of Exercise 13.1

NCERT Class 9th solution of Exercise 13.2

NCERT Class 9th solution of Exercise 13.4

NCERT Class 9th solution of Exercise 13.5

NCERT Class 9th solution of Exercise 13.6

NCERT Class 9th solution of Exercise 13.7

NCERT Class 9th solution of Exercise 13.8

NCERT Class 9th Projects

Exercise 13.3

Assume `pi=(22)/7`, unless stated otherwise.

Q1. Diameter of the base of a cone is `10.5cm` and its slant height is`10cm`. Find its curved surface area.

Sol. :

Given:

Radius of base is `r=(10.5)/2cm`

Slant height is `l=10cm`

To Find:

Curved Surface Area of Cone.

Solve:

Curved Surface Area of Cone`=pirl`

`=(22)/7times10.5/2times10`

`=11times1.5times10`

`=165.0cm^2`

Answer:

`text{The curved Surface Area of the Cone is 165.0cm}^2`. 

Q2. Find the total surface area of a cone, if its slant height is `21m` and diameter of its base is `24m`.

Sol. :

Given:

Radius of base is `r=(24)/2=12m`

The slant height is `l=21m`

To Find:

The total surface area of a cone.

Solve:

Total Surface Area of cone`=pir(r+l)`

`=(22)/7times12times(12+21)`

`=(22)/7times12times33`

`=(22)/7times396`

`=1244.57m^2`

Answer:

`text{The total Surface Area of the Cone is 1244.57m}^2`.

Q3. Curved surface area of a cone is `308cm^2`and its slant height is `14cm`. Find 

i) radius of the base and ii) total surface area of the cone.

Sol. :

Given:

The curved Surface Area of the Cone is `308cm^2`

Slant height is `l=14cm`

To Find:

i) radius of the base

ii) total surface area of the cone

Solve:

i) Curved Surface Area of Cone`=pirl`

`308=(22)/7times r times14`

`r=(308times7)/(22times14)`

`r=(154)/(11times2)`

`r=(77)/(11)`

`r=7`

Answer:

`text{The radius of the base is 7cm}`.

ii) Total Surface Area of the Cone`=pir(r+l)`

`=(22)/7times7(7+14)`

`=22times21`

`=462cm^2`

Answer:

`text{The total Surface Area of the Cone is 462cm}^2`.

Q4. A conical tent is `10m` high and the radius of its base is `24m`. Find

i) slant height of the tent.

ii) cost of the canvas required to make the tent if the cost of `1m^2` canvas is `₹70`.

Sol. :

Given:

The height of Cone is `h=10m`

The radius of the base is `r=24m`

Rate `₹70`per`m^2`

To Find:

i) Slant height of the tent.

ii) Cost of the canvas.

Solve:

Slant height of Cone`l=sqrt(h^2+r^2)`

`l=sqrt((10)^2+(24)^2)`

`l=sqrt(100+576)`

`l=sqrt(676)`

`l=26`

Answer:

The slant height of the tent is `26m`.

ii) Area of Cone`=pirl`

`=(22)/7times24times26`

`= 1961.14m^2`

Cost of the Canvas`=Areatimes Rate`

`=1961.14times70`

`=₹137280`

Answer:

`text{The cost of the canvas is ₹ 137280}`.

Q5. What length of tarpaulin `3m` wide will be required to make a conical tent of height `8m` and base radius `6m`? Assume that the extra length of material that will be required for stitching margins and stage in cutting is approximately `20cm` (Use `pi=3.14`).

Sol. :

Given:

The breadth of Tarpaulin `B=3m`

The height of Conical Tent `h=8m`

The radius of Base `r=6m`

Extra length `=20cm=0.2m`

To Find:

Length of Tarpaulin.

Solve:

Slant height `l=sqrt(r^2+h^2)`

`l=sqrt((6)^2+(8)^2)`

`l=sqrt(36+64)`

`l=sqrt100`

`l=10m`

Area of the Tarpaulin`=pirl`

`=3.14times6times10`

`=188.4m^2`

`text{Length of Tarpaulin} = (text{Area of Tarpaulin})/(text{Breadth of Tarpaulin})`

`=(188.4)/3`

`62.8m`

`text{Total Length of Tarpaulin =Length of Tarpaulin + Extra Lenght}`

`=62.8+0.20`

`=63m`

Answer:

`text{The Total Length of a Tarpaulin is 63m}`

Q6. The slant height and base diameter of a conical tomb are `25m` and `14m` respectively. Find the cost of white-washing its curved surface at the rate of `₹210` per `100m^2.

Sol. :

Given:

The slant height is `l=25m`

The radius of the Conical Tomb is `(14)/2=7m`

The rate of White-Washing is `₹210` per `100m^2`

To Find:

Cost of White-Washing.

Solve:

Curved Surface Area of Cone`=pirl`

`=(22)/7times25times7`

`=550m^2`

`text{Cost of White- Washing =Area}timestext{Rate}`

`=₹550times(210)/(100)`

`=₹1155`

Answer:

`text{Cost of White-Washing `₹ 1155}`.

Q7. A joker's cap is in the form of a right circular cone of base radius `7cm` and height `24cm`. Find the area of the sheet required to make `10`such caps.

Sol. :

Given:

The radius of Cone is `r=7cm`

Height of Cone is `h=24cm`

To Find:

Area of the sheet.

Solve:

Slant height `l=sqrt(r^2+h^2)`

`l=sqrt((7)^2+(24)^2)`

`l=sqrt(49+576)`

`l=sqrt(625)`

`l=25cm`

sheet required for one cap = Curved Surface Area of Cone

`=pirl`

`=(22)/7times7times25`

`=22times25`

`=550cm^2`

Sheet required for 10 such caps `=550times10`

`=5500cm^2`

Answer:

`text{Sheet required for 10 such caps 5500cm}^2`.

Q8. A bus stop is barricaded from the remaining part if the road, by using `50` hollow cones made of recycled cardboard. Each cone has a base diameter of `40cm` and height `1m`. If the outer side of each of the cones is to be painted and the cost of painting is `₹12`per`m^2`, what will be the cost of painting all these cones? (Use `pi=3.14` and takes `sqrt(1.04)=1.02`).

Sol. :

Given:

Radius of base is `r=(40)/2=20cm=0.2m`

Height of the Cone `h=1m`

The rate of paint is `₹12` per `m^2`

To Find:

Total Cost of paint.

Solve:

Slant height of Cone `l=sqrt(r^2+h^2)`

`l=sqrt((0.2)^2+(1)^2)`

`l=sqrt(0.04+1`

`l=sqrt(1.04)`

`l=1.02m`

Curved Surface Area of a Cone `=pirl`

`=(22)/7times0.2times1.02`

Curved Surface Area of `50` Cones `=50timespirl`

`=50times(22)/7times0.2times1.02`

Cost of Painting `=AreatimesRate`

`=₹50times(22)/7times0.2times1.02times12`

`=₹384.68`(approx.)

Answer:

`text{Total Cost of Paint ₹ 384.68 (approx).}`