A semi-circular sheet of metal of diameter 28 cm -Maths 9th

A semi-circular sheet of metal of diameter 28 cm -Maths 9th
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When semi-circular sheet is bent to form an open conical cup, the radius of the sheet becomes slant height of the cup and the semi-circular part of the sheet becomes the circumference of the base of the cone. ∴  Slant height of the conical cup (l) = 14 cm Let, r сm be the radius and h cm be the height of the conical cup. Then, circumference of the base of the conical cup = circumference of the semi-circular sheet 2 πr = 1/2 x 2 π x 14 ⇒ r = 7 cm  Now,    l2  = h2 + r2 ⇒  h = root under( √l2 - r2) = root under( √142 -72) = root under( √147) = 7 √3  Capacity of conical cup = 1/3 π2h = 1/3 x 22/7 x 7 x 7 x 7√3 = 1078/3.√3 = 359.3 x 1.732 = 622.31 cm3
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A semicircular thin sheet of a metal of diameter 28 cm is bent and an open conical cup is made. What is the capacity of the cup ? -Maths 9th

Description : A semicircular thin sheet of a metal of diameter 28 cm is bent and an open conical cup is made. What is the capacity of the cup ? -Maths 9th

Last Answer : answer:

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

A joker’s cap is in the form of right circular cone of base radius 7 cm and height 24cm. Find the area of the sheet required to make 10 such caps. -Maths 9th

Description : A joker’s cap is in the form of right circular cone of base radius 7 cm and height 24cm. Find the area of the sheet required to make 10 such caps. -Maths 9th

Last Answer : Radius of conical cap, r = 7 cm Height of conical cap, h = 24cm Slant height, l2 = (r2+h2) = (72+242) = (49+576) = (625) Or l = 25 cm CSA of 1 conical cap = πrl = (22/7)×7×24 = 550 CSA of 10 caps = (10×550) cm2 = 5500 cm2 Therefore, the area of the sheet required to make 10 such caps is 5500 cm2.

A semi-circle of diameter 14 cm has three chords of equal length connecting the two end points of the diameter so as -Maths 9th

Description : A semi-circle of diameter 14 cm has three chords of equal length connecting the two end points of the diameter so as -Maths 9th

Last Answer : (c)\(\bigg(rac{147 imes\sqrt3}{4}\bigg)\) cm2Take the trapezoid ABCD in the semi-circle with centre O such that AD = DC = CB. Now, complete the circle and draw an identical trapezoid in the other semicircle also. Then, ADCBEF ... {1}{2}\)x \(rac{3\sqrt3}{2}\) x 49 = \(rac{147 imes\sqrt3}{4}\) cm2.

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

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.

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.

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.

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!

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

Last Answer : answer:

If S denotes the area of the curved surface of a right circular cone of height h end semi-vertical angle a, then S equals -Maths 9th

Description : If S denotes the area of the curved surface of a right circular cone of height h end semi-vertical angle a, then S equals -Maths 9th

Last Answer : answer:

If a circular sheet of perimeter 2πr touching each side of a given quadrilateral sheet of perimeter 2p -Maths 9th

Description : If a circular sheet of perimeter 2πr touching each side of a given quadrilateral sheet of perimeter 2p -Maths 9th

Last Answer : Let ABCD be the quadrilateral from which a circular sheet is cut off touching each side of the quadrilateral. Also, given AB + BC + CD + DA = 2p ...(i) Circumference of the circle = 2πr ⇒ ... = pr.∴ Required remaining area = Area of quadrilateral - Area of circle = pr - πr2 = r(p - πr).

In figure, AOB is a diameter of the circle and C, D, E are any three points on the semi-circle. -Maths 9th

Description : In figure, AOB is a diameter of the circle and C, D, E are any three points on the semi-circle. -Maths 9th

Last Answer : Since, A, C, D and E are four point on a circle, then ACDE is a cyclic quadrilateral. ∠ACD+ ∠AED = 180° …(i) [sum of opposite angles in a cyclic quadrilateral is 180°]

In figure, AOB is a diameter of the circle and C, D, E are any three points on the semi-circle. -Maths 9th

Description : In figure, AOB is a diameter of the circle and C, D, E are any three points on the semi-circle. -Maths 9th

Last Answer : Since, A, C, D and E are four point on a circle, then ACDE is a cyclic quadrilateral. ∠ACD+ ∠AED = 180° …(i) [sum of opposite angles in a cyclic quadrilateral is 180°]

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.

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

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 sphere and a right circular cone of same radius have equal volumes. By what percentage does the height of the cone exceed its diameter ? -Maths 9th

Description : A sphere and a right circular cone of same radius have equal volumes. By what percentage does the height of the cone exceed its diameter ? -Maths 9th

Last Answer : answer:

The diameter of a solid mettalic right circular cylinder is equal to its height. After culting out the largest possible solid sphere -Maths 9th

Description : The diameter of a solid mettalic right circular cylinder is equal to its height. After culting out the largest possible solid sphere -Maths 9th

Last Answer : answer:

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

A rhombus shaped sheet with perimeter 40 cm and one diagonal 12 cm, is painted on both sides at the rate of Rs. -Maths 9th

Description : A rhombus shaped sheet with perimeter 40 cm and one diagonal 12 cm, is painted on both sides at the rate of Rs. -Maths 9th

Last Answer : Cost of painting =

A rhombus shaped sheet with perimeter 40 and digonals are 12 cm is painted on bith sides at the rate of rs 5 per metre square. Find the cost of painting -Maths 9th

Description : A rhombus shaped sheet with perimeter 40 and digonals are 12 cm is painted on bith sides at the rate of rs 5 per metre square. Find the cost of painting -Maths 9th

Last Answer : Let ABCD be a rhombus, then AB=BC=CD=DA=x Perimeter of rhombus =40cm ⇒4x=40cm⇒x=10cm ∴AB=BC=CD=DA=10cm In △ABC,S=2a+b+c​=210+10+12​=16cm ar△ABC=16(16−10)(16−10)(16−12)​=16×6×6×4​=48cm2ar.ABCD=2×48=96cm2 Cost of painting the sheet =Rs(5×96×2)=Rs960 [Both sides]

A rhombus shaped sheet with perimeter 40 cm and one diagonal 12 cm, is painted on both sides at the rate of Rs. -Maths 9th

Description : A rhombus shaped sheet with perimeter 40 cm and one diagonal 12 cm, is painted on both sides at the rate of Rs. -Maths 9th

Last Answer : Cost of painting =

A rhombus shaped sheet with perimeter 40 and digonals are 12 cm is painted on bith sides at the rate of rs 5 per metre square. Find the cost of painting -Maths 9th

Description : A rhombus shaped sheet with perimeter 40 and digonals are 12 cm is painted on bith sides at the rate of rs 5 per metre square. Find the cost of painting -Maths 9th

Last Answer : Let ABCD be a rhombus, then AB=BC=CD=DA=x Perimeter of rhombus =40cm ⇒4x=40cm⇒x=10cm ∴AB=BC=CD=DA=10cm In △ABC,S=2a+b+c​=210+10+12​=16cm ar△ABC=16(16−10)(16−10)(16−12)​=16×6×6×4​=48cm2ar.ABCD=2×48=96cm2 Cost of painting the sheet =Rs(5×96×2)=Rs960 [Both sides]

A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm (see figure). -Maths 9th

Description : A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm (see figure). -Maths 9th

Last Answer : NCERT Solutions for Class 9 Maths Chapter 12 Heron's Formula NCERT Solutions for Class 9 Maths Chapter 12 Heron's Formula Ex 12.1 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths ... = 48 m Sides ∆ABC are a = AB = 30m, b = AD = 30m, c = BD = 48m S

A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, -Maths 9th

Description : A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, -Maths 9th

Last Answer : For the given triangle, we have a = 28 cm, b = 30 cm, c = 26 cm Area of the given parallelogram = Area of the given triangle ∴ Area of the parallelogram = 336 cm2 ⇒ base x height = 336 ⇒ ... be the height of the parallelogram. ⇒ h = 33628 = 12 Thus, the required height of the parallelogram = 12 cm

30 circular plates, each of radius 14 cm -Maths 9th

Description : 30 circular plates, each of radius 14 cm -Maths 9th

Last Answer : Height of the cylinder formed (h) = 30 x 3 = 90 cm Radius of the base of the cylinder formed (r) = 14 cm (i) Total surface area of the cylinder = 2 πr(r + h) = 2 x 22/7 x 14(14 + 90) = 2 x 22/7 x 14 x 104 = 9152 cm2 (ii) Volume of the cylinder formed = πr2h = 22//7 x 14 x 14 x 90 = 55440 cm3

The area of the curved surface and the area of the base of a right circular cylinder are a square cm and b square cm respectively -Maths 9th

Description : The area of the curved surface and the area of the base of a right circular cylinder are a square cm and b square cm respectively -Maths 9th

Last Answer : answer:

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:

A solid right circular cylinder of radius 8 cm and height 2 cm is melted and cast into a right circular cone of height 3 times that of the cylinder. -Maths 9th

Description : A solid right circular cylinder of radius 8 cm and height 2 cm is melted and cast into a right circular cone of height 3 times that of the cylinder. -Maths 9th

Last Answer : Height of cone = 3 times height of cylinder = 3 3 = 9 cm Volume of cylinder = volume of cone r2 = 8 8 r = 8 cm l2 = h2 + r2 = (9)2 + (8)2 l = = 12 cm C.S.A (cone) = = 301.71 cm2

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.

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

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.

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 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 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.

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 .

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

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.

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 .

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

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.

Find the amount of water displaced by a solid spherical ball of diameter 4.2 cm, when it is completely immersed in water. -Maths 9th

Description : Find the amount of water displaced by a solid spherical ball of diameter 4.2 cm, when it is completely immersed in water. -Maths 9th

Last Answer : The amount of water displaced by a solid spherical ball when it is completely immersed in water is equal to its volume. Volume of a sphere of radius r is 34​πr3 As the diameter of the ball is 4.2 cm, its radius r=2.1 cm Hence, volume of water displaced =34​πr3=34​×722​×2.1×2.1×2.1=38.808 cm3

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

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

Last Answer : Volume of milk in 1 glass =πr2h =π×(3.5)2×12=461.58cm2​for 1600 students milk needed is =1600×461.58=738258litre​