CLASS-8 SUBJECT MATHS (FINALS-E) CHAPTER-11(EXERCISE11.4) MENSURATION.

EVENTS CONVENT HIGH SCHOOL

27/01/2022      CLASS- 8   SESSION 2021-22
SUBJECT : MATHS

CHAPTER-11 (EXERCISE 11.4)
MENSURATION

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 Question 1.Given a cylindrical tank, in which situation will you find the surface area and in which situation volume.

(a) To find how much it can hold.
(b) Number of cement bags required to plaster it.
(c) To find the number of smaller tanks that can be filled with water from it.
Solution:
(a) In this situation, we can find the volume.
(b) In this situation, we can find the surface area.
(c) In this situation, we can find the volume.

 Question 2.Diameter of cylinder A is 7 cm, and the height is 14 cm. Diameter of cylinder B is 14 cm and height is 7 cm. Without doing any calculations can you suggest whose volume is greater? Verify it by finding the volume of both the cylinders. Check whether the cylinder with greater volume also has greater surface area?

Solution:
Cylinder B has a greater volume.
Verification:
Volume of cylinder A = πr2h

Question 3.Find the height of a cuboid whose base area is 180 cm2 and volume is 900 cm3.

Solution:
Given: Area of base = lb = 180 cm2
V = 900 cm3
Volume of the cuboid = l × b × h
900 = 180 × h
h = 5 cm
Hence, the required height = 5 cm.

Question 4.A cuboid is of dimensions 60 cm × 54 cm × 30 cm. How many small cubes with side 6 cm can be placed in the given cuboid?

Solution:
Volume of the cuboid = l × b × h = 60 cm × 54 cm × 30 cm = 97200 cm3
Volume of the cube = (Side)3 = (6)3 = 216 cm3
Number of the cubes from the cuboid

Hence, the required number of cubes = 450.

 Question 5.Find the height of the cylinder whose volume is 1.54 m3 and the diameter of the base is 140 cm.

Solution:
V = 1.54 m3, d = 140 cm = 1.40 m
Volume of the cylinder = πr2h

Hence, the height of cylinder = 1 m.

Question 6.A milk tank is in the form of a cylinder whose radius is 1.5 m and length is 7 m. Find the quantity of milk in litres that can be stored in the tank.

Solution:
Here, r = 1.5 m
h = 7 m
.’. Volume of the milk tank = πr2h
= 227 × 1.5 × 1.5 × 7
= 22 × 2.25
= 49.50 m3
Volume of milk in litres = 49.50 × 1000 L (∵ 1 m3 = 1000 litres)
= 49500 L
Hence, the required volume = 49500 L.

 Question 7.If each edge of a cube is doubled,

(i) how many times will it be surface area increase?
(ii) how many times will its volume increase?
Solution:
Let the edge of the cube = x cm
If the edge is doubled, then the new edge = 2x cm
(i) Original surface area = 6x2 cm2
New surface area = 6(2x)2 = 6 × 4x2 = 24x2
Ratio = 6x2 : 24x2 = 1 : 4
Hence, the new surface area will be four times the original surface area.

(ii) Original volume of the cube = x3 cm3
New volume of the cube = (2x)3 = 8x3 cm3
Ratio = x3 : 8x3 = 1 : 8
Hence, the new volume will be eight times the original volume.

Question 8.Water is pouring into a cuboidal reservoir at the rate of 60 litres per minute. If the volume of the reservoir is 108 m3, find the number of hours it will take to fill the reservoir.

Solution:
Volume of the reservoir = 108 m3 = 108000 L [∵1 m3 = 1000 L]
Volume of water flowing into the reservoir in 1 minute = 60 L
Time taken to fill the reservoir

Hence, the required hour to fill the reservoir = 30 hours.

CLASS-8 SUBJECT MATHS (FINALS.E) CHAPTER-11 (EXERCISE11.3) MENSURATION

  EVENTS CONVENT HIGH SCHOOL

25/01/2022      CLASS- 8   SESSION 2021-22
SUBJECT :MATHS

CHAPTER-11 (EXERCISE 11.3)

MENSURATION

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 Question 1.There are two cuboidal boxes as shown in the figure. Which box requires the lesser amount of material to make?

Solution:
(a) Volume of the cuboid = l × b × h = 60 × 40 × 50 = 120000 cm3
(b) Volume of cube = (Side)3 = (50)3 = 50 × 50 × 50 = 125000 cm3
Cuboidal box (a) requires lesser amount of material.

Question 2.A suitcase with measures 80 cm × 48 cm × 24 cm is to be covered with a tarpaulin cloth. How many metres of tarpaulin of width 96 cm is required to cover 100 such suitcases?
Solution:
Measurement of the suitcase = 80 cm × 48 cm × 24 cm
l = 80 cm, b = 48 cm and h = 24 cm
Total surface area of the suitcase = 2[lb + bh + hl]
= 2 [80 × 48 + 48 × 24 + 24 × 80]
= 2 [3840 + 1152 + 1920]
= 2 × 6912
= 13824 cm2
Area of tarpaulin = length × breadth = l × 96 = 96l cm2
Area of tarpaulin = Area of 100 suitcase
96l = 100 × 13824
l = 100 × 144 = 14400 cm = 144 m
Hence, the required length of the cloth = 144 m.

Question 3.Find the side of a cube whose surface area is 600 cm2?
Solution:
Total surface area of a cube = 6l2
6l2 = 600
l2 = 100
l = √100 = 10 cm
Hence, the required length of side = 10 cm.

Question 4.Rukhsar painted the outside of the cabinet of measure 1 m × 2 m × 1.5 m. How much surface area did she cover if she painted all except the bottom of the cabinet?

Solution:
l = 2 m, b = 1.5 m, h = 1 m
Area of the surface to be painted = Total surface area of box – Area of base of box
= 2 [lb + bh + hl] – lb
= 2[2 × 1.5 + 1.5 × 1 + 1 × 2] – 2 × 1
= 2[3 + 1.5 + 2] – 2
= 2[6.5] – 2
= 13 – 2
= 11 m2
Hence, the required area = 11 m2.

Question 5.Daniel is painting the walls and ceiling of a cuboidal hall with length, breadth and height of 15 m, 10 m and 7 m respectively. From each can of paint 100 m2 of the area is painted. How many cans of paint will she need to paint the room?
Solution:
Surface area of a cuboidal hall without bottom = Total surface area – Area of base
= 2 [lb + bh + hl] – lb
= 2 [15 × 10 + 10 × 7 + 7 × 15] – 15 × 10
= 2[150 + 70 + 105] – 150
= 2 [325] – 150
= 650 – 150
= 500 m2
Area of the paint in one can = 100 m2
Number of cans required = 500100 = 5 cans.