A cubical box has each edge 10 cm and another cuboidal box is 12.5cm long, 10 cm wide and 8 cm high -Maths 9th

A cubical box has each edge 10 cm and another cuboidal box is 12.5cm long, 10 cm wide and 8 cm high -Maths 9th
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From the question statement, we have Edge of a cube = 10cm Length, l = 12.5 cm Breadth, b = 10cm Height, h = 8 cm (i) Find the lateral surface area for both the figures Lateral surface area of cubical box = 4 (edge)2 = 4(10)2 = 400 cm2 …(1) Lateral surface area of cuboidal box = 2[lh+bh] = [2(12.5×8+10×8)] = (2×180) = 360 Therefore, Lateral surface area of cuboidal box is 360 cm2. …(2) From (1) and (2), lateral surface area of the cubical box is more than the lateral surface area of the cuboidal box. The difference between both the lateral surfaces is, 40 cm2. (Lateral surface area of cubical box – Lateral surface area of cuboidal box=400cm2–360cm2 = 40 cm2) (ii) Find the total surface area for both the figures The total surface area of the cubical box = 6(edge)2 = 6(10 cm)2 = 600 cm2…(3) Total surface area of cuboidal box = 2[lh+bh+lb] = [2(12.5×8+10×8+12.5×100)] = 610 This implies, Total surface area of cuboidal box is 610 cm2..(4) From (3) and (4), the total surface area of the cubical box is smaller than that of the cuboidal box. And their difference is 10cm2. Therefore, the total surface area of the cubical box is smaller than that of the cuboidal box by 10 cm2
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A cubical box has each edge 10 cm -Maths 9th

Description : A cubical box has each edge 10 cm -Maths 9th

Last Answer : (i) Lateral surface area of cubical box = 4a2 = 4 x 102 = 400 cm2 Lateral surface area of cuboidal box = 2h (l + b) = 2 x 8(12.5 +10) = 16 X 22.5 = 360 cm2 Thus, lateral surface area of ... x 305cm2 = 610 cm2 Thus, total surface area of cuboidal box is greater by (610 - 600) cm2 = 10 cm2

Hameed has built a cubical water tank with lid for his house, with each other edge 1.5m long. -Maths 9th

Description : Hameed has built a cubical water tank with lid for his house, with each other edge 1.5m long. -Maths 9th

Last Answer : Edge of cubical tank = 1.5 m ∴ Area of 4 walls = 4 (side)² = 4(1.5)² m² = 4 x 225 = 9 m² Area of floor = (1.5)² = 2.25 m² ∴ Total surface area = 9 + 2.25 = 11.25 m² Edge of square tile = 25 m = 0.25 m² ∴ Area of 1 tile = (0.25)2 = .0625 m²

A cuboidal water tank is 6m long, 5m wide and 4.5m deep. How many litres of water can it hold? -Maths 9th

Description : A cuboidal water tank is 6m long, 5m wide and 4.5m deep. How many litres of water can it hold? -Maths 9th

Last Answer : Dimensions of a cuboidal water tank are: l = 6 m and b = 5 m and h = 4.5 m Formula to find volume of tank, V = l b h Put the values, we get V = (6 5 4.5) = 135 ... water, 135 m3volume hold = (135 1000) litres = 135000 litres Therefore, given cuboidal water tank can hold up to135000 litres of wate

A matchbox measures 4 cm×2.5cm×1.5cm. What will be the volume of a packet containing 12 such boxes? -Maths 9th

Description : A matchbox measures 4 cm×2.5cm×1.5cm. What will be the volume of a packet containing 12 such boxes? -Maths 9th

Last Answer : Dimensions of a matchbox (a cuboid) are l×b×h = 4 cm×2.5 cm×1.5 cm respectively Formula to find the volume of matchbox = l×b×h = (4×2.5×1.5) = 15 Volume of matchbox = 15 cm3 Now, volume of 12 such matchboxes = (15×12) cm3 = 180 cm3 Therefore, the volume of 12 matchboxes is 180cm3.

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

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%

If three cubes of copper, each with an edge 6 cm, 8 cm and 10 cm respectively are melted to form a single cube, -Maths 9th

Description : If three cubes of copper, each with an edge 6 cm, 8 cm and 10 cm respectively are melted to form a single cube, -Maths 9th

Last Answer : 20.8 cm Let the edge of the single cube be ‘a’ cm. Then, total volume melted = Volume of cube formed ⇒ (6)3 + (8)3 + (10)3 = a3 ⇒ a3 = 216 + 512 + 1000 = 1728 ... Diagonal of the new cube = 3–√a=(3–√×12)3a=(3×12) cm = 20.8 cm (approx.)

A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5cm. Find the outer curved surface of the bowl. -Maths 9th

Description : A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5cm. Find the outer curved surface of the bowl. -Maths 9th

Last Answer : Inner radius of hemispherical bowl = 5cm Thickness of the bowl = 0.25 cm Outer radius of hemispherical bowl = (5+0.25) cm = 5.25 cm Formula for outer CSA of hemispherical bowl = 2πr2, where r is radius of ... 22/7) (5.25)2 = 173.25 Therefore, the outer curved surface area of the bowl is 173.25 cm2.

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

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.

Find the cost of digging a cuboidal pit 8m long, 6m broad and 3m deep at the rate of Rs 30 per m3 -Maths 9th

Description : Find the cost of digging a cuboidal pit 8m long, 6m broad and 3m deep at the rate of Rs 30 per m3 -Maths 9th

Last Answer : The given pit has its length(l) as 8m, width (b)as 6m and depth (h)as 3 m. Volume of cuboidal pit = l×b×h = (8×6×3) = 144 (using formula) Required Volume is 144 m3 Now, Cost of digging per m3 volume = Rs 30 Cost of digging 144 m3 volume = Rs (144×30) = Rs 4320

A cuboidal water tank is 6 m long, -Maths 9th

Description : A cuboidal water tank is 6 m long, -Maths 9th

Last Answer : Volume of cuboidal tank = 1 x b x h = 6 m x 5 m x 4.5 m = 135 m3 = 135 x 1000 L = 135000 L

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

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

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

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

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

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=34​πr3=33.5238cm3 16 sphere=16(33.5238)=536.3808 Volume of liquid=1024−536.3808=488cm3

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

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 : According to question find the volume of this liquid.

A rectangular piece of paper is 22 cm long and 10 cm wide. -Maths 9th

Description : A rectangular piece of paper is 22 cm long and 10 cm wide. -Maths 9th

Last Answer : Since rectangular piece of paper of rolled along its length. ∴ 2πr = 22 r = 22 × 7 / 2 × 22 = 3.5 cm Height of cyclinder (h) = 10 cm ∴ Volume of cyclinder = πr2h = 22 / 7 × 3.5 × 3.5 × 10 = 385 cm3.

A rectangular piece of paper is 22 cm long and 10 cm wide. -Maths 9th

Description : A rectangular piece of paper is 22 cm long and 10 cm wide. -Maths 9th

Last Answer : Since rectangular piece of paper of rolled along its length. ∴ 2πr = 22 r = 22 × 7 / 2 × 22 = 3.5 cm Height of cyclinder (h) = 10 cm ∴ Volume of cyclinder = πr2h = 22 / 7 × 3.5 × 3.5 × 10 = 385 cm3.

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

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.

Shahid has built a cubical water tank with lid for his house, -Maths 9th

Description : Shahid has built a cubical water tank with lid for his house, -Maths 9th

Last Answer : Edge of tank = 2 m = 2 x 100 cm = 200 cm Area of five faces of the tank = 5a2 = 5 (200 cm)2 = 2,00,000 cm2 Area of a square tile = 25 cm x 25 cm = 625 cm2 Number of tiles required = Area of ... /12 do zen Cost of one dozen of tiles = ₹480 ∴ Cost of 320/12 dozen tiles = ₹480 x 320/12 = ₹12, 800

A hemispherical bowl made of brass has inner diameter 10.5cm. Find the cost of tin-plating it on the inside at the rate of Rs 16 per 100 cm2. -Maths 9th

Description : A hemispherical bowl made of brass has inner diameter 10.5cm. Find the cost of tin-plating it on the inside at the rate of Rs 16 per 100 cm2. -Maths 9th

Last Answer : Inner radius of hemispherical bowl, say r = diameter/2 = (10.5)/2 cm = 5.25 cm Formula for Surface area of hemispherical bowl = 2πr2 = 2 (22/7) (5.25)2 = 173.25 Surface area of hemispherical ... of tin-plating the inner side of the hemispherical bowl at the rate of Rs 16 per 100 cm2 is Rs 27.72.

Find the surface area of a sphere of radius: (i) 10.5cm (ii) 5.6cm (iii) 14cm -Maths 9th

Description : Find the surface area of a sphere of radius: (i) 10.5cm (ii) 5.6cm (iii) 14cm -Maths 9th

Last Answer : Formula: Surface area of sphere (SA) = 4πr2 (i) Radius of sphere, r = 10.5 cm SA = 4 (22/7) 10.52 = 1386 Surface area of sphere is 1386 cm2 (ii) Radius of sphere, r = 5.6cm Using formula, SA = 4 (22 ... 75 cm Surface area of sphere = 4πr2 = 4 (22/7) 1.752 = 38.5 Surface area of a sphere is 38.5 cm2

Find the lateral surface area and total surface area of a cube of edge 10 cm. -Maths 9th

Description : Find the lateral surface area and total surface area of a cube of edge 10 cm. -Maths 9th

Last Answer : Edge of cube (a) = 10 cm (i) ∴ Lateral surface area = 4a² = 4 x (10)² = 4 x 100 cm²= 400 cm² (ii) Total surface area = 6a² = 6 x(10)² cm² = 6 x 100 = 600 cm²

Rinku has built a cuboidal water tank in his house. The top of the water tank is covered with an iron lid. -Maths 9th

Description : Rinku has built a cuboidal water tank in his house. The top of the water tank is covered with an iron lid. -Maths 9th

Last Answer : Total inner surface area of the water tank including the base without top = 2(l + b) h + l b = 2(180 + 120) 60 + 180 120 = 36000 + 21600 = 57600 cm2 Area of each title = 10 ... 57600 / 80 = 720 Total amount required for 720 tiles at the rate of 480 per dozen = 480 / 12 720 = 28800

Rinku has built a cuboidal water tank in his house. The top of the water tank is covered with an iron lid. -Maths 9th

Description : Rinku has built a cuboidal water tank in his house. The top of the water tank is covered with an iron lid. -Maths 9th

Last Answer : Total inner surface area of the water tank including the base without top = 2(l + b) h + l b = 2(180 + 120) 60 + 180 120 = 36000 + 21600 = 57600 cm2 Area of each title = 10 ... 57600 / 80 = 720 Total amount required for 720 tiles at the rate of 480 per dozen = 480 / 12 720 = 28800

A cement concrete road is 1000 m long, 8 m wide and 15 cm thick over the sub-base of 10 cm thick gravel. The box cutting in road crust is (A) 500 m3 (B) 1000 m3 (C) 1500 m3 (D) 2000 m3

Description : A cement concrete road is 1000 m long, 8 m wide and 15 cm thick over the sub-base of 10 cm thick gravel. The box cutting in road crust is (A) 500 m3 (B) 1000 m3 (C) 1500 m3 (D) 2000 m3

Last Answer : (C) 1500 m3

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

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

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

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

Find the area of a triangle having perimeter 32cm. One side of its side is equal to 11cm and difference of the other two is 5cm. -Maths 9th

Description : Find the area of a triangle having perimeter 32cm. One side of its side is equal to 11cm and difference of the other two is 5cm. -Maths 9th

Last Answer : Solutions :- We have, Perimeter of triangle = 32 cm One of its side = 11 cm Let the second side be x And third side be x + 5 Perimeter of triangle = sum of three sides A/q => 11 + x + x + 5 ... 13 cm Now, By using heron's formula, Find the area of a triangle :- Answer : Area of triangle = 43.81 cm²

Is it possible to construct a triangle with lengths of its sides 5cm, 3cm and 8cm? Give reason for your answer. -Maths 9th

Description : Is it possible to construct a triangle with lengths of its sides 5cm, 3cm and 8cm? Give reason for your answer. -Maths 9th

Last Answer : Solution :- No, since sum of two sides is equal to third side. (5 cm + 3 cm = 8 cm)

If two adjacent sides of a kite are 5cm and 7cm, find its perimeter. -Maths 9th

Description : If two adjacent sides of a kite are 5cm and 7cm, find its perimeter. -Maths 9th

Last Answer : Solution :-

The area of a trapezium is 39cm2 . The distance between its parallel sides is 6cm. If one of the parallel sides is 5cm, then find the other parallel side. -Maths 9th

Description : The area of a trapezium is 39cm2 . The distance between its parallel sides is 6cm. If one of the parallel sides is 5cm, then find the other parallel side. -Maths 9th

Last Answer : area of trapezium = 39 cm sq height of trpezium = 6 cm one || side = 5 cm let other be x area of trapezium = 1/2 × ( sum of || sides) × h 39 = 1/2 × ( 5+x) × 6 39×2 = (5+x) × 6 78/6 = 5+x 13-5 = x 8 = x x = 8cm

Two cubes of edge 6 cm are joined to form a cuboid. Find the total surface area of the cuboid. -Maths 9th

Description : Two cubes of edge 6 cm are joined to form a cuboid. Find the total surface area of the cuboid. -Maths 9th

Last Answer : When two cubes are joined end to end, then Length of the cuboid = 6 + 6 = 12 cm Breadth of the cuboid = 6 cm Height of the cuboid = 6 cm Total surface area of the cuboid = 2 (lb + bh + hi) = 2(12 x6 + 6×6 + 6×12) = 2(72 + 36 + 72) = 2(180) = 360 cm2

Two cubes of edge 6 cm are joined to form a cuboid. Find the total surface area of the cuboid. -Maths 9th

Description : Two cubes of edge 6 cm are joined to form a cuboid. Find the total surface area of the cuboid. -Maths 9th

Last Answer : When two cubes are joined end to end, then Length of the cuboid = 6 + 6 = 12 cm Breadth of the cuboid = 6 cm Height of the cuboid = 6 cm Total surface area of the cuboid = 2 (lb + bh + hi) = 2(12 x6 + 6×6 + 6×12) = 2(72 + 36 + 72) = 2(180) = 360 cm2

How many bricks will be required to construct a wall 8 m long, 6 m high and 22.5 cm thick, if each brick measures 25 cm x 11.25 cm x 6 cm? -Maths 9th

Description : How many bricks will be required to construct a wall 8 m long, 6 m high and 22.5 cm thick, if each brick measures 25 cm x 11.25 cm x 6 cm? -Maths 9th

Last Answer : ∵ Volume of one brick = (25 × 11.25 × 6) cm3 and volume of the wall = (800 × 600 × 22.5) cm3 ∴ Number of bricks = Volume of the walls / Volume of one brick = 800 × 600 × 22.5 / 25 × 11.25 × 6 = 6400

How many bricks will be required to construct a wall 8 m long, 6 m high and 22.5 cm thick, if each brick measures 25 cm x 11.25 cm x 6 cm? -Maths 9th

Description : How many bricks will be required to construct a wall 8 m long, 6 m high and 22.5 cm thick, if each brick measures 25 cm x 11.25 cm x 6 cm? -Maths 9th

Last Answer : ∵ Volume of one brick = (25 × 11.25 × 6) cm3 and volume of the wall = (800 × 600 × 22.5) cm3 ∴ Number of bricks = Volume of the walls / Volume of one brick = 800 × 600 × 22.5 / 25 × 11.25 × 6 = 6400

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

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.

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

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 : Solution of this question

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

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...

A cylindrical drum of 15cm radius and 70cm height is full of water. The water is emptied into a rectangular tank 225cm long and 70cm wide. Find the height of water level in the tank. a) 3.5cm b) 2.25cm c) 3.142cm d) 1.57cm

Description : A cylindrical drum of 15cm radius and 70cm height is full of water. The water is emptied into a rectangular tank 225cm long and 70cm wide. Find the height of water level in the tank. a) 3.5cm b) 2.25cm c) 3.142cm d) 1.57cm

Last Answer : c) 3.142cm

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

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.

How can you work out the volume of a cube of edge 5cm?

Description : How can you work out the volume of a cube of edge 5cm?

Last Answer : 5 x 5 x 5 = 125 cubic cm

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

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

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

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

PQRS is a parallelogram, in which PQ = 12 cm and its perimeter is 40 cm. Find the length of each side of the parallelogram . -Maths 9th

Description : PQRS is a parallelogram, in which PQ = 12 cm and its perimeter is 40 cm. Find the length of each side of the parallelogram . -Maths 9th

Last Answer : Here, PQ = SR = 12cm Let PS = x and PS = QR ∴ x + 12 + x +12 = Perimeter 2x + 24 = 40 2x = 16 x = 8 Hence, length of each side of the parallelogram is 12cm , 8cm , 12cm and 8cm.

PQRS is a parallelogram, in which PQ = 12 cm and its perimeter is 40 cm. Find the length of each side of the parallelogram . -Maths 9th

Description : PQRS is a parallelogram, in which PQ = 12 cm and its perimeter is 40 cm. Find the length of each side of the parallelogram . -Maths 9th

Last Answer : Here, PQ = SR = 12cm Let PS = x and PS = QR ∴ x + 12 + x +12 = Perimeter 2x + 24 = 40 2x = 16 x = 8 Hence, length of each side of the parallelogram is 12cm , 8cm , 12cm and 8cm.

Find the area of an isosceles triangle, whose equal sides are of length 15 cm each and third side is 12 cm. -Maths 9th

Description : Find the area of an isosceles triangle, whose equal sides are of length 15 cm each and third side is 12 cm. -Maths 9th

Last Answer : We have, Three sides13cm,13cm and 20cm. By using Heron's formula We need to get the semi-perimeter s= 2 a+b+c​ = 2 13+13+20​ = 2 46​ =23 Now, put the heron's formula, s= s(s−a)(s−b)(s−c)​ = 23(23−13)(23−13)(23−20)​ = 23×10×10×3​ =10 23×3​ =83.07cm 2

The sides of a quadrilateral ABCD are 6 cm, 8 cm, 12 cm and 14 cm (taken in order), respectively and the angle between the first two sides is a right angle. -Maths 9th

Description : The sides of a quadrilateral ABCD are 6 cm, 8 cm, 12 cm and 14 cm (taken in order), respectively and the angle between the first two sides is a right angle. -Maths 9th

Last Answer : Given ABCD is a quadrilateral having sides AB=6cm, BC=8cm, CD=12cm and DA=14 cm. Now. Join AC. We have, ABC is a right angled triangle at B. Now, AC2=AB2+BC2 [by Pythagoras theorem]Now, AC2=AB2+BC2 ... =24(1+6-√)cm2=24+246=24(1+6)cm2 Hence, the area of quadrilateral is 241+6-√−−−−−−√cm2241+6cm2 .

The sides of a quadrilateral ABCD are 6 cm, 8 cm, 12 cm and 14 cm (taken in order), respectively and the angle between the first two sides is a right angle. -Maths 9th

Description : The sides of a quadrilateral ABCD are 6 cm, 8 cm, 12 cm and 14 cm (taken in order), respectively and the angle between the first two sides is a right angle. -Maths 9th

Last Answer : Given ABCD is a quadrilateral having sides AB = 6 cm, BC = 8 cm, CD = 12 cm and DA = 14 cm. Now, join AC.

If the difference in elevation of an edge of the pavement 9 m wide and its crown is 15 cm, the camber of the pavement, is (A) 1 in 60 (B) 1 in 45 (C) 1 in 30 (D) 1 in 15

Description : If the difference in elevation of an edge of the pavement 9 m wide and its crown is 15 cm, the camber of the pavement, is (A) 1 in 60 (B) 1 in 45 (C) 1 in 30 (D) 1 in 15

Last Answer : Answer: Option C

The outer and the inner radii of a hollow sphere are 12 cm and 10 cm. Find its volume. -Maths 9th

Description : The outer and the inner radii of a hollow sphere are 12 cm and 10 cm. Find its volume. -Maths 9th

Last Answer : This answer was deleted by our moderators...

The outer and the inner radii of a hollow sphere are 12 cm and 10 cm. Find its volume. -Maths 9th

Description : The outer and the inner radii of a hollow sphere are 12 cm and 10 cm. Find its volume. -Maths 9th

Last Answer : This answer was deleted by our moderators...