NCERT Solutions for Class 9 Maths Chapter 11 Exercise 11.1 - Surface Area and Volumes

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

Answer:

Given,

Base diameter of the cone = d=10.5 cm

Slant height = l=10 cm

We know, Curved surface area of a cone =πrl

∴ Required curved surface area of the cone=

=227×10.52×10=165 cm2

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

Answer:

Given,

Base diameter of the cone = d=24 m

Slant height = l=21 cm

We know, Total surface area of a cone = Curved surface area + Base area

=πrl+πr2=πr(l+r)

∴ Required total surface area of the cone=

=227×242×(21+12)=227×242×33=1244.57 m2

Q3 (i) Curved surface area of a cone is 308cm2 and its slant height is 14 cm. Find radius of the base .

Answer:

Given,

The curved surface area of a cone = 308cm2

Slant height =l=14 cm

(i) Let the radius of cone be r cm

We know, the curved surface area of a cone= πrl

∴ πrl=308⇒227×r×14=308⇒r=30844=7

Therefore, the radius of the cone is 7 cm

Q3 (ii) Curved surface area of a cone is 308cm2 and its slant height is 14cm . Find total surface area of the cone.

Answer:

Given,

The curved surface area of a cone = 308cm2

Slant height =l=14 cm

The radius of the cone is r= 7 cm

(ii) We know, Total surface area of a cone = Curved surface area + Base area

=πrl+πr2

=308+227×72=308+154=462 cm2

Therefore, the total surface area of the cone is 462 cm2

Q4 (i) A conical tent is 10 m high and the radius of its base is 24 m. Find slant height of the tent.

Answer:

Given,

Base radius of the conical tent = r=24 m

Height of the conical tent = h=10 m

∴ Slant height = l=h2+r2

=102+242=676=26 m

Therefore, the slant height of the conical tent is 26 m

Q4 (ii) A conical tent is 10 m high and the radius of its base is 24 m. Find cost of the canvas required to make the tent, if the cost of 1m2 canvas is Rs 70.

Answer:

Given,

Base radius of the conical tent = r=24 m

Height of the conical tent = h=10 m

∴ Slant height = l=h2+r2=26 m

We know, Curved surface area of a cone =πrl

∴ Curved surface area of the tent

=227×24×26=137287 m2

Cost of 1 m2 of canvas = Rs. 70

∴ Cost of 137287 m2 of canvas =

Rs. (137287×70)=Rs. 137280

Therefore, required cost of canvas to make tent is Rs. 137280

Q5 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 ).

Answer:

Given,

Base radius of the conical tent = r=6 m

Height of the tent = h=8 m

We know,

Curved surface area of a cone = πrl=πrh2+r2

∴ Area of tarpaulin required = Curved surface area of the tent

=3.14×6×82+62=3.14×6×10=188.4 m2

Now, let the length of the tarpaulin sheet be x m

Since 20 cm is wasted, effective length = x−20cm=(x−0.2) m

Breadth of tarpaulin = 3 m

∴[(x−0.2)×3]=188.4⇒x−0.2=62.8⇒x=63 m

Therefore, the length of the required tarpaulin sheet will be 63 m.

Q6 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 100m2 .

Answer:

Given, a conical tomb

The base diameter of the cone = d=14 m

Slant height =l=25 m

We know, Curved surface area of a cone =πrl

=227×142×25=22×25=550 m2

Now, Cost of whitewashing per 100m2 = Rs. 210

∴ Cost of whitewashing per 550m2 = Rs.(210100×550)

=Rs. (21×55)=Rs. 1155

Therefore, the cost of white-washing its curved surface of the tomb is Rs. 1155 .

Q7 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.

Answer:

Given, a right circular cone cap (which means no base)

Base radius of the cone = r=7 cm

Height =h=24 cm

∴l=h2+r2

We know, Curved surface area of a right circular cone =πrl

∴ The curved surface area of a cap =

=227×7×242+72=22×625=22×25 =550 cm2

∴ The curved surface area of 10 caps = 550×10=5500 cm2

Therefore, the area of the sheet required for 10 caps = 5500 cm2

Q8 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 )

Answer:

Given, hollow cone.

The base diameter of the cone = d=40 cm=0.4 m

Height of the cone = h=1 m

∴ Slant height = l=h2+r2 =12+0.22

We know, Curved surface area of a cone = πrl=πrh2+r2

∴ The curved surface area of 1 cone = 3.14×0.2×1.04=3.14×0.2×1.02

=0.64056 m2

∴ The curved surface area of 50 cones =(50×0.64056) m2

=32.028 m2

Now, the cost of painting 1 m2 area = Rs. 12

∴ Cost of the painting 32.028 m2 area =Rs. (32.028×12)

=Rs. 384.336

Therefore, the cost of painting 50 such hollow cones is Rs. 384.34 (approx)