Category: Maths Notes PPT

All of the sides of the object are the lateral surface of an object, except its base and top (when they exist). The lateral surface area is the lateral surface zone. This is to be distinguished from the overall surface area, which, along with the base and top areas, is the lateral surface area.

The Lateral Surface Area of the Cube:

A cube has six equal-area faces and 12 equal-length edges. Therefore, the Lateral surface area of the cube is 4 * side².

L.S.A = 4 (a)²

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Lateral Surface Area of Cuboid:

A cuboid is a 3-dimensional figure whose all sides are rectangles. It has 6 rectangular sides which are known as faces, 8 vertices, and 12 edges. The rectangle faces are at right angles to one another. Therefore formula of the lateral surface area of the cuboid is mentioned below:

L.S.A = 2(l + b)h

Where l = length of a cuboid, b is the breadth of cuboid, and h is the height of a cuboid.

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Lateral Surface Area of a Cylinder:

The lateral area of a cylinder is the area of the rectangle which wraps around the side of the cylinder. Therefore formula of the surface area of the cylinder is mentioned below:

L.S.A = 2 * π * r * h

Where π = 22/7 or 3.14, r = radius of a cylinder, and h is the height of a cylinder.

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Lateral Surface Area of Cone:

The lateral surface area of the cone is the area of the lateral or side surface only. The lateral surface area of the cone can be calculated as:

L.S.A = π * r * l

Where π = 22/7 or 3.14, r is the radius m, and l is the slant height. The slant height is defined as the shortest distance between the base to the apex along the surface of the solid.

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Lateral Surface Area of the Sphere:

The lateral surface area of the sphere is related to the lateral surface area of a cylinder.

L.S.A of cylinder = 2π * r * h

Where h = 2r

So, the Lateral Surface area of Sphere= 4 * π * r²

Lateral Surface Area of a Pyramid:

The lateral Surface area of the pyramid is the sum of the areas of its lateral faces.

L.S.A = 1/2 pl

Where p represents the perimeter of the base and l represents the slant height.

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The lateral surface area of a figure is a total surface area excluding the top and bottom of the figure, The formulas for calculating the surface area have been mentioned above with explanation.

While studying mathematics, we come across new concepts of comparing two more quantities with the same unit. In general terms, we consider the numerical values of the quantities and compare them. By performing a ratio between the terms, we can easily understand the outcome of the comparison. For this, we have framed a ratio formula for different techniques of comparison. In this section, we will study how a ratio between two quantities is determined and how we can use this concept to design various types of ratio formulas.

What is a Ratio?

A ratio is the comparison of two quantities of the same kind that helps to compare them easily. This is how we express a fraction where a numerator is compared to a denominator. The definition of numerator and denominator depends on the formula of a ratio used in different aspects of a mathematical calculation. For instance, a simple ratio between the profit margin and the cost price of a product is (profit : cost). When we multiply it by 100, we get the profit percentage.

In this way, the quick ratio formula is also formulated by using the concept and the terms associated with it. The formula is:

Quick Ratio = (Current assets – Inventory)/Current liabilities

This is how the rest of the ratio formulas are determined. The definition is properly studied and the relation between the terms is determined to form a ratio formula.

How a Ratio Formula is Determined?

The first method of comparison a student learns in mathematics is by forming a ratio. He is provided with a set of information where the same physical quantity is available in two different forms. For instance, Ram has 3 mangoes and Ashok has 5 mangoes. If we want to compare the number of mangoes owned by these two kids, we can form a ratio (3 : 5). In this way, other ratio formulas are also formed. You have already learned what the quick ratio formula stands for. Let us consider a few technical examples here.

Return on Equity Formula

This ratio formula is used by the commerce students in their advanced subjects. The return on equity formula is:

Return on Equity (ROE) = Net income/shareholder’s equity

Net Profit Margin Formula

This formula is one of the most common formulas you have studied in previous classes. Learn the definition of the terms used in the net profit margin formula.

Net Profit Margin = (Revenue – Cost)/Revenue

Debt Ratio Formula

The debt ratio formula is also called the debt-to-equity formula. It is used in the commerce subjects to calculate the ratio of debts to assets. The formula is

Debt Ratio = total debts/total assets

Gearing Ratio Formula

There are different kinds of gearing ratios used in the companies to understand the current financial conditions. In this section, you will study different types of gearing ratios such as debt to equity ratio, equity ratio, debt ratio, debt to capital ratio, etc.

In mathematics, a sequence is referred to as a systematic list of numbers. The numbers in the list are actually the terms of the sequence. A series is the summation of all the terms of a sequence. Sequence and series are like sets. However, the only difference between them is that in a sequence, individual terms can take place repeatedly in different positions. The length of a sequence is equivalent to the number of terms, which could either be finite or infinite. Below we will also learn the number series formula with which we are able to solve the number series math questions.

Number Series Formula

The sum of the first n terms of a sequence is what we call a series.

If a sequence is geometric or arithmetic then there are specified formulas to calculate the sum of the first n terms, represented as S[_{n}], without really adding all of the terms.

Formula for Sum of the Terms of an Arithmetic Series

In order to calculate the sum of the first n terms of an arithmetic sequence, we use the following formula,

S[_{n}] = n (a[_{1}] + a[_{2}])/2

Where,

n = number of terms

a[_{1}]  = the first term

a[_{n}] = the last term.

Note: In mathematical terms, a sequence can neither be arithmetic nor geometric.

Formula for Sum of the Terms of a Geometric Series

In order to calculate the sum of the first n terms of a geometric sequence, we use the following formula,

S[_{n}] = a[_{1}] (1−r[^{n}])/1−r, r≠1

Where,

n = number of terms

a[_{1}] = the first term

r = common ratio

Sequence and Series

A sequence is referred to as an ordered list of numbers. The numbers in that list are known as the terms of the sequence. The terms of a sequence are generally named as a[_{i}] or a[_{n}], having a subscripted letter i or n being the index. Thus, the second term of a sequence might be named a[_{2}], and a[_{12}] will be the twelfth term.

A series on the other hand is called the sum of all the terms in a sequence. But, there has to be a definite link between all the terms of the sequence.

S[_{N}] = a[_{1}] + a[_{2}] + a[_{3}] +… + a[_{n}]

Solved Examples

Example:

Determine the sum of the first 50 terms of the arithmetic sequence

Solution:

2, 5, 8, 11, 14, 19, 22, 25, 28, 31⋯

We will have to first find the 50th term:

A[_{50}] = a1+ (n−1) d        

=2+49(3)

=149

Then find the sum:

S[_{n}] = n(a[_{1}] + a[_{n}])/2

S[_{50}] = 50(2 + 149)/2

=3775

Example:

Identify S[_{10}] of the geometric series 24+12+6+……

Solution:

First, we will find r

r = r[_{2}]/r[_{1}] = 12/24

=12

Now, we will find the sum:

S[_{10}] = 24(1− (1/2)[^{10}])/1−1/2

=3069/64

A Fourier series is an enlargement of a periodic function f(x) with respect to an infinite sum of sines and cosines. The Fourier series considers the orthogonality links of the sine and cosine functions. The study and measure of Fourier series is referred to harmonic analysis and is tremendously useful to break up an arbitrary periodic function into a set of simple terms, which can be plugged in, solved separately, and then recombined to gain the solution to the actual problem or estimation to it to whatever appropriateness is desired or practical.

Fourier Series Formula

Following is the Fourier series formula:-

f(x) = 1/2a0 + ∑∞n = 1a_n cosnx + ∑∞n=1 b_n sinnx

Where,

a0 = 1/π∫^π_−pi f(x)dx

a_n = 1/π∫^π_−π f(x)sin nx dx

b_n = 1/π∫^π_−π f(x)sin nx dx

To know more about the Fourier series formula derivation as well as Fourier series notes, refer to — a top educational portal.

Finding the Coefficients Using Fourier Series Coefficients Formula

How did we know to implement sin (5x)/5, sin (7x)/7, etc?

There are formulas! Formulas like Fourier coefficients formula.

Below is a series of sines and cosines having a name for all coefficients:

f(x) = a0 + ∞n=1an cos(nxπ/L) + ∞n=1 bn sin(nxπ/L)

Where:

f(x) =  the function we desire (such as a square wave)

L = half of the period of the function

a0, an and bn = coefficients that we require to calculate!

To calculate the coefficients a0, an and bn we use the given below Fourier series formula list:

a0 = 1/2L ∫L−L f(x) dx

an = 1/L ∫L−L f(x) cos(nxπ/L) dx

bn = 1L ∫L−L f(x) sin(nxπ/L) dx

 

Solved Examples

Example:

Identify the Fourier series for f(x)=x f(x)=x on −L≤x≤L−L≤x≤L.

Solution:

Let’s begin with the integrals for An

A0=1/2L∫L−Lf(x)dx=1/2L∫L−Lxdx=0

An=1/L∫L−Lf(x)cos(nπx/L)dx=1/L∫L−Lxcos(nπx/L)dx=0

In both cases remember that we are integrating an odd function (x is odd and cosine is even, thus the product is odd) over the interval [−L, L] and thus we know that both of these integrals will be zero (0).

Next, is the integral for Bn

Bn=1/L∫L−Lf(x)sin(nπx/L)dx=1/L∫L−Lxsin(nπx/L)dx=2/L∫L0xsin(nπx/L)dx

In this case we are integrating an even function (x and sine are both odd thus the product is even) on the interval [−L,L] and thus we can also “simplify” the integral as above. Now, Using the previous result we get,

Bn=(−1)n+12L/nπ   n=1,2,3…

In the case the Fourier series is,

f(x) = ∞∑n =0A_ncos(nπx/L) + ∞∑n=1Bnsin(nπx/L) = ∞∑n=1(−1)n+12L/nπ sin(nπx/L)

A cube is a three-dimensional solid object with six square faces, facets, or sides, three of which meet at each vertex in geometry. The cube is one of the five Platonic solids and is the only regular hexahedron. It is made up of six faces, twelve edges, and eight vertices. The cube is also a right rhombohedron, a square parallelepiped, and an equilateral cuboid. In three orientations, it’s a regular square prism, and in four, it’s a trigonal trapezohedron. The only convex polyhedron with all square faces is the cube.

How to Find the Volume of a Cube Formula? 

The three-dimensional space enclosed by a boundary or filled by an object is known as volume. A cube generally has three fundamental dimensions: length, breadth and height. The volume of the cube can be determined by knowing all the fundamental dimensions.

Since the cube has the same dimensions so we can consider the length, breadth and height as equal. Here let us take length, breadth and height as equal to ‘s’.

Let ‘V’ be the volume of the cube and ‘s’ be the dimensions of the cube. So the volume of a cube formula will be as follows:

V = length × breadth × height

V = s × s × s = s3 cubic units

Steps to Find the Volume of a Cube Formula

• Take a square sheet of paper as an example.

• The area occupied by the square sheet will now be its surface area, which is equal to its length multiplied by its width.

• The surface area of the square will be ‘s2’ since the length and width will be equal.

• A cube is now created by piling several square sheets on top of one another until the height reaches ‘s’ units. The height or thickness of the cube is denoted by the letter ‘s’.

• The total area covered by the cube will now be equal to the area of the base multiplied by the height.

• So, the volume of a cube equation will be V = s × s × s = s3 cubic units.

Volume Formula of Cube and Cuboid

A cube is a three-dimensional figure with equal sides, i.e., all six faces are square. A cuboid, on the other hand, is a three-dimensional figure with non-equal sides and all six faces being rectangles.

The volume formula of cube and cuboid is the same : length × breadth × height. This formula is also known as the formula of cubic meters.

Volume of a Cube Formula When Diagonal is Given

When the three dimensions length, breadth and height are not given for a cube. But the diagonal side value is provided, then the formula to calculate the volume of a cube is as follows:

V = [sqrt{3}][frac{d^{3}}{9}]

Where V is the volume of the cube and d is the diagonal side of the cube.

Surface Area of the Cube

A solid object’s surface area is a measurement of the total area occupied by the object’s surface.

The general formula to calculate the surface area of a cube is 

A = 6s2

Where A is the surface area of the cube and ‘s’ is the dimension of the sides of the cube.

Solved Problems

1. Calculate the Surface Area and Volume of the Cube Having the Sides of Length Equal to 8 Cm.

Ans: The surface area of the cube is given by the formula

A = 6s2

Now substituting the values we get the surface area of a cube as follows:

A = 6 × 8 × 8

A = 384 cm2

The Volume of a cube formula is given as follows:

V = s × s × s = s3 

Substituting the values we get the volume of cube as follows:

V = 8 × 8 × 8 

V = 512 cm3.

Therefore, the surface area and volume of a cube having the sides of length equal to 8 cm is 384 cm2 and 512 cm3 respectively.

2. Calculate the Volume of a Cube if the Diagonal of the Cube is Given as 3 Cm.

Ans: The volume of a cube formula when diagonal is given is as follows:

V = [sqrt{3}][frac{d^{3}}{9}]

Now substituting the values we get the volume of the cube as follows:

V = [sqrt{3}][frac{3^{3}}{9}]

V = [sqrt{3}][frac{27}{9}]

V = [sqrt{3}] x 3

V = 3[sqrt{3}] cm3

Therefore the volume of the cube if the diagonal of the cube is given as 3cm is 3[sqrt{3}] cm3.

3. Find the Length of the Edges of the Cube if the Volume of the Cube is Given as 64 Cm3.

Ans: The volume of the cube formula is given as follows:

V = s × s × s = s3 

Since we are given the volume of a cube and we have to find the side of the cube. Let us substitute the values into the formula.

64 = s3

Taking cube roots on both sides of the equation we will get the length of the edges of the cube.

[sqrt[3]{64}] = s

We know that [sqrt[3]{64}] is equal to 4.

So, s = 4 cm.

Therefore, the length of the edges of the cube if the volume of the cube is given as 64 cm3 is 4 cm.

Conclusion

The 3-dimensional space enclosed by a boundary or filled by an object is known as volume. Finding the volume of an object will help us figure out how much water is needed to fill it, such as how much water is needed to fill a container, an aquarium, or a water tank. If we want to decide how much material is needed to cover an object’s surface, we must first determine its surface area.

Here let us have a look at all formula of coordinate geometry Class 10.

• Distance Formula

• Section Formula

• Area of a Triangle

A detailed explanation of coordinate geometry class 10 all formulas are given below.

Coordinate Geometry Class 10 Formulas – Distance Formula

To calculate the distance between two points the distance formula is used. Making a triangle by using the Pythagorean theorem to find the length of the hypotenuse gives the distance formula. The distance between the two points in a triangle is called the hypotenuse.

The distance formula can also be used to calculate the lengths of all the sides of a polygon, the perimeter of polygons on a coordinate plane, the area of polygons, and several other things.

The distance formula is denoted by ‘d’.

Distance Formula Between 2 Points in a 2D Plane:

Consider 2 points P and Q having the 2D coordinates as (x1,y1) and (x2,y2) respectively.

Now the distance between these 2 points in the 2D plane is

[d = PQ = sqrt{(x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2}}]

Distance Formula Between 2 Points in a 2D Plane in Polar Coordinates:

Consider 2 points P and Q having the 2D polar coordinates as (r1,θ1) and (r2,θ2) respectively.

Now the distance between these 2 points in the 2D plane is

[d = PQ = sqrt{r_{1}^{2} + r_{2}^{2} – 2r_{1}r_{2} cos(theta_{1} – theta_{2})}]

Distance Formula Between 2 Points in a 3D Plane:

Consider 2 points P and Q having the 3D coordinates as (x1,y1,z1) and (x2,y2,z2) respectively.

Now the distance between these 2 points in the 3D plane is

[d = PQ = sqrt{(x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2} + (z_{2} – z_{1})^{2}}]

Distance Formula Between a Point and a Line

Consider a straight line Ax + By + C = 0 and point P having the 2D coorrdinates as (x1, y1).

Now the distance between the line Ax + By + C = 0 and a point P (x1, y1) is

[d = frac{|Ax + By + C|}{sqrt{A^{2} + B^{2}}}]

Distance Formula Between Two Parallel Points

Consider two parallel lines y = mx + c1 and y = mx + c2.

Now the distance between these two parallel lines is given by

[d = frac{|c_{1} – c_{2}|}{sqrt{A^{2} + B^{2}}}]

Problems on Distance formula

1. Find the distance between two points A and B which are having the 2D coordinates as (4, 8) and (3, 6) respectively.

Ans: The distance between these 2 points in the 2D plane is

[d = AB = sqrt{(x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2}}]

Here x1 = 4, x2 = 8, y1 = 3 and y2 = 6. Now substituting these values in the distance formula we get,

[d = AB = sqrt{(8 – 4)^{2} + (6 – 3)^{2}}]

[d= AB = sqrt{(4)^{2} + (3)^{2}}]

[d = AB = sqrt{16 + 9}]

[d = AB = sqrt{25}]

d = AB = 5 units

So the distance between 2 points A and B is 5 units.

Coordinate Geometry Formulas Class 10 – Section Formula

The Section formula is used in coordinate geometry to find the ratio in which a line segment is separated by a point, either internally or externally. It is used to determine the triangle’s centroid, incenter, and excenters.

Section Formula When the Line Segment Separated Internally by a Point

Consider a point P (x, y) dividing the line segment joining the points A(x1,y1) and B(x2,y2) internally in the ratio m:n.

Now the section formula for a line segment separated internally by a point is:

[P(x, y) = (frac{mx_{2} + nx_{1}}{m + n}, frac{ny_{2} + ny_{1}}{m + n})]

Section Formula When the Line Segment Separated Externally by a Point

Consider a point P (x, y) dividing the line segment joining the points A(x1, y1) and B(x2, y2) externally in the ratio m:n.

Now the section formula for a line segment separated externally by a point is:

[P(x, y) = (frac{mx_{2} – nx_{1}}{m – n}, frac{ny_{2} – ny_{1}}{m – n})]

Section Formula When Line Segment Separated Internally by a Point in the Ratio K:1

Consider a point P (x, y) dividing the line segment joining the points A(x1,y1) and B(x2,y2) internally in the ratio K:1.

Now the section formula for a line segment separated internally by a point is:

[P(x, y) = (frac{Kx_{2} + x_{1}}{K + 1}, frac{Ky_{2} + y_{1}}{K + 1})]

Section Formula When Line Segment Separated Externally by a Point in the Ratio K:1

Consider a point P (x, y) dividing the line segment joining the points A(x1,y1) and B(x2,y2) externally in the ratio K:1.

Now the section formula for a line segment separated externally by a point is:

[P(x, y) = (frac{Kx_{2} – x_{1}}{K – 1}, frac{Ky_{2} – y_{1}}{K – 1})]

Problems on Section Formula

1. Two points A (5,8) and B (3,6) on the line segment are separating the point P (x,y) internally in the ratio 2:4. By using the section formula find the ratio in which a line segment is separated by a point.

Ans: The section formula for a line segment separated internally by a point is given by the formula:

[P(x, y) = (frac{mx_{2} + nx_{1}}{m + n}, frac{ny_{2} + ny_{1}}{m + n})]

Here x1 = 5, x2 = 8, y1 = 3 and y2 = 6, m=2, n=4. Now substituting these values in the section formula we get,

[P(x, y) = (frac{(2 times 8) + (4 times 5)}{2 + 4}, frac{(2 times 6) + (4 times 3)}{2 + 4})]

[P(x, y) = (frac{16 + 20}{6}, frac{12 + 12}{6}]

[P(x, y) = (frac{36}{6}, frac{24}{6})]

Px, y = 6, 4

The ratio in which the line segment separated by a point internally is (6,4).

2. Two points A (5,8) and B (3,5) on the line segment are separating the point P (x,y) externally in the ratio 2:4. By using the section formula find the ratio in which a line segment is separated by a point.

Ans: the section formula for a line segment separated externally by a point is 

[P(x, y) = (frac{mx_{2} – nx_{1}}{m – n}, frac{ny_{2} – ny_{1}}{m – n})]

Here x1 = 5, x2 = 8, y1 = 3 and y2 = 5, m=2, n=4. Now substituting these values in the section formula we get,

[P(x, y) = (frac{(2 times 8) – (4 times 5)}{2 – 4}, frac{(2 times 5) – (4 times 3)}{2 – 4})]

[P(x, y) = (frac{16 – 20}{-2}, frac{10 – 12}{-2}]

[P(x, y) = (frac{-4}{-2}, frac{-2}{-2})]

P(x, y) = (2, 1)

The ratio in which the line segment separated by a point externally is (2, 1).

Formulas of Coordinate Geometry Class 10 – Area of a Triangle

The area of a triangle is defined as the total area enclosed by the triangle’s three sides.

Here in this coordinate geometry all formula Class 10 we will discuss the area of the triangle of three different triangles namely right-angled triangle, equilateral triangle, and isosceles triangle.

Area of a Right Angled Triangle

A right-angled triangle is one in which one of the angles is the right angle. The hypotenuse is the hand opposite the right angle. Legs are the sides that are adjacent to the right angle.

The area of the right-angled triangle is:

A = ½ × Base × Height

Area of an Equilateral Triangle

An equilateral triangle is one in which all three sides are the same length and all three internal angles are congruent and 60°each.

The area of an equilateral triangle is:

[A = frac{sqrt{3} times a^{2}}{4}]

Area of an Isosceles Triangle

A triangle with two sides of equal length is called an isosceles triangle.

The area of an isosceles triangle is:

A = ½ × Base × Height

Area of Triangle With all 3 Sides Different

When all 3 sides of a triangle are different, then to find the area of the triangle we will use Heron’s formula.

When the lengths of all three sides are known, Heron’s formula, named after Hero of Alexandria, measures the area of a triangle. Unlike other triangle area formulas, no angles or other distances in the triangle must be determined first.

Heron’s formula to calculate the area of the triangle is:

[A = sqrt{s(s – a)(s – b)(s – c)}]

Where ‘s’ is the semi perimeter of the triangle and a, b and c are the sides of the triangle.

Semi perimeter of the triangle is given by:

S = (a + b + c)/2

Problems on Area of Triangle

1. If the base of the right-angled triangle is 4 cm and height is 6 cm. Find the area of the triangle.

Ans: The area of the right-angled triangle is:

A = ½ × Base × Height

Here Base = 4 cm and Height = 6 cm. Substituting these values in the formula we get,

A = ½ × 4 × 6

A = ½ × 24

A = 12 cm2.

Therefore the area of the right-angled triangle is 12 cm2.

2. If the three sides of a triangle are having the same length equal to 6 cm. Find the area of the triangle and name the type of triangle.

Ans: If all three sides are the same length then the triangle is an equilateral triangle.

The area of an equilateral triangle is:

[A = frac{sqrt{3} times a^{2}}{4}]

Here ‘a’ is the sides of the triangle given as 6cm. So substituting this value we get 

[A = frac{sqrt{3} times 6^{2}}{4}]

[A = frac{sqrt{3} times 36}{4}]

[A = sqrt{3} times 9]

[A = 9sqrt{3}] cm2

Therefore the area of the equilateral triangle is [9sqrt{3}] cm2.

Conclusion

One of the most important and exciting concepts in mathematics is coordinate geometry. By the use of graphs of lines and curves, it connects algebra and geometry. This allows algebraic solutions to geometric problems and offers geometric insights into algebra. These coordinate geometry Class 10 formulas will help students to understand the basic concepts of geometrical structures and to apply them in our daily uses too.