Important Questions Class 8 Mathematics Chapter 11

Important Questions Class 8 Mathematics Chapter 11 – Mensuration

Class 7 Mathematics introduced students to important topics like unit conversion, perimeter, and area of plane shapes, including squares, rectangles, parallelograms, triangles, and circles. 

Class 8 Mathematics Chapter 11 Mensuration is the continuation of previous classes where you will learn to calculate the perimeter and area of other closed plane figures like quadrilaterals and trapeziums. You will also learn to find the area of the pathway in rectangular shapes and areas of regular and irregular polygons. The surface area, length, volume, and capacity of three-dimensional solid objects such as cubes, cuboids, and cylinders are also covered in this Chapter 11. 

You will study the below key topics and important formulas involving 2D and 3D shapes in the Mensuration chapter:

• Perimetre is the distance around the boundary of a plane-closed figure(2D shapes). The mensuration formula to find the perimeter is shown below,

              Rectangle = 2 (length + breadth)

              Square = 4 × side

              Circle = 2 π r, where r is the radius

              Triangle = a + b + c

              Rhombus =   4 × side

              Parallelogram = 2(l+b)

              Trapezium = a + b + c + d          

• The area is the region enclosed in a plane-closed figure(2D shapes). The mensuration formula to find the area is shown below,

              Rectangle = length x breadth

              Square = side x side

              Circle =   π r²

              Triangle = 1/2 × base × corresponding height

              Rhombus = 1/2 x  product of diagonals

              Parallelogram = base × corresponding height

              Trapezium = 1/2 x (Sum of parallel sides) × Height

• Volume or capacity is the amount of space occupied by a solid(3D shapes). The mensuration formula to calculate the volume is shown below,

              Cube = a³

              Sphere = (4/3) π r³

              Cuboid = l × b × h

              Hemisphere = (⅔) π r³

              Cylinder = π r² h

              Cone = (⅓) π r² h            

• The lateral and curved surface area of solids(3D shapes),

              Cube = 4 (side)²

Cuboid = 2 × height × (length + breadth).

Sphere = 4 π r²

              Hemisphere = 2 π r²

              Cylinder = 2πrh.

              Cone = π r l

• Total surface area is the sum of the  structure’s curved and lateral surface areas.

              Cube = 6 (side)²

              Cuboid = 2 (lb + bh + hl)

              Sphere = 4 π r²

              Hemisphere = 3 π r²

              Cylinder = 2πr (r + h)

              Cone = πr (r + l)

Thorough knowledge of geometrical shapes and algebraic equations is required  to solve the problems related to the perimeter, area, and volume of the shapes. Learning NCERT solutions, CBSE revision notes, and solving Important Questions Class 8 Mathematics Chapter 11 on Extramarks, will help students have a thorough knowledge of all the concepts and score well in their examination. Chapter 11 Class 8 Mathematics Important Questions with Step-by-Step Solutions will help you improve critical thinking and solve problems easily. 

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Important Questions Class 8 Mathematics Chapter 11 – With Solutions

A full list of Important Questions Class 8 Mathematics Chapter 11 is collated from the NCERT textbook, NCERT exemplars, CBSE sample papers, CBSE past years’ question papers, and other genuine sources for students to help practise all types of questions from the chapter before facing their final examination. 

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Below is a list of a few questions and their solutions from our question bank of Mathematics Class 8 Chapter 11 Important Questions

Question 1. What will be the change in the volume of a cube when its side becomes ten times the original side?

(a) Volume becomes 1000 times.

(b) Volume becomes 10 times.

(c) Volume becomes 100 times.

(d) Volume becomes 1 / 1000 times

Answer 1. Consider the side of the cube is x, then its volume = x³.

                    Given that if the side becomes ten times the original side, then the new volume is 

                    Volume = (10x)³ = 1000x³

Hence the correct answer is (a) Volume becomes 1000 times.

Question 2. When a polygon ABCDE is divided into parts, as shown below. Find its area if

AD = 8 cm, AH = 6 cm, AG = 4 cm, AF = 3 cm and perpendiculars BF = 2 cm,

CH = 3 cm, EG = 2.5 cm.

Answer 2. Area of Polygon ABCDE = area of (▲AFB + ▲CHD + ▲ADE) + area of trapezium FBCH

We know that the area of a triangle is 1/2 × base × corresponding height, and the area of the trapezium is 1/2 x (Sum of parallel sides) × height.

Hence,

Area of ▲AFB = 1/2 × AF × BF = 1 / 2 x (3 x 2) = 3 cm²

Area of ▲CHD = 1/2 × CH × HD = 1 / 2 x (3 x ( AD – AH ) ) = 1 / 2 x (3 x ( 8 – 6 ) ) = 3 cm²

Area of ▲ADE = 1/2 x AD x GE = 1 / 2 x ( 8 x 2.5 ) = 10cm²

Area of trapezium FBCH  = 1/2 x ( BF + CH ) x FH = 1/2 x ( 2 + 3 ) x ( AH – AF ) = 1/2 x ( 2 + 3 ) x ( 6 – 3 ) = 7.5 cm²

So the total area of the polygon ABCDE = 3 + 3 + 10 + 7.5 =  23.5 cm²

Question 3. Fill in the blanks.

(a) Area of a rhombus is equal to __________ of its diagonals.

(b) The area of the cube’s face is 10 cm², then the total surface area of the cube is __________.

(c) A cube of side 4 cm is painted on all the sides. Further, if it is sliced into 1 cubic cm cubes, then the number of such small cubes that will have exactly two of their faces painted is __________.

(d) The volume of a cylinder becomes __________ the original volume if its radius becomes half of the original radius.

(e) If the diagonal d of a quadrilateral is doubled and the heights h1 and h2 falling on d is halved, then the area of the quadrilateral is __________.

(f) Two cylinders, A and B, are made by folding a rectangular sheet with dimensions 20 cm × 10 cm along its length and breadth. The volume of A is ________ of the volume of B.

(g) A trapezium has three equal sides, and one side is double the equal side. The trapezium can be divided into __________ equilateral triangles of _______ area.

Answer 3.

(a) Half the product.

(b) 60 cm². 

As we know, the cube’s total surface area = 6 (side)². Given that side² = 10 cm², the total surface area of the cube = 6 x 10cm² = 60 cm²

(c) 24 cubes. 

When the cube side = 4 cm, then the volume of that cube = 4³ = 64 cm³

If the bigger cube of 64 cm³ is sliced into 1 cm³ cube, then we get 64 smaller cubes.

As we know that a cube consists of 12 edges, the number of small cubes with two faces painted  

will be 12 x 2 = 24 cubes, as shown in the image below.

(d) 1/4 

The volume of the cylinder  = π r² h

Given that if the radius is reduced to half of the original radius, then the new volume of the cylinder becomes,

=  π ( 1/2r)² h = 1/4 πr²h

(e) 1/2 (h1 +h2) d

Area of quadrilateral = 1/2 (sum of the heights) x diagonal 

If the diagonal d is doubled and the heights h1 and h2 are halved, then the new area of the quadrilateral is,

A = 1/2 ( h1 + h2 )/2 x 2d

    = 1/2 (h1 + h2 )d .

we get the same area as the original area of the quadrilateral.

(f) Twice

Given that the rectangular sheet dimension is 20 cm(length) × 10 cm(breadth)

using the formulas radius r = circumference/2π   and volume of cylinder = π r² h, let us calculate the volume in both cases ,

When the rectangle is folded along its length, cylinder A will have a circumference of 20 cm and a height of 10 cms.

h = 10, r = 20/2π = 10/ π

The volume of cylinder A = π(10/π)²h

= (100/π) × 10 = 1000/π

When the rectangle is folded along its breadth, cylinder B will have a circumference of 10 cms and a height of 20 cms. 

h = 20, r = 10/2π = 5/π

Volume of cylinder B = π (5/ π)²h 

= (25/π) x 20 = 500/π

Volume of cylinder A/Volume of cylinder B = [1000/π]/[500/π] = 2

Hence the volume of cylinder A is twice of volume B.

(g) Three, same

Consider the above trapezium  Q2Q1AB  

Let Q2Q1 = Q1A = AB = a and Q2B = 2a

Draw medians through the vertices Q1 and A on the side Q2B, now 

Q2Q3 = Q3B = a

We get 2 parallelograms Q2Q1AQ3 and Q1Q3BA

Since a parallelogram has equal opposite sides,

In Q2Q1AQ3 

Q1Q2 = AQ3 and Q1A = Q2Q3

so ▲Q1Q2Q3 = ▲Q1Q3A

In Q1Q3BA

Q1Q3 = AB and Q1A = Q3B

so ▲Q1Q3A = ▲Q3AB

Hence, the trapezium can be divided into three equilateral triangles of the same area.

Question 4. Give whether the statements are true or false.

(a) 1L = 1000 cm3

(b) Amount of region occupied by a solid is called it’s surface area.

(c) The surface area of a cuboid formed by joining face-to-face three cubes of side x is three times the surface area of a cube of side x.

(d) The surface area of a cube formed by cutting a cuboid of dimensions 2 × 1 × 1 in 2 equal parts is 2 sq. units.

(e) The areas of any two faces of a cube are equal.

(f) Two cuboids with equal volumes will always have equal surface areas.

(g) The areas of any two faces of a cuboid are equal.

Answer :

 (a) True

 (b) False. The amount of region occupied by a solid is called its volume.

 (c) False.

When three cubes with side x are joined face-to-face, the resulting cuboid will have the same height and breadth as the cubes, but its length is three times more.  

The cuboid height, breadth and length are x, x, and 3x, respectively. 

Total surface area = 2 (lb + bh + hl) = 2( (3x x x) + (x x x) + (x x 3x ) = 2(3x² + x² + 3x²) = 2 x 7x² = 14x² 

Now, the total surface area of the cube of side x = 6 (Side)² = 6x², now three times the surface area of cube = 3 x 6x² = 18x² ≠ 14x² Hence, the statement is false.

(d) False. 

The cuboid dimensions are 2 x 1 x 1 (l x b x h).

When it is sliced into two equal cubes, the cube’s dimensions will be 1 x 1 x 1. 

The surface area of the cube = 6 (Side)² = 6 x (1)² = 6 sq units 

Hence, the statement is false.

(e) True

(f) False. 

We can prove this with the following example.

Consider two cuboids, A and B, with different dimensions. 

Cuboid A : l x b x h = 10 x 4 x 3

Cuboid B : l x b x h = 12 x 5 x 2

The volume of A = 120 cm³ and volume of B = 120 cm³ 

Both cuboids have the same volume.

Now, 

The total surface area of A = 2 ( lb x bh x lh ) = 2 (10 × 4 + 4 × 3 + 10 × 3) = 164 cm³

The total surface area of B = 2 ( lb x bh x lh ) = 2 (12 × 5 + 5 × 2 + 12× 2) = 188 cm³

Hence, the statement two cuboids with equal volumes will always have equal surface areas is false.

Question 5. Prashant has got three containers with him,

(a) Cylindrical container P has radius r and height h,

(b) Cylindrical container Q has a radius of 2r and height of 1/2 h, and

(c) Cuboidal container R has dimensions r × r × h

Please choose the correct option below to arrange the containers in the increasing order of their volumes.

(a) P, Q, R

(b) Q, R, P

(c) R, P, Q

(d) cannot be arranged

Answer 5. (c) R, P, Q

Explanation: 

The volume of cylindrical container P having radius r and height h = π r² h.

 The volume of the cylindrical container Q having radius 2r and height 1/2 h = π x (2r)² x 1/2 h = 2π r² h.

 The volume of cuboidal container R having dimensions r × r × h = r²h

 Hence, the arrangement of containers in increasing order will be R, P, and Q

Question 6. The cylinder’s height becomes 1/4 of the original height, and the radius is doubled; then state the correct option from below.

(a) Cylinder curved surface area will remain unchanged.

(b) Cylinder curved surface area will be doubled.

(c) Cylinder curved surface area will be halved.

(d) Cylinder curved surface area will be 1/4 of the original curved surface.

Answer 6. (c) Cylinder curved surface area will be halved.

Explanation : 

Given that the height of a cylinder is h/4 and the radius is 2r, then 

The curved surface area of the cylinder = 2πrh, 

The new curved surface area of cylinder = 2π x (2r) x h/4 = 4πr x h/4 = πrh 

From the above calculation, we can conclude that the curved surface area of the cylinder will be halved.

Question 7. The cylinder’s height becomes 1/4 of the original height, and the radius is doubled; then state the correct option from below.

(a) Cylinder total surface area will be doubled.

(b) Cylinder total surface area will remain unchanged.

(c) Cylinder total surface area will be halved.

(d) None of the above.

Answer 7. (d) None of the above.

Explanation: 

Total surface area of a cylinder = 2πr (r + h).

If the height of the cylinder is h/4 and the radius is 2r, then

Total surface area of a new cylinder = 2 x π x 2r ( 2r + 1/4 h )

= 4 πr(2r+h/4)

​= 8πr² + 4πrh/4

= 8πr² + πrh = πr(8r+h)

Hence, option (d) is correct.

Question 8. The circumference of the front wheel of a cart is 3 m long, and that of the back wheel is 

4 m long. What is the distance travelled by the cart when the front wheel makes five more revolutions than the rear wheel?

Answer 8. Circumference of the front wheel = 3 m and circumference of the rear wheel = 4 m.

Let us consider the distance as Y

we know that number of revolutions of the wheel = Distance travelled/circumference

Number of revolutions of the front wheel = Y / 3 and the number of revolutions of the rear wheel = Y / 4                   

If the front wheel makes five revolutions more than the rear wheel,

∴ Y/3 – Y/ 4 = 5

= 4Y – 3Y/12 = 5

= Y / 12 = 5

Y = 5 x 12 = 60 m

The distance traveled by the cart is 60 m.

Question 9. A circular pond has a footpath running along its boundary. If a person completes one round around the pond with exactly 400 steps, his step is 66 cm long. Find the diameter of the pond.

Answer 9. We know that the diameter d = 2r.

Given that the person takes 400 steps 66 cm long to go around the pond,

then the circumference of the pond = 66 x 400 = 26400 cm = 264 m 

we now know that circumference of the pond 2πr = 264

r = 264/2π

r = 264/ 2x 3.14 = 42 m

hence, the diameter of the pond = 2 x 42 = 84 m.

Question 10. The cylinder’s radius is r, and the height is h. Find the change in the volume if the

(a) Height is doubled.

(b) Height is doubled, and the radius is halved.

(c) Height remains the same, and the radius is halved.

Answer 10. Original volume V0 of the cylinder = πr²h

1. When height is doubled.

              New volume V1 = 2πr²h 

The ratio of new volume to the old volume = V1/V0 = 2πr²h / πr²h = 2 : 1

               The new volume is doubled 

1. When the height is doubled, and the radius is halved

              New volume V1 = 2π(r/2)²h = 1/2 × πr²h

The ratio of new volume to the old volume = V1/V0 = ( 1/2 × πr²h) / (πr²h ) = 1:2

               The new volume is halved.

       (c) When the height remains the same, and the radius is halved.

              New volume V1 = π(r/2)²h = πr²h/4

              The ratio of new volume to the old volume = V1/V0 = (1/4 × πr²h)/πr²h = 1/4 = 1:4

              The new volume is one-fourth of the original volume.

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