8th Mensuration Solved Questions

8^th Mensuration CBSE Enrichment questions

1. A rectangular tank can hold 650 litres of milk . if it is 130cm long & 250cm wide , find the height of the tank.

Solution: Volume of rectangular tank = 650 litres = 650 × 1000 cm ³ = 650000 cm ³

We know that Volume of a cuboid = l × b × h

650000 = 130 × 250 × h

⇒ h = (650000)/( 130 × 250 )= 20 cm

Therefore, height of the tank = 20 cm

2. a hall is of length 16m , breadth 14m & height 5m . Calculate the no: of persons that can be accommodated in the hall , assuming 3.5m³ of air is required for each person.

Solution: ∴Volume of the hall = length × breadth × height

                                                  = 16 × 14 × 5

                                                    = 1120 m³

Volume occupied my 1 man = 3.5 m³

3.5 m³  of air occupied my 1 man

1120 m³  of air occupied my ( 1/3.5)x 1120= 320 man

Therefore, number of men that can be accommodated = 32 0 men

2. A cylinder container with diameter 48cm contains sufficient water to submerge a rectangular solid of iron with dimensions 33cm x 18cm x 12cm.find the rise in the level of water when solid is completely submerged.

Solution:  Diameter of cylinder = 48 cm     ⇒ Radius of cylinder (r) = 24 cm

Now, when the Cuboid  is completely submerged

Then, Volume of cylindrical portion formed by original water level and after increase in water level = volume of cuboid

⇒ πr ² h = 33 cm × 18 cm × 12 cm

22/7 x 24 x 24 x h = 33 cm × 18 cm × 12 cm

h = (33 cm × 18 cm × 12 cm x7)/ 22 x 24 x 24=3.94cm

Hence, the required increase in water level is 3.94 cm

3. The difference between the circumference and radius of a circle is 74 cm. Find the circle diameter.

Solution: circumference - radius = 74 Þ      pd- d/2 =74 Þ  22d/7  - d/2 =74

Þ  (44d-7d)/14 =74 Þ 37d = 74x14 Þ d= (74x14)/37 Þ d  = 28cm

4. Find the area of the rhombus if the length of each side be 7 cm. and  corresponding altitude is 8 cm

Solution:  area of rhombus = base × height = 7 cm × 8 cm = 56 sq cm

5. A solid cubical block of fine wood costs rupees 256 at rupees 500/m³. Find its volume and the length of each side.

Solution: Let the edge of the cubical box be “a” m.

Rs. 500 is the cost of 1 m³ wood

Rs. 256. is the cost of 1x256/500 m³ wood=0.512m³

So, Volume of total wood = 0.512m³

a x a x a = 0.512m³

a x a x a = 0.8 x 0.8 x0.8x

The length of each side = 0.8m=80cm

6. What will be the volume of the cylinder in which area of the base is 45 sq.cm and height is 6 cm?

Solution: The volume of the cylinder = Area of base x height =45 sq.cm x 6 cm =270cm³

7. Area of 4 walls of a room is 108 m² .If height and length of room is in ratio 2: 5 and height and breadth in ratio 4: 5. Find the area of floor of the room.

Solution: Given: L : H = 5 : 2 Þ  H : B = 4 : 5

L : H : B = 20 : 8 : 10

Area of four walls of a room is =2(l + b) h = 108 Þ (l + b) h = 54

(20 x + 10x) x 8x = 54

240x2 = 54

x² = 54 / 240

x² = 0.225

The area of floor of the room = l x b =l × b = 20x × 10x =200x² =200x 225= 45m²

8. If Area of the square inscribed in a semicircle is 2cm² .Find the area of the Square Inscribed In a full circle.

Solution: Area of square = 2cm²

Side of square = square root of 2

Let r be the radius of the circle and adjust it like that it forms a right angle triangle

then

r² = (square root of 2)² + (square root of 2 / 2)²

r² = 2 + ( 2 / 4)

r² = 5 / 2

r = square root of ( 5 / 2)

In a full square the diagonal of square will be the diameter of the cirlcle.

r = square root of ( 5 / 2) is this so diameter = 2 * square root of (5 / 2)

Diagonal of square = side * square root of 2

side * square root of 2 = 2 * square root of 5 / square root of 2

side * 2 = 2 * square root of 5

side = square root of 5

Area of square = r²

= (square root of 5)²

= 5 cm²

9. A cylinder tube open at boyh ends is made of metal. The internal diameter of the tube is 11.2 cm and its length is 21 cm .The metal everywhere is 0.4 cm thick . Calculate the volume of the metal correct to 1 place of decimal.

Solution: Internal diameter = 11.2 cm

Internal radius= 11.2/2 = 5.6 cm  Length of the tube = 21 cm

Volume of the internal cylinder π r ² h = 22x 5.6x5.6x21/7 =2069.76cm ²

External radius, R = (5.6 + 0.4) cm = 6 cm

Volume of the external cylinder = πR² h =22x6x6x21/7=2376cm²

Now, volume of the metal = Volume of the external diameter – Volume of the internal diameter  = (2376 – 2069.76) cm³ = 306.24 cm³ = 306.2 cm³

10.  The dimension of an oil tin are 26 m x 26 cm x 45 cm .Find the area of the tin sheet required for making 20 such tins. If 1 square meter sheet costs Rs. 10 , find the total cost for making 20 such tins.

Solution : Area of the tin sheet required =  Area of 20 oil tins

= 20 × 2 (lb + bh + hl)   = 20x2(26x26+26x45+45x26)=12.06m²

Cost of 1 m² tin sheet = Rs.10

Cost of  12.064 m² tin sheet = Rs.10 2.064  = Rs.120.64

So, total cost for making 20 such tins = Rs.120.64

Practice problems for class 8th Mathematics