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MATHS CBSE VIII| Chapter 11 - Mensuration (Math) | Exercise 11.4 Page 194 Q 8

Question 8:

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

Answer

Volume of cuboidal reservoir = 108 m3 = (108 × 1000) L = 108000 L

It is given that water is being poured at the rate of 60 L per minute.

That is, (60 × 60) L = 3600 L per hour

Required number of hours = 30 hours

Thus, it will take 30 hours to fill the reservoir.

MATHS CBSE VIII| Chapter 11 - Mensuration (Math) | Exercise 11.4 Page 191 Q 1 Q2, Q3, Q4,Q5, Q6, Q7

Question 1:

Given a cylindrical tank, in which situation will you find 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.

Answer

(a) In this situation, we will find the volume.

(b) In this situation, we will find the surface area.

(c) In this situation, we will 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?

Answer

The heights and diameters of these cylinders A and B are interchanged.

We know that,

Volume of cylinder

If measures of r and h are same, then the cylinder with greater radius will have greater area.

MATHS CBSE VIII| Chapter 11 - Mensuration (Math) | Exercise 11.2 Page 186 Q 1, Q2, Q3, Q4, Q5, Q6, Q7, Q8, Q9, Q10

Question 1:

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

 

Answer

We know that,

Total surface area of the cuboid = 2 (lh + bh + lb)

Total surface area of the cube = 6 (l)2

Total surface area of cuboid (a) = [2{(60) (40) + (40) (50) + (50) (60)}] cm2

= [2(2400 + 2000 + 3000)] cm2

= (2 × 7400) cm2

= 14800 cm2

Total surface area of cube (b) = 6 (50 cm)2 = 15000 cm2

Thus, the cuboidal box (a) will require 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?


Answer

Total surface area of suitcase = 2[(80) (48) + (48) (24) + (24) (80)]

= 2[3840 + 1152 + 1920]

= 13824 cm2

Total surface area of 100 suitcases = (13824 × 100) cm2 = 1382400 cm2

MATHS CBSE VIII| Chapter 11 - Mensuration (Math) | Exercise 11.2 Page 178 Q 2, Q3, Q4, Q5,Q6, Q7,Q8, Q9, Q10, Q11

Question 2:

The area of a trapezium is 34 cm2 and the length of one of the parallel sides is 10 cm and its height is 4 cm. Find the length of the other parallel side.

Answer

It is given that,area of trapezium = 34 cm2 and height = 4 cm

Let the length of one parallel side be a. We know that,

Area of trapezium = (Sum of parallel sides) × (Distances between parallel sides)

Thus, the length of the other parallel side is 7 cm.

 

 

 

Question 3:

Length of the fence of a trapezium shaped field ABCD is 120 m. If BC = 48 m, CD = 17 m and AD = 40 m, find the area of this field. Side AB is perpendicular to the parallel sides AD and BC.

 

Answer

Length of the fence of trapezium ABCD = AB + BC + CD + DA

120 m = AB + 48 m + 17 m + 40 m

AB = 120 m − 105 m = 15 m