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

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

1 Answer

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 m Curved surface area of the conical tent = πrl = 22/7 x 7 x 25 = 550 m2 ∴ Area of the cloth used = 550 m2   Length of 5 m wide cloth used  = Area/Width = 550 m2/5 m  = 110 m
by

Related questions

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

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

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 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

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

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

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.

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 cloth having an area of 165 m sq. is -Maths 9th

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

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...

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 river 3 m deep and 40 m wide -Maths 9th

Description : A river 3 m deep and 40 m wide -Maths 9th

Last Answer : Length of water canal in one minute = 2 x 1000 m/60 = 100/3 m Volume of water flowing into the sea in one minute = l x b x h = (100 x 40 x 3)/3 = 4000 m 3 = 4000 x 1000 L = 4000000 L ( Since,1 m3 = 1000 L)

An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see figure), each piece measuring 20 cm, 50 cm and 50 cm. -Maths 9th

Description : An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see figure), each piece measuring 20 cm, 50 cm and 50 cm. -Maths 9th

Last Answer : Let the sides of each triangular piece be a = 20 cm, b = 50 cm, c = 50 cm

An umbrella is made by stitching 10 triangular pieces of cloth.... -Maths 9th

Description : An umbrella is made by stitching 10 triangular pieces of cloth.... -Maths 9th

Last Answer : The sides of triangular pieces are 20 cm, 50 cm and 50 cm. Let a = 20 cm, b = 50 cm, c = 50 cm ∴ Semi - perimeter, s = (a + b + c)/2 = (20 + 50 + 50)/2 s = 60 cm ∴ Area of ... 40 x 10 x 10) = 200√6 cm2 Cloth of each design required = Area of 5 triangular pieces = 5 x 200√6 = 1000√6 cm2

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 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.

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

Description : 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

Last Answer : 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 ... . Therefore, the total surface area of the cubical box is smaller than that of the cuboidal box by 10 cm2

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.

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

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.

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

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 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.

Poke your fingers in my eyes and I will open wide my jaws. Linen cloth, quills, or paper, my greedy lust devours them all. What am I? -Riddles

Description : Poke your fingers in my eyes and I will open wide my jaws. Linen cloth, quills, or paper, my greedy lust devours them all. What am I? -Riddles

Last Answer : Shears, or scissors.

Which mechanical device was used for weaving with ropes and pullies, which helped to weave wide pieces of cloth? -SST 10th

Description : Which mechanical device was used for weaving with ropes and pullies, which helped to weave wide pieces of cloth? -SST 10th

Last Answer : Fly shuttle.

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

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

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

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

Three measuring rods are 64 cm,80 cm and 96 cm in length . What is the least length of cloth that can can be measured exact number of times using any one of these rods? (a) 9.60 m (b) 8 m (c) 9.60 cm (d) 96 m

Description : Three measuring rods are 64 cm,80 cm and 96 cm in length . What is the least length of cloth that can can be measured exact number of times using any one of these rods? (a) 9.60 m (b) 8 m (c) 9.60 cm (d) 96 m

Last Answer : Ans 4. 9.60 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

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).

If two sides of a triangle are of lengths 5 cm and 1.5 cm, then the length of third side of the triangle cannot be -Maths 9th

Description : If two sides of a triangle are of lengths 5 cm and 1.5 cm, then the length of third side of the triangle cannot be -Maths 9th

Last Answer : (d) Given, the length of two sides of a triangle are 5 cm and 1.5 cm, respectively. Let sides AB = 5 cm and CA = 1.5 cm We know that, a closed figure formed by three intersecting lines ( ... options (a), (b) and (c) satisfy the above inequality but option (d) does not satisfy the above inequality.

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

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).

If two sides of a triangle are of lengths 5 cm and 1.5 cm, then the length of third side of the triangle cannot be -Maths 9th

Description : If two sides of a triangle are of lengths 5 cm and 1.5 cm, then the length of third side of the triangle cannot be -Maths 9th

Last Answer : (d) Given, the length of two sides of a triangle are 5 cm and 1.5 cm, respectively. Let sides AB = 5 cm and CA = 1.5 cm We know that, a closed figure formed by three intersecting lines ( ... options (a), (b) and (c) satisfy the above inequality but option (d) does not satisfy the above inequality.

If the length of hypotenuse of a right angled triangle is 5 cm and its area is 6 sq cm, then what are the lengths of the remaining sides? -Maths 9th

Description : If the length of hypotenuse of a right angled triangle is 5 cm and its area is 6 sq cm, then what are the lengths of the remaining sides? -Maths 9th

Last Answer : Let one of the remaining sides be x cm.Then, other side = \(\sqrt{5^2-x^2}\) cm∴ Area = \(rac{1}{2} imes{x} imes\sqrt{25-x^2}\) = 6⇒ \(x\sqrt{25-x^2}\) = 12 ⇒ x2(25 - x2) = 144⇒ 25x2 - x4 = 144 ⇒ x4 - 25x2 ... (x2 - 16) (x2 - 9) = 0 ⇒ x2 = 16 or x2 = 9 ⇒ x = 4 or 3∴ The two sides are 4 cm and 3 cm.

WXYZ is a square of side length 30. V is a point on XY and P is a point inside the square with PV perpendicular to XY. PW = PZ = PV – 5. Find PV. -Maths 9th

Description : WXYZ is a square of side length 30. V is a point on XY and P is a point inside the square with PV perpendicular to XY. PW = PZ = PV – 5. Find PV. -Maths 9th

Last Answer : answer:

Prove that the points (2, –2), (–2, 1) and (5, 2) are the vertices of a right angled triangle. Also find the length of the hypotenuse -Maths 9th

Description : Prove that the points (2, –2), (–2, 1) and (5, 2) are the vertices of a right angled triangle. Also find the length of the hypotenuse -Maths 9th

Last Answer : Let the co-ordinates of any point on the x-axis be (x, 0). Then distance between (x, 0) and (– 4, 8) is 10 units.⇒ \(\sqrt{(x+4)^2+(0-8)^2}\) = 10 ⇒ x2 + 8x + 16 + 64 = 100 ⇒ x2 + 8x – 20 = 0 ⇒ (x + 10) (x – 2) = 0 ⇒ x = –10 or 2 ∴ The required points are (– 10, 0) and (2, 0).

Point A(5, 1) is the centre of the circle with radius 13 units. AB ⊥ chord PQ. B is (2, –3). The length of chord PQ is -Maths 9th

Description : Point A(5, 1) is the centre of the circle with radius 13 units. AB ⊥ chord PQ. B is (2, –3). The length of chord PQ is -Maths 9th

Last Answer : (c) ParallelogramAB = \(\sqrt{(4-7)^2+(5-6)^2}\) = \(\sqrt{9+1}\) = \(\sqrt{10}\)BC = \(\sqrt{(7-4)^2+(6-3)^2}\) = \(\sqrt{9+9}\) = \(3\sqrt2\)CD =\(\sqrt{(4-1)^2+(3-2) ... = \(2\sqrt{13}\)AB = CD, BC = AD and AC ≠ BD ⇒ opposite sides are equal and diagonals are not equal. ⇒ ABCD is a parallelogram.

The point A(0, 0), B(1, 7) and C(5, 1) are the vertices of a triangle. Find the length of the perpendicular from -Maths 9th

Description : The point A(0, 0), B(1, 7) and C(5, 1) are the vertices of a triangle. Find the length of the perpendicular from -Maths 9th

Last Answer : (b) 3x - y = 0 Given lines are 3x - y - 3 = 0 and 3x - y + 5 = 0. Line parallel to the given lines can be written as 3x - y + c = 0 ...(i) Let us taken a point, say ... 5c + 15 = - 3c + 15 ⇒ 8c = 0 ⇒ c = 0. Substituting c = 0 in (i), the required equation is 3x - y = 0.

A straight line passes through the points (5, 0) and (0, 3). The length of the perpendicular from the point (4, 4) on the line is: -Maths 9th

Description : A straight line passes through the points (5, 0) and (0, 3). The length of the perpendicular from the point (4, 4) on the line is: -Maths 9th

Last Answer : (b) \(rac{\sqrt{17}}{2}\)Equation of the line through the points (5, 0) and (0, 3) y - 0 = \(rac{3-0}{0-5}\) (x - 5)⇒ y = \(rac{-3}{5}\)(x - 5)⇒ 5y + 3x - 15 = 0 ∴ Distance of perpendicular from ... (rac{|20+12-15|}{\sqrt{25+9}{}}\) = \(rac{17}{\sqrt{34}}\) units. = \(rac{\sqrt{17}}{2}\) units.

(a) The length of Ramesh’s notebook is 9 cm 5 mm. What will be its length in cm? (b) The length of a young gram plant is 65 mm. Express its length in cm. -Maths 9th

Description : (a) The length of Ramesh’s notebook is 9 cm 5 mm. What will be its length in cm? (b) The length of a young gram plant is 65 mm. Express its length in cm. -Maths 9th

Last Answer : (a) The length of Ramesh notebook is 9 cm 5 mm The length in cm is [(9 + 5 / 10)] cm = 9.5 cm (b) The length of a gram plant is 65 mm Hence, the length in cm is 65 / 10 = 6.5 cm

Draw a line segment AB of length 5.8 cm. Draw the perpendicular bisector of this line segment. -Maths 9th

Description : Draw a line segment AB of length 5.8 cm. Draw the perpendicular bisector of this line segment. -Maths 9th

Last Answer : Steps of Construction (i) Draw a line segment AB = 5.8 cm. (ii) Taking A as centre and radius more than 1/2AB, draw two arcs, one on either side of AB. (iii) Taking B as centre and ... . (iv) Join CD, intersecting AB at point P. Then, line CPD is the required perpendicular bisector of AB.

If the rate of change of the super-elevation along a curved portion of a 7 metre wide road is 1 in 150 and the maximum super-elevation allowed is 1 in 15, the maximum length of the transition curve to be provided at either end, is (A) 65 m (B) 70 m (C) 75 m (D) 80 m

Description : If the rate of change of the super-elevation along a curved portion of a 7 metre wide road is 1 in 150 and the maximum super-elevation allowed is 1 in 15, the maximum length of the transition curve to be provided at either end, is (A) 65 m (B) 70 m (C) 75 m (D) 80 m

Last Answer : Answer: Option B