Cbse 8, Math, Cbse- Mensuration, Ncert Solutions

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CBSE 8, Math, CBSE­ Mensuration, NCERT Solutions

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Mensuration-NCERT Solutions NCERT Solutions

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Class VIII Math NCERT Solution For Mensuration

NCERT QUESTIONS WITH SOLUTIONS EXERCISE: 11.1 1.    A square and a rectangular eld with measurements as given in the following gure have the same perimeter. Which eld has a larger area?

        

Sol. Let x be the breadth of the rectangle. It is given that the perimeter of a rectangle = perimeter of the square.         ∴ 2(80 + x) = 4 × 60         ⇒80 + x = 120         ⇒x = 120 – 80 = 40         i.e. breadth of the rectangle = 40 m         Now, Area of the square = (60 × 60) m2         = 3600 m2         and the area of the rectangle = (40 × 80) m2         = 3200 m2         Hence, the square eld has a larger area. 2.   Mrs. Kaushik has a square plot with the measurement as shown in the gure. She wants to construct a house in the middle of the plot. A garden is developed around the house. And the total cost of developing a garden around the house at the rate of Rs. 55 per m2.

Sol. Area of the garden = Area of the outer square         � Area of the inner rectangle         = 25 × 25 m2 � 20 × 15 m2         (625 � 300) m2 = 325 m2         Cost of developing a garden @ Rs. 55 per sq. meter         = Rs. (55 × 325) = Rs. 17875 3.   The shape of a garden is rectangular in the middle and semi circular at the ends as shown in the diagram. Find the area and the perimeter of this garden [Length of rectangle is 20 � (3.5 + 3.5) metres].

Sol. Total area of the garden         = Area of the rectangular portion + The sum of the areas of the pair of semi circles                  Perimeter of the garden = 2 × length of rectangular portion + circumference of circle          4.   A ooring tile has the shape of a parallelogram whose base is 24 cm and the corresponding height is 10 cm. How many such tiles are required to cover a oor of area 1080 m2? Sol. Area of one tile = base × height 2

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        = (24 × 10)         = 240 cm2         Number of tiles required to cover the oor          

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5.   An ant is moving around a few food pieces of different shapes scattered on the oor. For which food piece would the ant have to take a longer round? Remember, circumference of a circle can be obtained by using the expression C = 2πr, where r is the radius of the circle.

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Sol. Let us mark points A, B, C and D in the given gures as shown. Let A be the point in each gure from where the ant start moving on the food pieces. She is to reach the initial point after moving around the boundary of each food piece.         For food piece (a)         Distance moved = Arc AB + BA

EXERCISE: 11.2 6.   The shape of the top surface of a table is trapezium. Find its area if its parallel sides are lm and 1.2 m and perpendicular distance between then is 0.8 m.

Sol. Area of top surface of a table         = Area of the trapezium

        

7.   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 other parallel side. Sol. Let the required side be x cm.         Then, area of the trapezium                  But, the area of the trapezium = 34 cm2 (given) ∴ 2 (10 + x) = 34         ⇒10 + × = 17         ⇒x = 17 – 10 = 7         Hence, the outer side = 7 cm 8.   Length of the fence of a trapezium shaped eld ABCD is 120 m. If BC = 48 m, CD = 17 rn and AD = 40 m, nd the area of this eld. Side AB is perpendicular to the parallel sides AD and BC.

Sol. Let ABCD be the given trapezium in which BC 48 m, CD = 17 m and AD = 40 m

        Through D, draw DL ⊥ BC.

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        Now, BL = AD = 40 m         and LC = BC – BL = (48 – 40) m = 8 m         Applying pythagoras theorem in right ΔDLC, we have         DL2 = DC2 – LC2 = 172 – 82 = 289 – 64 = 225 ⇒ DL         Now, area of the trapezium ABCD

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        = (44 × 15) m2 = 660 m2. 9.   The diagonal of a quadrilateral shaped eld is 24 m and the perpendiculars dropped on it from the remaining opposite vertices are 8 m and 13 m. Find the area of the eld.

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Sol. Let ABCD be the given quadrilateral in which BE ⊥ AC and DF ⊥ AC.         It is given that         AC = 24 m, BE = 8 m and DF= 13 m.

        Now, area of quad. ABCD         = area of ΔABC + area of ΔACD                  = (12 × 8 + 12 × 13) m2         = (96 + 156) m2 = 252 m2 10.   The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area. Sol. 11.   Find the area of a rhombus whose side is 6 cm and whose altitude is 4 cm. if one of its diagonals is 8 cm long, nd the length of the other diagonal. Sol. Let ABCD be a rhombus of side 6 cm and whose altitude DE = 4 cm. Also, one of its diagonals, BD = 8 cm.         Area of the rhombus ABCD

        

        Hence, the other diagonal is 6 cm. 12.   The oor of a building consists of 3000 tiles which are rhombus shaped end each of its diagonals are 45 cm and 30 cm in len. Find the total cost of polishing the oor, if the cost per m2 is Rs. 4. Sol. Area of the oor = 3000 � Area of one tile

        

13.   Mohan wants to b a trapezium shaped eld. It side along the river is parallel to and twice the side along the road. If the area of this eld is 10500 m2 and the perpendicular distance between the two parallel sides is 100 m, nd the length of the side along the river.

Sol. Let the parallel sides of the trapezium shaped eld be x m and 2x m. Then, its area.

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                   ∴The length of the side along the river is 2 × 70, i.e., 140 metres. 14.   Top surface of a raised platform is in the shape of a regular octagon as shown in the gure. Find the area of the octagonal surface.

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Sol. Area of the octagonal surface ABCDEFGH = Area (trap. ARCH) + Area (rect. HCDG) + Area (trap. GDEF)

15.   There is a Pentagonal shaped park as shown in the gure. For nding its area Jyoti and Kavita divided it in two different ways.

        Find the area of this park using both ways. Can you suggest some other way of nding its area? Sol. Taking Jyoti’s diagram:         Area of the pentagonal shaped park         = 2 × Area of trapezium ABEF

        

        Taking Kavita’s diagram:         Area of the pentagonal shaped park

        

16.   Diagram of the adjacent picture frame has outer dimensions = 24 cm × 28 cm and inner dimensions 16 cm × 20 cm. Find the area of each section of the frame, if the width of each section is same.

EXERCISE: 11.3 1   There are two cuboidal boxes as shown in the gure below. Which box requires the lesser amount of material to make?

Sol. Total surface area of rst box         = 2(lb + bh + lh)         = 2(60 × 40 + 40 × 50 + 60 × 50)cm2         = 200(24 + 20 + 30) cm2         = 200 × 74 cm2 = 14800 cm2         Total surface area of second box         = 6 (Edge)2 = 6 × 50 × 50 cm2         = 15000 cm2

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        Since the total surface area of rst box is less than that of the second, therefore the rst box i.e., (a) requires the least amount of material to make. 2.   A suitcase of rneasures 80 cm x48 cm x24 cm is to be covered with a tarpaulin cloth. How many metres of tarpaulin of width 96 cm is required to cover 100 such suitacases? Sol. Total surface area of suitcase         = 2[(80) (48) + (48) (24) + (24) (80))         = 2[3840 + 1152 + 1920]         = 13824 cm2         Total surface area of 100 suitcase         = (13824 × 100) cm2 =1382400 cm2         Required tarpaulin = Length × Breadth         1382400 cm2 = Length x96 cm

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                 Thus 144 m of tarpaulin is required to cover 100 suitcases. 3.   Find the side of a cube whose surface area is 600 cmz. Sol. Let a be the side of the cube having surface area 600 cm2.         ∴ 6a2 = 600 ⇒ a2 = 100 ⇒ a = 10         Hence, the side of the cube = 10 cm. 4.   Rukhsar painted the outside of the cabinet of measure 1 m x2m x1.5 m. How much surface area did she cover if she painted all except the bottom of the cabinate.

Sol. Here l = 2m, b = 1m and h = 1.5 m         Area to be painted         = 2bh + 2lh + lb         (2 × 1 × 1.5 + 2 × 2 × 1.5 + 2 × 1)m2         = (3 + 6 + 2) m2 = 11m2 5.   Deniel is painting the walls and ceiling of a cuboidal hall with length, breadth and height of 15 m, 10 m and 7 m respectively. From each can of paint 100 m2 of area is painted.         How many cans of paint will she need to paint the room? Sol. Here l = 15m, b = 10 m and h = 7 m         Area to be painted         = 2bh + 2lh + lb         = 2 × 10 × 7 + 2 × 15 × 7 + 15 × 10)m2         = (140 + 210 150) m2 = 500m2         Since each can of paint covers 100 m2, therefore number of cans required 6.   Describe how the two gures at the right are alike and how they are different. Which box has larger lateral surface area.

Sol. Try yourself 7.   A closed cylindrical tank of radius 7 m and height 3 m is made from a sheet of metal. How much sheet of metal is required? Sol. Here, r = 7m and h = 3 m.         Sheet of metal required to make a closed cylinder = Total surface area of the cylinder. = (2πrh + 2πr2) sq. units.          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 33 cm. Find the perimetre of rectangular sheet. Sol. A hollow cylind :r is cut alcng its height to form a rectangular sheet.         Area of cylinder = Area of rectangular sheet 4224 cm2 = 33 cm × Length                  Thus, the length of the rectangular sheet is 128 cm. Perimeter of the rectangular sheet         = 2 (Length + Width)         = [2(128 + 33)] cm         =(2 × 161)cm         = 322 cm 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 84 cm and length is 1 m.

Sol. In one revolution, the roller will cover an area equal to its lateral surface area. Thus, in 1 revolution, area of the road covered = 27πrh

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           10.   A company packages its milk powder in cylindrical container whose base has a diameter of 14 cm and height 20 cm. Company places a label around the surface of the container (as shown in the gure). if the label is placed 2 cm from top and bottom, what is the surface area of the label.

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Sol. Since the company places a label around the surface of the cylindrical container of radius 7 cm and height 20 cm such that it is placed 2 cm from top and bottom.         We have to nd the curved surface of a cylinder of radius 7 cm and height (20 � 4) cm i.e., 16 cm.         This curved surface area          EXERCISE: 11.4 1.   Given a cylindrical tank, in which situation will you nd surface area and in which situation volume.         (a) To nd how much it can hold.         (b) Number of cement bags required to plaster it.         (c) To nd the number of smaller tanks that can be lled with water from it. Sol. (a) volume (b) Surface area (c) Volume 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 by nding the volume of both the cylinders. Check whether the cylinder with greater volume also has greater surface area?

Sol. The heights and diameters of these cylinders A and B are interchanged.         We know that,         Volume of cylinder = πr2h         If measures of r and h are same, then the cylinder with greater radius will have greater area.                  As the radius of cylinder B is greater, therefore, the volume of cylinder B will be greater.         Let us verify by calculating the volume of both the cylinders.

        

        Thus, the surface area of cylinder B is also greater than the surface area of cylinder A. 3.   Find the height of a cuboid whose base area is 180 cm2 and volume is 900 cm3? Sol. Volume of the cuboid = 900 cm3         ⇒ (Area of the base) × Height = 900 cm3 180 × Height = 900                  Hence, the height of the cuboid is 5 cm. 4.   A cuboid is of dimensions 60 cm × 54cm × 30 cm. How many small cubes with side 6 cm can be placed in the given cuboid? Sol. Volume of cuboid = (60 × 54 × 30)cm3         = 97200 cm3         Volume of cube = (6 × 6 × 6) cm3         = 216 cm3

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         5.   Find the height of the cylinder whose volume is 1.54 m3 Sol. Let the h be the height of cylinder whose radius,

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        Hence, the height of cylinder is 1 metre. 6.   A milk tank is in the form of cylinder whose radius is 1.5 m and length is 7 m. nd the quantity of milk in litres that can be stored in the tank?

Sol. Quantity of milk that can be stored in the tank         = Volume of the tank         = πr2h, where r = 1.5 and h = 7 m                  = (49.5 × 1000) litres [∵ 1 m3 = 1000 Itr.]         = 49500 litres. 7.   If each edge of a cube is doubled,         (i) How many times Al its surface area increase?         (ii) How many times will its volume increase? Sol. Let x units be the edge of the cube. Then, its surface area = 6x2 and its volume = x3.         When its edge is doubled,         (i) Its surf ace area = 6 (2x)2 = 6 × 4x2 = 24x2                 ⇒ The surface area of the new cube will be 4 times that of the original cube.         (ii) Its volume = (2x)3 = 8x3                 ⇒ The volume of the new cube will be 8 times that of the original cube. 8.   Water is pouring into a cuboidal reservoir at the rate of 60 litres per minute. If the volume of reservoi: is 108 m3, nd the number of hours it will take to ll the reservoir.

Sol. Volume of the reservoir = 108 m3         = 108 × 1000 litres         = 108000 litres         Since water is pouring into reservoir @ 60 litres per minute.         ∴ Time taken to ll the reservoir         

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  STUDY MATERIAL FOR CBSE CLASS 8 MATH

Chapter 1 - Algebraic Expressions and Identities (/cbse-ncert/class-8/Math/CBSE-AlgebraicExpressionsandIdentities.html) Chapter 2 - Comparing Quantities (/cbse-ncert/class-8/Math/CBSE-ComparingQuantities.html) Chapter 3 - Cubes and Cube Roots (/cbse-ncert/class-8/Math/CBSE-CubesandCubeRoots.html) Chapter 4 - Data handling (/cbse-ncert/class-8/Math/CBSE-Datahandling.html) Chapter 5 - Direct and Inverse Proportions (/cbse-ncert/class-8/Math/CBSE-DirectandInverseProportions.html) Chapter 6 - Exponents and Powers (/cbse-ncert/class-8/Math/CBSE-ExponentsandPowers.html) Chapter 7 - Factorization (/cbse-ncert/class-8/Math/CBSE-Factorization.html) Chapter 8 - Introduction to Graphs (/cbse-ncert/class-8/Math/CBSE-IntroductiontoGraphs.html) Chapter 9 - Mensuration (/cbse-ncert/class-8/Math/CBSE-Mensuration.html) Chapter 10 - Playing with Numbers (/cbse-ncert/class-8/Math/CBSE-PlayingwithNumbers.html) Chapter 11 - Practical Geometry (/cbse-ncert/class-8/Math/CBSE-PracticalGeometry.html) Chapter 12 - Squares and Square Roots (/cbse-ncert/class-8/Math/CBSE-SquaresandSquareRoots.html) Chapter 13 - Visualizing Solid Shapes (/cbse-ncert/class-8/Math/CBSE-VisualizingSolidShapes.html) Chapter 14 - Linear Equations in One Variable (/cbse-ncert/class-8/Math/CBSE-LinearEquationsinOneVariable.html) Chapter 15 - Rational Numbers (/cbse-ncert/class-8/Math/CBSE-RationalNumbers.html) Chapter 16 - Understanding Quadrilaterals (/cbse-ncert/class-8/Math/CBSE-UnderstandingQuadrilaterals.html)

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