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Exercise 13.3 - Chapter 13 Surface Areas & Volumes class 9 ncert solutions Maths - SaraNextGen [2024]


Question 1:

Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area. https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/8773/Chapter%2013_html_4110777b.gif

Answer:

Radius (r) of the base of cone =https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/8773/Chapter%2013_html_m46bb855e.gif= 5.25 cm

Slant height (l) of cone = 10 cm

CSA of cone = πrl

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/8773/Chapter%2013_html_2f03dd1d.gif

Therefore, the curved surface area of the cone is 165 cm2.

 

Question 2:

Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m. https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/8774/Chapter%2013_html_4110777b.gif

Answer:

Radius (r) of the base of cone =https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/8774/Chapter%2013_html_15f3dd9.gif= 12 m

Slant height (l) of cone = 21 m

Total surface area of cone = πr(r + l)

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/8774/Chapter%2013_html_3bc5a671.gif

 

Question 3:

Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find

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

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/8775/Chapter%2013_html_4110777b.gif

Answer:

(i) Slant height (l) of cone = 14 cm

Let the radius of the circular end of the cone be r.

We know, CSA of cone = πrl

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/8775/Chapter%2013_html_m43e8e8c6.gif

Therefore, the radius of the circular end of the cone is 7 cm.

(ii) Total surface area of cone = CSA of cone + Area of base

= πrl + πr2

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/8775/Chapter%2013_html_26a09e10.gif

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

 

Question 4:

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

(i) slant height of the tent

(ii) cost of the canvas required to make the tent, if the cost of 1 m2 canvas is Rs 70.

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/2883/Chapter%2013_html_4110777b.gif

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/2883/Chapter%2013_html_m424ded9.jpg

(i) Let ABC be a conical tent.

Height (h) of conical tent = 10 m

Radius (r) of conical tent = 24 m

Let the slant height of the tent be l.

In ΔABO,

AB2 = AO2 + BO2

l2 = h2 + r2

= (10 m)2 + (24 m)2

= 676 m2

∴ l = 26 m

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

(ii) CSA of tent = πrl

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/2883/Chapter%2013_html_5229ec23.gif

Cost of 1 m2 canvas = Rs 70

Cost of https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/2883/Chapter%2013_html_m17d67e8e.gif canvas =https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/2883/Chapter%2013_html_m5c237ad0.gif

= Rs 137280

Therefore, the cost of the canvas required to make such a tent is

Rs 137280.

 

Question 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]

Answer:

Height (h) of conical tent = 8 m

Radius (r) of base of tent = 6 m

Slant height (l) of tent =https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/2886/Chapter%2013_html_7a18fc84.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/2886/Chapter%2013_html_423ef8c9.gif

CSA of conical tent = πrl

= (3.14 × 6 × 10) m2

= 188.4 m2

Let the length of tarpaulin sheet required be l.

As 20 cm will be wasted, therefore, the effective length will be (l − 0.2 m).

Breadth of tarpaulin = 3 m

Area of sheet = CSA of tent

[(l − 0.2 m) × 3] m = 188.4 m2

l − 0.2 m = 62.8 m

l = 63 m

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

 

Question 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 m2https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/2891/Chapter%2013_html_4110777b.gif

Answer:

Slant height (l) of conical tomb = 25 m

Base radius (r) of tomb = https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/2891/Chapter%2013_html_m487cd9da.gif7 m

CSA of conical tomb = πrl

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/2891/Chapter%2013_html_5a388553.gif

= 550 m2

Cost of white-washing 100 m2 area = Rs 210

Cost of white-washing 550 m2 area =https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/2891/Chapter%2013_html_m441be263.gif

= Rs 1155

Therefore, it will cost Rs 1155 while white-washing such a conical tomb.

 

Question 7:

A joker’s cap is in the form of right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps. https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/2895/Chapter%2013_html_4110777b.gif

Answer:

Radius (r) of conical cap = 7 cm

Height (h) of conical cap = 24 cm

Slant height (l) of conical cap =https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/2895/Chapter%2013_html_7a18fc84.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/2895/Chapter%2013_html_mc1cde60.gif

CSA of 1 conical cap = πrl

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/2895/Chapter%2013_html_78b37562.gif

CSA of 10 such conical caps = (10 × 550) cm2 = 5500 cm2

Therefore, 5500 cm2 sheet will be required.

 

Question 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 takehttps://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/2897/Chapter%2013_html_m5d98932b.gif= 1.02).

Answer:

Radius (r) of cone =https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/2897/Chapter%2013_html_m325483e9.gif = 0.2 m

Height (h) of cone = 1 m

Slant height (l) of cone =https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/2897/Chapter%2013_html_e83d6ff.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/2897/Chapter%2013_html_2369389f.gif

CSA of each cone = πrl

= (3.14 × 0.2 × 1.02) m2 = 0.64056 m2

CSA of 50 such cones = (50 × 0.64056) m2

= 32.028 m2

Cost of painting 1 m2 area = Rs 12

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

= Rs 384.336

= Rs 384.34 (approximately)

Therefore, it will cost Rs 384.34 in painting 50 such hollow cones.

Also Read : Exercise-13.4-Chapter-13-Surface-Areas-&-Volumes-class-9-ncert-solutions-Maths

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