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51. Radius of the base of a right circular cone is 3cm and the height of the cone is 4cm.Find the curved surface area of the cone.
  A.  47
5 / 7
sq cm.
  B.  47
1 / 7
sq.cm
  C.  47
3 / 5
sq. cm
  D.  49
5 / 4
sq.cm
     
   
View Answer

Shortcut:
To find the curved surface area of the cone,
(i) if its slant height and radius of its base are given.
Curved surface area of the cone = πrl sq units.

(ii) if its height and radius of its base are given.
Curved surface area of the cone = πr[√(r2 + h2)] sq units.
Here, r = 3, h = 4
Using these values in the shortcut, we get:
Curved surface area of the cone =
22 / 7
x 3[√(32 + 42)]
=
22 / 7
x 3[√(9 + 16)]
=
22 / 7
x 3[√(25)]
=
22 / 7
x 3 x 5
=
22 x 3 x 5 / 7
=
330 / 7
= 47
1 / 7

Hence, the curved surface area of the cone is 47
1 / 7
sq cm.

52. The area of the base of a right circular cone is 154cm2 and its height is 14cm.The curved surface of cone is:
  A.  153√2cm2
  B.  154√6cm2
  C.  154√5cm2
  D.  153√3cm2
     
   
View Answer

Shortcut:
To find the curved surface area of the cone,
(i) if its slant height and radius of its base are given.
Curved surface area of the cone = πrl sq units.

(ii) if its height and radius of its base are given.
Curved surface area of the cone = πr[√(r2 + h2)] sq units.
Here, Area = 154, h = 14
πr2 = 154 r2 =
154 / 22
x 7
r = 7
slant height = √(h2 + r2) = √(142 + 72) = 7√5 cm
Curved surface =
22 / 7
x 7 x 7√5 = 154√5
Hence, the curved surface area of the cone is 154√5 sq cm.

53. The slant hegiht of a conical tomb is 17
1 / 2
metres. If its diameter be 28 metres,find the cost of construction in at Rs135 per cubic metre and also find the cost of whitewashing its slant surface at Rs 3.30 per square metre.
  A.  Rs 261020 ; Rs 2450
  B.  Rs 26260 ; Rs 2731
  C.  Rs 26150 ; Rs 2521
  D.  Rs 291060 ; Rs 2541
     
   
View Answer

Shortcut:
To find the curved surface area of the cone,
(i) if its slant height and radius of its base are given.
Curved surface area of the cone = πrl sq units.

(ii) if its height and radius of its base are given.
Curved surface area of the cone = πr[√(r2 + h2)] sq units.
Here, Height of the cone = √[(
35 / 2
)2 − 142]
=
21 / 2
m
Volume of the cone =
1 / 3
x
22 / 7
x 14 x 14 x
21 / 2
= 2156 cubic m.
∴ Cost of constructing the cone = 2156 x 135 = Rs 291060
Curved Surface area of the cone = πrl
=
22 / 7
x 14 x
35 / 2
= 77 cubic m.
∴ Cost of white washing = 770 x 3.30 = Rs 2541.

54. Radius of he base of a right circular cone is 3cm and the height of the cone is 4cm. Find the total surface area of the cone.
  A.  75
3 / 7
sq cm
  B.  25
3 / 7
sq cm
  C.  75
6 / 7
sq cm
  D.  85
3 / 7
sq cm
     
   
View Answer

Shortcut:
To find the total surface area of the right circular cone,
(i) if its slant height and radius of its base are given.
Total surface area of the cone = πrl + πr2 = πr(l + r) sq units.

(ii) if its height and radius of its base are given.
Total surface area of the cone = πr[√(h2 + r2)] + r sq units.
Here, h = 4, r = 3
Total surface area of the cone = πr[√(h2 + r2)] + r sq units.
=
22 / 7
x 3 [√(42 + 32)] + 3
=
22 / 7
x 3 [√(16 + 9)] + 3
=
22 / 7
x 3[√(25) + 3]
=
22 / 7
x 3 x (5 + 3)
=
22 / 7
x 3 x 8
=
528 / 7
= 75
3 / 7

Hence, the total surface area of the cone is 75
3 / 7
sq cm.

55. A frustum of a right ciruclar cone has a diameter of base 10 cm of top 6 cm and a height of 5 cm. Find
(i)slant height
(ii) curved surface area
(iii) total surface area
(iv) volume of the frustum
  A.  (i) 6.385 cm; (ii) 135.4 sq cm; (iii)142.25 sq cm; (iv) 156.67 cu cm
  B.  (i) 5.385 cm; (ii) 235.4 sq cm; (iii)282.25 sq cm; (iv) 256.67 cu cm
  C.  (i) 5.385 cm; (ii) 135.4 sq cm; (iii)242.25 sq cm; (iv) 256.67 cu cm
  D.  (i) 6.385 cm; (ii) 235.4 sq cm; (iii)342.25 sq cm; (iv) 256.67 cu cm
     
   
View Answer

Shortcut:
Frustum: If a cone is cut by a plane parallel to the base so as to divide the cone into two parts as shown in the figure, lower part is called the frustum of the cone.


Let the radius of the base of the frustum = R,
the radius of the top of the frustum = r
slant height of the frustum = l
(i) Slant height (l) = √[h2 + (R − r)2] units.
(ii) Curved Surface Area = π(R + r)l sq units.
(iii) Total Surface Area = π[(R + r)l + r2 + R2] sq units.
(iv) Volume of the frustum =
πh / 3
(r2 + R2 +rR) cubic units.
Here, r =
6 / 2
= 3, R =
10 / 2
= 5, h = 5
(i) Slant height (l) = √[52 + (5 − 3)2]
= √[25 + (2)2] = √[25 + 4] = √(29) = 5.385 sq cm
(ii) Curved Surface Area = π(5 + 3)x 5.385 sq
=
22 / 7
x 8 x 5.385 = 135.4 sq cm
(iii) Total Surface Area =
22 / 7
[(5 + 3) x 5.385 + 32 + 52]
=
22 / 7
[8 x 5.385 + 9 + 25]
=
22 / 7
[43.08 + 34] = 242.25 sq cm
(iv) Volume of the frustum =
22/7 x 5 / 3
(32 + 52 + 3 x 5)
=
22 x 5 / 7 x 3
(9 + 25 + 15)
=
22 x 5 / 7 x 3
x 49 = 256.67 cubic cm.

56. If the radii of the ends of a bucket 45 cm high are 28 cm and 7 cm, determine its capacity and the surface area.
  A.  49510 cu cm, 5410 sq cm
  B.  48510 cu cm, 5610 sq cm
  C.  48510 cu cm, 5510 sq cm
  D.  46510 cu cm, 5610 sq cm
     
   
View Answer

Shortcut:
Frustum: If a cone is cut by a plane parallel to the base so as to divide the cone into two parts as shown in the figure, lower part is called the frustum of the cone.



Let the radius of the base of the frustum = R,
the radius of the top of the frustum = r
slant height of the frustum = l
(i) Slant height (l) = √[h2 + (R − r)2] units.
(ii) Curved Surface Area = π(R + r)l sq units.
(iii) Total Surface Area = π[(R + r)l + r2 + R2] sq units.
(iv) Volume of the frustum =
πh / 3
(r2 + R2 +rR) cubic units.
Here, r = 7, h = 45, R = 28
Slant height (l) = √[452 + (28 − 7)2]
= √[2025 + 21 x 21]
= √[2025 + 441]
= √(2466) = 49.6
Hence, the slant height is 49.6 cm. Now, Volume of the frustum =
22/7 x 45 / 3
(282 + 72 + 28 x 7)
=
22 x 15 / 7
(784 + 49 + 196)
=
22 x 15 / 7
x 1029 = 22 x 15 x 147 = 48510
Hence, the volume of bucket is 48510 cubic cm.
Surface area of the bucket = π(R + r)l + πr2] = π(28 + 7)49.6 + π72] =
22 / 7
[(35)49.6 +
22 / 7
x 7 x 7
=
22 / 7
(1736) + 22 x 7 = 22 x 248 + 154 = 5610
Hence, the surface area of the bucket is 5610 sq cm.

57. A reservior is in the shape of a frustum of a right circular cone. It is 8 m across at the top and 4 m across the bottom. It is 6 m deep. Its capacity is:
  A.  186 m
  B.  376 m
  C.  276 m
  D.  176 m
     
   
View Answer

Shortcut:
Frustum: If a cone is cut by a plane parallel to the base so as to divide the cone into two parts as shown in the figure, lower part is called the frustum of the cone.



Let the radius of the base of the frustum = R,
the radius of the top of the frustum = r
slant height of the frustum = l
(i) Slant height (l) = √[h2 + (R − r)2] units.
(ii) Curved Surface Area = π(R + r)l sq units.
(iii) Total Surface Area = π[(R + r)l + r2 + R2] sq units.
(iv) Volume of the frustum =
πh / 3
(r2 + R2 +rR) cubic units.
Here, r = 2, h = 6, R = 4
Volume of the frustum =
22/7 x 6 / 3
(42 + 22 + 4 x 2)
=
22 x 2 / 7
(16 + 4 + 8)
=
22 x 2 / 7
x 28 = 22 x 2 x 4 = 176
Hence, the volume of frustum is 176 cubic cm.

58. A brick measures 20 cm by 10 cm by7
1 / 2
cm . How many bricks will be required for a all 25 m long, 2 m high and 3/4 m thick?
  A.  15000
  B.  35000
  C.  25000
  D.  30000
     
   
View Answer

Shortcut:
To find number of bricks when the dimensions of brick and wall are given.
Required no. of bricks =
Volume of wall / Volume of 1 brick

Here for wall, l = 25, b = 2, h = 3/4
Volume of wall = 25 x 2 x
3 / 4
=
25 x 3 / 2
Here for brick, l = 20/100, b = 10/100, h = 15/200
Volume of brick =
20 / 100
x
10 / 100
x
15 / 200

=
20 x 10 x 15 / 100 x 100 x 200

=
15 / 10000

No. of brick =
75/2 / 15/10000
=
75 x 10000 / 15 x 2
= 2.5 x 10000 = 25000
Hence, the required number of bricks is 25000.

59. A closed wooden box measures externally cm long, 7 cm broad, 6 cm high. if the thickness of the wood is half a cm, find
(i) the capacity of the box
(ii) the weight supposing that one cubic cm. of wood weights 0.9gm.
  A.   (i)240 cm3,(ii) 124.2 gm
  B.   (i)140 cm3,(ii) 124.2 gm
  C.   (i)240 cm3,(ii) 114.2 gm
  D.   (i)260 cm3,(ii) 144.2 gm
     
   
View Answer

Shortcut:
To find capacity, volume of material and weight of material of a closed box, when external dimensions (l x b x h) and thickness of material of which box is made, are given.
(i) Capacity of box = (External length − 2 x thickness) x (External breadth − 2 x thickness) x (External height − 2 x thickness)
(ii) Volume of material = External Volume − Capacity
(iii) Weight of wood = Volume of wood x Density of wood.

Here, l = 9, b = 7, h = 6, thickness = 1/2
(i) Required Capacity = (9 − 2 x 1/2) x (7 − 2 x 1/2) x (6 − 2 x 1/2)
= (9 − 1) x (7 − 1) x (6 − 1) = 8 x 6 x 5 = 240
Hence, the required capacity is 240 cubic cm.
Vol. of wood = external vol − capacity
= 9 x 7 x 6 − 240 = 138 cubic cm.
∴ Weight of wood = Vol. of wood x density of wood = 138 x 0.9 = 124.2 g

60. The volume of a cube is 125 cm3. The surface are of the cube is:
  A.   140 sq cm.
  B.   150 sq cm.
  C.   120 sq cm.
  D.   180 sq cm.
     
   
View Answer

Shortcut:
To find the volume of a cube if the surface area of the cube is given
Vol. of cube = √(
surface area of cube / 6
)3
Here, vol of cube = 125, Surface area = ?
Vol. of cube = √(
surface area of cube / 6
)3
125 = √(
SA / 6
)3
or, 5 = √(
SA / 6
)
Squaring both sides, we get:
25 =
SA / 6

SA = 25 x 6 = 150
Hence, surface area of the cube is 150 sq cm.

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