define (a+b)2 -Maths 9th

define (a+b)2 -Maths 9th
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It is a^2+2ab+b^2
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define (a+b)2 -Maths 9th

Description : define (a+b)2 -Maths 9th

Last Answer : It is a^2+2ab+b^2

Define the term of Relation with example. -Maths 9th

Description : Define the term of Relation with example. -Maths 9th

Last Answer : Relation: If A and B are any two non-empty sets, then any subset of A B is defined as a relation from A to B. For example, Suppose A = {1, 2, 3} and B = {1, 2, 3, 4}. Then {(2, 3), ... 3)} is a relation in A B. Many more relations (subsets) can be selected at random from our product set A B.

Define the term of Domain, codomain and range of a relation: -Maths 9th

Description : Define the term of Domain, codomain and range of a relation: -Maths 9th

Last Answer : Let R be a relation from set A to set B. Then, the set of first element of the ordered pairs in R is called the domain and the set of second elements of the ordered pairs in R is called the range. The second set B is called ... 16, 25, 36}, Range of R = {4, 5, 6} and Codomain of R = {1, 4, 5, 6}.

Define the term of Inverse of a relation. -Maths 9th

Description : Define the term of Inverse of a relation. -Maths 9th

Last Answer : For any binary relation R, a second relation can be constructed by merely interchanging first and second components in every ordered pair. The relation thus obtained is called the inverse of the first one and designated as R-1. Thus, R-1 = {(y, x ... {(1, 2), (2, 3), (3, 4), (5, 4)}. So (R-1)-1 = R.

Define the types of relations and their examples. -Maths 9th

Description : Define the types of relations and their examples. -Maths 9th

Last Answer : There are various types of relations:Let A be a non-empty set. Then, a relation R on A is said to be Reflexive if (a, a) ∈ R for each a ∈ A, i.e., if a R a for each a ∈ A. For example, the relation is as strong as is reflexive ... a || b ⇒ b || a a || b, b || c ⇒ a || c.

Define: Trail. -Maths 9th

Description : Define: Trail. -Maths 9th

Last Answer : It is the performance of an experiment, such as throwing a dice or tossing a coin.

Define : Algebra of Events. -Maths 9th

Description : Define : Algebra of Events. -Maths 9th

Last Answer : Let A, B and C be any two events associated with a random experiment whose sample space is S. Then, (i) A ∪ B. (Union of A and B) is the event that occurs if A occurs or B occurs or both A and B occur ... a mutually exclusive and exhaustive set of events. ∴ A∩B = ϕ, B∩C = ϕ, A ∩C = ϕ and A∪B∪C = S.

Define : Addition Theorem of Probability. -Maths 9th

Description : Define : Addition Theorem of Probability. -Maths 9th

Last Answer : (a) For Two Events. If A and B are two events associated with a random experiment, then P(A ∪ B) = P(A) + P(B) - P(A ∩ B) ⇒ P(A or B) = P(A) + P(B) - P(A and B) Corollary 1: If A and B are ... that A ⊆ B, then P(A) ≤ P(B) (ii) If E is an event associated with a random experiment, then 0 P(E) ≤ 1

Define : Multiplication Theorem on Probability. -Maths 9th

Description : Define : Multiplication Theorem on Probability. -Maths 9th

Last Answer : Statement I. If two events A and B are independent, then probability that they will both occur is equal to the product of their individual probabilities. i.e. P (A and B) = P (A) P (B) ... occurrence of two independent events. Method. Use the relation P (A ∩ B) = P (A) . P (B).

Define : Right circular cylinder. -Maths 9th

Description : Define : Right circular cylinder. -Maths 9th

Last Answer : A right circular cylinder is a cylinder that has a closed circular surface having two parallel bases on both the ends and whose elements are perpendicular to its base. It is also called a right ... circular surface is at a fixed distance from a straight line known as the axis of the cylinder.

Define : Prism. -Maths 9th

Description : Define : Prism. -Maths 9th

Last Answer : answer:

Define : Parallelopiped. -Maths 9th

Description : Define : Parallelopiped. -Maths 9th

Last Answer : answer:

Define :Volume and surface area of a hollow cylinder. -Maths 9th

Description : Define :Volume and surface area of a hollow cylinder. -Maths 9th

Last Answer : answer:

Define : Right circular cone. -Maths 9th

Description : Define : Right circular cone. -Maths 9th

Last Answer : A right circular cone is a cone where the axis of the cone is the line meeting the vertex to the midpoint of the circular base. That is, the centre point of the circular base is joined with ... cone is a three-dimensional shape having a circular base and narrowing smoothly to a point above the base.

Define : Sphere. -Maths 9th

Description : Define : Sphere. -Maths 9th

Last Answer : answer:

Define : Hemisphere. -Maths 9th

Description : Define : Hemisphere. -Maths 9th

Last Answer : answer:

Define latent heat of fusion. Also write its SI unit. -Maths 9th

Description : Define latent heat of fusion. Also write its SI unit. -Maths 9th

Last Answer : The SI unit of latent heat is Joule. ... Latent heat is defined as the heat required for converting solid into a liquid or to vapour without any change in the temperature.

the curved surface area of a cylinder is 154 cm. the total surface area of the cylinder is three times its curved surface area. find the volume of the cylinder. -Maths 9th

Description : the curved surface area of a cylinder is 154 cm. the total surface area of the cylinder is three times its curved surface area. find the volume of the cylinder. -Maths 9th

Last Answer : T.S.A = 3*154 = 462 cm² C.S.A = 154 cm² C.S.A = 2πrh T.S.A = 2πr(r+h) Now, In T.S.A = 2πrr + 2πrh 462 = 2πrr + 2πrh 462 = 2*22/7*r*r + 154 462 - 154 = 2*22/7*r*r 308*7/2*22 = r*r 49 = r*r R = 7 cm ... 7*h 154/44 = h 7/2 =h H = 3.5 cm or 7/2 cm Now volume = πrrh = 22/7 * 7* 7 *7/2 = 11*49 = 539 cm³

case study questions class 9 maths surface area and volume -Maths 9th

Description : case study questions class 9 maths surface area and volume -Maths 9th

Last Answer : Q. Read the source or text given below and answer the following questions: A conical circus tent has to be made with a cloth that is 5m wide, whose height is 24m, and the radius of the base is 7m. ... 4. Find the Curved Surface Area Answers: 1. Curved Surface Area 2. 550 m 3. Rs.3850 4.110m²

Sohan wants to show gratitude towards his teacher by giving her a card made by him. He has three pieces of trapezium pasted one above the other as shown in fig. These pieces are arranged in a way that AB||HC || GD || FE. Also BC=CD=DE, and GF=6 cm... -Maths 9th

Description : Sohan wants to show gratitude towards his teacher by giving her a card made by him. He has three pieces of trapezium pasted one above the other as shown in fig. These pieces are arranged in a way that AB||HC || GD || FE. Also BC=CD=DE, and GF=6 cm... -Maths 9th

Last Answer : Given : Sohan wants to show gratitude towards his teacher by giving her a card made by him. He has three pieces of trapezium pasted one above the other as shown in the fig. These pieces are arranged ... length of coloured tape required = 30 cm (b) The values are : Happiness, beauty, Knowledge.

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

Hameed has built a cubical water tank with lid for his house, with each other edge 1.5m long. -Maths 9th

Description : Hameed has built a cubical water tank with lid for his house, with each other edge 1.5m long. -Maths 9th

Last Answer : Edge of cubical tank = 1.5 m ∴ Area of 4 walls = 4 (side)² = 4(1.5)² m² = 4 x 225 = 9 m² Area of floor = (1.5)² = 2.25 m² ∴ Total surface area = 9 + 2.25 = 11.25 m² Edge of square tile = 25 m = 0.25 m² ∴ Area of 1 tile = (0.25)2 = .0625 m²

A 4 cm cube is cut into 1 cm cubes. Calculate the total surface area of all the small cubes. -Maths 9th

Description : A 4 cm cube is cut into 1 cm cubes. Calculate the total surface area of all the small cubes. -Maths 9th

Last Answer : Side of cube = 4 cm But cutting into 1 cm cubes, we get = 4 x 4 x 4 = 64 Now surface area of one cube = 6 x (1)² = 6 x 1=6 cm² and surface area of 64 cubes = 6 x 64 cm² = 384 cm²

Three equal cubes are placed adjacently in a row. Find the ratio of total surface area of the new cuboid to that of the sum of the surface areas of the three cubes. -Maths 9th

Description : Three equal cubes are placed adjacently in a row. Find the ratio of total surface area of the new cuboid to that of the sum of the surface areas of the three cubes. -Maths 9th

Last Answer : Let each side of a cube = a cm Then surface area = 6a² cm² and surface area of 3 such cubes = 3 x 6a² = 18a² cm² By placing three cubes side by side we get a cuboid whose ... + 3a²] = 14 a² ∴ Ratio between their surface areas = 14a² : 18a² = 7 : 9

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

Mary wants to decorate her ********* tree. She wants to place the tree on a wooden block covered with coloured paper with picture of Santa Claus on it. -Maths 9th

Description : Mary wants to decorate her Christmas tree. She wants to place the tree on a wooden block covered with coloured paper with picture of Santa Claus on it. -Maths 9th

Last Answer : Length of box (l) = 80 cm Breadth (b) = 40 cm and height (h) = 20 cm ∴ Total surface area = 2(lb + bh + hl) = 2[80 x 40 + 40 x 20 + 20 x 80] cm² = 2[3200 + 800 + 1600] ... of one sheet = (40 cm)² = 1600 cm² ∴ No. of sheets required for the box = 11200 = 1600 = 7 sheets

Find the ratio of the total surface area and lateral surface area of a cube. -Maths 9th

Description : Find the ratio of the total surface area and lateral surface area of a cube. -Maths 9th

Last Answer : Let a be the edge of the cube, then Total surface area = 6a2² and lateral surface area = 4a² Now ratio between total surface area and lateral surface area = 6a² : 4a² = 3 : 2

Find the lateral surface area and total surface area of a cube of edge 10 cm. -Maths 9th

Description : Find the lateral surface area and total surface area of a cube of edge 10 cm. -Maths 9th

Last Answer : Edge of cube (a) = 10 cm (i) ∴ Lateral surface area = 4a² = 4 x (10)² = 4 x 100 cm²= 400 cm² (ii) Total surface area = 6a² = 6 x(10)² cm² = 6 x 100 = 600 cm²

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

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

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

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

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

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

A 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 kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in figure. -Maths 9th

Description : A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in figure. -Maths 9th

Last Answer : Each shade of paper is divided into 3 triangles i.e., I, II, III 8 cm For triangle I: ABCD is a square [Given] ∵ Diagonals of a square are equal and bisect each other. ∴ AC = BD = 32 cm Height of AABD ... are: Area of shade I = 256 cm2 Area of shade II = 256 cm2 and area of shade III = 17.92 cm2

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

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

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

Radha made a picture of an aeroplane with coloured paper as shown in figure. Find the total area of the paper used. -Maths 9th

Description : Radha made a picture of an aeroplane with coloured paper as shown in figure. Find the total area of the paper used. -Maths 9th

Last Answer : For surface I: It is an isosceles triangle whose sides are a = 5 cm, b = 5 cm, c = 1 cm = (0.75 x 3.3) cm2 = 2.475 cm2 (approx.) For surface II: It is a rectangle with length 6.5 cm and breadth 1 cm. ∴ Area of ... surface V) = [2.475 + 6.5 + 1.3 + 4.5 + 4.5] cm2 = 19.275 cm2 = 19.3 cm2 (approx.)

Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm. -Maths 9th

Description : Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm. -Maths 9th

Last Answer : Given a quadrilateral ABCD with AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm. For ∆ABC, a = AB = 3 cm, b = BC = 4 cm and c = AC = 5 cm Now, area of quadrilateral ABCD = area of ∆ABC + area of ∆ACD = 6 cm2 + 9.2 cm2 = 15.2 cm2 (approx.)

Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540 cm. Find its area. -Maths 9th

Description : Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540 cm. Find its area. -Maths 9th

Last Answer : Let the sides of the triangle be a = 12x cm, b = 17x cm, c = 25x cm Perimeter of the triangle = 540 cm Now, 12x + 17x + 25x = 540 ⇒ 54x = 54 ⇒ x = 10 ∴ a = (12 x10)cm = 120cm, b = (17 x 10) cm = 170 cm and c = (25 x 10)cm = 250 cm Now, semi-perimeter, s = 5402cm = 270 cm

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 triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m (see figure). The advertisements yield an earning of ₹5000 per m² per year. -Maths 9th

Description : The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m (see figure). The advertisements yield an earning of ₹5000 per m² per year. -Maths 9th

Last Answer : Let the sides of the triangular will be a = 122m, b = 12cm, c = 22m Semi-perimeter, s = a+b+c2 (122+120+224)m = 2642 m = 132m The area of the triangular side wall Rent for 1 year (i.e. 12 months) per m2 = ... = Rs. 5000 x 312 = Rent for 3 months for 1320 m2 = Rs. 5000 x 312 x 1320 = Rs. 16,50,000.

A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with side a. Find the area of the signal board, using Heron’s formula. -Maths 9th

Description : A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with side a. Find the area of the signal board, using Heron’s formula. -Maths 9th

Last Answer : Let each side of the equilateral triangle be a. Semi-perimeter of the triangle,

p(x)=x3+3x2+3x+1, g(x) = x+2 -Maths 9th

Description : p(x)=x3+3x2+3x+1, g(x) = x+2 -Maths 9th

Last Answer : p(x) = x3+3x2+3x+1, g(x) = x+2 g(x) = 0 ⇒ x+2 = 0 ⇒ x = −2 ∴ Zero of g(x) is -2. Now, p(−2) = (−2)3+3(−2)2+3(−2)+1 = −8+12−6+1 = −1 ≠ 0 ∴By factor theorem, g(x) is not a factor of p(x

Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases: (i) p(x) = 2x3+x2–2x–1, g(x) = x+1 -Maths 9th

Description : Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases: (i) p(x) = 2x3+x2–2x–1, g(x) = x+1 -Maths 9th

Last Answer : Solution: p(x) = 2x3+x2–2x–1, g(x) = x+1 g(x) = 0 ⇒ x+1 = 0 ⇒ x = −1 ∴Zero of g(x) is -1. Now, p(−1) = 2(−1)3+(−1)2–2(−1)–1 = −2+1+2−1 = 0 ∴By factor theorem, g(x) is a factor of p(x

x4+3x3+3x2+x+1 -Maths 9th

Description : x4+3x3+3x2+x+1 -Maths 9th

Last Answer : Solution: Let p(x)= x4+3x3+3x2+x+1 The zero of x+1 is -1. p(−1)=(−1)4+3(−1)3+3(−1)2+(−1)+1 =1−3+3−1+1 =1 ≠ 0 ∴By factor theorem, x+1 is not a factor of x4+3x3+3x2+x+1

Determine which of the following polynomials has (x + 1) a factor: (i) x3+x2+x+1 -Maths 9th

Description : Determine which of the following polynomials has (x + 1) a factor: (i) x3+x2+x+1 -Maths 9th

Last Answer : Solution: Let p(x) = x3+x2+x+1 The zero of x+1 is -1. [x+1 = 0 means x = -1] p(−1) = (−1)3+(−1)2+(−1)+1 = −1+1−1+1 = 0 ∴By factor theorem, x+1 is a factor of x3+x2+x+1

3. Check whether 7+3x is a factor of 3x3+7x. -Maths 9th

Description : 3. Check whether 7+3x is a factor of 3x3+7x. -Maths 9th

Last Answer : Solution: 7+3x = 0 ⇒ 3x = −7 ⇒ x = -7/3 ∴Remainder: 3(-7/3)3+7(-7/3) = -(343/9)+(-49/3) = (-343-(49)3)/9 = (-343-147)/9 = -490/9 ≠ 0 ∴7+3x is not a factor of 3x3+7x

Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer. (i) 4x2–3x+7 -Maths 9th

Description : Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer. (i) 4x2–3x+7 -Maths 9th

Last Answer : Solution: The equation 4x2–3x+7 can be written as 4x2–3x1+7x0 Since x is the only variable in the given equation and the powers of x (i.e., 2, 1 and 0) are whole numbers, we can say that the expression 4x2–3x+7 is a polynomial in one variable

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.