Surface Areas and Volumes: 9 Math

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

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

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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