\nQuestion 1.
\nFill in the blanks:
\n(i) Area of a parallelogram = base \u00d7 …….
\n(ii) Area of a trapezium = \\(\\frac{1}{2}\\) \u00d7 ……….. \u00d7 distance between parallel sides.
\n(iii) Area of a rhombus = \\(\\frac{1}{2}\\) \u00d7 product of ……..
\n(iv) Area is measured in ……….. units.
\n(v) Volume of a solid is the measurement of ………… occupied by it.
\n(vi) Volume is measured in ………… units.
\n(vii) The volume of a unit cube is ……….
\n(viii) 1 litre = …………… cm3<\/sup>
\n(ix) 1 m3<\/sup> = ………… litres
\n(x) Volume of a cuboid = ……….. \u00d7 height.
\n(xi) Cylinders in which line segment joining the centres of the circular faces is perpendicular to the base are called ……….
\n(xii) Volume of a cylinder = area of base \u00d7 ………..
\n(xiii) Area of four walls = perimeter of floor \u00d7 …….
\n(xiv) Lateral surface area of a cube = 4 \u00d7 (…………)2<\/sup>
\n(xv) Total surface area of a cylinder of radius r and height h is ………..
\nSolution:
\n(i) Area of a parallelogram = base \u00d7 height.
\n(ii) Area of a trapezium = \\(\\frac{1}{2}\\) \u00d7 (sum of parallel sides)
\n\u00d7 distance between parallel sides.
\n(iii) Area of a rhombus = \\(\\frac{1}{2}\\) \u00d7 product of its diagonals.
\n(iv) Area is measured in square units.
\n(v) Volume of a solid is the measurement of the space occupied by it.
\n(vi) Volume is measured in cubic units.
\n(vii) The volume of a unit cube is 1 cubic unit.
\n(viii) l litre = 1000 cm3<\/sup>
\n(ix) 1 m3<\/sup> = 1000 litres.
\n(x) Volume of a cuboid = length \u00d7 breadth \u00d7 height.
\nOR
\nVolume of a cuboid = area of the base \u00d7 height.
\n(xi) Cylinders in which line segment joining the centres of the circular faces
\nis perpendicular to the base are called right circular cylinders.
\n(xii) Volume of a cylinder = area of base \u00d7 height.
\n(xiii) Area of four walls = perimeter of floor \u00d7 height of the room.
\n(xiv) Lateral surface area of a cube = 4 \u00d7 (edge)2<\/sup>
\n(xv) Total surface area of a cylinder of radius r and height h is 2\u03c0r(h + r).<\/p>\n
Question 2. Multiple Choice Questions<\/strong> Question 4. Question 5. Question 6. Question 7. Question 8. Question 9. Question 10. Question 11. Question 12. Question 13. Question 14. Question 15. Question 16. Question 17. Value Based Questions<\/strong> Question 2. Higher Order Thinking Skills (Hots)<\/strong> Question 2. Question 3. Question 4.
\nState which of the following statements are true (T) or false (F):
\n(i) Perimeter of a rectangle is the sum of lengths of its four sides.
\n(ii) Area of a quadrilateral can be found by splitting it into two triangles.
\n(iiii) Perimeter of a circle of radius r = \u03c0r2<\/sup>.
\n(iv) Volume of a cube = 6 \u00d7 (side)2<\/sup>
\n(v) 1 m3<\/sup> = 100000 cm3<\/sup>
\n(vi) Total surface area of a cuboid
\n= 2 (lb + bh + hl)
\n(vii) There is no difference between volume and capacity.
\n(viii)Total surface area of a cylinder = lateral surface area + area of two circular ends.
\n(ix) Surface area of a cube = 4 \u00d7 (side)2<\/sup>
\n(x) Lateral surface area of a cuboid = perimeter of base \u00d7 height.
\nSolution:
\n(i) Perimeter of a rectangle is the sum of lengths of its four sides. True
\n(ii) Area of a quadrilateral can be found by splitting it into two triangles. True
\n(iii) Perimeter of a circle of radius r = \u03c0r2<\/sup>. False
\nCorrect :
\nIt is area of a circle perimeter is 2\u03c0r.
\n(iv) Volume of a cube = 6 \u00d7 (side)2<\/sup> False Correct :
\nIt is surface area not volume, volume is (side)3<\/sup>.
\n(v) 1 m3<\/sup> = 100000 cm3<\/sup> False
\nCorrect:
\n1 m3<\/sup> = 1000000 cm3<\/sup>
\n(vi) Total surface area of a cuboid
\n= 2 (lb + bh +hl) True
\n(vii) There is no difference between volume and capacity. False
\nCorrect :
\nVolume refers to the amount of space occupied by an object
\nwhereas capacity refers to the quantity that a container holds.
\n(viii)Total surface area of a cylinder = lateral surface area
\n+ area of two circular ends. True
\n(ix) Surface area of a cube = 4 \u00d7 (side)2<\/sup> False Correct :
\nIt is 6 \u00d7 (side)2<\/sup>
\n(x) Lateral surface area of a cuboid = perimeter of base \u00d7 height. True<\/p>\n
\nChoose the correct answer from the given four options (3 to 17):<\/strong>
\nQuestion 3.
\nArea of a triangle is 30 cm2<\/sup>. If its base is 10 cm, then its height is
\n(a) 5 cm
\n(b) 6 cm
\n(c) 7 cm
\n(d) 8 cm
\nSolution:
\nArea of a triangle = 30 cm2<\/sup>
\nBase = 10 cm
\nArea = \\(\\frac{1}{2}\\) \u00d7 Base \u00d7 Height
\nHeight = \\(\\frac{A \\times 2}{B}=\\frac{30 \\times 2}{10}\\) =6 cm (b)<\/p>\n
\nIf the perimeter of a square is 80 cm, then its area is
\n(a) 800 cm2<\/sup>
\n(b) 600 cm2<\/sup>
\n(c) 400 cm2<\/sup>
\n(d) 200 cm2<\/sup>
\nSolution:
\nPerimeter of a square = 80 cm
\nPerimeter of square = 4(Side)
\n\u2234 Side = \\(\\frac{\\text { Perimeter }}{4}\\)
\n\u2234 Side = \\(\\frac{80}{4}\\) = 20 cm
\nArea = (side)2<\/sup> = (20)2<\/sup> = 400 cm2<\/sup> (c)<\/p>\n
\nArea of a parallelogram is 48 cm2<\/sup>. If its height is 6 cm then its base is
\n(a) 8 cm
\n(b) 4 cm
\n(c) 16 cm
\n(d) none of these
\nSolution:
\nArea of parallelogram = 48 cm2<\/sup>
\nHeight = 6 cm
\nArea of ||gm = Base \u00d7 Height
\nBase = \\(\\frac{\\mathrm{A}}{h}=\\frac{48}{6}\\) = 8 cm (a)<\/p>\n
\nIf d is the diameter of a circle, then its area is
\n(a) \u03c0d2<\/sup>
\n(b) \\(\\frac{\\pi d^{2}}{2}\\)
\n(c) \\(\\frac{\\pi d^{2}}{4}\\)
\n(d) 2\u03c0d2<\/sup>
\nSolution:
\nd is the diameter a circle
\n\u2234 Radius = r =\\(\\frac{d}{2}\\)
\nArea = \u03c0r2<\/sup> = \u03c0\\(\\left(\\frac{d}{2}\\right)^{2}\\) = \u03c0\\(\\frac{d^{2}}{4}\\) (c)<\/p>\n
\nIf the area of a trapezium is 64 cm2<\/sup> and the distance between parallel sides is 8 cm, then sum of its parallel sides is
\n(a) 8 cm
\n(b) 4 cm
\n(c) 32 cm
\n(d) 16 cm
\nSolution:
\nArea of trapezium = 64 cm2<\/sup>
\nDistance between parallelogram (h) = 8 cm
\nArea of trapezium = \\(\\frac{1}{2}\\) \u00d7 (Sum of ||gm sides) \u00d7 h
\nSum of parallel lines = \\(\\frac{\\mathrm{A} \\times 2}{h}\\)
\n= \\(\\frac{64 \\times 2}{8}\\) = 16 cm (d)<\/p>\n
\nArea of a rhombus whose diagonals are 8 cm and 6 cm is
\n(a) 48 cm2<\/sup>
\n(b) 24 cm2<\/sup>
\n(c) 12 cm2<\/sup>
\n(d) 96 cm2<\/sup>
\nSolution:
\nArea of rhombus =\\(\\frac{d_{1} \\times d_{2}}{2}=\\frac{8 \\times 6}{2}\\)
\n= \\(\\frac{48}{2}\\) = 24 cm2<\/sup> (b)<\/p>\n
\nIf the lengths of diagonals of a rhombus is doubled, then area of rhombus will be
\n(a) doubled
\n(b) tripled
\n(c) four times
\n(d) remains same
\nSolution:
\nArea of rhombus1<\/sub> = \\(\\frac{1}{2}\\) \u00d7 d1<\/sub> \u00d7 d2<\/sub>
\nNow the diagonals are doubled
\nArea of rhombus = \\(\\frac{1}{2}\\) \u00d7 2d1<\/sub> \u00d7 2d2<\/sub> = 2d1<\/sub>d2<\/sub>
\nwith doubled diagonals.
\nLengths of diagonals are doubled, then the area will be four times. (c)<\/p>\n
\nIf the length of a diagonal of a quadrilateral is 10 cm and lengths of the perpendiculars on it from opposite vertices are 4 cm and 6 cm, then area of quadrilateral is
\n(a) 100 cm2<\/sup>
\n(b) 200 cm2<\/sup>
\n(c) 50 cm2<\/sup>
\n(d) none of these
\nSolution:
\nLength of a diagonal of a quadrilateral = 10 cm
\nand lengths of perpendicular on it from the
\nopposite vertices = 4 cm and 6 cm
\n\u2234 Area = \\(\\frac{1}{2}\\) (4 + 6) \u00d7 10 cm2<\/sup>
\n= \\(\\frac{1}{2}\\) \u00d7 10 \u00d7 10 = 50 cm2<\/sup> (c)<\/p>\n
\nArea of a rhombus is 90 cm2<\/sup>. If the length of one diagonal is 10 cm then the length of other diagonal is
\n(a) 18 cm
\n(b) 9 cm
\n(c) 36 cm
\n(d) 4.5 cm
\nSolution:
\nArea of a rhombus = 90 cm2<\/sup>
\nLength of one diagonal = 10 cm
\nArea of rhombus = \\(\\frac{1}{2}\\) \u00d7 d1<\/sub> \u00d7 d2<\/sub>
\n\u2234 Length of second diagonal = \\(\\frac{A \\times 2}{d_{1}}\\)
\n= \\(\\frac{90 \\times 2}{10}\\) = 18cm (a)<\/p>\n
\nIf the volume of a cube is 729 cm3<\/sup>, then its surface area is
\n(a) 486 cm2<\/sup>
\n(b) 324 cm2<\/sup>
\n(c) 162 cm2<\/sup>
\n(d) none of these
\nSolution:
\nVolume of a cube = 729 cm3<\/sup>
\nV = (Side)3<\/sup>
\n\u2234 Side (edge) = \\(\\sqrt[3]{729}=\\sqrt[3]{9 \\times 9 \\times 9}\\) = 9 cm
\nThen surface area = 6 \u00d7 (side)2<\/sup>
\n= 6 \u00d7 (9)2<\/sup> = 6 \u00d7 81 cm2<\/sup>
\n= 486 cm2<\/sup> (a)<\/p>\n
\nIf the lateral surface area of a cube is 100 cm2<\/sup>, then its volume is
\n(a) 25 cm3<\/sup>
\n(b) 125 cm3<\/sup>
\n(c) 625 cm3<\/sup>
\n(d) none of these
\nSolution:
\nLateral surface area of a cube = 4(Edge)2<\/sup> = 100 cm2<\/sup>
\n\u2234 4 \u00d7 (edge)2<\/sup> = 100
\n\u21d2 (edge)2<\/sup> = \\(\\frac{100}{4}\\) = 25 = (5)2<\/sup>
\n\u2234 Edge of cube = 5 cm
\nVolume = (edge)3\u00a0<\/sup>= (5)3<\/sup> = 125 cm3<\/sup> (b)<\/p>\n
\nIf the length of side of a cube is doubled, then the ratio of volumes of new cube and original cube is
\n(a) 1 : 2
\n(b) 2 : 1
\n(c) 4 : 1
\n(d) 8 : 1
\nSolution:
\nLet original side of a cube = x
\nThen volume = x3<\/sup>
\nIf edge is doubled i.e. edge = 2x
\nThen volume = (2x)3<\/sup> = 8x3<\/sup>
\n\u2234 Ratio between new cube and original cube
\n= 8x3<\/sup> : x = 8 : 1 (d)<\/p>\n
\nIf the dimensions of a rectangular room are 10m\u00a0\u00d7 12m \u00d7 9m, then the cost of painting its four walls at the rate of \u20b98 per m2<\/sup> is
\n(a) \u20b93186
\n(b) \u20b93618
\n(c) \u20b93168
\n(d) none of these
\nSolution:
\nDimensions of a room = 10m\u00d7 12 \u00d7 9m
\nArea of 4 walls = 2(l + b)h
\n= 2(10 + 12) \u00d7 9 = 2 \u00d7 22 \u00d7 9 m2<\/sup>
\n= 396 cm2<\/sup>
\nCost of painting = \u20b98 per m2<\/sup>
\n\u2234 Total cost = 396 \u00d7 8 = \u20b93168 (c)<\/p>\n
\nVolume of a cylinder is 1848 cm2<\/sup>. If the diameter of its base is 14 cm, then the height of the cylinder is
\n(a) 12 cm
\n(b) 6 cm
\n(c) 3 cm
\n(d) none of these
\nSolution:
\nVolume of a cylinder = 1848 cm2<\/sup>
\nDiameter of base = 14 cm
\n\u2234 Radius (r) = \\(\\frac{14}{2}\\) = 7 cm
\nV = \u03c0r2<\/sup>h
\n\u2234 Height = \\(\\frac{\\mathrm{V}}{\\pi r^{2}}\\)
\n= \\(\\frac{1848 \\times 7}{22 \\times 7 \\times 7}\\) = 12 cm<\/p>\n
\nIf the radius of a cylinder is doubled and height is halved, then new volume is
\n(a) same
\n(b) 2 times
\n(c) 4 times
\n(d) 8 times
\nSolution:
\nLet radius = r
\nand height = h
\nThen volume = \u03c0r2<\/sup>h
\nIf radius is doubled i. e. 2 r and height is halved
\ni.e. \\(\\frac{h}{2}\\), then
\nVolume = \u03c0(2r)2<\/sup> \u00d7 \\(\\frac{h}{2}\\)
\n= \u03c0 \u00d7 4r2<\/sup> \u00d7 \\(\\frac{h}{2}\\) = 2\u03c0r2<\/sup>h
\n\u2234 Its volume is doubled (2 times) (b)<\/p>\n
\nQuestion 1.
\nPulkit painted four walls and roof of a rectangular room of size 10m \u00d7 12m \u00d7 12m. He got \u20b910 per m2<\/sup> for his work. How much money he earned? He always give one fourth of his income to an orphanage. Find how much money he gave to orphanage? What values are being promoted?
\nSolution:
\nDimensions of a room = 10m \u00d7 12m \u00d7 10m
\n\u2234 Area of 4-walls = 2(l + b) \u00d7 h
\n= 2(10 + 12) \u00d7 10 m2<\/sup>
\n= 2 \u00d7 22 \u00d7 10 = 440 m3<\/sup>
\nand area of cielings = l x b = 10 \u00d7 12 = 120 m2<\/sup>
\nTotal area = 440 + 120 = 560 m2<\/sup>
\nRate of painting = \u20b910 per m2<\/sup>
\nTotal changes for painting = \u20b9560 \u00d7 10 = \u20b95600
\nMoney gave to an orphanage = \\(\\frac{1}{4}\\) of \u20b95600 = \u20b91400
\nRemaining money = \u20b95600 – 1400 = \u20b94200
\nAmount given to an orphanage is a good and noble deed.
\nHe help the poor and needy.<\/p>\n
\nIn a slogan writing competition in a school, Rama wrote the slogan ‘Truth pays, never betrays’ on a trapezium shaped cardboard. If the lengths of parallel sides of trapezium are 60 cm and 80 cm and the distance between them is 50 cm, find the area of trapezium. What are the advantages of speaking truth?
\nSolution:
\nParallel sides of a trapezium = 60 cm and 80 cm
\nand distance between then = 50 cm
\n\u2234 Area of trapezium = \\(\\frac{1}{2}\\) (sum of parallel sides) \u00d7 height
\n= \\(\\frac{1}{2}\\)(60 + 80) \u00d7 50
\n= \\(\\frac{1}{2}\\) \u00d7 140 \u00d7 50 cm3<\/sup>
\n= 3500 cm3<\/sup>
\nRama wrote on it a slogan: Truth pays, never betrays’
\nAlways speek the truth. It pays in the long run on speaking truth,
\npeople will believe you, as truth is like a God.<\/p>\n
\nQuestion 1.
\nThe length of a room is 50% more than its breadth. The cost of carpeting the room at the rate of \u20b938.50 m2<\/sup> is \u20b9924 and the cost of papering the walls at \u20b93.30 m2<\/sup> is \u20b9214.50. If the room has one door of dimensions 1 m \u00d7 2 m and two windows each of dimensions 1 m \u00d7 1.5 m, find the dimensions of the room.
\nSolution:
\nLength of a room is 50% more than its breadth
\nLet breadth (b) = xm
\nThen length (l) = x + \\(\\frac{150}{100}=\\frac{3}{2} x \\mathrm{m}\\)
\nCost of carpeting the room at the rate of \u20b938.50 = \u20b9924
\nArea of floor = \u20b9 \\(\\frac{924}{38.50}\\)
\n= \\(\\frac{924 \\times 100}{3850}\\) = 24 m2<\/sup>
\n\u2234 l \u00d7 b = 24 m2<\/sup> …(i)
\nCost of papering the walls at the rate of \u20b933.30 per m2<\/sup> = \u20b93214.50
\n\u2234 Area of paper \\(\\frac{214.50}{3.30}=\\frac{21450}{330}=65 \\mathrm{m}^{2}\\)
\nArea of one door of dimension 1 m \u00d7 2 m = 2 m2<\/sup>
\nand area of two windows of size
\n= 1 \u00d7 1.5 m = 1 \u00d7 1.5 \u00d7 2 = 3 m2<\/sup>
\n\u2234 Area of 4-walls = 65 + 2 + 3 = 70 m2<\/sup>
\nNow, l \u00d7 b = 24 \u21d2 \\(\\frac{3}{2}\\)x \u00d7 x = 24
\n\u21d2 x2<\/sup> = \\(\\frac{24 \\times 2}{3}\\) = 16 = (4)2<\/sup>
\n\u2234 x = 4
\n\u2234 Length = 4 \u00d7 \\(\\frac{3}{2}\\) = 6 m
\nand breadth = x = 4 m
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-18-Mensuration-Objective-Type-Questions-hots-Q1.1.png\")
\n\u2234 Length = 6 m, breadth = 4 m and height = 3.5 m<\/p>\n
\nThe barrel of a fountain pen, cylindrical in shape, is 7 cm long and 5 mm in diameter. A full barrel of ink in the pen will be used up when writing 310 words on an average. How many words would use up a bottle of ink containing one-fifth of a litre? Answer correct to the nearest 100 words.
\nSolution:
\nHeight of cylindrical shaped barrel (h) = 7 cm
\nDiameter = 5 mm
\n\u2234 Radius (r) = \\(\\frac{5}{2}\\) mm
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-18-Mensuration-Objective-Type-Questions-hots-Q2.1.png\")
\nOne-fifth of litre = 200 ml
\n\u2234 In 200 ml, words will be written
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-18-Mensuration-Objective-Type-Questions-hots-Q2.2.png\")
<\/p>\n
\nA cylindrical jar is 20 cm high with internal diameter 7 cm. An iron cube of edge 5 cm is immersed in the jar completely in the water which was originally 12 cm high. Find the rise in the level of water.
\nSolution:
\nHeight of cylindrical jar = 20 cm
\nand diameter = 7 cm
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-18-Mensuration-Objective-Type-Questions-hots-Q3.1.png\")
<\/p>\n
\nSquares each of side 6 cm are cut off from the four comers of a sheet of tin measuring 42 cm by 30 cm. The remaining portion of the tin sheet is made into an open box by folding up the flaps. Find the capacity of the box.
\nSolution:
\nFrom a sheet, squares of 6 cm sides are cut and cutout
\nLength of sheet = 32 cm
\nand breadth = 30 cm
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-18-Mensuration-Objective-Type-Questions-hots-Q4.1.png\")
\nRemaining sheet is folded into a box whose length
\n= 42 – 6 \u00d7 2 = 42 – 12 = 30 cm
\nBreadth = 30 – 6 \u00d7 2 = 30 – 12 = 18 cm
\nand height = 6 cm
\nCapacity of the box = 30 \u00d7 18 \u00d7 6 = 3240 cm3<\/sup><\/p>\nML Aggarwal Class 8 Solutions for ICSE Maths<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"