Two hemispherical domes are to be painted -Maths 9th

Two hemispherical domes are to be painted -Maths 9th
by

1 Answer

Answer :

Let radii of the bases of two domes be r and R. ∴   2πr = 17.6  ⇒  2 x 22/7 x r = 17.6 ⇒ r = 17.6 x 7/2 x 22 = 2.8 cm and 2 πR = 70.4  ⇒ 2 x 22/7 x R = 70.4 ⇒ R = 70.4 x 7/2 x 22 = 11.2 cm Now, area of two hemispherical domes = 2πr2  + 2πR2 = 2 x 22/7 x 2.8 x 2.8 x 2 x 22/7 x 11.2 x 11.2 = 49.28 + 788.48 cm2  = 837.76 cm2  Cost of painting at the rate of  ₹10 per cm2 = 837.76 x 10  = ₹8377.6
by

Related questions

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

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

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

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.

Calculate the surface area of a hemispherical dome of a temple with radius 14 m to be whitewashed from outside. -Maths 9th

Description : Calculate the surface area of a hemispherical dome of a temple with radius 14 m to be whitewashed from outside. -Maths 9th

Last Answer : Here, radius of hemispherical dome (r) = 14 m Surface area of dome = 2πr2 = 2 × 22 / 7 × 14 × 14 = 1232 m2 Hence , total surface area to be whitewashed from outside is 1232 m2

Calculate the surface area of a hemispherical dome of a temple with radius 14 m to be whitewashed from outside. -Maths 9th

Description : Calculate the surface area of a hemispherical dome of a temple with radius 14 m to be whitewashed from outside. -Maths 9th

Last Answer : Here, radius of hemispherical dome (r) = 14 m Surface area of dome = 2πr2 = 2 × 22 / 7 × 14 × 14 = 1232 m2 Hence , total surface area to be whitewashed from outside is 1232 m2

The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is being pumped into it. -Maths 9th

Description : The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is being pumped into it. -Maths 9th

Last Answer : NEED ANSWER

The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is being pumped into it. -Maths 9th

Description : The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is being pumped into it. -Maths 9th

Last Answer : Find the ratios of the surface areas of the balloon in the two cases.

A hemispherical bowl is made of steel 0.25 cm thick. -Maths 9th

Description : A hemispherical bowl is made of steel 0.25 cm thick. -Maths 9th

Last Answer : Outer radius of hemispherical bowl = R = r + 0.25 = 5 + 0.25 = 5.25 cm Outer curved surface area of bowl = 2 πR2 = 2 x 22/7 x (5.25)2 = 2 x 22 x 5.25 x 0.75 = 173.25 cm2

A hemispherical tank is made up of an iron -Maths 9th

Description : A hemispherical tank is made up of an iron -Maths 9th

Last Answer : Inner radius of the hemispherical tank (r) = 1 m Outer radius of the hemispherical tank (R) = 1 + 0.01 = 1.01 m Volume of iron used to make the hemispherical tank = 2/3 πR3 - 2/3πr3 = 2/3π(R3 - r3) = 2/3 x 22/7 [(1.01)3 - 13] = 44/21(1.0303 - 1) = 44/21 x 0.0303 = 0.06349 m3

A hemispherical bowl is made of steel 0.5 cm thick. The inside radius of the bowl is 4 cm. What is the volume of steel used in making the bowl ? -Maths 9th

Description : A hemispherical bowl is made of steel 0.5 cm thick. The inside radius of the bowl is 4 cm. What is the volume of steel used in making the bowl ? -Maths 9th

Last Answer : answer:

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

A hemispherical bowl is 176 cm round the brim. Supposing it to be half full, how many persons may be served from it in hemispherical -Maths 9th

Description : A hemispherical bowl is 176 cm round the brim. Supposing it to be half full, how many persons may be served from it in hemispherical -Maths 9th

Last Answer : Let the radius of the hemispherical bowl be r cm Then, 2πr=176⇒r=176×72×22⇒r=282πr=176⇒r=176×72×22⇒r=28 Volume of liquid in the bowl : =12×( ... : =Volume of liquid in the bowlVolume of 1 glass=(14×28×282×2×2)= 1372

A sphere is cut into two equal halves and both the halves are painted from all the sides. The radius of the sphere is r unit and the -Maths 9th

Description : A sphere is cut into two equal halves and both the halves are painted from all the sides. The radius of the sphere is r unit and the -Maths 9th

Last Answer : answer:

There is a slide in a park. One of its side Company hired one of its walls for 3 months.walls has been painted in some colour with a message “KEEP THE PARK GREEN AND CLEAN -Maths 9th

Description : There is a slide in a park. One of its side Company hired one of its walls for 3 months.walls has been painted in some colour with a message “KEEP THE PARK GREEN AND CLEAN -Maths 9th

Last Answer : Let the sides of the wall be a = 15m, b = 11m, c = 6m Semi-perimeter, Thus, the required area painted in colour = 20√2 m2

The paint in a certain container is sufficient to paint an area equal to 9.375 m2. How many bricks of dimensions 22.5 cm×10 cm×7.5 cm can be painted out of this container? -Maths 9th

Description : The paint in a certain container is sufficient to paint an area equal to 9.375 m2. How many bricks of dimensions 22.5 cm×10 cm×7.5 cm can be painted out of this container? -Maths 9th

Last Answer : Total surface area of one brick = 2(lb +bh+lb) = [2(22.5 10+10 7.5+22.5 7.5)] cm2 = 2(225+75+168.75) cm2 = (2 468.75) cm2 = 937.5 cm2 Let n bricks can be painted out by the ... 93750 cm2 So, we have, 93750 = 937.5n n = 100 Therefore, 100 bricks can be painted out by the paint of the container.

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]

There is a slide in a park,one of its side walls has been painted.... -Maths 9th

Description : There is a slide in a park,one of its side walls has been painted.... -Maths 9th

Last Answer : Let the dimensions of triangular shape wall be a = 15 m, b = 6 m, c = 11 m ∴ s = (a + b + c)/2 = (15 + 6 + 11)/2 = 32/2 = 16 m Area painted = Area of triangle = root under(√s(s - a)(s - b)(s - c) = root under (√16(16 - 15)(16 - 6)(16 - 11)) = root under (√16 x 1 x 10 x 5) = 20√2 m2

What is the number of domes in sixty dome mosque ?

Last Answer : There are 81 domes in sixty domed mosques .

Why are salt domes important?

Description : Why are salt domes important?

Last Answer : Need answer

From the following which type of work envelope is made in jointed arm robot a.Rectangular work envelope b.Cylindrical work envelope c.Spherical or hemispherical work envelope d.C

Description : From the following which type of work envelope is made in jointed arm robot a.Rectangular work envelope b.Cylindrical work envelope c.Spherical or hemispherical work envelope d.C

Last Answer : b.Cylindrical work envelope

One end of a cylindrical rod is grounded to a hemispherical surface of radius`R=20mm`.It is immersed in water`(mu=4//3)`.If the refractive index of th

Description : One end of a cylindrical rod is grounded to a hemispherical surface of radius`R=20mm`.It is immersed in water`(mu=4//3)`.If the refractive index of th

Last Answer : One end of a cylindrical rod is grounded to a hemispherical surface of radius`R=20mm`.It is immersed in water`(mu ... is ................(determine)

What is the radiation characteristic of a dipole antenna? A. Omnidirectional B. Bidirectional C. Unidirectional D. Hemispherical

Description : What is the radiation characteristic of a dipole antenna? A. Omnidirectional B. Bidirectional C. Unidirectional D. Hemispherical

Last Answer : B. Bidirectional

From the following which type of work envelope is made in jointed arm robot. (A) Rectangular work envelope (B) Cylindrical work envelope (C) Spherical or hemispherical work envelope (D) None of the above

Description : From the following which type of work envelope is made in jointed arm robot. (A) Rectangular work envelope (B) Cylindrical work envelope (C) Spherical or hemispherical work envelope (D) None of the above

Last Answer : C

Abrupt uplift and gentle subsidence are indicative of: (A) Hat microatoll (B) Hemispherical coral head (C) Cup microatoll (D) Cylindrical coral head

Description : Abrupt uplift and gentle subsidence are indicative of: (A) Hat microatoll (B) Hemispherical coral head (C) Cup microatoll (D) Cylindrical coral head

Last Answer : (C) Cup microatoll  

While discussing the properties of a parallelogram teacher asked about the relation between two angles x and y of a parallelogram as shown ... -Maths 9th

Description : While discussing the properties of a parallelogram teacher asked about the relation between two angles x and y of a parallelogram as shown ... -Maths 9th

Last Answer : (a) Yes , x < y is correct (b) Ð ADB =Ð DBC = y (alternate int. angles) since BC < CD (angle opp. to smaller side is smaller) there for, x < y (c) Truth value

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

The radius of a spherical balloon increases from 7cm to 14cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases. -Maths 9th

Description : The radius of a spherical balloon increases from 7cm to 14cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases. -Maths 9th

Last Answer : Let r1 and r2 be the radii of spherical balloon and spherical balloon when air is pumped into it respectively. So r1 = 7cm r2 = 14 cm Now, Required ratio = (initial surface area)/(Surface area after pumping air into ... = (7/14)2 = (1/2)2 = ¼ Therefore, the ratio between the surface areas is 1:4.

The perimeter of a right triangle is 30 cm. If its hypotenuse is 13 cm, then what are two sides? -Maths 9th

Description : The perimeter of a right triangle is 30 cm. If its hypotenuse is 13 cm, then what are two sides? -Maths 9th

Last Answer : The other two sides of the triangle are 12 cm and 5 cm Explanation: Let the other two sides of triangle be x and y It's hypotenuse is 13 cm Perimeter of triangle = Sum of all sides ... When y = 12 x=17-y = 17-12 =5 So, the other two sides of the triangle are 12 cm and 5 cm

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

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

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

What is the general form of linear equation in two variables ? -Maths 9th

Description : What is the general form of linear equation in two variables ? -Maths 9th

Last Answer : The general form of linear equation in two variables can be written as: ax+by+c=0

1. Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres. -Maths 9th

Description : 1. Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres. -Maths 9th

Last Answer : To recall, a circle is a collection of points whose every point is equidistant from its centre. So, two circles can be congruent only when the distance of every point of both the circles are equal from the centre ... ) So, by SSS congruency, ΔAOB ΔCOD ∴ By CPCT we have, AOB = COD. (Hence proved).

Two coin are tossed 400 times and we get a. Two Heads : 112 times b. One Head : 160 times c. No Head : 128 times. When two coins are tossed at random, what is the probability of getting a. Two Heads b. One Head c. No Head -Maths 9th

Description : Two coin are tossed 400 times and we get a. Two Heads : 112 times b. One Head : 160 times c. No Head : 128 times. When two coins are tossed at random, what is the probability of getting a. Two Heads b. One Head c. No Head -Maths 9th

Last Answer : Given, Total number of events = 400 (a) No. of times two heads occur = 112 Probability of getting two heads = 112/400 = 7/25 (b) No. of times one heads occur = 160 Probability of getting one heads = 160/400 = 2/5 (c) No. of times no heads occur = 128 Probability of getting no heads = 128/400 = 8/25

The mean of 200 items was 50. Later on, it was discovered that the two items were misread as 92 and 8 instead of 192 and 88.find the correct mean. -Maths 9th

Description : The mean of 200 items was 50. Later on, it was discovered that the two items were misread as 92 and 8 instead of 192 and 88.find the correct mean. -Maths 9th

Last Answer : Here's ur answer..

find two solution of 4 x + Y is equal to 2 -Maths 9th

Description : find two solution of 4 x + Y is equal to 2 -Maths 9th

Last Answer : When solving for y in this problem(4x+y=2) we need to get y by itself in the equation. To do this we need to move 4x over to the right side of the equation. To do this we subtract, 4x from ... we do to one side of the equation we have to do the same to the other side of the equation. So,..

Find two irrational numbers between ROOT 2 and ROOT 7. -Maths 9th

Description : Find two irrational numbers between ROOT 2 and ROOT 7. -Maths 9th

Last Answer : Root 3 root 5

An exterior angle of a triangle is 110° and the two interior opposite angles are equal find the interior opposite angels -Maths 9th

Description : An exterior angle of a triangle is 110° and the two interior opposite angles are equal find the interior opposite angels -Maths 9th

Last Answer : each interior opposite angles are 55

EF is the transversal to two parallel lines AB and CD. GM and HL are the bisector of the corresponding angles EGB and EHD.Prove that GL parallel to HL. -Maths 9th

Description : EF is the transversal to two parallel lines AB and CD. GM and HL are the bisector of the corresponding angles EGB and EHD.Prove that GL parallel to HL. -Maths 9th

Last Answer : AB || CD and a transversal EF intersects them ∴ ∠EGB = ∠GHD ( Corresponding Angles) ⇒ 2 ∠EGM = 2 ∠GHL ∵ GM and HL are the bisectors of ∠EGB and ∠EHD respectively. ⇒ ∠EGM = ∠GHL But these angles form a pair of equal corresponding angles for lines GM and HL and transversal EF. ∴ GM || HL.

ABC and ADC are two right triangles with common hypotenuse AC. Prove that angle CAD = angle CAB -Maths 9th

Description : ABC and ADC are two right triangles with common hypotenuse AC. Prove that angle CAD = angle CAB -Maths 9th

Last Answer : Given, AC is the common hypotenuse. ∠B = ∠D = 90°. To prove, ∠CAD = ∠CBD Proof: Since, ∠ABC and ∠ADC are 90°. These angles are in the semi circle. Thus, both the triangles are lying in the semi ... D are concyclic. Thus, CD is the chord. ⇒ ∠CAD = ∠CBD (Angles in the same segment of the circle)

Two circles intersect at A and B. AC and AD are respectively the diameters of the circles. Prove that C, B and D are collinear. -Maths 9th

Description : Two circles intersect at A and B. AC and AD are respectively the diameters of the circles. Prove that C, B and D are collinear. -Maths 9th

Last Answer : Join CB, BD and AB, Since, AC is a diameter of the circle with centre O. ∴ ∠ABC = 90° [angle in semi circle] ---- (i) Also, AD is a diameter of the circle with center O . ∴ ∠ABD = 90° [angle in ... ⇒ ∠ABC + ∠ABD = 180° So. CBD is a straight line. Hence C, B and D are collinear . Hence proved.

Prove that the line joining the mid-points of two parallel chords of a circle passes through the centre. -Maths 9th

Description : Prove that the line joining the mid-points of two parallel chords of a circle passes through the centre. -Maths 9th

Last Answer : Let AB and CD be two parallel chords having P and Q as their mid-points, respectively. Let O be the centre of the circle. Join OP and OQ and draw OX | | AB | | CD. Since, Pis the mid-point of AB. ⇒ OP ... 90° Now, ∠POX + ∠XOQ = 90° + 90° = 180° so, POQ is a straight line . Hence proved

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.

Two cubes of side 2 cm each are joined end to end. Find the volume of the cuboid so formed. -Maths 9th

Description : Two cubes of side 2 cm each are joined end to end. Find the volume of the cuboid so formed. -Maths 9th

Last Answer : When two cubes of side 2 cm each are joined end to end then, Length (l) = (2 + 2) = 4cm Breadth (b) = 2 cm; Height (h) = 2 cm ∴ Volume of cuboid = lbh = 4 x 2 x 2 = 16 cm3

State and prove-line joining the midpoint of any two sides of a triangle is parallel to throw side and is equal to 1/2 of it -Maths 9th

Description : State and prove-line joining the midpoint of any two sides of a triangle is parallel to throw side and is equal to 1/2 of it -Maths 9th

Last Answer : Here, In △△ ABC, D and E are the midpoints of sides AB and AC respectively. D and E are joined. Given: AD = DB and AE = EC. To Prove: DE ∥∥ BC and DE = 1212 BC. Construction: Extend line segment DE to ... we have DF ∥∥ BC and DF = BC DE ∥∥ BC and DE = 1212BC (DE = EF by construction) Hence proved.

If a and b are two rational numbers, prove that a + b, a - b, ab are rational numbers. -Maths 9th

Description : If a and b are two rational numbers, prove that a + b, a - b, ab are rational numbers. -Maths 9th

Last Answer : In this way a / b is also a rational number.

Find two irrational numbers between 2 and 2.5 . -Maths 9th

Description : Find two irrational numbers between 2 and 2.5 . -Maths 9th

Last Answer : The two irrational numbers between 2 and 2.5 are 2.101001000100001----- and 2.201 001 0001 00001-----

Find two irrational numbers between 0.1 and 0.12. -Maths 9th

Description : Find two irrational numbers between 0.1 and 0.12. -Maths 9th

Last Answer : The two irrational numbers between 0.1 and 0.12 are 0.1 010010001--- and 0.1101001000100001 -----

Give two examples to show that the product of two irrational numbers may be a rational number . -Maths 9th

Description : Give two examples to show that the product of two irrational numbers may be a rational number . -Maths 9th

Last Answer : Take a = (2+ √3) and b =(2 - √3 ); a and b are irrational numbers, but their product = 4-3 = 1, is a rational number. Take c = √3 and d = -√3; c and d are irrational numbers. but their product = -3, is a rational number.

Give two rational numbers lying between 0.232332333233332---- and 0.21211211121111---- -Maths 9th

Description : Give two rational numbers lying between 0.232332333233332---- and 0.21211211121111---- -Maths 9th

Last Answer : The two rational numbers are 0.222. and 0.221