A wall of length 10 m is to be built across an open ground. -Maths 9th

A wall of length 10 m is to be built across an open ground. -Maths 9th
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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
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A wall of length 10 m is to be built across an open ground. -Maths 9th

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

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 length of the longest pole that can be put in a room of dimensions 10 m x 10 m x 5 m. -Maths 9th

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

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

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

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

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

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

A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree. -Maths 9th

Description : A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree. -Maths 9th

Last Answer : The broken part bends so that the top of the tree touches the ground making an angle 30 °. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Original height of ... After rationalising Now we have , A`BC = AB + BC Take LCM So , original height of tree is

The shadow of a tower standing on level ground is found to be 40 m longer when the Sun’s altitude is 30° than when it is 60°. Find the height of the tower. -Maths 9th

Description : The shadow of a tower standing on level ground is found to be 40 m longer when the Sun’s altitude is 30° than when it is 60°. Find the height of the tower. -Maths 9th

Last Answer : this is the answer

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

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

What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6m -Maths 9th

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.

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

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.

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

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

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

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

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

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 : Solution of the question

The length, breadth and height of a room are 5 m, -Maths 9th

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.

What length of 5 m wide cloth will be -Maths 9th

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

What length of tarpaulin 3 m wide -Maths 9th

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

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

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:

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

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

7 persons enter an elevator on the ground floor of a 11 storey hotel. Any one of them can leave the elevator at any of the 10 floors. -Maths 9th

Description : 7 persons enter an elevator on the ground floor of a 11 storey hotel. Any one of them can leave the elevator at any of the 10 floors. -Maths 9th

Last Answer : answer:

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

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

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

The outer and inner diameters of a circular pipe are 6 cm and 4 cm respectively. If its length is 10 cm, then what is the total surface -Maths 9th

Description : The outer and inner diameters of a circular pipe are 6 cm and 4 cm respectively. If its length is 10 cm, then what is the total surface -Maths 9th

Last Answer : answer:

The length and width of a swimming pool are 50 metres and 15 metres respectively. If the depth of the swimming pool at one end is 10 -Maths 9th

Description : The length and width of a swimming pool are 50 metres and 15 metres respectively. If the depth of the swimming pool at one end is 10 -Maths 9th

Last Answer : 11250 m3 If we take the vertical crossection of the face of the swimming pool, then it is a trapezium ABCD, with parallel sides AD and BC respectively of lengths 10 m and 20 m and distance between parallel sides as 50 m ... 50]×15[12(10+20)×50]×15 m3 = (15 x 50 x 15) m3 = 11250 m3.

The heat flux (from outside to inside) across an insulating wall with thermal conductivity, K = 0.04 W/m.°K and thickness 0.16m is 10 W/m2 . The temperature of the inside wall is - 5°C. The outside wall temperature is (A) 25°C (B) 30°C (C) 35°C (D) 40°C

Description : The heat flux (from outside to inside) across an insulating wall with thermal conductivity, K = 0.04 W/m.°K and thickness 0.16m is 10 W/m2 . The temperature of the inside wall is - 5°C. The outside wall temperature is (A) 25°C (B) 30°C (C) 35°C (D) 40°C

Last Answer : (C) 35°C

The cost of levelling a ground in the form .... -Maths 9th

Description : The cost of levelling a ground in the form .... -Maths 9th

Last Answer : True, Let a = 51m, b = 37m, c = 20m s = (a + b + c)/2 = (51 + 37 + 20)/2 = 108/2 = 54 m ∴ Area of triangular ground = root under(√s(s - a)(s - b)(s - c)) = root under√54(54 - 51)(54 - 37)(54 - 20) = root under (√54 x 3 x 17 x 34) = 306 m2 Cost of levelling the ground = Rs 3 x 306 = Rs 918

A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field. -Maths 9th

Description : A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field. -Maths 9th

Last Answer : Let the given field is in the form of a trapezium ABCD such that parallel sides are AB = 10 m and DC = 25 m Non-parallel sides are AD = 13 m and BC = 14 m. We draw BE || AD, such that BE = 13 m. ... = 112 m2 So, area of the field = area of ∆BCE + area of parallelogram ABED = 84 m2 + 112 m2 = 196 m2

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

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.

The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of Rs.10 per m2 is Rs.15000, find the height of the hall. -Maths 9th

Description : The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of Rs.10 per m2 is Rs.15000, find the height of the hall. -Maths 9th

Last Answer : Let length, breadth, and height of the rectangular hall be l, b, and h respectively. Area of four walls = 2lh+2bh = 2(l+b)h Perimeter of the floor of hall = 2(l+b) = 250 m Area of four walls = 2( ... of paining the walls is Rs. 15000. 15000 = 2500h Or h = 6 Therefore, the height of the hall is 6 m.

A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. -Maths 9th

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

A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. -Maths 9th

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

In Fig. 6.10, if m|n, then find the value of x. -Maths 9th

Description : In Fig. 6.10, if m|n, then find the value of x. -Maths 9th

Last Answer : Solution :-

The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find -Maths 9th

Description : The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find -Maths 9th

Last Answer : Radius of well = (r) = 3.5/2 m Depth of well = (h) = 10 m (i) Inner curved surface area of well = 2 πrh = 2 x 22/7 x 3.5/2 x 10 = 110 m2 (ii) Cost of plastering 1 m2 = ₹ 40 ∴ Cost of plastering 110 m2 = ₹110 X 40 = ₹4400

A godown measures 45 m x 25 m x 10 m. -Maths 9th

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

A heap of wheat is in the form of a cone whose diameter is 10.5 m -Maths 9th

Description : A heap of wheat is in the form of a cone whose diameter is 10.5 m -Maths 9th

Last Answer : Radius of the conical heap of wheat (r) = 10.5/2 m Height of the conical heap of wheat (h) = 3 m Volume of the conical heap of wheat = 1/3 πr2h = 1/3 x 22/7 x (10.5/2)2 x 3 = 173.25/2 = 86.625 ... = 6.05 m Area of canvas required = curved surface area of cone πrl = 22/7 x 10.5/2 x 6.05 = 99.825 m2

A conical tent is 10 m high -Maths 9th

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

If 450 volts AC were measured across the load as shown in the illustration it would indicate a/an _______________. EL-0083 A. open winding between 'H1' and 'X1' B. properly operating circuit C. open winding between 'X1' and 'X4' D. ground on one side of 'Ep'

Description : If 450 volts AC were measured across the load as shown in the illustration it would indicate a/an _______________. EL-0083 A. open winding between 'H1' and 'X1' B. properly operating circuit C. open winding between 'X1' and 'X4' D. ground on one side of 'Ep'

Last Answer : Answer: C

A straight wire of length 0.20 m moves at a steady speed of 3.0 m s-1 at right angles to the magnetic field of flux density 0.10 T. The e.m.f induced across ends of wire is A. 0.5 V B. 0.06 V C. 0.05 V D. 0.04 V

Description : A straight wire of length 0.20 m moves at a steady speed of 3.0 m s-1 at right angles to the magnetic field of flux density 0.10 T. The e.m.f induced across ends of wire is A. 0.5 V B. 0.06 V C. 0.05 V D. 0.04 V

Last Answer : 0.06 V

Find the lateral surface area and total surface area of a cuboid of length 80 cm, breadth 40 cm and height 20 cm. -Maths 9th

Description : Find the lateral surface area and total surface area of a cuboid of length 80 cm, breadth 40 cm and height 20 cm. -Maths 9th

Last Answer : Length of cuboid (l) = 80 cm Breadth (b) = 40 cm Height (h) = 20 cm (i) ∴ Lateral surface area = 2h(l + b) = 2 x 20(80 + 40) cm² = 40 x 120 = 4800 cm² (ii) Total surface area = 2(lb ... x 40 + 40 x 20 + 20 x 80) cm² = 2(3200 + 800 + 1600) cm² = 5600 x 2 = 11200 cm²

The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m2. -Maths 9th

Description : The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m2. -Maths 9th

Last Answer : The roller is in the form of a cylinder of diameter = 84 cm ⇒ Radius of the roller(r) = 842 cm = 42 cm Length of the roller (h) = 120 cm Curved surface area of the ... roller = 31680 cm2 = 3168010000m2 ∴ Area of the playground levelled in 500 revolutions = 500 x 3168010000m2 = 1584m2

The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m2? -Maths 9th

Description : The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m2? -Maths 9th

Last Answer : A roller is shaped like a cylinder. Let h be the height of the roller and r be the radius. h = Length of roller = 120 cm Radius of the circular end of roller = r = (84/2) cm = 42 cm Now, CSA of roller = 2πrh = ... , we have 2 (22/7) 0.7 h = 4.4 Or h = 1 Therefore, the height of the cylinder is 1 m.

A chord of a circle of radius 7.5 cm with centre 0 is of length 9 cm. Find its distance from the centre. -Maths 9th

Description : A chord of a circle of radius 7.5 cm with centre 0 is of length 9 cm. Find its distance from the centre. -Maths 9th

Last Answer : ∵ PM = MQ = 1/2 = PQ = 45 cm and OP = 7.5 cm In right angled ΔOMP, using phthagoras theorem OM2 = OP2 - PM2 ⇒OM2 = 7.52 - 4.52 ⇒OM2 = 56.25 - 20.25 ⇒OM2 = 36 ∴ OM = √36 = 6 cm

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.

The total surface area of a cube is 726 cm2. Find the length of its edge . -Maths 9th

Description : The total surface area of a cube is 726 cm2. Find the length of its edge . -Maths 9th

Last Answer : Total surface area of a cube = 726 cm2 6 × (side)2 = 726 (side)2 = 121 side = 11 cm Hence, the length of the edge of cube is 11 cm.

The diameter of a roller is 42 cm and its length is 120 cm. -Maths 9th

Description : The diameter of a roller is 42 cm and its length is 120 cm. -Maths 9th

Last Answer : We have the diameter of a cyclindrial roller = 42 cm ⇒ The radius of cyclindrical roller (r) = 42 / 2 = 21 cm Length of a cyclindrical roller (h) = 120 cm Curved surface of the roller = 2πrh = ... of the playground = Area covered by the roller in 500 complete revolutions = 500 1.584 = 792 m2

Give possible expression for the length and breadth of the rectangle whose area is given by 4a2 +4a - 3. -Maths 9th

Description : Give possible expression for the length and breadth of the rectangle whose area is given by 4a2 +4a - 3. -Maths 9th

Last Answer : Given, area of rectangle = 4a2 + 6a-2a-3 = 4a2 + 4a – 3 [by splitting middle term] = 2a(2a + 3) -1 (2a + 3) = (2a – 1)(2a + 3) Hence, possible length = 2a -1 and breadth = 2a + 3

Write the coordinates of the vertices of a rectangle whose length and breadth are 5 and 3 units respectively, -Maths 9th

Description : Write the coordinates of the vertices of a rectangle whose length and breadth are 5 and 3 units respectively, -Maths 9th

Last Answer : Given, length of a rectangle = 5 units and breadth of a rectangle = 3 units One vertex is at origin i.e., (0, 0) and one of the other vertices lies in III quadrant. So, the length of the rectangle is 5 ... negative,direction of y-axis and then vertex is C(0, -3). The fourth vertex B is (-5, - 3).