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Mensuration - 8th math

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NCERT Exercise 11.3 solution q6 to q10


Mensuration Class Eighth Math NCERT Exercise 11.3 Question (6) Describe how the two figures at the right are alike and how they are different. Which box has larger lateral surface area?

Solution

The dimensions of both the figures are same but one is cylindrical while other is cubical.

Calculation of Lateral surface area of Cylinder

Given, Height of the cylinder = 7cm

And Diameter of the cylinder = 7cm

Thus, radius of the cylinder = diameter/2

=7/2=3.5cm

Thus, radius of the cylinder = 3.5cm

Thus, lateral surface area of the given cylinder = ?

We know that, lateral surface area or curved surface area is the surface area except top and base

And we know that, Lateral surface area or curved surface area of a cylinder = 2πrh

Thus, lateral surface area of the given cylinder = 2 π × 3.5cm × 7cm

= 2 × 22 × 3.5 cm2

= 154 cm2

Thus, lateral surface area of given cylinder = 154cm2

Calculation of lateral surface area or curved surface area of cube

We know that, Lateral surface area of a cube = 2h(l+b)

Thus, lateral surface area of the given cube = 2 × 7cm (7cm +7cm)

= 14 cm × 14

Thus, lateral surface area of the given cube = 196 cm2

Both of the figure are alike in dimension but area different in shape as one is cylinder while other is cube. And the lateral surface area of the cube is larger. Answer

Mensuration Class Eighth Math NCERT Exercise 11.3 Question (7) A closed cylindrical tank of radius 7m and height 3m is made from a sheet of metal. How much sheet of metal is required?

Solution

The metal sheet required is equal to the total surface area of the cylindrical tank.

Given, Radius of cylindrical tank = 7m

And height of the cylindrical tank = 3m

Thus, total surface area of the cylindrical tank = ?

Calculation of total surface area of the cylindrical tank

We know that, total surface area of a cylinder = 2πr(r+h)

Thus, total surface area of the given cylinder

= 2 × 22 × 10m2

= 440 m2

Thus, total surface area of the given cylinder = 440m2

Thus, area of metal will be required to make the given cylindrical tank = 440m2 Answer

Mensuration Class Eighth Math NCERT Exercise 11.3 Question (8) The lateral surface area of a hollow cylinder is 4224 cm2. It is cut along its height and formed a rectangular sheet of width 33cm. Find the perimeter of rectangular sheet.

Solution

The lateral surface area of the cylinder will be equal to the area of the rectangular sheet. Since the hollow cylinder is cut along its height thus given width of the sheet will be equal to the height of the cylinder. And area of rectangular sheet divided by the given width will give the length of the sheet. Using length and width of the sheet the perimeter can be calculated.

Given, Lateral surface area of the hollow cylinder = 4224cm2

Since, rectangular sheet is made after the cutting of the given hollow cylinder, thus area of rectangular sheet = lateral surface area of the hollow cylinder = 4224cm2

And width of the sheet = 33cm

Now, we know that, Area of a rectangle = length × breadth

⇒ 4224cm2 = Length × 33cm

Thus, Length

⇒ Length = 128 cm

Calculation of perimeter of the rectangular sheet

We know that, Perimeter of a rectangular sheet = 2(Length + Breadth)

= 2 (128cm + 33cm)

= 2 × 161 cm

= 322 cm

Thus, perimeter of the given rectangular sheet = 322 cm Answer

Mensuration Class Eighth Math NCERT Exercise 11.3 Question (9) A road roller takes 750 complete revolutions to move once over to level a road. Find the area of the road if the diameter of a road roller is 84cm and length is 1m.

Solution

Given, to move once over to level a road = 750 complete revolution

Diameter of the road roller = 84cm

= 84/100m = 0.84m

Thus, radius of wheel of road roller = 0.84/2 = 0.42m

And length of the road roller = 1m

Thus, area of the road = ?

The curved surface area of the wheel of the road roller will be equal to the area covered in one revolution of road roller.

We know that, lateral surface area of a cylinder = 2πrh

Thus, lateral surface area of the wheel of given road roller

= 2 × 22 × 0.6 × 1 m2

= 26.4 m2

Thus, lateral surface area of wheel of given road roller = 26.4 m2

Now, in one complete revolution of the road roller the area of road leveled = 26.4m2

Thus, in 750 complete revolution of the road roller the area of road leveled = 26.4m2 × 750

= 1980 m2

Thus, area of road = 1980 m2 Answer

Mensuration Class Eighth Math NCERT Exercise 11.3 Question (10) A company packages its milk powder in cylindrical container whose base has a diameter of 14cm and height 20cm. Company places a label around the surface of the container (as shown in the figure). If the label is placed 2cm from top and bottom, what is the area of the label.

Solution

The base of the cylindrical milk powder can= Diameter = 14cm

Thus, radius of the can = diameter/2 = 14/2

Thus, radius of the can = 7 cm

And height of the can (h) = 20cm

Label to be placed 2cm from top and bottom.

Thus area of label = ?

Now, since label is to be placed 2cm from top and bottom

Thus, height of the label = height of the can – (2cm + 2cm)

= 20 cm – 4cm = 16cm

Thus, height of the label = 16cm

And, radius of the label = radius of the can = 7cm

Thus, lateral surface area of the label = area of the label

Now, we know that, Lateral surface area of a cylinder = 2πrh

Thus, lateral surface area of the cylindrical shaped label

= 2 × 22 × 16 cm2

= 704 cm2

Thus, area of the label = 704 cm2 Answer




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