Volumes and Surface Area of Solids - Exercise 15C | Extra Documents & Tests for Class 9 PDF Download

 Page 1


                                   
                
             
   
 
                     
                       
 
   
 
                         
 
   
 
       
               
Question:62
Note Use p =
22
7
, unless stated otherwise.
Find the curved surface area of a cone with base radius 5.25 cm and slant height 10 cm.
Solution:
Radius of the base, r = 5.25 cm
Slant height, l = 10 cm
? Curved surface area of the cone = prl =
22
7
×5. 25 ×10 = 165 cm
2
Thus, the curved surface area of the cone is 165 cm
2
.
Question:63
Note Use p =
22
7
, unless stated otherwise.
Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
Solution:
Slant height, l = 21 m
Radius of the base, r = 
24
2
 = 12 m
? Total surface area of the cone = pr(r+l) =
22
7
×12 ×(12 +21) =
22
7
×12 ×33 =
8712
7
 cm
2
Thus, the total surface area of the cone is 
8712
7
 cm
2
.
Question:64
Note Use p =
22
7
, unless stated otherwise.
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.
Solution:
Radius of the cone, r = 7 cm
Height of the cone, h = 24 cm
? Slant height of the cone, l =
v
r
2
+h
2
=
v
7
2
+24
2
= v 49 +576 = v 625 = 25 cm
Area of the sheet required to make one cap = Curved surface area of the cone = prl =
22
7
×7 ×25 = 550 cm
2
? Area of the sheet required to make 10 such caps = 550 × 10 = 5500 cm
2
Thus, the area of the sheet required to make 10 such jocker caps is 5500 cm
2
.
Page 2


                                   
                
             
   
 
                     
                       
 
   
 
                         
 
   
 
       
               
Question:62
Note Use p =
22
7
, unless stated otherwise.
Find the curved surface area of a cone with base radius 5.25 cm and slant height 10 cm.
Solution:
Radius of the base, r = 5.25 cm
Slant height, l = 10 cm
? Curved surface area of the cone = prl =
22
7
×5. 25 ×10 = 165 cm
2
Thus, the curved surface area of the cone is 165 cm
2
.
Question:63
Note Use p =
22
7
, unless stated otherwise.
Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
Solution:
Slant height, l = 21 m
Radius of the base, r = 
24
2
 = 12 m
? Total surface area of the cone = pr(r+l) =
22
7
×12 ×(12 +21) =
22
7
×12 ×33 =
8712
7
 cm
2
Thus, the total surface area of the cone is 
8712
7
 cm
2
.
Question:64
Note Use p =
22
7
, unless stated otherwise.
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.
Solution:
Radius of the cone, r = 7 cm
Height of the cone, h = 24 cm
? Slant height of the cone, l =
v
r
2
+h
2
=
v
7
2
+24
2
= v 49 +576 = v 625 = 25 cm
Area of the sheet required to make one cap = Curved surface area of the cone = prl =
22
7
×7 ×25 = 550 cm
2
? Area of the sheet required to make 10 such caps = 550 × 10 = 5500 cm
2
Thus, the area of the sheet required to make 10 such jocker caps is 5500 cm
2
.
Question:65
Note Use p =
22
7
, unless stated otherwise.
The curved surface area of a cone is 308 cm
2
 and its slant height is 14 cm. Find the radius of the base and total surface area of the cone.
Solution:
Slant height, l = 14 cm
Let the radius of the base be r cm.
Curved surface area of the cone = 308 cm
2 
        Given
? prl = 308 ?
22
7
×r×14 = 308 ? r =
308×7
22×14
? r = 7 cm
? Total surface area of the cone = pr(r+l) =
22
7
×7 ×(7 +14) =
22
7
×7 ×21 = 462 cm
2
Thus, the radius of the base is 7 cm and the total surface area of the cone is 462 cm
2
.
Question:66
Note Use p =
22
7
, unless stated otherwise.
The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of whitewashing its curved surface at the rate of  12 per m
2
.
Solution:
Slant height of the conical tomb, l = 25 m
Radius of the conical tomb, r = 
14
2
 = 7 m
? Curved surface area of the conical tomb = prl =
22
7
×7 ×25 = 550 m
2
Rate of whitewashing =  12 per m
2
? Cost of whitewashing the conical tomb
= Curved surface area of the conical tomb × Rate of whitewashing
= 550 × 12
=  6,600
Thus, the cost of whitewashing the conical tomb is  6,600.
Question:67
Note Use p =
22
7
, unless stated otherwise.
A conical tent is 10 m high and the radius of its base is 24 m. Find the slant height of the tent. If the cost of 1 m
2
 canvas is  70, find the cost of canvas required to make the tent.
Solution:
Radius of the conical tent, r = 24 m
Height of the conical tent, h = 10 m
 
? Slant height of the conical tent, l =
v
r
2
+h
2
=
v
24
2
+10
2
= v 576 +100 = v 676 = 26 m
Curved surface area of the conical tent = prl =
22
7
×24 ×26 m
2
The cost of 1 m
2
 canvas is  70.
? Cost of canvas required to make the tent
= Curved surface area of the conical tent ×  70
=
22
7
×24 ×26 ×70
=  1,37,280
Thus, the cost of canvas required to make the tent is  1,37,280.
Question:68
A bus stop is barricated from the remaining part of the road by using 50 hollow cones made of recycled cardboard. Each one 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  25 per m
2
, what will be the cost of painting all these cones? (Use p = 3.14 and v 1. 04 = 1.02).
Solution:
Radius of each cone, r = 
40
2
 = 20 cm = 0.2 m         1m = 100cm
Height of each cone, h = 1 m
 
? Slant height of each cone, l =
v
r
2
+h
2
=
v
(0. 2)
2
+1
2
= v 0. 04 +1 = v 1. 04 = 1.02 m
Curved surface area of each cone = prl = 3. 14 ×0. 2 ×1. 02 = 0.64056 m
2
? Curved surface area of 50 cones = 0.64056 × 50 = 32.028 m
2
Cost of painting =  25 per m
2
Page 3


                                   
                
             
   
 
                     
                       
 
   
 
                         
 
   
 
       
               
Question:62
Note Use p =
22
7
, unless stated otherwise.
Find the curved surface area of a cone with base radius 5.25 cm and slant height 10 cm.
Solution:
Radius of the base, r = 5.25 cm
Slant height, l = 10 cm
? Curved surface area of the cone = prl =
22
7
×5. 25 ×10 = 165 cm
2
Thus, the curved surface area of the cone is 165 cm
2
.
Question:63
Note Use p =
22
7
, unless stated otherwise.
Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
Solution:
Slant height, l = 21 m
Radius of the base, r = 
24
2
 = 12 m
? Total surface area of the cone = pr(r+l) =
22
7
×12 ×(12 +21) =
22
7
×12 ×33 =
8712
7
 cm
2
Thus, the total surface area of the cone is 
8712
7
 cm
2
.
Question:64
Note Use p =
22
7
, unless stated otherwise.
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.
Solution:
Radius of the cone, r = 7 cm
Height of the cone, h = 24 cm
? Slant height of the cone, l =
v
r
2
+h
2
=
v
7
2
+24
2
= v 49 +576 = v 625 = 25 cm
Area of the sheet required to make one cap = Curved surface area of the cone = prl =
22
7
×7 ×25 = 550 cm
2
? Area of the sheet required to make 10 such caps = 550 × 10 = 5500 cm
2
Thus, the area of the sheet required to make 10 such jocker caps is 5500 cm
2
.
Question:65
Note Use p =
22
7
, unless stated otherwise.
The curved surface area of a cone is 308 cm
2
 and its slant height is 14 cm. Find the radius of the base and total surface area of the cone.
Solution:
Slant height, l = 14 cm
Let the radius of the base be r cm.
Curved surface area of the cone = 308 cm
2 
        Given
? prl = 308 ?
22
7
×r×14 = 308 ? r =
308×7
22×14
? r = 7 cm
? Total surface area of the cone = pr(r+l) =
22
7
×7 ×(7 +14) =
22
7
×7 ×21 = 462 cm
2
Thus, the radius of the base is 7 cm and the total surface area of the cone is 462 cm
2
.
Question:66
Note Use p =
22
7
, unless stated otherwise.
The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of whitewashing its curved surface at the rate of  12 per m
2
.
Solution:
Slant height of the conical tomb, l = 25 m
Radius of the conical tomb, r = 
14
2
 = 7 m
? Curved surface area of the conical tomb = prl =
22
7
×7 ×25 = 550 m
2
Rate of whitewashing =  12 per m
2
? Cost of whitewashing the conical tomb
= Curved surface area of the conical tomb × Rate of whitewashing
= 550 × 12
=  6,600
Thus, the cost of whitewashing the conical tomb is  6,600.
Question:67
Note Use p =
22
7
, unless stated otherwise.
A conical tent is 10 m high and the radius of its base is 24 m. Find the slant height of the tent. If the cost of 1 m
2
 canvas is  70, find the cost of canvas required to make the tent.
Solution:
Radius of the conical tent, r = 24 m
Height of the conical tent, h = 10 m
 
? Slant height of the conical tent, l =
v
r
2
+h
2
=
v
24
2
+10
2
= v 576 +100 = v 676 = 26 m
Curved surface area of the conical tent = prl =
22
7
×24 ×26 m
2
The cost of 1 m
2
 canvas is  70.
? Cost of canvas required to make the tent
= Curved surface area of the conical tent ×  70
=
22
7
×24 ×26 ×70
=  1,37,280
Thus, the cost of canvas required to make the tent is  1,37,280.
Question:68
A bus stop is barricated from the remaining part of the road by using 50 hollow cones made of recycled cardboard. Each one 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  25 per m
2
, what will be the cost of painting all these cones? (Use p = 3.14 and v 1. 04 = 1.02).
Solution:
Radius of each cone, r = 
40
2
 = 20 cm = 0.2 m         1m = 100cm
Height of each cone, h = 1 m
 
? Slant height of each cone, l =
v
r
2
+h
2
=
v
(0. 2)
2
+1
2
= v 0. 04 +1 = v 1. 04 = 1.02 m
Curved surface area of each cone = prl = 3. 14 ×0. 2 ×1. 02 = 0.64056 m
2
? Curved surface area of 50 cones = 0.64056 × 50 = 32.028 m
2
Cost of painting =  25 per m
2
? Total cost of painting all the cones
= Curved surface area of 50 cones ×  25
= 32.028 × 25
=  800.70
Thus, the cost of painting all the cones is  800.70.
Question:69
Note Use p =
22
7
, unless stated otherwise.
Find the volume, curved surface area and the total surface area of a cone having base radius 35 cm and height is 12 cm.
Solution:
Radius of the cone, r = 35 cm
Height of the cone, h = 12 cm
 
? Slant height of the cone, l =
v
r
2
+h
2
=
v
35
2
+12
2
= v 1225 +144 = v 1369 = 37 cm
i Volume of the cone = 
1
3
pr
2
h =
1
3
×
22
7
×(35)
2
×12 = 15400 cm
3
ii Curved surface area of the cone = prl =
22
7
×35 ×37 = 4070 cm
2
iii Total surface area of the cone = pr(r+l) =
22
7
×35 ×(35 +37) =
22
7
×35 ×72 = 7920 cm
2
Question:70
Find the volume, curved surface area and the total surface area of a cone whose height and slant height are 6 cm and 10 cm respectively.
Solution:
Height of the cone, h = 6 cm
Slant height of the cone, l = 10 cm
Radius, r = 
v
l
2
-h
2
= v 100 -36 = v 64 = 8 cm
Volume of the cone = pr
2
h 
                              =
1
3
×3. 14 ×8
2
×6 = 401. 92 cm
3
Curved surface area of the cone = prl 
                                                = 3. 14 ×8 ×10 = 251. 2 cm
2
? Total surface area = prl + pr
2
 
                               = 251. 2 + 3. 14 ×8
2
= 452. 16 cm
2
Question:71
Note Use p =
22
7
, unless stated otherwise.
A conical pit of diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?
HINT 1 m
3
 = 1 kilolitre.
Solution:
Radius of the conical pit, r = 
3.5
2
 m 
Depth of the conical pit, h = 12 m
 
? Capacity of the conical pit = 
1
3
pr
2
h =
1
3
×
22
7
×
3.5
2
2
×12 = 38.5 m
3
 = 38.5 kL         (1 m
3
 = 1 kilolitre)
Thus, the capacity of the conical pit is 38.5 kL.
Question:72
A heap of wheat is in the form of a cone of diameter 9 m and height 3.5 m. Find its volume. How much canvas cloth is required to just cover the heap? Use p = 3.14.
Solution:
Radius of the heap, r = 
9
2
 m = 4.5 m
Height of the heap, h = 3.5 m
 
?? Volume of the heap of wheat = 
1
3
pr
2
h =
1
3
×3. 14 ×(4. 5)
2
×3. 5 = 74.1825 m
3
Now, 
Slant height of the heap, l =
v
r
2
+h
2
=
v
(4. 5)
2
+(3. 5)
2
= v 20. 25 +12. 25 = v 32. 5 ˜ 5.7 m
? Area of the canvas cloth required to just cover the heap of wheat
= Curved surface area of the heap of wheat
= prl ˜ 3. 14 ×4. 5 ×5. 7 ˜ 80. 541 m
2
Thus, the area of canvas cloth required to just cover the heap is approximately 80.541 m
2 
.
Question:73
( )
Page 4


                                   
                
             
   
 
                     
                       
 
   
 
                         
 
   
 
       
               
Question:62
Note Use p =
22
7
, unless stated otherwise.
Find the curved surface area of a cone with base radius 5.25 cm and slant height 10 cm.
Solution:
Radius of the base, r = 5.25 cm
Slant height, l = 10 cm
? Curved surface area of the cone = prl =
22
7
×5. 25 ×10 = 165 cm
2
Thus, the curved surface area of the cone is 165 cm
2
.
Question:63
Note Use p =
22
7
, unless stated otherwise.
Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
Solution:
Slant height, l = 21 m
Radius of the base, r = 
24
2
 = 12 m
? Total surface area of the cone = pr(r+l) =
22
7
×12 ×(12 +21) =
22
7
×12 ×33 =
8712
7
 cm
2
Thus, the total surface area of the cone is 
8712
7
 cm
2
.
Question:64
Note Use p =
22
7
, unless stated otherwise.
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.
Solution:
Radius of the cone, r = 7 cm
Height of the cone, h = 24 cm
? Slant height of the cone, l =
v
r
2
+h
2
=
v
7
2
+24
2
= v 49 +576 = v 625 = 25 cm
Area of the sheet required to make one cap = Curved surface area of the cone = prl =
22
7
×7 ×25 = 550 cm
2
? Area of the sheet required to make 10 such caps = 550 × 10 = 5500 cm
2
Thus, the area of the sheet required to make 10 such jocker caps is 5500 cm
2
.
Question:65
Note Use p =
22
7
, unless stated otherwise.
The curved surface area of a cone is 308 cm
2
 and its slant height is 14 cm. Find the radius of the base and total surface area of the cone.
Solution:
Slant height, l = 14 cm
Let the radius of the base be r cm.
Curved surface area of the cone = 308 cm
2 
        Given
? prl = 308 ?
22
7
×r×14 = 308 ? r =
308×7
22×14
? r = 7 cm
? Total surface area of the cone = pr(r+l) =
22
7
×7 ×(7 +14) =
22
7
×7 ×21 = 462 cm
2
Thus, the radius of the base is 7 cm and the total surface area of the cone is 462 cm
2
.
Question:66
Note Use p =
22
7
, unless stated otherwise.
The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of whitewashing its curved surface at the rate of  12 per m
2
.
Solution:
Slant height of the conical tomb, l = 25 m
Radius of the conical tomb, r = 
14
2
 = 7 m
? Curved surface area of the conical tomb = prl =
22
7
×7 ×25 = 550 m
2
Rate of whitewashing =  12 per m
2
? Cost of whitewashing the conical tomb
= Curved surface area of the conical tomb × Rate of whitewashing
= 550 × 12
=  6,600
Thus, the cost of whitewashing the conical tomb is  6,600.
Question:67
Note Use p =
22
7
, unless stated otherwise.
A conical tent is 10 m high and the radius of its base is 24 m. Find the slant height of the tent. If the cost of 1 m
2
 canvas is  70, find the cost of canvas required to make the tent.
Solution:
Radius of the conical tent, r = 24 m
Height of the conical tent, h = 10 m
 
? Slant height of the conical tent, l =
v
r
2
+h
2
=
v
24
2
+10
2
= v 576 +100 = v 676 = 26 m
Curved surface area of the conical tent = prl =
22
7
×24 ×26 m
2
The cost of 1 m
2
 canvas is  70.
? Cost of canvas required to make the tent
= Curved surface area of the conical tent ×  70
=
22
7
×24 ×26 ×70
=  1,37,280
Thus, the cost of canvas required to make the tent is  1,37,280.
Question:68
A bus stop is barricated from the remaining part of the road by using 50 hollow cones made of recycled cardboard. Each one 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  25 per m
2
, what will be the cost of painting all these cones? (Use p = 3.14 and v 1. 04 = 1.02).
Solution:
Radius of each cone, r = 
40
2
 = 20 cm = 0.2 m         1m = 100cm
Height of each cone, h = 1 m
 
? Slant height of each cone, l =
v
r
2
+h
2
=
v
(0. 2)
2
+1
2
= v 0. 04 +1 = v 1. 04 = 1.02 m
Curved surface area of each cone = prl = 3. 14 ×0. 2 ×1. 02 = 0.64056 m
2
? Curved surface area of 50 cones = 0.64056 × 50 = 32.028 m
2
Cost of painting =  25 per m
2
? Total cost of painting all the cones
= Curved surface area of 50 cones ×  25
= 32.028 × 25
=  800.70
Thus, the cost of painting all the cones is  800.70.
Question:69
Note Use p =
22
7
, unless stated otherwise.
Find the volume, curved surface area and the total surface area of a cone having base radius 35 cm and height is 12 cm.
Solution:
Radius of the cone, r = 35 cm
Height of the cone, h = 12 cm
 
? Slant height of the cone, l =
v
r
2
+h
2
=
v
35
2
+12
2
= v 1225 +144 = v 1369 = 37 cm
i Volume of the cone = 
1
3
pr
2
h =
1
3
×
22
7
×(35)
2
×12 = 15400 cm
3
ii Curved surface area of the cone = prl =
22
7
×35 ×37 = 4070 cm
2
iii Total surface area of the cone = pr(r+l) =
22
7
×35 ×(35 +37) =
22
7
×35 ×72 = 7920 cm
2
Question:70
Find the volume, curved surface area and the total surface area of a cone whose height and slant height are 6 cm and 10 cm respectively.
Solution:
Height of the cone, h = 6 cm
Slant height of the cone, l = 10 cm
Radius, r = 
v
l
2
-h
2
= v 100 -36 = v 64 = 8 cm
Volume of the cone = pr
2
h 
                              =
1
3
×3. 14 ×8
2
×6 = 401. 92 cm
3
Curved surface area of the cone = prl 
                                                = 3. 14 ×8 ×10 = 251. 2 cm
2
? Total surface area = prl + pr
2
 
                               = 251. 2 + 3. 14 ×8
2
= 452. 16 cm
2
Question:71
Note Use p =
22
7
, unless stated otherwise.
A conical pit of diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?
HINT 1 m
3
 = 1 kilolitre.
Solution:
Radius of the conical pit, r = 
3.5
2
 m 
Depth of the conical pit, h = 12 m
 
? Capacity of the conical pit = 
1
3
pr
2
h =
1
3
×
22
7
×
3.5
2
2
×12 = 38.5 m
3
 = 38.5 kL         (1 m
3
 = 1 kilolitre)
Thus, the capacity of the conical pit is 38.5 kL.
Question:72
A heap of wheat is in the form of a cone of diameter 9 m and height 3.5 m. Find its volume. How much canvas cloth is required to just cover the heap? Use p = 3.14.
Solution:
Radius of the heap, r = 
9
2
 m = 4.5 m
Height of the heap, h = 3.5 m
 
?? Volume of the heap of wheat = 
1
3
pr
2
h =
1
3
×3. 14 ×(4. 5)
2
×3. 5 = 74.1825 m
3
Now, 
Slant height of the heap, l =
v
r
2
+h
2
=
v
(4. 5)
2
+(3. 5)
2
= v 20. 25 +12. 25 = v 32. 5 ˜ 5.7 m
? Area of the canvas cloth required to just cover the heap of wheat
= Curved surface area of the heap of wheat
= prl ˜ 3. 14 ×4. 5 ×5. 7 ˜ 80. 541 m
2
Thus, the area of canvas cloth required to just cover the heap is approximately 80.541 m
2 
.
Question:73
( )
Note Use p =
22
7
, unless stated otherwise.
A man uses a piece of canvas having an area of 551 m
2
, to make a conical tent of base radius 7 m. Assuming that all the stitching margins and wastage incurred while cutting, amount
to approximately 1 m
2
, find the volume of the tent that can be made with it.
Solution:
 
Area of the canvas = 551 m
2
Area of the canvas used in stitching margins and wastage incurred while cutting = 1 m
2
 
? Area of the canvas used in making the tent = 551 - 1 = 550 m
2
Radius of the tent, r = 7 m
Let the slant height and height of the tent be l m and h m, respectively.
Area of the canvas used in making the tent = 550 m
2
? prl = 550 ?
22
7
×7 ×l = 550 ? l =
550
22
= 25 m
Now,
Height of the tent, h =
v
l
2
-r
2
=
v
25
2
-7
2
= v 625 -49 = v 576 = 24 m
? Volume of the tent = 
1
3
pr
2
h =
1
3
×
22
7
×(7)
2
×24 = 1232 m
3
Thus, the volume of the tent that can be made with the given canvas is 1232 m
3
.
Question:74
How many metres of cloth, 2.5 m wide, will be required to make a conical tent whose base radius is 7 m and height 24 metres?
Solution:
Radius of the conical tent, r = 7 m
Height of the conical tent, h = 24 m
 Now, l =
v
r
2
+h
2
 = v 49 +576  = v 625   = 25 m
 
Curved surface area of the cone = prl                                                       =
22
7
×7 ×25                                                 = 550 m
2
Here, area of the cloth = curved surface area of the cone = 550
Width of the cloth = 2.5 m
? Length of the cloth = 
area of the cloth
width of the cloth
=
550
2.5
= 220 m
Question:75
Two cones have their heights in the ratio 1 : 3 and the radii of their bases in the ratio 3 : 1. Show that their volumes are in the ratio 3 : 1.
Solution:
Let the heights of the first and second cones be h  and 3h, respectively .
Also, let the radius of the first and second cones be 3r and r,  respectively.
? Ratio of volumes of the cones = 
volume of the first cone
volume of the second cone
                            
                                               =
1
3
p ×(3r)
2
×h
1
3
p ×r
2
×3h
=
9r
2
h
3r
2
h
= 3 : 1 
Question:76
A cylinder and a cone have equal radii of their bases and equal heights. If their curved surface areas are in the ratio 8 : 5., show that the radius and height of each has the ratio 3 : 4.
Solution:
Suppose that the respective radii and height of the cone and the cylinder are r and h.
Then ratio of curved surface areas = 8 : 5
Let the curved surfaces areas be 8x and 5x.
   
i. e. , 2 prh = 8x and prl = 5x ? pr
v
h
2
+r
2
= 5xHence 4 p
2
r
2
h
2
= 64x
2
 and p
2
r
2
h
2
+r
2
= 25x
2
? Ratio of curved surface areas =
4 p
2
r
2
h
2
p
2
r
2
h
2
+r
2
 
=
64
25
 ? 
4 p
2
r
2
h
2
p
2
r
2
h
2
+r
2
 
=
64
25
?
4h
2
h
2
+r
2
=
64
25
     ? 25h
2
 =
? The ratio of the radius and height of the cone is 3 : 4.
Question:77
A right circular cone is 3.6 cm high and the radius of its base is 1.6 cm. It is melted and recast into a right circular cone having base radius 1.2 cm. Find its height.
Solution:
Let the cone which is being melted be denoted by cone 1 and let the cone into which cone 1 is being melted be denoted by cone 2.
Height of cone 1 = 3.6 cm
Radius of the base of cone 1 = 1.6 cm
Radius of the base of cone 2 = 1.2 cm
Let h cm be the height of cone 2.
The volumes of both the cones should be equal.
 i. e. , 
1
3
p ×1. 6
2
×3. 6 =
1
3
p ×1. 2
2
×h ? h  = 
1.6×1.6×3.6
1.2×1.2
= 6. 4 cm
? Height of cone 2 = 6.4 cm
( )
( ) ( ) ( )
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Question:62
Note Use p =
22
7
, unless stated otherwise.
Find the curved surface area of a cone with base radius 5.25 cm and slant height 10 cm.
Solution:
Radius of the base, r = 5.25 cm
Slant height, l = 10 cm
? Curved surface area of the cone = prl =
22
7
×5. 25 ×10 = 165 cm
2
Thus, the curved surface area of the cone is 165 cm
2
.
Question:63
Note Use p =
22
7
, unless stated otherwise.
Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
Solution:
Slant height, l = 21 m
Radius of the base, r = 
24
2
 = 12 m
? Total surface area of the cone = pr(r+l) =
22
7
×12 ×(12 +21) =
22
7
×12 ×33 =
8712
7
 cm
2
Thus, the total surface area of the cone is 
8712
7
 cm
2
.
Question:64
Note Use p =
22
7
, unless stated otherwise.
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.
Solution:
Radius of the cone, r = 7 cm
Height of the cone, h = 24 cm
? Slant height of the cone, l =
v
r
2
+h
2
=
v
7
2
+24
2
= v 49 +576 = v 625 = 25 cm
Area of the sheet required to make one cap = Curved surface area of the cone = prl =
22
7
×7 ×25 = 550 cm
2
? Area of the sheet required to make 10 such caps = 550 × 10 = 5500 cm
2
Thus, the area of the sheet required to make 10 such jocker caps is 5500 cm
2
.
Question:65
Note Use p =
22
7
, unless stated otherwise.
The curved surface area of a cone is 308 cm
2
 and its slant height is 14 cm. Find the radius of the base and total surface area of the cone.
Solution:
Slant height, l = 14 cm
Let the radius of the base be r cm.
Curved surface area of the cone = 308 cm
2 
        Given
? prl = 308 ?
22
7
×r×14 = 308 ? r =
308×7
22×14
? r = 7 cm
? Total surface area of the cone = pr(r+l) =
22
7
×7 ×(7 +14) =
22
7
×7 ×21 = 462 cm
2
Thus, the radius of the base is 7 cm and the total surface area of the cone is 462 cm
2
.
Question:66
Note Use p =
22
7
, unless stated otherwise.
The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of whitewashing its curved surface at the rate of  12 per m
2
.
Solution:
Slant height of the conical tomb, l = 25 m
Radius of the conical tomb, r = 
14
2
 = 7 m
? Curved surface area of the conical tomb = prl =
22
7
×7 ×25 = 550 m
2
Rate of whitewashing =  12 per m
2
? Cost of whitewashing the conical tomb
= Curved surface area of the conical tomb × Rate of whitewashing
= 550 × 12
=  6,600
Thus, the cost of whitewashing the conical tomb is  6,600.
Question:67
Note Use p =
22
7
, unless stated otherwise.
A conical tent is 10 m high and the radius of its base is 24 m. Find the slant height of the tent. If the cost of 1 m
2
 canvas is  70, find the cost of canvas required to make the tent.
Solution:
Radius of the conical tent, r = 24 m
Height of the conical tent, h = 10 m
 
? Slant height of the conical tent, l =
v
r
2
+h
2
=
v
24
2
+10
2
= v 576 +100 = v 676 = 26 m
Curved surface area of the conical tent = prl =
22
7
×24 ×26 m
2
The cost of 1 m
2
 canvas is  70.
? Cost of canvas required to make the tent
= Curved surface area of the conical tent ×  70
=
22
7
×24 ×26 ×70
=  1,37,280
Thus, the cost of canvas required to make the tent is  1,37,280.
Question:68
A bus stop is barricated from the remaining part of the road by using 50 hollow cones made of recycled cardboard. Each one 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  25 per m
2
, what will be the cost of painting all these cones? (Use p = 3.14 and v 1. 04 = 1.02).
Solution:
Radius of each cone, r = 
40
2
 = 20 cm = 0.2 m         1m = 100cm
Height of each cone, h = 1 m
 
? Slant height of each cone, l =
v
r
2
+h
2
=
v
(0. 2)
2
+1
2
= v 0. 04 +1 = v 1. 04 = 1.02 m
Curved surface area of each cone = prl = 3. 14 ×0. 2 ×1. 02 = 0.64056 m
2
? Curved surface area of 50 cones = 0.64056 × 50 = 32.028 m
2
Cost of painting =  25 per m
2
? Total cost of painting all the cones
= Curved surface area of 50 cones ×  25
= 32.028 × 25
=  800.70
Thus, the cost of painting all the cones is  800.70.
Question:69
Note Use p =
22
7
, unless stated otherwise.
Find the volume, curved surface area and the total surface area of a cone having base radius 35 cm and height is 12 cm.
Solution:
Radius of the cone, r = 35 cm
Height of the cone, h = 12 cm
 
? Slant height of the cone, l =
v
r
2
+h
2
=
v
35
2
+12
2
= v 1225 +144 = v 1369 = 37 cm
i Volume of the cone = 
1
3
pr
2
h =
1
3
×
22
7
×(35)
2
×12 = 15400 cm
3
ii Curved surface area of the cone = prl =
22
7
×35 ×37 = 4070 cm
2
iii Total surface area of the cone = pr(r+l) =
22
7
×35 ×(35 +37) =
22
7
×35 ×72 = 7920 cm
2
Question:70
Find the volume, curved surface area and the total surface area of a cone whose height and slant height are 6 cm and 10 cm respectively.
Solution:
Height of the cone, h = 6 cm
Slant height of the cone, l = 10 cm
Radius, r = 
v
l
2
-h
2
= v 100 -36 = v 64 = 8 cm
Volume of the cone = pr
2
h 
                              =
1
3
×3. 14 ×8
2
×6 = 401. 92 cm
3
Curved surface area of the cone = prl 
                                                = 3. 14 ×8 ×10 = 251. 2 cm
2
? Total surface area = prl + pr
2
 
                               = 251. 2 + 3. 14 ×8
2
= 452. 16 cm
2
Question:71
Note Use p =
22
7
, unless stated otherwise.
A conical pit of diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?
HINT 1 m
3
 = 1 kilolitre.
Solution:
Radius of the conical pit, r = 
3.5
2
 m 
Depth of the conical pit, h = 12 m
 
? Capacity of the conical pit = 
1
3
pr
2
h =
1
3
×
22
7
×
3.5
2
2
×12 = 38.5 m
3
 = 38.5 kL         (1 m
3
 = 1 kilolitre)
Thus, the capacity of the conical pit is 38.5 kL.
Question:72
A heap of wheat is in the form of a cone of diameter 9 m and height 3.5 m. Find its volume. How much canvas cloth is required to just cover the heap? Use p = 3.14.
Solution:
Radius of the heap, r = 
9
2
 m = 4.5 m
Height of the heap, h = 3.5 m
 
?? Volume of the heap of wheat = 
1
3
pr
2
h =
1
3
×3. 14 ×(4. 5)
2
×3. 5 = 74.1825 m
3
Now, 
Slant height of the heap, l =
v
r
2
+h
2
=
v
(4. 5)
2
+(3. 5)
2
= v 20. 25 +12. 25 = v 32. 5 ˜ 5.7 m
? Area of the canvas cloth required to just cover the heap of wheat
= Curved surface area of the heap of wheat
= prl ˜ 3. 14 ×4. 5 ×5. 7 ˜ 80. 541 m
2
Thus, the area of canvas cloth required to just cover the heap is approximately 80.541 m
2 
.
Question:73
( )
Note Use p =
22
7
, unless stated otherwise.
A man uses a piece of canvas having an area of 551 m
2
, to make a conical tent of base radius 7 m. Assuming that all the stitching margins and wastage incurred while cutting, amount
to approximately 1 m
2
, find the volume of the tent that can be made with it.
Solution:
 
Area of the canvas = 551 m
2
Area of the canvas used in stitching margins and wastage incurred while cutting = 1 m
2
 
? Area of the canvas used in making the tent = 551 - 1 = 550 m
2
Radius of the tent, r = 7 m
Let the slant height and height of the tent be l m and h m, respectively.
Area of the canvas used in making the tent = 550 m
2
? prl = 550 ?
22
7
×7 ×l = 550 ? l =
550
22
= 25 m
Now,
Height of the tent, h =
v
l
2
-r
2
=
v
25
2
-7
2
= v 625 -49 = v 576 = 24 m
? Volume of the tent = 
1
3
pr
2
h =
1
3
×
22
7
×(7)
2
×24 = 1232 m
3
Thus, the volume of the tent that can be made with the given canvas is 1232 m
3
.
Question:74
How many metres of cloth, 2.5 m wide, will be required to make a conical tent whose base radius is 7 m and height 24 metres?
Solution:
Radius of the conical tent, r = 7 m
Height of the conical tent, h = 24 m
 Now, l =
v
r
2
+h
2
 = v 49 +576  = v 625   = 25 m
 
Curved surface area of the cone = prl                                                       =
22
7
×7 ×25                                                 = 550 m
2
Here, area of the cloth = curved surface area of the cone = 550
Width of the cloth = 2.5 m
? Length of the cloth = 
area of the cloth
width of the cloth
=
550
2.5
= 220 m
Question:75
Two cones have their heights in the ratio 1 : 3 and the radii of their bases in the ratio 3 : 1. Show that their volumes are in the ratio 3 : 1.
Solution:
Let the heights of the first and second cones be h  and 3h, respectively .
Also, let the radius of the first and second cones be 3r and r,  respectively.
? Ratio of volumes of the cones = 
volume of the first cone
volume of the second cone
                            
                                               =
1
3
p ×(3r)
2
×h
1
3
p ×r
2
×3h
=
9r
2
h
3r
2
h
= 3 : 1 
Question:76
A cylinder and a cone have equal radii of their bases and equal heights. If their curved surface areas are in the ratio 8 : 5., show that the radius and height of each has the ratio 3 : 4.
Solution:
Suppose that the respective radii and height of the cone and the cylinder are r and h.
Then ratio of curved surface areas = 8 : 5
Let the curved surfaces areas be 8x and 5x.
   
i. e. , 2 prh = 8x and prl = 5x ? pr
v
h
2
+r
2
= 5xHence 4 p
2
r
2
h
2
= 64x
2
 and p
2
r
2
h
2
+r
2
= 25x
2
? Ratio of curved surface areas =
4 p
2
r
2
h
2
p
2
r
2
h
2
+r
2
 
=
64
25
 ? 
4 p
2
r
2
h
2
p
2
r
2
h
2
+r
2
 
=
64
25
?
4h
2
h
2
+r
2
=
64
25
     ? 25h
2
 =
? The ratio of the radius and height of the cone is 3 : 4.
Question:77
A right circular cone is 3.6 cm high and the radius of its base is 1.6 cm. It is melted and recast into a right circular cone having base radius 1.2 cm. Find its height.
Solution:
Let the cone which is being melted be denoted by cone 1 and let the cone into which cone 1 is being melted be denoted by cone 2.
Height of cone 1 = 3.6 cm
Radius of the base of cone 1 = 1.6 cm
Radius of the base of cone 2 = 1.2 cm
Let h cm be the height of cone 2.
The volumes of both the cones should be equal.
 i. e. , 
1
3
p ×1. 6
2
×3. 6 =
1
3
p ×1. 2
2
×h ? h  = 
1.6×1.6×3.6
1.2×1.2
= 6. 4 cm
? Height of cone 2 = 6.4 cm
( )
( ) ( ) ( )
Question:78
A circus tent is cylindrical to a height of 3 metres and conical above it. If its diameter is 105 m and the slant height of the conical portion is 53 m, calculate the length of the canvas 5 m
wide to make the required tent.
Solution:
Radius of the tent, r = 52.5 m
Height of the cylindrical portion of the tent, H = 3 m
Slant height of the conical portion of the tent, l = 53 m
The tent is a combination of a cylindrical and a conical portion.
i.e., area of the canvas = curved surface area of the cone + curved surface area of the cylinder
                                   = prl +2 prH  
                                  =
22
7
×
105
2
(53 +2 ×3 ) =
22
7
×
105
2
×(53 +6) = 9735 m
2
But area of the canvas = length × breadth
? Length of the canvas = 
area
breadth
=
9735
5
= 1947 m
                       
Question:79
An iron pillar consists of a cylindrical portion 2.8 m high and 20 cm in diameter and a cone 42 cm high is surmounting it. Find the weight of the pillar, given that 1 cm
3
 of iron weights 7.5
g.
Solution:
Height of the cylindrical portion of the iron pillar, h = 2.8 m = 280 cm
Radius of the cylindrical portion of the iron pillar, r = 20 cm
Height of the cone which is surmounted on the cylindrical portion, H = 42 cm
Now, volume of the pillar = volume of the cylindrical portion + volume of the conical portion
                                       = pr
2
h +
1
3
pr
2
H   
                                      =
22
7
×10
2
280 +
1
3
×42 =
22
7
×(280 +14) =
22
7
×100 ×294 = 92400 cm
3
? Weight of the pillar = volume of the pillar × weight per cubic cm
                                 = 92400 ×
7.5
1000
= 693 kg
Question:80
From a solid right circular cylinder with height 10 cm and radius of the base 6 cm, a right circular cone of the same height and base is removed. Find the volume of the remaining solid.
Solution:
Height of the cylinder = 10 cm
Radius of the cylinder = 6 cm
The respective heights and radii of the cone and the cylinder are the same.
? Volume of the remaining solid = volume of the cylinder - volume of the cone
                                                 = pr
2
h -
1
3
pr
2
h  =
2
3
pr
2
h =
2
3
×3. 14 ×6
2
×10 = 753. 6 cm
3
Question:81
Water flows at the rate of 10 metres per minute through a cylindrical pipe 5 mm in diameter. How long would it take to fill a conical vessel whose diameter art the surface 40 cm and
depth 24 cm?
Solution:
Radius of the cylindrical pipe = 2.5 mm = 0.25 cm
Water flowing per minute = 10 m = 1000 cm
Volume of water flowing per minute through the cylindrical pipe = p 0. 25)
2
1000 cm
3
 = 196.4 cm
3
Radius of the the conical vessel = 40 cm
Depth of the vessel = 24 cm
Volume of the vessel = 
1
3
p 20)
2
24 = 10057. 1 cm
3
Let the time taken to fill the conical vessel be x min.
Volume of water flowing per minute through the cylindrical pipe × x = volume of the conical vessel
? x =
10057.1
196.4
= 51 min 12 sec
? The cylindrical pipe would take 51 min 12 sec to fill the conical vessel.
Question:82
Note Use p =
22
7
, unless stated otherwise.
A cloth having an area of 165 m
2
 is shaped into the form of a conical tent of radius 5 m. i How many students can sit in the tent if a student, on an average, iccupies 
5
7
 m
2
 on the ground? 
ii Find the volume of the cone.
Solution:
Radius of the conical tent, r = 5 m
Area of the base of the conical tent = pr
2
=
22
7
×5
2
=
550
7
 m
2
Average area occupied by a student on the ground = 
5
7
 m
2
? Number of students who can sit in the tent
=
Area of the base of the conical tent
Average area occupied by a student on the ground
=
550
7
5
7
= 110
Thus, the number of students who can sit in the tent is 110.
Let the slant height of the tent be l m.
Curved surface area of the tent = 165 m
2
( )
(
(