Ex 13.3 NCERT Solutions Class 9 Chapter 13
The solutions to exercise 13.3 (from class 9th's maths NCERT book) are given below. The exercise has 8 questions which have been explained thoroughly with proper reasoning.
l= 10 cm
r= 10.5/2 = 5.25 cm
CSA of the cone
= πrl
= (22/7) x 5.25 x 10
= 22 x 0. 75 x 10
= 22 x 7.5
= 11 x 15
= 165 cm2
TSA of the cone
= πr (l+ 2r)
= (22/7) x 12 x (21 + 24)
= (22/7) x 12 x 35
= 22 x 12 x 5
= 22 x 60
= 1320 cm2
CSA of cone = 308 cm2
Slant height = 14 cm
(i) CSA = πrl
308 = (22/7) x r x 14
308 = 22 x r x 2
308/22 = 2r
14 = 2r
r = 7 cm
(ii) TSA of cone
= CSA + πr2
= 308 + {(22/7) x 7 x 7}
= 308 + (22 x 7)
= 308 + 154
= 462
(i) Slant height of the cone:
l2=h2+r 2
l2=102+242
l2=100 + 576
l2=676
l= 26 m
(ii) For the tent no cloth will be used for the base, so we will find the CSA
= πrl
= (22/7) x 24 x 26 m2
Cost of 1 m2 of canvas = Rs 70
Cost of (22/7)x24 x 26 m2 of canvas
= Rs 70 x (22/7)x24 x 26
= Rs 10 x 22 x 24 x 26
= Rs 137280
The cloth is in the form of the rectangle, and its area is equal to the CSA of the conical tent.
l2=h2+r 2
l2=82+6 2
l2= 64 + 36
l2= 100
l=10
Let the length of the cloth be n
CSA = Area of the cloth
Ï€rl = (n) x 3
3.14 x 6 x 10 = 3n
188.4 = 3n
n= 62.8
n= 63 m
The slant height of the cone = 25 m
Radius = 14/2 = 7 m
CSA of the conical tomb
= πrl
= (22/7) x 7 x 25
= 22 x 25
= 550 m2
Cost of whitewashing 100 m2 CSA of the conical tomb
= Rs 210
Cost of whitewashing 550 m2 CSA of the conical tomb
= (210 x 550)/100= Rs 1155
CSA of 1 cap
= πrl
= (22/7) x 7 x 24
= 22 x 24
= 528 m2
Area of the sheet required to make 10 such conical caps
= 10 x 528
= 5280 m2
r= 20 cm = 0.20 m
h= 1 m
l2=h2+r 2
l2=12+0.20 2
l2= 1+ 0.04
l2=1.04
l=1.02
CSA of 50 hollow cones
= 50 x πrl
= 50 x (22/7) x 0.20 x 1.02
= 3.14 x 50 x 0.20 x 1.02
= 32.028 m2
Cost of painting 1 m2 CSA of the hollow cone
= Rs 12
Cost of painting 32.028 m2 CSA of the hollow cone
= Rs 12 x 32.028
= 384.336 m2 = 384.34 m2
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