The diameter of a garden roller is 1.4 m and it is 2 m long. How much area will it level in 5 revolutions? -Maths 9th

The diameter of a garden roller is 1.4 m and it is 2 m long. How much area will it level in 5 revolutions? -Maths 9th
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1 Answer

Answer :

 Radius of a garden roller (r) = 1.4 / 2 = 0.7 m  and length of the garden roller (h) = 2 m ∴  Curved surface area of garden roller = 2πrh .  So, the area levelled of a garden roller in one revolution is 8.8 m2.  Now, area levelled of a garden in 5 revolutions = 8.8 x 5= 44m2
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The diameter of a garden roller is 1.4 m and it is 2 m long. How much area will it level in 5 revolutions? -Maths 9th

Description : The diameter of a garden roller is 1.4 m and it is 2 m long. How much area will it level in 5 revolutions? -Maths 9th

Last Answer : Radius of a garden roller (r) = 1.4 / 2 = 0.7 m and length of the garden roller (h) = 2 m ∴ Curved surface area of garden roller = 2πrh . So, the area levelled of a garden roller in one revolution is 8.8 m2. Now, area levelled of a garden in 5 revolutions = 8.8 x 5= 44m2

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 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 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

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

The diameter of a roller is 84 cm.. -Maths 9th

Description : The diameter of a roller is 84 cm.. -Maths 9th

Last Answer : Radius of roller = 84/2 cm = 42 cm Length of roller = 120 cm ∴ Curved surface area of roller = 2 πrh = 2 x 22/7 x 42 x 120 = 31680 cm2 As the roller takes 500 complete revolutions to level the playground ∴ Area of play ground = 500 x 31680 cm2 = 15840000 cm2 = 15840000/10000 m2 = 1584 m2

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

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

Find (i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5m high. -Maths 9th

Description : Find (i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5m high. -Maths 9th

Last Answer : Height of cylindrical tank, h = 4.5m Radius of the circular end , r = (4.2/2)m = 2.1m (i) the lateral or curved surface area of cylindrical tank is 2πrh = 2 (22/7) 2.1 4.5 m2 = (44 0.3 ... ) = 87.12 m2 This implies, S = 95.04 m2 Therefore, 95.04m2 steel was used in actual while making such a tank.

Find the area of the sheet required to make closed cylindrical vessel of height 1 m and diameter 140 cm. -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

Find the area of the sheet required to make closed cylindrical vessel of height 1 m and diameter 140 cm. -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

What is the critical rotation speed in revolutions per second, for a ball mill of 1.2 m diameter charged with 70 mm dia balls? (A) 0.5 (B) 1.0 (C) 2.76 (D) 0.66

Description : What is the critical rotation speed in revolutions per second, for a ball mill of 1.2 m diameter charged with 70 mm dia balls? (A) 0.5 (B) 1.0 (C) 2.76 (D) 0.66

Last Answer : (D) 0.66

If diameter of a wheel is 1.26 m, what the distance covered in 500 revolutions? (a) 1.38 km (b) 4.64 km (c) 2.46 km (d) 1.98 km

Description : If diameter of a wheel is 1.26 m, what the distance covered in 500 revolutions? (a) 1.38 km (b) 4.64 km (c) 2.46 km (d) 1.98 km

Last Answer : (d) 1.98 km

Water is following at the rate of 5 km/hr through a pipe of diameter 14 cm into a rectangular tank which is 50 m long -Maths 9th

Description : Water is following at the rate of 5 km/hr through a pipe of diameter 14 cm into a rectangular tank which is 50 m long -Maths 9th

Last Answer : Convert all to metres: 5 km = 5000 m 14 cm = 0.14 m 7 cm = 0.07 m Find the radius: Radius = Diameter 2 Radius = 0.14 2 = 0.07 m Find the amount of water that flowed out in an hour: Volume ... hours needed: Number of hours = 154 77 = 2 hours It takes 2 hours to fill up the tank to rise by 7 cm

A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of ₹12.50 per m2 . -Maths 9th

Description : A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of ₹12.50 per m2 . -Maths 9th

Last Answer : Diameter of the pillar = 50 cm ∴ Radius (r) = 502m = 25 m = 14m and height (h) = 3.5m Curved surface area of a pillar = 2πrh ∴ Curved surface area to be painted = 112m2 ∴ Cost of painting of 1 m2 pillar = Rs. 12.50 ∴ Cost of painting of 112 m2 pillar = Rs. ( 112 x 12.50 ) = Rs. 68.75.

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.

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

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 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 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

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.

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

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.

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

It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. -Maths 9th

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

Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area -Maths 9th

Description : Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area -Maths 9th

Last Answer : Radius of the base of cone = diameter/ 2 = (10.5/2)cm = 5.25cm Slant height of cone, say l = 10 cm CSA of cone is = πrl = (22/7)×5.25×10 = 165

The inner diameter of a circular well is 3.5m. It is 10m deep. Find (i) its inner curved surface area, (ii) the cost of plastering this curved surface at the rate of Rs. 40 per m2. -Maths 9th

Description : The inner diameter of a circular well is 3.5m. It is 10m deep. Find (i) its inner curved surface area, (ii) the cost of plastering this curved surface at the rate of Rs. 40 per m2. -Maths 9th

Last Answer : Inner radius of circular well, r = 3.5/2m = 1.75m Depth of circular well, say h = 10m (i) Inner curved surface area = 2πrh = (2 (22/7 ) 1.75 10) = 110 Therefore, the inner curved surface ... area = Rs (110 40) = Rs.4400 Therefore, the cost of plastering the curved surface of the well is Rs. 4400.

The curved surface area of a right circular cylinder of height 14 cm is 88 cm2. Find the diameter of the base of the cylinder. (Assume π =22/7 ) -Maths 9th

Description : The curved surface area of a right circular cylinder of height 14 cm is 88 cm2. Find the diameter of the base of the cylinder. (Assume π =22/7 ) -Maths 9th

Last Answer : Height of cylinder, h = 14cm Let the diameter of the cylinder be d Curved surface area of cylinder = 88 cm2 We know that, formula to find Curved surface area of cylinder is 2πrh. So 2πrh =88 cm2 (r is the ... 88 cm2 2r = 2 cm d =2 cm Therefore, the diameter of the base of the cylinder is 2 cm.

If the volume of a sphere is numerically equal to its surface area, then find the diameter of the sphere. -Maths 9th

Description : If the volume of a sphere is numerically equal to its surface area, then find the diameter of the sphere. -Maths 9th

Last Answer : Let r be the radius of the sphere. and Volume of a sphere = surface area of the sphere ⇒ 4 / 3πr3 = 4πr2 ⇒ r = 3 cm ∴ Diameter of the sphere = 2r = 2 × 3 = 6 cm

If the volume of a sphere is numerically equal to its surface area, then find the diameter of the sphere. -Maths 9th

Description : If the volume of a sphere is numerically equal to its surface area, then find the diameter of the sphere. -Maths 9th

Last Answer : Let r be the radius of the sphere. and Volume of a sphere = surface area of the sphere ⇒ 4 / 3πr3 = 4πr2 ⇒ r = 3 cm ∴ Diameter of the sphere = 2r = 2 × 3 = 6 cm

Nidhi has to find the area of a sphere whose diameter was 14 cm. -Maths 9th

Description : Nidhi has to find the area of a sphere whose diameter was 14 cm. -Maths 9th

Last Answer : Area is two-dimensional while 4 πr represents a length.

A hemispherical bowl has its external diameter equal to 10 cm and its thickness is 1 cm. What is the whole surface area of the bowl ? -Maths 9th

Description : A hemispherical bowl has its external diameter equal to 10 cm and its thickness is 1 cm. What is the whole surface area of the bowl ? -Maths 9th

Last Answer : External radius of hemispherical bowl = 5 cm Internal radius of the bowl = (5 – 1) cm = 4 cm Surface area of external portion = 2π(5)2 = 50 p sq. cm Surface area of internal portion = 2π(4)2 = ... = 91π sq. cm = (91×227)(91×227) sq. cm = 13 × 22 sq. cm = 286 cm2

While Lilo was playing with a toy tractor, she rolled it over an ink drop. The image on the right shows the ink drop and the marks on the floor left by tractor’s wheels. The diameter of the back wheel of the tractor is 1.5 times that of its front wheel. What is the number of revolutions made by the front wheel between the ink drop and the last mark?

Description : While Lilo was playing with a toy tractor, she rolled it over an ink drop. The image on the right shows the ink drop and the marks on the floor left by tractor's wheels. The diameter of the back ... is the number of revolutions made by the front wheel between the ink drop and the last mark?

Last Answer : 6

A wheel has a diameter of 21 inches. How far does it travel if it makes 30 complete revolutions?

Description : A wheel has a diameter of 21 inches. How far does it travel if it makes 30 complete revolutions?

Last Answer : I need help

A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4cm. -Maths 9th

Description : A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4cm. -Maths 9th

Last Answer : Let r1 and r2 Inner and outer radii of cylindrical pipe r1 = 4/2 cm = 2 cm r2 = 4.4/2 cm = 2.2 cm Height of cylindrical pipe, h = length of cylindrical pipe = 77 cm (i) curved surface ... CSA of roller = (500 31680) cm2 = 15840000 cm2 = 1584 m2. Therefore, area of playground is 1584 m2. Answer!

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

Half the perimeter of a rectangular garden in 36m. Write a linear equation which satisfies this data. Draw the graph for the same. -Maths 9th

Description : Half the perimeter of a rectangular garden in 36m. Write a linear equation which satisfies this data. Draw the graph for the same. -Maths 9th

Last Answer : Solution :-

A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting? -Maths 9th

Description : A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting? -Maths 9th

Last Answer : Here, each side of the rhombus = 30 m. Let ABCD be the given rhombus and the diagonal, BD = 48 m Sides ∆ABC are a = AB = 30m, b = AD = 30m, c = BD = 48m Since, a diagonal divides the rhombus into ... Area of grass for 18 cows to graze = 864 m2 ⇒ Area of grass for 1 cow to graze = 86418 m2 = 48 m2

he frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. -Maths 9th

Description : he frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. -Maths 9th

Last Answer : Say h = height of the frame of lampshade, looks like cylindrical shape r = radius Total height is h = (2.5+30+2.5) cm = 35cm and r = (20/2) cm = 10cm Use curved surface area formula to find the ... 2πrh = (2 (22/7) 10 35) cm2 = 2200 cm2 Hence, 2200 cm2 cloth is required for covering the lampshade.

Diameter of the base of a cone is 10.5 cm -Maths 9th

Description : Diameter of the base of a cone is 10.5 cm -Maths 9th

Last Answer : Radius of cone (r) = 10.5/2 cm Slant height of cone (l) = 10 cm Curved surface area of cone = πrl = 22/7 x 10.5/2 x 10 = 165 cm2

A spherical iron shell with external diameter 21 cm weighs 22775 x 5/21 grams. Find the thickness of the shell if the metal weighs 10 gms per cu cm. -Maths 9th

Description : A spherical iron shell with external diameter 21 cm weighs 22775 x 5/21 grams. Find the thickness of the shell if the metal weighs 10 gms per cu cm. -Maths 9th

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The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas -Maths 9th

Description : The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas -Maths 9th

Last Answer : If diameter of earth is said d, then the diameter of moon will be d/4 (as per given statement) Radius of earth = d/2 Radius of moon = ½×d/4 = d/8 Surface area of moon = 4π(d/8)2 Surface area of earth = 4π(d/2)2 Ncert solutions class 9 chapter 13-6

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.

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

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.

If two chords of a circle with a common end-point are inclined equally to the diameter through this common end-point, then prove that chords are equal. -Maths 9th

Description : If two chords of a circle with a common end-point are inclined equally to the diameter through this common end-point, then prove that chords are equal. -Maths 9th

Last Answer : Let AB and AC be two chords and AD be the diameter of the circle . Draw OL ⊥ AB and OM ⊥ AC . In ΔOLA and ΔOMA ∠ OLA = ∠ OMA = 90° OA = OA [common sides] ∠ OAL = ∠ OAM [given] . OLA ≅ OMA [by ASA congruency] OL =OM (by CPCT) Chords AM and AC are equidistant from O. ∴ AB = AC Hence proved.

A circle with centre O and diameter COB is given. If AB and CD are parallel, then show that chord AC is equal to chord BD. -Maths 9th

Description : A circle with centre O and diameter COB is given. If AB and CD are parallel, then show that chord AC is equal to chord BD. -Maths 9th

Last Answer : O Join AC and BD. Given, COB is the diameter of circle. ∠CAB = ∠BDC = 90° [angle in a semi-circle] Also, AB II CD ∠ABC = ∠DCB (alternate angles] Now, ∠ACB = 90° - ∠ABC and ∠DBC = 90° - ∠DCB = ... = ∠DBC BC = BC [common sides] ΔABC = ΔDCB [by ASA congruency] ∴ AC = BD [by CPCT] Hence Proved.

The given figure shows a circle with centre O in which a diameter AB bisects the chord PQ at the point R. If PR = RQ = 8 cm and RB = 4 cm, then find the radius of the circle. -Maths 9th

Description : The given figure shows a circle with centre O in which a diameter AB bisects the chord PQ at the point R. If PR = RQ = 8 cm and RB = 4 cm, then find the radius of the circle. -Maths 9th

Last Answer : Let r be the radius, then OQ = OB = r and OR = (r - 4) ∴ OQ2 = OR2 + RO2 ⇒ r2 = 64 + (r-4)2 ⇒ r2 = 64 + r2 + 16 - 8r ⇒ 8r = 80 ⇒ r = 10 cm

AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, then find the distance of AB from the centre of the circle. -Maths 9th

Description : AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, then find the distance of AB from the centre of the circle. -Maths 9th

Last Answer : ∵ The perpendicular drawn from the centre to the chord bisects it. ∴ AM = 1/2 AB = 1/2 × 30 cm = 15 cm Also, OA = 1/2 AD = 1/2 × 34 cm = 17 cm In rt. △OAM, we have OA2 = OM2 + AM2 172 = OM2 + 152 ⇒ 289 = OM2 + 225 ⇒ OM2 = 289 - 225 ⇒ OM2 = 64 ⇒ OM = √64 = 8 cm

A school provides milk to the students daily in cylindrical glasses of diameter 7 cm. -Maths 9th

Description : A school provides milk to the students daily in cylindrical glasses of diameter 7 cm. -Maths 9th

Last Answer : Diameter d = 7 cm Radius r = 7 / 2 cm and h = 12 cm ∴ V = πr2h = 22 / 7 × 7 / 2 × 7 / 2 × 12 = 462 Total milk for 1600 students = 462 × 1600 = 739200 cm3 = 739200 / 1000 litres = 739.2 litres .

AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, the distance of AB from the centre of the circle is -Maths 9th

Description : AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, the distance of AB from the centre of the circle is -Maths 9th

Last Answer : (d) Given, AD = 34 cm and AB = 30 cm In figure, draw OL ⊥ AB. Since, the perpendicular from the centre of a circle to a chord bisects the chord.