Description : Find the length of the longest pole that can be put in a room of dimensions 10 m x 10 m x 5 m. -Maths 9th
Last Answer : Here, we have a cuboid with dimensions l = length = 10 m, b = breadth = 10 m and h = height = 5 m Now, length of longest pole = diagonal of cuboid ∴ Required length = √(l2 + b2 + h2) = √(100 + 100 + 25) = √225 = 15 cm
Description : The length of the longest pole that can be put in a room of dimensions (10m x 10m x 5m) is -Maths 9th
Last Answer : longest pole can be faced along the diagonal and its length =102+102+52=225=15m
Last Answer : Solution of this question
Description : A city has a park shaped as a right angled triangle. The length of the longest side of this park is 80 m. The Mayor of the city -Maths 9th
Last Answer : answer:
Description : If a wooden box of dimensions 8 m x 7 m x 6 m is to carry boxes of dimensions 8 cm x 7 cm x 6 cm, then find the maximum number of boxes that can be carried in the wooden box. -Maths 9th
Last Answer : Volume of wooden box = 800 cm × 700 cm × 600 cm Volume of box = 8 cm × 7 cm × 6 cm ∴ Number of boxes = volume of wooden box / volume of each box = 800 cm × 700 cm × 600 cm / 8 cm × 7 cm × 6 cm = 1000000
Description : The number of planks of dimensions (4 m x 50cm x 20cm) that can be stored in a pit which is 16 m long, 12 m wide and 40 m deep is -Maths 9th
Last Answer : Length of the plank=4m=400cm Breadth=50cm Height=20cm Volume of the plank=L*B*H =400*50*20 =400000cm^3 Length of the pit=16m=1600cm Breadth=12m=1200cm Height=4m=400cm Volume of the pit= L ... *1200*400 =768000000cm^3 Number of planks that can be fitted= 768000000/400000 =1920 planks is the answer.
Description : The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of ₹7.50 m². -Maths 9th
Last Answer : Length of a room (l) = 5m Breadth (b) = 4 m and height (h) = 3 m ∴ Area of 4 walls = 2(l + b) x h = 2(5 + 4) x 3 = 6 x 9 = 54 m² and area of ceiling = l x b = 5 x ... ∴ Total area = 54 + 20 = 74 m2 Rate of white washing = 7.50 per m² ∴ Total cost = ₹74 x 7.50 = ₹555
Description : The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and ceiling at the rate of Rs 7.50 per m2. -Maths 9th
Last Answer : Length (l) of room = 5m Breadth (b) of room = 4m Height (h) of room = 3m It can be observed that four walls and the ceiling of the room are to be white washed. Total area to be white washed = Area of walls + ... m2 area = Rs.7.50 (Given) Cost of white washing 74 m2 area = Rs. (74 7.50) = Rs. 555
Description : The length, breadth and height of a room are 5 m, -Maths 9th
Last Answer : Area of four walls = 2h (l + b) Here, l = 5 m, b = 4 m and h = 3 m Area of four walls = 2 x 3(5 + 4) = 54 m2 Area of ceiling = l x b = 5 x 4 = 20 m2 Total area to be white-washed = 54 + 20 = 74 m2 Cost of white-washing of 1 square metre = ₹ 7.50 ∴ Cost of white-washing = ₹74 x 7.50 = ₹ 555.
Description : The paint in a certain container is sufficient to paint an area equal to 9.375 m2. How many bricks of dimensions 22.5 cm×10 cm×7.5 cm can be painted out of this container? -Maths 9th
Last Answer : Total surface area of one brick = 2(lb +bh+lb) = [2(22.5 10+10 7.5+22.5 7.5)] cm2 = 2(225+75+168.75) cm2 = (2 468.75) cm2 = 937.5 cm2 Let n bricks can be painted out by the ... 93750 cm2 So, we have, 93750 = 937.5n n = 100 Therefore, 100 bricks can be painted out by the paint of the container.
Description : The sides of a triangle are 35 cm, 54 cm and 61 cm, respectively. The length of its longest altitude -Maths 9th
Last Answer : s= 2 a+b+c = 2 35+54+61 =75 Area, A= s(s−a)(s−b)(s−c) = 75(75−35)(75−54)(75−61) =420 5 cm 2 Now, Area of the triangle is also given as A= 2 1 ×a×h Where, h is the longest altitude. Therefore, 2 1 ×a×h=420 5 Hence, h=24 5 cm
Last Answer : The length of its longest altitude
Description : Find the length of the longest rod that -Maths 9th
Last Answer : Length of the longest rod = Diagonal of the room = root under( √a2 + b2 + c2) = root under( √122 + 92 + 82) = root under( √144 + 81 + 64) = √289 =17 m
Description : A metallic sheet is of rectangular shape with dimensions 48 cm x 36 cm. From each of its corners, a square of 8 cm is cut-off and an open box is made of the remaining sheet. Find the volume of the box. -Maths 9th
Last Answer : When squares of 8 cm is cutt-off from rectangulare sheet then, Length of box (l) = (98 - 8 - 8) = 32 cm Breadth of box (b) = (36 - 8 - 8) = 20 cm Height of box (h) = 8cm ∴ Volume of box = lbh = 32 x 20 x 8 = 5120 cm3
Description : The dimensions of a rectangle ABCD are 51 cm x 25 cm. -Maths 9th
Last Answer : According to question find the lengths QC and PD.
Description : A design is made on a rectangular tile of dimensions 50 cm x 17 cm as shown in figure. -Maths 9th
Last Answer : According to question find the the total area of the design and the remaining area of the tiles.
Description : Metal spheres, each of radius 2 cm, are packed into a rectangular box of internal dimensions 16 cm x 8 cm x 8 cm. -Maths 9th
Last Answer : Volume of rectangular box=lbh=16(64)=1024cm3 Volume of sphere=34πr3=33.5238cm3 16 sphere=16(33.5238)=536.3808 Volume of liquid=1024−536.3808=488cm3
Last Answer : According to question find the volume of this liquid.
Description : A rectangular box has dimensions x, y and z units, where x < y < z. If one dimension is only increased by one unit, -Maths 9th
Description : The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank, if its length and depth are respectively 2.5 m and 10 m -Maths 9th
Last Answer : Length (l) and depth (h) of tank is 2.5 m and 10 m respectively. To find: The value of breadth, say b. Formula to find the volume of a tank = l b h = (2.5 b 10) m3= 25b m3 Capacity ... of water (Given) Therefore, 25000 b = 50000 This implies, b = 2 Therefore, the breadth of the tank is 2 m.
Description : A wall of length 10 m is to be built across an open ground. -Maths 9th
Last Answer : Length of the wall = 10 m = 1000 cm Thickness of the wall = 42 cm Height of the wall = 5 m = 500 cm ∴ Volume of the wall = 1000 × 500 × 42 cm3 Volume of each brick = 42 × 12 × 10 cm3 No. of bricks = Volume of the wall / Volume of each brick = 1000 × 500 × 42 / 42 × 12 × 10 = 4166.67 = 4167
Description : A storage tank consists of a circular cylinder with a hemisphere adjoined on either end. If the external diameter of the cylinder be 1.4 m and its length be 8 m, find the cost of painting it on the outside at the rate of Rs. 10 per m -Maths 9th
Last Answer : Answer We have, r=0.7m, h=8m ∴ Total surface area = 2πr2+2πrh=2πr(r+h)=2×722×0.7×8.7m2 Required cost = Rs. {2×722×0.7×8.7×10}=Rs.382.80
Description : The whole surface area of a rectangular block is 1300 cm2. Find its volume, if their dimensions are in the ratio of 4 : 3 : 2. -Maths 9th
Last Answer : Let the length, breadth and height of the rectangular box be 4x, 3x and 2x, respectively. ∵ Total surface area = 1300 cm2 2(4x × 3x + 3x × 2x + 4x × 2x) = 1300 52x2 =1300x2 = 25x = 5 ∴ Volume of rectangular box = 4x × 3x × 2x = 24(5)2 = 3000 cm3
Description : The dimensions of a rectangle ABCD are 51 cm × 25 cm. -Maths 9th
Last Answer : Area of rectangle ABCD = AB x BC = 51 x 25 = 1275 cm2 Area of trapezium PBCQ = 5/6 x 1275 = 6375/6 cm2 Let QC = 9x cm and PB = 8x cm ∴ Area of trapezium PBCQ = 1/2(QC + PB) x BC ⇒ 6375/6 = 1/2(9x + 8x) x 25 ⇒ 17x ... 6375/6 x 2/17 x 25 ⇒ x = 5 ∴ QC = 9 x 5 cm = 45 cm and PB = 8 x 5 cm = 40 cm
Description : If V is the volume of a cuboid of dimensions l, -Maths 9th
Last Answer : V = lbh and S = 2(lb + bh + hl) Now RHS = 2/S(1/l + 1/b + 1/h) = 2/S(bh + hl + lb/lbh) = 2/S x S/2 x 1/V = 1/V = LHS.
Description : A wooden bookshelf has external dimensions as follows: -Maths 9th
Last Answer : Area of the bookshelf to be polished = Area of the five complete surfaces + Area of 2 rectangles of dimensions 110 cm x 5 cm in the front + Area of 4 rectangles of dimensions 75 cm x 5 cm in the ... of painting = ₹ 10/100 x 19350 = ₹1935 Hence, total expenses required = ₹(4340 + 1935) = ₹6275
Description : A metallic sheet is of rectangular shape with dimensions 28m × 36m. From each of its corners, a square is cut off so as to make an open box. -Maths 9th
Last Answer : R.E.F image Volume of box =l×b×h From the diagram l=48−2(8) ∵ Two square formed side =32m b=36−2(8) =20m Also h=8m from question ∴ Volume =32×20×8 =5120m3
Description : If two rectangular sheets each of dimensions 2x and 2y form the curved surfaces of two different cylinders, then the ratio -Maths 9th
Description : The volume of a certain rectangular solid is 8 cm^3. Its total surface area is 32 cm^2 and its three dimensions are in geometric progression. -Maths 9th
Last Answer : (b) 32 Let the edges of the solid be a, ar, ar2. Then, Volume = a x ar x ar2 = a3r3 = (ar)3. Given (ar)3 = 8 ⇒ ar = 2 Also, surface area = 2(a x ar + ar x ar2 + a × ar2) = 2(a2r + ... Given, 2ar (a + ar + ar2) = 32 ⇒ 4(a + ar + ar2) = 32 ; Sum of lengths of all edges = 32.
Description : In a hot water heating system, there is cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system. -Maths 9th
Last Answer : Height of cylindrical pipe = Length of cylindrical pipe = 28m Radius of circular end of pipe = diameter/ 2 = 5/2 cm = 2.5cm = 0.025m Now, CSA of cylindrical pipe = 2πrh, where r = radius and h = height of ... = 2 (22/7) 0.025 28 m2 = 4.4m2 The area of the radiating surface of the system is 4.4m2.
Description : A cylindrical roller 2.5 m in length, 1.75 m in radius when rolled on a road was found to cover the area of 5500 m2. -Maths 9th
Last Answer : NEED ANSWER
Last Answer : Solution of the question
Description : What length of 5 m wide cloth will be -Maths 9th
Last Answer : Radius of the base of the conical tent (r) = 7 m Height of the conical tent (h) = 24 m Let 'l' be the slant height of the cone then l = root under ( √r2 + h2) = root under ( √72 + 242 ) = √625 = 25 ... of the cloth used = 550 m2 Length of 5 m wide cloth used = Area/Width = 550 m2/5 m = 110 m
Description : What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6m -Maths 9th
Last Answer : Solution: Height of conical tent, h = 8m Radius of base of tent, r = 6m Slant height of tent, l2 = (r2+h2) l2 = (62+82) = (36+64) = (100) or l = 10 Again, CSA of conical tent = πrl = (3.14 6 ... .2) 3] = 188.4 L-0.2 = 62.8 L = 63 Therefore, the length of the required tarpaulin sheet will be 63 m.
Description : What length of tarpaulin 3 m wide -Maths 9th
Last Answer : Radius of the base of cone (r) = 6 m Height of the cone (h) = 8 m Let 'l' be the slant height of the cone. Then l = root under(√r2 + h2) = root under(√62 + 82) = root under(√100) = 10 m Surface area of ... make a conical tent of width 3 m = 188.4/3 = 62.8 m Wastage = 20 cm = 20/100 m = 0.2 m
Description : The area of parallelogram PQRS is 88 cm sq. A perpendicular from S is drawn to intersect PQ at M. If SM = 8 cm, then find the length of PQ. -Maths 9th
Last Answer : Given area of parallelogram = 88 cm² And SM = 8cm Area of a parallelogram = height × base (Height is the measurement of a perpendicular drawn from one side to other) Here, Area of PQRS = SM × PQ 88cm² = 8cm × PQ 11cm = PQ
Description : In Fig. 6.10, if m|n, then find the value of x. -Maths 9th
Last Answer : Solution :-
Description : A godown measures 45 m x 25 m x 10 m. -Maths 9th
Last Answer : Total number of crates that can be accommodated in the godown = Volume of godown/Volume of one crate = 45 x 25 x 10/ 1.5 x 1.25 x 0.5 = 12000 Number of crates already stored = 8000 Remaining crates that can be accommodated = 12000 - 8000 = 4000
Description : In a right angle triangle, prove that the hypotenuse is the longest side. -Maths 9th
Last Answer : The sum of angles of a triangle is180° If one aangke is of 90° then the sum of two angles is 90° It means that the angle forming 90° is biggest angle We know , Angle opposite to the longest side is largest. It means hypotenuse is the biggest side of right angled triangle
Description : If the lengths of the sides of a triangle are in the ratio 6:11:15 and it's perimeter is 96cm , then the height corresponding to the longest side is -Maths 9th
Last Answer : LET EACH SIDE BE X 6X+11X+15X=96 32X=96 X=3 SIDES=6 3=18 11 3=33 15 3=45 AREA OF TRIANGLE BY HERONS FORMULA=S=96/2=48 WHOLE UNDERROOT 48 48-18 48-33 48-45 UNDERROOT=12 4 30 15 3 4 3 15ROOT2 180 ... bh/2 180root2=18 h/2 360root2=18h h=20 root2 But root 2=1.4(approx) h=20 1.4(approx) h=28cm(approx).
Description : Prove that in a triangle,other than an euilateral triangle, angle opposite the longest side is greater than 2/3 of a right angle. -Maths 9th