{"id":47055,"date":"2022-05-31T23:30:10","date_gmt":"2022-05-31T18:00:10","guid":{"rendered":"https:\/\/cbselibrary.com\/?p=47055"},"modified":"2023-01-25T11:18:38","modified_gmt":"2023-01-25T05:48:38","slug":"ml-aggarwal-class-9-solutions-for-icse-maths-chapter-16-chapter-test","status":"publish","type":"post","link":"https:\/\/cbselibrary.com\/ml-aggarwal-class-9-solutions-for-icse-maths-chapter-16-chapter-test\/","title":{"rendered":"ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 16 Mensuration Chapter Test"},"content":{"rendered":"

ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 16 Mensuration Chapter Test<\/h2>\n

Question 1.
\n(a) Calculate the area of the shaded region.
\n\"ML
\n(b) If the sides of a square are lengthened by 3 cm, the area becomes 121 cm2<\/sup>. Find the perimeter of the original square.
\nAnswer:
\n(a) In the figure,
\nOA \u22a5 BC
\nAC = 15 cm, AO = 12 cm, BO = 5 cm,
\nBC = 14 cm
\n\u2234 OC = BC – BO = 14 – 5 = 9 cm
\nArea of right \u2206AOC
\n= \\(\\frac{1}{2}\\) base \u00d7 altitude
\n= \\(\\frac{1}{2}\\) \u00d7 9 \u00d7 12 cm2<\/sup> = 54 cm2<\/sup><\/p>\n

(b) Let the side of original square = x cm
\nThen length of given square = (x + 3) cm
\nArea = side \u00d7 side
\n\u21d2 121 = (x + 3)(x + 3)
\n\u21d2 (11)2<\/sup> = (x + 3)2<\/sup>
\n\u21d2 11 = x + 3 \u21d2 x + 3 = 11
\n\u21d2 x = 11 – 3 cm \u21d2 x = 8 cm<\/p>\n

Question P.Q.
\nThe given figure shows a kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and side 6 cm each. How much paper is used in making the kite ? Ignore the wastage of the paper is making the kite.
\n\"ML
\nAnswer:
\nLength of each diagonal of the square ABCD = 32 cm
\ni.e. AC = BD = 32 cm
\nBase EF of isosceles \u2206BEF = 8 cm
\nand each side = 6 cm
\nDraw BL \u22a5 EF
\nArea of square = \\(\\frac{(\\text { Diagonal })^{2}}{2}\\)
\n= \\(\\frac{(32)^{2}}{2}=\\frac{1024}{2} \\mathrm{cm}^{2}\\) = 512 cm2<\/sup>
\nIn \u2206BEL, \u2220L = 90\u00b0
\nBL2<\/sup> = BE2<\/sup> – EL2<\/sup> (Pythagoras Theorem)
\n= (6)2<\/sup> – (4)2<\/sup> = 36 – 16 = 20
\n\u2234BL = \\( \\sqrt{{20}} \\) = \\(\\sqrt{4 \\times 5}\\) = 2\\( \\sqrt{{5}} \\) cm
\nArea of isosceles triangle = \\(\\frac{1}{2}\\) Base \u00d7 alt.
\n= \\(\\frac{1}{2}\\) \u00d7 8 \u00d7 2\\( \\sqrt{{5}} \\) cm2<\/sup>
\n= 8\\( \\sqrt{{5}} \\) cm2<\/sup>
\n\u2234 Total area of the kite = (512 + 8\\( \\sqrt{{5}} \\)) cm2<\/sup>
\n= 512 + 8 (2.236) cm2<\/sup>
\n= 512 + 17.89 = 529.89 cm2<\/sup><\/p>\n

Question 2.
\n(a) Find the area enclosed by the figure (i) given below. All measurements are in centimetres:
\n(b) Find the area of the quadrilateral ABCD shown in figure (ii) given below. All measurements are in centimetres.
\n(c) Calculate the area of the shaded region shown in figure (iii) given below. All measurements are in metres.
\n\"ML
\nAnswer:
\n(a) Area of figure (i) = Area of ABCD – Area of both triangles
\n\"ML
\n= (9 \u00d7 9) – \\(\\left(\\frac{1}{2} \\times 5 \\times 6\\right)\\) \u00d7 2 cm2<\/sup>
\n= (81 – 15 \u00d7 2)
\n= (81 – 30) cm2<\/sup> = 51 cm2<\/sup><\/p>\n

(b) In \u2206 ABD By Pythagoras theorem,
\n\u21d2 BD2<\/sup> = AB2<\/sup> + AD2\u00a0<\/sup>\u21d2 BD2<\/sup> =(6)2<\/sup> + (8)2<\/sup>
\n\u21d2 BD2<\/sup> = 36 + 64 \u21d2 BD2<\/sup> = 100 \u21d2 BD = 10cm
\nIn \u2206BCD
\nBy Pythagoras theorem
\n\"ML
\nBC2<\/sup> = BD2<\/sup> + CD2<\/sup>
\n\u21d2 (26)2<\/sup> = (10)2<\/sup> + (CD)2<\/sup>
\n\u21d2 676 = 100 + CD2<\/sup>
\n\u21d2 CD2<\/sup> + 100 = 676 \u21d2 CD2<\/sup> = 676 – 100
\n\u21d2 CD2<\/sup> = 576 \u21d2 CD = \\( \\sqrt{{576}} \\) cm \u21d2 CD = 24 cm
\nArea of given fig.= Area of \u2206 ABD + Area of \u2206 BCD
\n= \\(\\frac{1}{2}\\) \u00d7 Base \u00d7 height + \\(\\frac{1}{2}\\) \u00d7 Base \u00d7 height
\n= \\(\\frac{1}{2}\\) \u00d7 AB \u00d7 AD + \\(\\frac{1}{2}\\) \u00d7 CD \u00d7 BD
\n= [\\(\\frac{1}{2}\\) \u00d7 6 \u00d7 8 + \\(\\frac{1}{2}\\) \u00d7 24 \u00d7 10]cm2<\/sup> = [3 \u00d7 8 + 12 \u00d7 10] cm2<\/sup>
\n= (24 + 120) cm2<\/sup> = 144 cm2<\/sup><\/p>\n

(c) Area of figure (iii) = Area of ABCD – (Area of Jst part + Area of 2nd part + Area of 3rd part)
\n\"ML
\n\"ML
\nHence, required area of given figure = 54 m2<\/sup><\/p>\n

Question 3.
\nAsifa cut an aeroplane from a coloured chart paper (as shown in the adjoining figure). Find the total area of the chart paper used, correct to 1 decimal place.
\nAnswer:
\nThe whole figure of an aeroplane consists 5 figures, three triangles, one rectangle and one trapezium
\nTake midpoint M of AB
\n\u2234 AM = MB = 1 cm
\n\"ML
\n\"ML
\nJoin MN and CN
\nTheir \u2206 AMN, \u2206NCB and \u2206 MNC are equilateral traingles having 1 cm side each
\nNow area of \u2206 GHF
\n\"ML
\n\"ML<\/p>\n

Question 4.
\nIf the area of a circle is 78.5 cm2<\/sup>, find its circumference. (Take \u03c0 = 3.14)
\nAnswer:
\nArea of a circle = 78.5 cm2<\/sup>
\nLet r be the radius
\n\u2234 r2<\/sup> = \\(\\frac{\\text { Area }}{\\pi}=\\frac{78.50}{3.14}\\) = 25 = (5)2<\/sup>
\n\u2234 r = 5 cm
\nNow circumference = 2\u03c0r
\n= 2 \u00d7 3.14 \u00d7 5 cm = 31.4 cm<\/p>\n

Question 5.
\nFrom a square cardboard, a circle of biggest area was cut out. If the area of the circle is 154 cm2<\/sup>, calculate the original area of the cardboard.
\nAnswer:
\nArea of circle cut out from the square board = 154 cm2<\/sup>
\nLet r be the radius
\n\"ML
\n\u21d2 r2<\/sup> = \\(\\frac{154 \\times 7}{22}\\) = 49 = (7)2<\/sup>
\n\u21d2 r = 7 cm
\nNow side of the square = 7 \u00d7 2 = 14 cm
\n\u2234 Area of the original cardboard
\n= a2<\/sup> = (14)2<\/sup> = 196 cm2<\/sup><\/p>\n

Question 6.
\n(a) From a sheet of paper of dimensions = 2m \u00d7 1\u00b75m, how many circles can you cut of radius 5cm. Also find the area of the paper wasted. Take \u03c0 = 3\u00b714.
\n(b) If the diameter of a semicircular protractor is 14cm, then find its perimeter.
\nAnswer:
\nLength of sheet of paper = 2m = 200cm
\nBreadth of sheet = 1\u00b75 m = 150 cm
\n\"ML
\nArea = l \u00d7 b = 200 \u00d7 150 cm2<\/sup>
\nRadius of circle = 5cm.
\n\u2234 No. of circles in lengthwise
\n= \\(\\frac{200}{5 \\times 2}\\) = 20
\nand widthwise = \\(\\frac{150}{10}\\) = 15
\n\u2234 No. of circles = 20 \u00d7 15 = 300
\nArea of one circle = \u03c0r2<\/sup>
\n= 3\u00b714 \u00d7 5 \u00d7 5 cm2<\/sup>
\nArea of 300 circles
\n= 300 \u00d7 \\(\\frac{314}{100}\\) \u00d7 25 cm2<\/sup> = 23550 cm2<\/sup>
\n\u2234 Area of the remaining portion = Area of square – area of 300 circles
\n= (30000 – 23550) cm2<\/sup>
\n= 6450 cm2<\/sup>
\n(b)Diameter of semicircular protractor = 14cm.
\n\"ML
\n\u2234 Its perimeter = \\(\\frac{1}{2}\\)\u03c0d + d
\n= \\(\\frac{1}{2}\\) \u00d7 \\(\\frac{22}{7}\\) \u00d7 14 + 14 = 22 + 14 = 36 cm<\/p>\n

Question 7.
\nA road 3.5 m wide surrounds a circular park whose circumference is 88 m. Find the cost of paving the road at the rate of Rs. 60 per square metre.
\nAnswer:
\nWidth of the road = 3.5 m
\nCircumference of the circular park = 88 m
\nLet r be the radius of the park
\n\u2234 2\u03c0r = 88
\n\"ML
\n= \\(\\frac{22}{7}\\) \u00d7 3.15 \u00d7 3.5 m2<\/sup> = 346.5 m2<\/sup>
\nRate of paving the road = Rs. 60 per m2<\/sup>
\n\u2234 Total cost = Rs. 60 \u00d7 346.5
\n= Rs. 20790<\/p>\n

Question 8.
\nThe adjoining sketch shows a running tract 3.5 m wide all around which consists of two straight paths and two semicircular rings. Find the area of the track.
\n\"ML
\nAnswer:
\nWidth of track = 3.5 m.
\nInner length of rectangular base = 140 m
\nand width = 42 m
\nOuter length = 140 + 2 \u00d7 3.5 = 140 + 7
\n= 147 m
\nand width = 42 + 2 \u00d7 3.5 = 42 + 7 = 49 m
\nRadius of inner semicircle (r) = \\(\\frac{42}{2}\\) = 21m
\nand outer radius (R) = 21 + 3.5 = 24.5 m
\nNow area of track = 2(140 \u00d7 3.5) + 2 \u00d7 \\(\\frac{1}{2}\\)\u03c0 (R2<\/sup> – r2<\/sup>)
\n\"ML<\/p>\n

Question 9.
\nIn the adjoining figure, O is the centre of a circular arc and AOB is a line segment. Find the perimeter and the area of the shaded region correct to one decimal place. (Take \u03c0 = 3.142)
\n\"ML
\nAnswer:
\nIn a semicircle, \u2220ACB = 90\u00b0
\n\u2234 \u2206 ABC is a right angled triangle
\nNow AB2<\/sup> = AC2<\/sup> + BC2<\/sup> (Pythagoras Theorem)
\n= 122<\/sup> + 162<\/sup>
\n= 144 + 256 = 400
\n= (20)2<\/sup>
\n\u2234 AB = 20 cm
\n\u2234 Radius of semicircle = \\(\\frac{20}{2}\\) = 10 cm<\/p>\n

(i) Area of shaded portion
\n= Area of semicircle – area of \u2206 ABC
\n\"ML<\/p>\n

(ii) Perimeter of shaded portion
\n= circumference of semicircle + AC + BC
\n= \u03c0r + 12 + 16 = 3.142 \u00d7 10 + 28
\n= 31.42 + 28 = 59.42 cm = 59.4 cm<\/p>\n

Question 10.
\n(a) In the figure (i) given below, the radius is 3\u00b75 cm. Find the perimeter of the quarter of the circle.
\n(b) In the figure (ii) given below, there are five squares each of side 2 cm.
\n(i) Find the radius of the circle.
\n(ii) Find the area of the shaded region. (Take \u03c0 = 3.14).
\n\"ML
\nSolution:
\n(a) Radius of quadrant = 3\u00b75 cm
\n\"ML
\nArea of 5 square of side 2 cm each
\n= (2)2<\/sup> \u00d7 5 = 4 \u00d7 5 = 20 cm2<\/sup>
\n\u2234 Area of shaded portion = 31.4 – 20
\n= 11.4 cm2<\/sup><\/p>\n

Question 11.
\n(a) In the figure (i) given below, a piece of cardboard in the shape of a quadrant of a circle of radius 7 cm is bounded by perpendicular radii OX and OY. Points A and B lie on OX and OY respectively such that OA = 3 cm and OB = 4 cm. The triangular part OAB is removed. Calculate the area and the perimeter of the remaining piece.
\n\"ML
\n(b) In the figure (ii) given below, ABCD is a square. Points A, B, C and D are centres of quadrants of circles of the same radius. If the area of the shaded portion is \\(21 \\frac{3}{7}\\) cm2<\/sup>, find the radius of the quadrants. Take \u03c0 = \\(\\frac{22}{7}\\).
\nSolution:
\n(a) Radius of quadrant = 7 cm
\nOA = 3 cm, OB = 4 cm
\n\u2234 AX = 7 – 3 = 4 cm and
\nBY = 7 – 4 = 3 cm
\n\u2234 AB2<\/sup> = OA2<\/sup> + OB2<\/sup> = (3)2<\/sup> + (4)2<\/sup> = 9 + 16 = 25
\n\u21d2 AB = \\( \\sqrt{{25}} \\) = 5cm
\nNow (i) Area of shaded portion
\n\"ML
\n(ii) Perimeter of shaded portion
\n= \\(\\frac{1}{4}\\)(2\u03c0r) + AX + BY + AB
\n= \\(\\frac{1}{2}\\) \u00d7 \\(\\frac{22}{7}\\) \u00d7 7 + 4 + 3 + 5
\n= 11 + 12 = 23 cm.<\/p>\n

(b) ABCD is a square and with centres A, B, C and D quadrants are drawn.
\nLet side of square = a
\n\u2234 Radius of each quadrant = \\(\\frac{a}{2}\\)
\n\u2234 Area of shaded portion
\n\"ML<\/p>\n

Question 12.
\nIn the adjoining figure, ABC is a right angled triangle right angled at B. Semicircle are drawn on AB, BC and CA as diameter. Show that the sum of areas of semi circles drawn on AB and BC as diameter is equal to the area of the semicircle drawn on CA as diameter.
\n\"ML
\nAnswer:
\n\u2206 ABC is a right angled triangle right angled at B
\n\u2234 AC2<\/sup> = AB2<\/sup> + BC2<\/sup> …(i) (Pythagoras theorem)
\nNow area of semicircle on AC as diameter
\n\"ML
\n\"ML
\n\"ML<\/p>\n

Question 13.
\nThe length of minute hand of a clock is 14 cm. Find the area swept by the minute hand in 15 minutes.
\nSolution:
\nRadius of hand = 14 cm
\n\u2234 Area swept in 15 minutes
\n\"ML<\/p>\n

Question 14.
\nFind the radius of a circle if a 90\u00b0 arc has a length of 3\u00b75 \u03c0 cm. Hence, find the area of sector formed by this arc.
\nSolution:
\nLength of arc of the sector of a circle = 3\u00b75 \u03c0 cm
\nand angle at the centre = 90\u00b0
\n\"ML<\/p>\n

Question 15.
\nA cube whose each edge is 28 cm long has a circle of maximum radius on each of its face painted red. Find the total area of the unpainted surface of the cube.
\nSolution:
\nEdge of cube = 28 cm
\n\u2234 Surface area = 6 a2<\/sup> = 6 \u00d7 (28)2<\/sup> cm2<\/sup>
\n= 6 \u00d7 28 \u00d7 28 = 4704 cm2<\/sup>
\nNow diameter of each circle = 28 cm
\n\u2234 Radius = \\(\\frac{28}{2}\\) = 14 cm
\n\u2234 Area of each circle
\n= \u03c0r2<\/sup> = \\(\\frac{22}{7}\\) \u00d7 14 \u00d7 14cm = 616 cm2<\/sup>
\nand area of such 6 circles drawn on 6 faces of cube = 616 \u00d7 6 = 3696 cm2<\/sup>
\n\u2234 Area of remaining portion of the cube
\n= 4704 – 3696 = 1008 cm2<\/sup><\/p>\n

Question 16.
\nCan a pole 6.5 m long fit into the body of a truck with internal dimensions of 3.5m, 3 m and 4m?
\nAnswer:
\nNo,
\nBecause length of pole = 6.5 m
\nBut internal dimensions of truck are 3.5 m, 3 m and 4 m all of these dimensions are less than that of 6.5 m. So that pole cannot fit into the body of truck with given dimensions.<\/p>\n

Question 17.
\nA car has a petrol tank 40 cm long, 28 cm wide and 25 cm deep. If the fuel consumption of the car averages 13.5 km per litre, how far can the car travel with a full tank of petrol ?
\nAnswer:
\nCapacity of car tank = 40cm \u00d7 28cm \u00d7 25cm
\n= (40 \u00d7 28 \u00d7 25) cm3<\/sup> = \\(\\frac{40 \\times 28 \\times 25}{1000}\\) litre (\u2235 1000cm3<\/sup> = 1 litre)
\nAverage of car = 13.5 km per litre
\nThen, distance travelled by car
\n= \\(\\frac{40 \\times 28 \\times 25}{1000}\\) \u00d7 135km
\n\"ML
\nHence, The car can travel 378 km with a full tank of petrol.<\/p>\n

Question 18.
\nAn aquarium took 96 minutes to completely fill with water. Water was filling the aquarium at a rate of 25 litres every 2 minutes. Given that the aquarium was 2 m long and 80 cm wide, compute the height of the aquarium.
\nAnswer:
\nWater fill in 2 minutes = 25 litres
\nWater fill in 1 minutes = \\(\\frac{25}{2}\\)
\nWater fill in 96 minutes = \\(\\frac{25}{2}\\) \u00d7 96 litres
\n= 25 \u00d7 48 litres = 1200 litres
\ni.e. Capacity of aquarium = 1200 litres ……(1)
\nBut, Length of aquarium = 2m = 2 \u00d7 100 cm = 200 cm
\nBreadth of aquarium = 80 cm
\nLet height of aquarium = h cm
\nThen, capacity of aquarium = 200 \u00d7 80 \u00d7 h cm3<\/sup>
\n= \\(\\frac{200 \\times 80 \\times h}{1000}\\) litre = \\(\\frac{1}{5}\\) \u00d7 80 \u00d7 h litre
\n= 16 h litre
\nFrom (1) and (2)
\n16h = 1200 \u21d2 h = \\(\\frac{1200}{16}\\) cm \u21d2 h = 75 cm
\nHence, height of aquarium = 75 cm<\/p>\n

Question 19.
\nThe lateral surface area of a cubiod is 224 cm2<\/sup>. Its height is 7 cm and the base is a square. Find :
\n(i) a side of the square, and
\n(ii) the volume of the cubiod.
\nAnswer:
\nGiven that lateral surface Area of a cubiod = 224 cm2<\/sup>
\nHeight of cubiod = 7 cm
\nAlso base is square
\nLet length of cubiod = x cm
\nThen Breadth of cubiod = x cm (\u2235 Base is square so length and breadth are same)
\nLateral Surface Area = 2 (l + b) \u00d7 h
\n\u21d2 224 = 2 (x + x) \u00d7 7
\n\u21d2 224 = 2 \u00d7 2x \u00d7 7 \u21d2 224 = 28x
\n\u21d2 28x = 224 \u21d2 x = \\(\\frac{224}{28}\\) cm = 8 cm
\n(i) Hence, side of the square = 8 cm
\n(ii) volume of the cuboid = = l \u00d7 b \u00d7 h
\n= 8 \u00d7 8 \u00d7 7 cm3<\/sup> = 448 cm3<\/sup><\/p>\n

Question 20.
\nIf the volume of a cube is V m3<\/sup>, its surface area is S m2<\/sup> and the length of a diagonal is d metres, prove that 6\\( \\sqrt{{3}} \\) V = S d.
\nSolution:
\nVolume of cube = (V) = (Side)3<\/sup>
\nLet a be the side of the cube, then
\nV = a3<\/sup> and S = 6a2<\/sup>
\nDiagonal (d) = \\( \\sqrt{{3}} \\). a.
\nNow Sd = 6a2<\/sup> \u00d7 \\( \\sqrt{{3}} \\) a = 6\\( \\sqrt{{3}} \\) a3<\/sup>
\n= 6\\( \\sqrt{{3}} \\) V (\u2235 V = a3<\/sup>)
\nHence 6\\( \\sqrt{{3}} \\) V = Sd.<\/p>\n

Question 21.
\nThe adjoining figure shows a victory stand, each face is rectangular. All measurement are in centimetres. Find its volume and surface area (the bottom of the stand is open).
\n\"ML
\nSolution:
\nIn the figure, it has three parts as indicated by 3, 1 and 2.
\n\u2234 Volume of part (3) = 50 \u00d7 40 \u00d7 12 cm3<\/sup> = 24000 cm3<\/sup>
\nVolume of part (1) = 50 \u00d7 40 \u00d7 (16 + 24) cm3<\/sup>
\n= 50 \u00d7 40 \u00d7 40 cm3<\/sup> = 80000 cm3<\/sup>
\nand volume of part (2) = 50 \u00d7 40 \u00d7 24 cm3<\/sup>
\n= 48000 cm3<\/sup>
\nTotal volume = (24000 + 8000 + 48000) cm3<\/sup> = 153000 cm3<\/sup>
\nNow total surface area = Area of front and back + area of vertical faces + area of top faces
\n= 2 (50 \u00d7 12 + 50 \u00d7 40 + 50 \u00d7 24) cm2<\/sup> + (12 \u00d7 40 + 28 \u00d7 40 + 16 \u00d7 40 + 24 \u00d7 40) cm2<\/sup> + 3 (50 \u00d7 40) cm2<\/sup>
\n= 2 (600 + 2000 + 1200)cm2<\/sup> + (480 + 1120 + 640 + 960) cm2<\/sup> + 3 \u00d7 2000 cm2<\/sup>
\n= 2 (3800) + 3200 + 6000 cm2<\/sup>
\n= 7600 + 3200 + 6000 = 16800 cm2<\/sup><\/p>\n

Question 22.
\nThe external dimensions of an open rectangular wooden box are 98 cm by 84 cm by 77 cm. If the wood is 2 cm thick all around, find :
\n(i) the capacity of the box
\n(ii) the volume of the wood used in making the box, and
\n(iii) the weight of the box in kilograms correct to one decimal place, given that 1 cm3 of wood weighs 0.8 gm.
\nAnswer:
\nGiven that external dimensions of open rectangular wooden box = 98 cm, 84 cm, and 77 cm.
\nThickness = 2 cm
\nThen internal dimensions of open rectangular wooden box (98 – 2 \u00d7 2 )cm, (84 – 2 \u00d7 2) cm (77 – 2) cm
\n= (98 – 4) cm, (84 – 4) cm, 75 cm = 94 cm, 80 cm, 75 cm
\n(i) Capacity of the box = 94cm \u00d7 80cm \u00d7 75cm = 564000 cm3<\/sup>
\n(ii) Internal volume of box = 564000 cm3<\/sup>
\nExternal volume of box = 98cm \u00d7 84cm \u00d7 77 cm = 633864 cm3<\/sup>
\nVolume of wood used in making the
\nBox = 633864 cm3<\/sup> – 564000 cm3<\/sup>3 = 69864 cm3<\/sup>
\n(iii) Weight of 1 cm3<\/sup> wood = 0.8 gm
\nWeight of 69864 cm3<\/sup> wood = 0. 8 \u00d7 69864 gm
\n= \\(\\frac{0.8 \\times 69864}{1000} \\mathrm{kg}=\\frac{55891.2}{1000} \\mathrm{kg}\\)
\n= 55.9 kg (Correct to one decimal)<\/p>\n

Question 23.
\nA cuboidal block of metal has dimensions 36 cm by 32 cm by 0.25 m. It is melted and recast into cubes with an edge of 4 cm.
\n(i) How many such cubes can be made ?
\n(ii) What is the cost of silver coating the surfaces of the cubes at the rate of Rs. 1.25 per square centimetre ?
\nAnswer:
\nGiven, dimensions of cuboidal block are 36 cm, 32 cm, 0.25 m.
\nVolume of cuboidal block = 36 cm \u00d7 32 cm \u00d7 0.25 m
\n= 36 cm \u00d7 32 cm \u00d7 (0.25 \u00d7 100) cm = (36 \u00d7 32 \u00d7 25) cm3<\/sup>
\nVolume of cube having edge is 4 cm
\n= 4 cm \u00d7 4 cm \u00d7 4 cm = 64 cm3<\/sup>
\n(i) Number of cubes
\n\"ML
\n(ii) Total surface area of one cube
\n= 6 (a)2<\/sup> = 6 (4)2<\/sup> cm2<\/sup> = 6 \u00d7 4 \u00d7 4 cm2<\/sup> = 96 cm2<\/sup>
\nTotal surface area of 450 cube = 450 \u00d7 96 cm2<\/sup> = 43200 cm2<\/sup>
\nCost of silver coating the surface for 1 cm2<\/sup> = Rs. 1.25
\ncost of silver coating the surface for 43200 cm2<\/sup>
\n= 43200 \u00d7 1.25 = Rs. 54000<\/p>\n

Question 24.
\nThree cubes of silver with edges 3 cm, 4 cm and 5 cm are melted and recast into a single cube. Find the cost of coating the surface of the new cube with gold at the rate of Rs. 3.50 per square centimetre.
\nSolution:
\nVolume of First cube = (edge)3<\/sup>
\n= (3 cm)3<\/sup> = 3 \u00d7 3 \u00d7 3 cm3<\/sup> = 27 cm3<\/sup>
\nVolume of second cube = (edge)3<\/sup>
\n= (4 cm)3<\/sup> = 4 \u00d7 4 \u00d7 4 cm3<\/sup> = 64 cm3<\/sup>
\nVolume of third cube = (edge)3<\/sup>
\n= (5 cm)3<\/sup> = 5 \u00d7 5 \u00d7 5 cm3<\/sup> = 125 cm3<\/sup>
\nTotal volume = (27 + 64 + 125) cm3<\/sup> = 216 cm3<\/sup>
\nMade new cube whose volume = 216 cm3<\/sup>
\n(edge)3<\/sup> = 216 cm3<\/sup> \u21d2 (edge)3<\/sup> = (6cm)3<\/sup>
\n\u21d2 edge = 6 cm.
\nSurface area of new cube = 6 (edge)2<\/sup>
\n= 6 (6)2<\/sup> cm2<\/sup> = 6 \u00d7 6 \u00d7 6 cm2<\/sup> = 216 cm2<\/sup>
\nCost of coating the surface for 1 cm2<\/sup>
\n= Rs. 3.50
\nCost of coating the surface for 216 cm2<\/sup>
\n= Rs. 3. 50 \u00d7 216 = Rs. 756<\/p>\n

ML Aggarwal Class 9 Solutions for ICSE Maths<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"

ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 16 Mensuration Chapter Test Question 1. (a) Calculate the area of the shaded region. (b) If the sides of a square are lengthened by 3 cm, the area becomes 121 cm2. Find the perimeter of the original square. Answer: (a) In the figure, OA \u22a5 BC … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[3034],"tags":[],"yoast_head":"\nML Aggarwal Class 9 Solutions for ICSE Maths Chapter 16 Mensuration Chapter Test - CBSE Library<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/cbselibrary.com\/ml-aggarwal-class-9-solutions-for-icse-maths-chapter-16-chapter-test\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 16 Mensuration Chapter Test\" \/>\n<meta property=\"og:description\" content=\"ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 16 Mensuration Chapter Test Question 1. 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