Description : How many metres of 5 m wide cloth will be required to make a conical tent, the radius of whose base is 3.5 m and height is 12 m ? -Maths 9th
Last Answer : l = √h2 + r2 = √(3.5)2 + (12)2 = √12.25 + 144 = √156.25 = 12.5 m Curved surface area = πrl = 22 / 7 × 3.5 × 12.5 = 137.5 m2 Area of cloth = 137.5 m2 Length of cloth required = C.S.A. / Width l = 137.5 / 5 = 27.5 m
Description : How many metres of cloth 5 m wide will be required to make a conical tent, the radius of whose base is 7 m and whose height is 24 m? -Maths 9th
Last Answer : Given, radius (r) = 7 m and height (h) = 24m ∴ Slant height (l) = √h2 + r2 = √242 + 72 = √625 = 25 m ∴ Length of canvas required
Description : How many square metres of canvas is required for a conical tent whose height is 3.5 m and the radius of the base is 12 m? -Maths 9th
Last Answer : Slantheight5l = h2+r2 =(3.5)2+(12)2 total canvas required = πrl =π×12×12.5 =471 sq m
Last Answer : Solution of this question
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 : A conical tent is 10 m high and the radius of its base is 24 m. Find (i) slant height of the tent. -Maths 9th
Last Answer : : Ncert solutions class 9 chapter 13-5 Let ABC be a conical tent Height of conical tent, h = 10 m Radius of conical tent, r = 24m Let the slant height of the tent be l. (i) In right triangle ... (13728/7) 70 = Rs 137280 Therefore, the cost of the canvas required to make such a tent is Rs 137280.
Description : A conical tent is 10 m high -Maths 9th
Last Answer : Radius of conical tent = r = 24 m Height of conical tent = h = 10 m (i) Let l be the slant height of the cone. Then l = root under(√r2 + h2) ⇒ l = root under(√242 + 102) = root under(√576 + 100) = root ... 1 m2 canvas = ₹70 ∴ Cost of 22/7 x 24 x 26 cm2 canvas = ₹ 70 x 22/7 x 24 x 26 = ₹137280
Description : The slant height and base diameter of conical tomb are 25m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of Rs. 210 per 100 m2 -Maths 9th
Last Answer : Slant height of conical tomb, l = 25m Base radius, r = diameter/2 = 14/2 m = 7m CSA of conical tomb = πrl = (22/7)×7×25 = 550 CSA of conical tomb= 550m2 Cost of white-washing 550 m2 area, which is Rs (210×550)/100 = Rs. 1155 Therefore, cost will be Rs. 1155 while white-washing tomb.
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 : A joker’s cap is in the form of right circular cone of base radius 7 cm and height 24cm. Find the area of the sheet required to make 10 such caps. -Maths 9th
Last Answer : Radius of conical cap, r = 7 cm Height of conical cap, h = 24cm Slant height, l2 = (r2+h2) = (72+242) = (49+576) = (625) Or l = 25 cm CSA of 1 conical cap = πrl = (22/7)×7×24 = 550 CSA of 10 caps = (10×550) cm2 = 5500 cm2 Therefore, the area of the sheet required to make 10 such caps is 5500 cm2.
Description : How much ice-cream can be put into a cone with base radius 3.5 cm and height 12 cm ? -Maths 9th
Last Answer : Here, radius (r) = 3.5 cm and height (h) = 12 cm ∴ Amount of ice cream = 1 / 3 πr2h = 1 / 3 × 22 / 7 × 3.5 × 3.5 × 12 = 154 cm3
Description : Find the capacity in litres of a conical vessel having height 8 cm and slant height 10 cm. -Maths 9th
Last Answer : Height of conical vessel (h) = 8 cm Slant height of conical vessel (l) = 10 cm ∴ r2 + h2 = l2 r2 + 82 = 102 r2 = 100 - 64 = 36 r = 6 cm Now, volume of conical vessel = 1 / 3 πr2h = 1 / 3 × 22 / 7 × 6 × 6 × 8 = 301.71 cm3 = 0.30171 litre
Description : It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. -Maths 9th
Last Answer : Radius of the closed cylindrical tank = 140/2 cm = 70 cm = 0.7 m Height of the closed cylindrical tank = 1 m Area of metal sheet required = 2πr(r + h) = 2 x 22/7 x 0.7 (1 + 0.7) = 7.48 m2
Description : A cone of height 7 cm and base radius 1 cm is carved from a cuboidal block of wood 10 cm × 5 cm × 2 cm -Maths 9th
Last Answer : 92239223% Volume of cone = 1313πr2h = 13×227×1×7=22313×227×1×7=223 cu. cm Volume of cubical block = (10 × 5 × 2) cm3 = 100 cm3 ∴ Wastage of wood = (100−227)100×100(100−227)100×100 = 27832783% = 92239223%
Description : A cylindrical container of height 14 m and base 12 m contains oil. The oil is to be transferred to one cylindrical can, -Maths 9th
Last Answer : answer:
Description : Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find (i) radius of the base -Maths 9th
Last Answer : Slant height of cone, l = 14 cm Let the radius of the cone be r. (i) We know, CSA of cone = πrl Given: Curved surface area of a cone is 308 cm2 (308 ) = (22/7) r 14 308 = 44 r r = 308 ... Total surface area of cone = 308+(22/7) 72 = 308+154 Therefore, the total surface area of the cone is 462 cm2.
Description : The total surface area of a cone whose radius is r/2 and slant height 2l is -Maths 9th
Last Answer : Total surface area of cone = πr(r+l) Given, radius = r/2 and slant height = 2l Therefore, new total surface area of cone = πr/2(r/2+2l) = π(r/4^2+rl) = πr(l+r/4)
Last Answer : Radius (r)=r/2 & slant height=2l TSA (S)=PIE R (l+r) =22/7×r/2(2l+r/2) =11/7×r(2l+r/2)
Description : A cylindrical rod of iron whose height is eight times its radius is melted and cast into spherical balls each of half the radius of the cylinder. -Maths 9th
Last Answer : Let radius of iron rod = r ∴∴ Height = 8r ∴∴ Volume of iron rod =π×(r)2×8r⇒8πr3=π×(r)2×8r⇒8πr3 ⇒⇒ Radius of spherical ball =r2=r2 Volume of spherical ball =43π(r2)3=43π(r2)3 Let n balls are casted ∴n×43π(r38)=8πr3∴n×43π(r38)=8πr3 ⇒n6=8⇒n=48
Description : A cylindrical rod of iron whose radius is one-fourth of its height is melted and cast into spherical balls of the same radius as that of the cylinder. -Maths 9th
Last Answer : Let radius of cylindrical rod =r ⇒ height =4r Volume of cylindrical rod =πr2h =πr2(4r) =4πr3 Volume of spherical balls of radius r=34πr3 No. of balls =34πr34πr3=
Description : A child consumed an ice-cream of inverted right-circular conical shape from the top and left only 12.5% of the cone for her mother. -Maths 9th
Description : Two cans have the same height equal to 21 cm. One can is cylindrical, the diameter of whose base is 10 cm. -Maths 9th
Last Answer : (c) 450 cm3. Required difference in capacities = 227227 x (5)2 x 21~ (10)2 x 21 = (1650 ~ 2100) cm3 = 450 cm3
Description : It is required to make a closed cylindrical tank of height 1m and base diameter 140cm from a metal sheet. How many square meters of the sheet are required for the same? -Maths 9th
Last Answer : Let h be the height and r be the radius of a cylindrical tank. Height of cylindrical tank, h = 1m Radius = half of diameter = (140/2) cm = 70cm = 0.7m Area of sheet required = Total surface are of tank = 2πr( ... [2 (22/7) 0.7(0.7+1)] = 7.48 Therefore, 7.48 square meters of the sheet are required.
Description : A closed iron tank 12 m long 9 m wide and 4 m deep is to be made . Determine the cost of iron sheet used at the rate of rs 5 per meter , sheet being 2 m wide. -Maths 9th
Last Answer : This answer was deleted by our moderators...
Description : A right triangular prism of height 18 cm and of base sides 5 cm, 12 cm and 13 cm is transformed into another right triangular prism on a base -Maths 9th
Last Answer : Vol. of △ ular prism = Area of △ ular base × height. ∴ Area of triangular base = area of triangle PQR By heron's formula. S=S(s−a)(s−b)(s−c)where S=2a+b+c∴Areaof△PQR= S=23+4+5=6 S=6(6−3)(6−4)(6−5)=3×2×3×2×1=6cm2 ∴ vol. of Prism =6×10 =60cm3Answer.
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 : A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. -Maths 9th
Last Answer : Diameter of cone = 10.5 m Radius of cone (r) = 5.25 m Height of cone (h) = 3 m Volume of cone = 1 / 3 πr2h = 1 / 3 × 22 / 7 × 5.25 × 5.25 × 3 = 86.625m3 Cost of 1m3 of wheat = 10 ∴ Cost of 86.625 m3 of wheat = 10 × 86.625 = 86.625
Description : A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. -Maths 9th
Last Answer : Given: Radius of cone, r = diameter/2 = 40/2 cm = 20cm = 0.2 m Height of cone, h = 1m Slant height of cone is l, and l2 = (r2+h2) Using given values, l2 = (0.22+12) = (1.04) Or l ... (32.028 12) = Rs.384.336 = Rs.384.34 (approximately) Therefore, the cost of painting all these cones is Rs. 384.34.
Description : Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m -Maths 9th
Last Answer : Radius of cone, r = 24/2 m = 12m Slant height, l = 21 m Formula: Total Surface area of the cone = πr(l+r) Total Surface area of the cone = (22/7)×12×(21+12) m2 = 1244.57m2
Description : The circumference of the base of 9 m high wooden solid cone is 44 m. Find the slant height of the cone. -Maths 9th
Last Answer : Circumference of the base of a cone = 2πr
Description : Water flows in a tank 150 m × 100 m at the base, through a pipe whose cross-section is 2 dm by 1.5 dm, at a speed of 15 km per hour. -Maths 9th
Last Answer : Volume of water discharged through the pipe = Volume increase of the tank First, consider the pipe. Volume discharge through pipe = length breadth speed time Let the time taken to fill the tank to 3 m depth be t. ... the tank V=150 100 3 cu. m ⇒ V=450 100 Therefore, 450t=450 100 ⇒ t=100 hours
Description : A cloth having an area of 165 m sq. is -Maths 9th
Last Answer : Let l m be the height of the conical tent. Radius of the base of conical tent (r) = 5 m (i) Area of the circular base of the cone = πr2 = 22/7 x 52 m2 Number of students = Area of the base/ Area occupied by one ... ~ 9.23 cm Volume of conical tent = 1/3πr2h = 1/3 x 22/7 x 52 x 9.23 m3 = 241.74 m3
Description : Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm. -Maths 9th
Last Answer : Steps of Construction (i) Draw BC = 12 cm. (ii) Construct ÐCBY = 90°. (iii) From ray BY, cut-off line segment BD = 18 cm. (iv) Join CD. (v) Draw the perpendicular bisector of CD intersecting BD at A. (vi ... = AC Now, BD = BA + AD ⇒ BD = AB + AC Hence, △ABC is the required triangle.
Description : A semicircular thin sheet of a metal of diameter 28 cm is bent and an open conical cup is made. What is the capacity of the cup ? -Maths 9th
Description : Find the area of the sheet required to make closed cylindrical vessel of height 1 m and diameter 140 cm. -Maths 9th
Last Answer : Required sheet = T.S.A. of cyclinder = 2πr (h+r) = 2 × 22 / 7 × 70 / 100(1 + 70 / 100) = 2 × 22 × 0.1 × 1.7 = 7.48 m2
Description : A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30cm long, 25 cm wide and 25 cm high. -Maths 9th
Last Answer : Length of greenhouse, say l = 30cm Breadth of greenhouse, say b = 25 cm Height of greenhouse, say h = 25 cm (i) Total surface area of greenhouse = Area of the glass = 2[lb+lh+bh] = [2(30 ... (30+25+25)] (after substituting the values) = 320 Therefore, 320 cm tape is required for all the 12 edges.
Description : A cylindrical vessel can hold 154 g of water. If the radius of its base is 3.5 cm, and 1 cm3 of water weighs lg,find the depth of water. -Maths 9th
Last Answer : Since 1 cm3 of water weighs 1 g. ∴ Volume of cyclinder vessel = 154 cm3 πr2h = 154 h = 154 × 7 / 22 × 3.5 × 3.5 h ;= 4 cm Hence, the depth of water is 4 cm.
Description : If in a cylinder, radius is doubled and height is halved, then find its curved surface area. -Maths 9th
Last Answer : Let r and h be radius and height of the cyclinder, then C.S.A. = 2πrh Now, radius is doubled and height is halved. ∴ New radius = 2r and new height = h / 2 New C.S.A. = 2π × 2r × h / 2 = 2πrh .
Description : Find the volume of cone of radius r/2 and height ‘2h’. -Maths 9th
Last Answer : Volume of cone = 1 / 3π × (r / 2)2 × 2h = 1 / 3π × r2 / 4 × 2h = 1 / 6 πr2h cu. units.
Description : A spherical ball is divided into two equal halves. If the curved surface area of each half is 56.57 cm?, find the volume of the spherical ball.11531/cylinder-radius-halved-and-height-doubled-then-find-volume-with-respect-original-volume -Maths 9th
Last Answer : since curved surface of half of the spherical ball = 56.57 cm2 ∴ 2πr2 = 56.57 ⇒ r2 = 56.57 / 2 × 3.14 = 9 ⇒ r = 3 cm Now, volume of spherical ball = 4 / 3 πr3 = 4 / 3 × 3.14 × 3 × 3 × 3 = 113.04 cm3