Surface Areas and Volumes: 9 Math


mathematics Class Nine

NCERT Exercise 13.3 part-2: 9th math

Surface Areas And Volumes Class nine Math NCERT Exercise 13.3Question (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)

Solution

9 math surface areas and volumes ncert exercise 13.3 question 5 9 math surface areas and volumes ncert exercise 13.3 question 5

Given, Height of tent = 8 m

And, Base radius of tent = 6 m

And, Width of tarpaulin to cover the given conical tent = 3 m

And, length of tarpaulin for margins and wastage = 20 cm

= 20/100 = 0.2 m

And, ℿ = 3.14

Thus, Length of tarpaulin required to cover the tent = ?

Here, the area of tarpaulin required to cover the given tent = curved surface area of conical tent

Now, we know that, Curved Surface Area of a Cone `=pi\r\l`

Thus, to find the curved surface area of given conical tent, first slant height will be calculated.

Now, since height, radius and slant height of a cone together make a right angled triangle.

Thus, [slant height(`l`)]2 = (h)2 + (r)2

Thus, [slant height of the given conical tent (`l`)]2 =

= (8)2 + (6) 2

⇒ `l^2` = 64 m2 + 36 m2

⇒ `l^2` = 100 m2

⇒ `l = sqrt(100\ m^2)`

⇒ `l`=10 m

Thus, slant height (`l`) of the tent = 10 m

Now, We know that, Curved surface area of a cone `= pi\ r\ l`

Thus, Curved Surface Area of the given conical tent

= 3.14 × 6 m × 10 m

= 3.14 × 60 m2

= 188.40 m2

Thus, Curved Surface Area of given conical tent = 188.40 m2

Now, we know that, curved surface area of conical tent = Area of rectangular piece of tarpaulin to cover the tent

[Here, Let length of the tent = `l` and width of the tent = b]

⇒ 188.4 m2 = `l` × b

⇒ 188.4 m2= `l` × 3 m

⇒ `l = (188.4\ m^2)/(3\ m)`

⇒ `l` = 62.8 m

Thus, length of tarpaulin to cover the given tent = 62.8 m

Now, Total required length of tarpaulin = Length of tarpaulin + 20 cm Extra for margins and wastage

As given in question extra length of material required = 20 cm = 0.2m

Therefore, total length of tarpaulin required to cover the given tent

= 62.8 m + 0.2 m

= 63 m

Thus, length of tarpaulin required to cover the given tent = 63 m Answer

Surface Areas And Volumes Class nine Math NCERT Exercise 13.3Question (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 100 m2.

Solution

Given, Slant height of the conical tomb = 25 m

And, Base diameter of the conical tomb = 14 m

Therefore, Radius of the conical tomb (r) = 14/2 = 7 m

Rate of white washing = Rs 210 per 100 m2

Therefore, total cost of white washing of the given conical tomb = ?

Now, we know that, Curved Surface Area of a Cone `=pi\ r\ l`

Therefore, Curved Surface Area of given conical tomb

=`22/7` × 7 m × 25 m

= 22 × 25 m2

= 550 m2

Thus, curved surface area of given tomb = 550 m2 which is to be white washed.

Calculation of Cost of White Washing of given Conical Tomb

Now, as per question, rate of white washing for 100 m2 = Rs 210

Therefore, cost white washing of 1 m2 `=210/100 = 2.1`

Therefore, cost of white washing of 550 m2 = 2.1 × 550

= Rs 1155

Thus, cost of white washing of the given conical tomb = Rs 1155.00 Answer

Surface Areas And Volumes Class nine Math NCERT Exercise 13.3Question (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 a caps.

Solution

Given, Radius of jockers's conical cap = 7 cm

Height of jocker's conical cap = 24 cm

Therefore, Area of sheet required to make 10 such jocker's cap = ?

Here, Area of Sheet required to make a given conical cap = Curved surface area of given conical cap

Now, we know that, Curved Surface Area of A Cone `=pi\ r\ l`

Thus, before calculating the curved surface area of a cone, its slant height will required to calculate first.

Calculation of Slant Height of the joker's conical cap

9 math surface areas and volumes ncert exercise 13.3 question 7

Here, height, base radius and slant height of a cone together make a right angled triangle,

Therefor, [slant height (`l`)]2= (r)2+ (h)2

⇒ `l^2` = (7 cm)2 + (24 cm)2

⇒ `l^2` = 49 cm2 + 576 cm2

⇒ `l^2` = 625 cm2

`=>l=sqrt(625\ cm^2)`

⇒ Slant height (`l`) = 25 cm

Now, Curved surface area of given conical cap = π r l

= `22/7` × 7 cm × 25 cm

= 22 × 25 cm2

= 550 cm2

Thus, curved surface area of the given one conical joker's cap = 550 cm2

Thus, area of sheet of required to make given one joker's cap = 550 cm2

Therefore, area of sheet required for 10 similar caps = 550 ×10

= 5500 cm2

Thus, total area of sheet required to make 10 given caps = 5500 cm2 Answer

Surface Areas And Volumes Class nine Math NCERT Exercise 13.3Question (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 `sqrt(1.04)` = 1.02)

Solution :

Given, Diameter of cone = 40 cm

Therefore, radius of cone = 40/2 cm

=20 cm = 20/100 = 0.2 m

And, Height of cone = 1 m

Cost of painting = Rs 12 per m2

ℿ = 3.14

And, `sqrt(1.04)` = 1.02

And total number of cones = 50

Therefore, cost of painting of all these 50 cones = ?

Here, Total area to be painted = curved surface area of one cone × 50

Now, we know that, Curved Surface Area of A Cone `=pi\ r\ l`

Thus, before calculating the curved surface area of a cone, its slant height will required to calculate first.

Calculation of Slant Height of the given cone

Now, [Slant height (`l`)]2 = (Height)2 + (Radius)2

⇒ `l^2` = ( 1 m )2 + ( 0.2 m )2

⇒ `l^2` = 1 m2 + 0.04 m2

⇒ `l^2` = 1.04 m2

`=>l=sqrt(1.04\ m^2)`

⇒ Slant height (`l`) = 1.02 m

Now, Curved surface are of one given cone `=pi\ r\ l`

= 3.14 × × 0.2 m × 1.02 m

= 3.14 × 0.204

= 0.64056 m2

Thus, curved surface area of one cone = 0.64056 m2

Now as per question cost of painting for 1 m2 = Rs 12

Therefore, cost of painting of area of one cone, i.e. of 0.64056 m2

= 0.64056 × 12

= Rs 7.68672

Thus, cost of painting of one given cone = Rs 7.68672

Thus, cost of painting of similar 50 cones = Rs 7.68672 × 50

= Rs 384.336 = ≃ Rs 384.34

Thus, cost of painting of given all 50 cones = Rs 384.34 Answer

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