Study MaterialsCBSE NotesMensuration Class 8 Extra Questions Maths Chapter 11

Mensuration Class 8 Extra Questions Maths Chapter 11

Mensuration Class 8 Extra Questions Maths Chapter 11

Extra Questions for Class 8 Maths Chapter 11 Mensuration

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    Mensuration Class 8 Extra Questions Very Short Answer Type

    Question 1.
    Find the perimeter of the following figures:
    Mensuration NCERT Extra Questions for Class 8 Maths Q1
    Solution:
    (i) Perimeter of the rectangle = 2(l + b) = 2(8 + 6) = 2 × 14 = 28 cm
    (ii) Perimeter of the square = 4 × side = 4 × 6 = 24 cm
    (iii) Perimeter of the circle = 2πr = 2 × \(\frac { 22 }{ 7 }\) × 7 = 44 cm.

    Question 2.
    The length and breadth of a rectangle are 10 cm and 8 cm respectively. Find its perimeter if the length and breadth are (i) doubled (ii) halved.
    Solution:
    Length of the rectangle = 10 cm
    Breadth of the rectangle = 8 cm
    (i) When they are doubled,
    l = 10 × 2 = 20 cm
    and b = 8 × 2 = 16 cm
    Perimeter = 2(l + b) = 2(20 + 16) = 2 × 36 = 72 cm
    (ii) When they are halved,
    l = \(\frac { 10 }{ 2 }\) = 5 cm
    b = \(\frac { 8 }{ 2 }\) = 4 cm
    Perimeter = 2(l + b) = 2(5 + 4) = 2 × 9 = 18 cm

    Question 3.
    A copper wire of length 44 cm is to be bent into a square and a circle. Which will have a larger area?
    Solution:
    (i) When the wire is bent into a square.
    Side = \(\frac { 44 }{ 4 }\) = 11 cm
    Area of the square = (side)2 = (11)2 = 121 cm2
    (ii) When the wire is bent into a circle.
    Circumference = 2πr
    44 = 2πr
    Mensuration NCERT Extra Questions for Class 8 Maths Q3
    So, the circle will have a larger area.

    Question 4.
    The length and breadth of a rectangle are in the ratio 4 : 3. If its perimeter is 154 cm, find its length and breadth.
    Solution:
    Let the length of the rectangle be 4x cm and that of breadth = 3x cm
    Perimeter = 2(l + b) = 2(4x + 3x) = 2 × 7x = 14x cm
    14x = 154
    x = 11
    Length = 4 × 11 = 44 cm
    and breadth = 3 × 11 = 33 cm

    Question 5.
    The area of a rectangle is 544 cm2. If its length is 32 cm, find its breadth.
    Solution:
    Area = 544 cm2
    Length = 32 cm
    Breadth of the rectangle = \(\frac { Area }{ Length }\)
    = \(\frac { 544 }{ 32 }\)
    = 17 cm
    Hence, the required breadth = 17 cm

    Question 6.
    If the side of a square is doubled then how much time its area becomes?
    Solution:
    Let the side of the square be x cm.
    Area = (side)2 = x2 sq. cm
    If its side becomes 2x cm then area = (2x)2 = 4x2 sq. cm
    Ratio is x2 : 4x2 = 1 : 4
    Hence, the area would become four times.

    Question 7.
    The areas of a rectangle and a square are equal. If the length of the rectangle is 16 cm and breadth is 9 cm, find the side of the square.
    Solution:
    Area of the square = Area of the rectangle = 16 × 9 = 144 cm2
    Side of the square = √Area of the square = √144 = 12 cm
    Hence, the side of square = 12 cm.

    Question 8.
    If the lengths of the diagonals of a rhombus are 16 cm and 12 cm, find its area.
    Solution:
    Given:
    First diagonal d1 = 16 cm
    Second diagonal d2= 12 cm
    Mensuration NCERT Extra Questions for Class 8 Maths Q8
    Hence, the required area = 96 cm2.

    Question 9.
    The area of a rhombus is 16 cm2. If the length of one diagonal is 4 cm, find the length of the other diagonal.
    Solution:
    Given: Area of the rhombus = 16 cm2
    Length of one diagonal = 4 cm
    Mensuration NCERT Extra Questions for Class 8 Maths Q9
    Hence, the required length = 8 cm.

    Question 10.
    If the diagonals of a rhombus are 12 cm and 5 cm, find the perimeter of the rhombus.
    Solution:
    Given: d1 = 12 cm, d2 = 5 cm
    Mensuration NCERT Extra Questions for Class 8 Maths Q10
    Mensuration NCERT Extra Questions for Class 8 Maths Q10.1
    The perimeter = 4 × side = 4 × 6.5 = 26 cm
    Hence, the perimeter = 26 cm.

    Mensuration Class 8 Extra Questions Short Answer Type

    Question 11.
    The volume of a box is 13400 cm3. The area of its base is 670 cm2. Find the height of the box.
    Solution:
    Volume of the box = 13400 cm3
    Area of the box = 670 cm2
    Mensuration NCERT Extra Questions for Class 8 Maths Q11
    Hence, the required height = 20 cm.

    Question 12.
    Complete the following table; measurement in centimetres.
    Mensuration NCERT Extra Questions for Class 8 Maths Q12
    Solution:
    Mensuration NCERT Extra Questions for Class 8 Maths Q12.1
    Mensuration NCERT Extra Questions for Class 8 Maths Q12.2

    Question 13.
    Two cubes are joined end to end. Find the volume of the resulting cuboid, if each side of the cubes is 6 cm.
    Solution:
    Mensuration NCERT Extra Questions for Class 8 Maths Q13
    Length of the resulting cuboid = 6 + 6 = 12 cm
    Breadth = 6 cm
    Height = 6 cm
    Volume of the cuboid = l × b × h = 12 × 6 × 6 = 432 cm3

    Question 14.
    How many bricks each 25 cm by 15 cm by 8 cm, are required for a wall 32 m long, 3 m high and 40 cm thick?
    Solution:
    Converting into same units, we have,
    Length of the wall = 32 m = 32 × 100 = 3200 cm
    Breadth of the wall = 3 m = 3 × 100 = 300 cm
    and the height = 40 cm
    v, length of the brick = 25 cm
    breadth = 15 cm
    and height = 8 cm
    Number of bricks required
    Mensuration NCERT Extra Questions for Class 8 Maths Q14
    Hence, the required number of bricks = 12800.

    Question 15.
    MNOPQR is a hexagon of side 6 cm each. Find the area of the given hexagon in two different methods.
    Mensuration NCERT Extra Questions for Class 8 Maths Q15
    Solution:
    Method I: Divide the given hexagon into two similar trapezia by joining QN.
    Mensuration NCERT Extra Questions for Class 8 Maths Q15.1
    Area of the hexagon MNOPQR = 2 × area of trapezium MNQR
    = 2 × \(\frac { 1 }{ 2 }\) (6 + 11) × 4
    = 17 × 4
    = 68 cm2
    Method II: The hexagon MNOPQR is divided into three parts, 2 similar triangles and 1 rectangle by joining MO, RP.
    Mensuration NCERT Extra Questions for Class 8 Maths Q15.2

    Question 16.
    The area of a trapezium is 400 cm2, the distance between the parallel sides is 16 cm. If one of the parallel sides is 20 cm, find the length of the other side.
    Solution:
    Given: Area of trapezium = 400 cm2
    Height = 16 cm
    Mensuration NCERT Extra Questions for Class 8 Maths Q16
    Hence, the required length = 30 cm.

    Question 17.
    Find the area of the hexagon ABCDEF given below. Given that: AD = 8 cm, AJ = 6 cm, AI – 5 cm, AH = 3 cm, AG = 2.5 cm and FG, BH, EI and CJ are perpendiculars on diagonal AD from the vertices F, B, E and C respectively.
    Mensuration NCERT Extra Questions for Class 8 Maths Q17
    Solution:
    Given:
    AD = 8 cm
    FG = 3 cm
    AJ = 6 cm
    EI = 4 cm
    AI = 5 cm
    BH = 3 cm
    AH = 3 cm
    CJ = 2 cm
    AG = 2.5 cm
    Mensuration NCERT Extra Questions for Class 8 Maths Q17.1
    Mensuration NCERT Extra Questions for Class 8 Maths Q17.2
    Mensuration NCERT Extra Questions for Class 8 Maths Q17.3
    Area of hexagon ABCDEF = Area of ΔAGF + Area of trapezium FGIE + Area of ΔEID + Area of ΔCJD + Area of trapezium HBCJ + Area of ΔAHB
    = 3.75 cm2 + 8.75 cm2 + 6 cm2 + 2 cm2 + 7.5 cm2 + 4.5 cm2
    = 32.50 cm2.

    Question 18.
    Three metal cubes of sides 6 cm, 8 cm and 10 cm are melted and recast into a big cube. Find its total surface area.
    Solution:
    Volume of the cube with side 6 cm = (side)3 = (6)3 = 216 cm3
    Volume of the cube with side 8 cm = (side)3 = (8)3 = 512 cm3
    Volume of the cube with side 10 cm = (side)3 = (10)3 = 1000 cm3
    Volume of the big cube = 216 cm3 + 512 cm3 + 1000 cm3 = 1728 cm3
    Side of the resulting cube = \(\sqrt [ 3 ]{ 1728 }\) = 12 cm
    Total surface area = 6 (side)2 = 6(12)2 = 6 × 144 cm2 = 864 cm2.

    Question 19.
    The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m2.
    Solution:
    Given: Diameter of the roller = 84 cm
    Radius = \(\frac { 84 }{ 2 }\) = 42 cm
    Height = 120 cm
    Curved surface area of the roller = 2πrh
    Mensuration NCERT Extra Questions for Class 8 Maths Q19
    Area covered by the roller in one complete revolution = 3.168 m2
    Area covered in 500 complete revolutions = 500 × 3.168 = 1584 m2
    Hence, the required area = 1584 m2.

    Question 20.
    A rectangular metal sheet of length 44 cm and breadth 11 cm is folded along its length to form a cylinder. Find its volume.
    Solution:
    Circumference of the base = 2πr
    Mensuration NCERT Extra Questions for Class 8 Maths Q20
    = 423.5 cm3
    Hence, the required volume = 423.5 cm3.

    Question 21.
    160 m3 of water is to be used to irrigate a rectangular field whose area is 800 m2. What will be the height of the water level in the field? (NCERT Exemplar)
    Solution:
    Volume of water = 160 m3
    Area of rectangular field = 800 m2
    Let h be the height of water level in the field.
    Now, the volume of water = volume of cuboid formed on the field by water.
    160 = Area of base × height = 800 × h
    ⇒ h = 0.2
    So, required height = 0.2 m

    Question 22.
    Find the area of a rhombus whose one side measures 5 cm and one diagonal as 8 cm. (NCERT Exemplar)
    Solution:
    Let ABCD be the rhombus as shown below.
    Mensuration NCERT Extra Questions for Class 8 Maths Q22
    DO = OB = 4 cm, since diagonals of a rhombus are perpendicular bisectors of each other.
    Therefore, using Pythagoras theorem in ΔAOB, AO2 + OB2 = AB2
    Mensuration NCERT Extra Questions for Class 8 Maths Q22.1

    Question 23.
    The parallel sides of a trapezium are 40 cm and 20 cm. If its non-parallel sides are both equal, each being 26 cm, find the area of the trapezium.
    Solution:
    Let ABCD be the trapezium such that
    AB = 40 cm and CD 20 cm and AD = BC = 26 cm.
    Mensuration NCERT Extra Questions for Class 8 Maths Q23
    Now, draw CL || AD
    Then ALCD is a parallelogram.
    So AL = CD = 20 cm
    and CL = AD = 26 cm.
    In ΔCLB, we have CL = CB = 26 cm
    Therefore, ΔCLB is an isosceles triangle.
    Draw altitude CM of ΔCLB.
    Since ΔCLB is an isosceles triangle. So, CM is also the median.
    Then LM = MB = \(\frac { 1 }{ 2 }\) BL = \(\frac { 1 }{ 2 }\) × 20 cm = 10 cm
    [as BL = AB – AL = (40 – 20) cm = 20 cm].
    Applying Pythagoras theorem in ΔCLM, we have
    CL2 = CM2 + LM2
    262 = CM2 + 102
    CM2 = 262 – 102 = (26 – 10) (26 + 10) = 16 × 36 = 576
    CM = √576 = 24 cm
    Hence, the area of the trapezium = \(\frac { 1 }{ 2 }\) (sum of parallel sides) × height
    = \(\frac { 1 }{ 2 }\) (20 + 40) × 24
    = 30 × 24
    = 720 cm2.

    Question 24.
    Find the area of polygon ABCDEF, if AD = 18 cm, AQ = 14 cm, AP = 12 cm, AN = 8 cm, AM = 4 cm, and FM, EP, QC and BN are perpendiculars to diagonal AD. (NCERT Exemplar)
    Solution:
    Mensuration NCERT Extra Questions for Class 8 Maths Q24
    In the figure
    MP = AP – AM = (12 – 4) cm = 8 cm
    PD = AD – AP = (18 – 12) cm = 6 cm
    NQ = AQ – AN = (14 – 8) cm = 6 cm
    QD = AD – AQ = (18 – 14) cm = 4 cm
    Area of the polygon ABCDEF = area of ∆AFM + area of trapezium FMPE + area of ∆EPD + area of ∆ANB + area of trapezium NBCQ + area of ∆QCD.
    Mensuration NCERT Extra Questions for Class 8 Maths Q24.1

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