Category: Class 10

Students can go through AP SSC 10th Class Maths Notes Chapter 14 Statistics to understand and remember the concepts easily.

AP State Syllabus SSC 10th Class Maths Notes Chapter 14 Statistics

→ Statistics is a branch of mathematics which deals with collection, organisation, presentation, analysis and interpretation of numerical data.

→ Data is a collection of actual information which is used to make logical inferences.

→ Arithmetic Mean of raw data:
The Arithmetic Mean (A.M.) of a raw data viz. x₁, x₂, x₃, ……., x_n is the sum of values of all observations divided by the number of observations.
Arithmetic Mean (A.M.) =
Eg.: Sita secured 23, 24, 24, 22 and 20 marks in a test. Her mean marks are
A.M. = \(\frac{23+24+24+22+20}{5}\) = \(\frac{113}{5}\) = 22.6

→ A.M. by direct method:
Let x₁, x₂, x₃, ……., x_n be observations with respective frequencies f₁, f₂, ……, f_n
i.e., x₁ occurs for f₁ times, x₂ occurs for f₂ times, ….., x_n occurs for f_n times.

→ For a grouped data, it is assumed that the frequency of each class interval is centered around its mid-point and the A.M. is given by A.M. = \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)

→ A.M. by deviation method, \(\overline{\mathbf{x}}=\mathbf{a}+\frac{\Sigma \mathbf{f}_{\mathbf{i}} \mathbf{d}_{\mathbf{i}}}{\Sigma \mathbf{f}_{\mathbf{i}}}\)
where, a – assumed mean
d_i – deviation = x_i – a.
Step – 1: Choose ‘a’ from the central values.
Step – 2: Obtain d_i by subtracting a from x_i.
Step – 3: Multiply f_i and d_i.
Step – 4: Find ∑f_id_i and ∑f_i .
Step – 5: Find \(\overline{\mathbf{x}}=\mathbf{a}+\frac{\Sigma \mathbf{f}_{\mathbf{i}} \mathbf{d}_{\mathbf{i}}}{\Sigma \mathbf{f}_{\mathbf{i}}}\)

→ A.M. by step-deviation method:

Step – 1: Choose ‘a’ from mid values.
Step – 2: Obtain u_i = \(\frac{x_{i}-a}{h}\).
Step – 3: Multiply f_i and u_i.
Step – 4: Find Ef_iu_i and Sf_i.
Step – 5: Find \(\overline{\mathrm{x}}=\mathrm{a}+\left(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\right) \times \mathrm{h}\)

→ Mode : Mode is the size of variable which occurs most frequently.

→ Mode of a grouped data:

Where, l – lower boundary of the modal class,
h – size of the modal class interval,
f₁ – frequency of modal class.
f₀ – frequency of the class preceding the modal class.
f₂ – frequency of the class succeeding the modal class.

→ Median: Median is defined as the measure of the central items when they are in descending or ascending order of magnitude.

→ Median for a grouped data:

where,
l – lower boundary of median class,
n – number of observations.
cf – cumulative frequency of class preceding the median class.
f – frequency of median class.
h – size of the median class.

→ Cumulative frequency curve or an ogive:
First we prepare the cumulative frequency table, then the cumulative frequencies are plotted against the upper or lower limits of the corresponding class intervals. By joining the points the curve so obtained is called a cumulative frequency or ogive.
Ogives are of two types.

1. Less than ogive: Plot the points with the upper limits of the classes as abscissa and the corresponding less than cumulative frequencies as ordinates. The points are joined by free hand smooth curve to give less than cumulative frequency curve or the less than ogive. It is a rising curve.
2. Greater than ogive: Plot the points with the lower limits of the classes as abscissa and the corresponding greater than cumulative frequencies as ordinates. Join the points by a free hand smooth curve to get the greater than ogive. It is a falling curve.

When the points are joined by straight lines, the figure obtained is called cumulative frequency polygon.

→ Median can be obtained from cumulative frequency curve: From \(\frac{n}{2}\) frequency draw a line parallel to X-axis cutting the curve at a point. From this point draw a perpendicular to the axis. The point at which the perpendicular meets the X – axis determines the median.

Less than type and greater than type curves intersects at a point. From this point of intersection if we draw a perpendicular on the X-axis then this cuts X-axis at some point. This point gives the median.

Students can go through AP SSC 10th Class Maths Notes Chapter 13 Probability to understand and remember the concepts easily.

AP State Syllabus SSC 10th Class Maths Notes Chapter 13 Probability

→ Theory of probability has its origin date back to 16th century.

→ J. Cardan, an Italian physician and mathematician wrote the first book on probability named “The Book of Games of Chance”.

→ James Bernoulli (1654 – 1705), A.De Moivre (1667 – 1754) and Pierre Simon Laplace (1749 – 1827) made significant contribution to the theory of probability.

→ Experimental or empirical probability : The probability estimated on the basis of results of an actual experiment is called experimental probability of empirical probability.
Eg : An unbiased coin is tossed for 1000 times, head turned up for 455 times and tail turned up 545 times, then the probability or likelyhood of getting a head is = \(\frac{455}{1000}\) = 0.455.

Thus experimental probability = \(\frac{\text { No. of trials in which the event happened }}{\text { Total no. of trials }}\)

→ Classical or Theoretical probability: Classical probability of an event (E) is defined Number as

P(E) = \(\frac{\text { Number of outcomes favourable to E }}{\text { No. of all possible outcomes of the experiment }}\)

This definition was given by ‘Pierre Simon Laplace’.
Eg: The probability of getting a head when a coin is tossed is given by Number of outcomes favourable to this event getting a head = 1 Number of all possible outcomes of this experiment = 2 (Head, Tail)

∴ P(E) = \(\frac{\text { No. of favourable outcomes }}{\text { Total events }}\) = \(\frac{1}{2}\)

Note: If an experiment is conducted for many number of times, then the experimental probability may become closer and closer to theoretical probability.

→ The probability of a sure event is 1.

→ The probability of an impossible event is zero.

→ The probability of an event E is a number P(E) such that 0 ≤ P(E) ≤ 1.

→ An event having only one outcome is called an elementary event. The sum of the probabilities of all the elementary events of an experiment is 1.

→ For any event E, P(E) + P(\(\overline{\mathrm{E}}\)) = 1, where E and \(\overline{\mathrm{E}}\) are complementary events.

→ Playing cards and their probability : A deck of playing cards consists of 52 cards which are divided into four suits of 13 cards each.
They are:

→ The cards in each suit are:

Eg : When a card is drawn at random from a deck of cards then

• Getting a black or red card – equally likely exhaustive events.
• Getting an ace or king – mutually exclusive.
  
• Getting an ace or a hearts – not mutually exclusive since the hearts contain an ace.
  

→ When a coin is tossed, the outcomes are H, T (Head, Tail).

→ When a dice is thrown the outcomes are 1, 2, 3, 4, 5 and 6.

→ When two dice are thrown, the outcomes are

→ If a coin is tossed n-times or n – coins are tossed simultaneously, then the number of total outcomes = 2ⁿ.

→ If a dice is thrown for n – times or n – dice are thrown simultaneously then the number of total outcomes = 6ⁿ.

Students can go through AP SSC 10th Class Maths Notes Chapter 12 Applications of Trigonometry to understand and remember the concepts easily.

AP State Syllabus SSC 10th Class Maths Notes Chapter 12 Applications of Trigonometry

→ If a person is looking at an object then the imaginary line joining the object and the eye of the observer is called the line of sight or ray of view.

→ An imaginary line parallel to earth surface and passing through the point of observation is called the horizontal.

→ If the line of sight is above the horizontal then the angle between them is called “angle of elevation”.

→ If the line of sight is below the horizontal then the angle between them is called the angle of depression.

→ Useful hints to solve the problems:

1. Draw a neat diagram of a right triangle or a combination of right triangles if necessary.
2. Represent the data given on the triangle.
3. Find the relation between known values and unknown values.
4. Choose appropriate trigonometric ratio and solve for the unknown.

→ The height or length of an object or the distance between two distant objects can be determined with the help of trigonometric ratios.

→ To use this application of trigonometry, we should know the following terms.

→ The terms are Horizontal line, Line of Sight, Angle of Elevation and Angle of Depression.

→ Horizontal line: A line which is parallel to earth from observation point to object is called “horizontal line”.

→ Line of Sight (or) Ray of Vision: The line of sight is the line drawn from the eye of an observer to the point in the object viewed by the observer.

→ Angle of Elevation: The line of sight is above the horizontal line then angle between the line of sight and the horizontal line is called “angle of elevation”.

Note:

1. If the angle of observer moves towards the perpendicular line (pole/tree/ building), then angle of elevation increases and if the observer moves away from the perpendicular line (pole/tree/building), then angle of elevation decreases.
2. If height of tower is doubled and the distance between the observer and foot of the tower is also doubled, then the angle of elevation remains same.
3. If the angle of elevation of sun above a tower decreases, then the length of shadow of a tower increases.

→ Angle of Depression: The line of sight is below the horizontal line then angle between the line of sight and the horizontal line is called angle of depression.

Note:

1. The angle of elevation and depression are always acute angles.
2. The angle of elevation of a point P as seen from a point ‘O’ is always equal to the angle of depression of ‘O’ as seen from P.

→ Points to be kept in mind:
I. Trigonometric ratios in a right triangle:

II. Trigonometric ratios of some specific angles:

→ Solving Procedure:
When we want to solve the problems of height and distances, we should consider the following :

1. All the objects such as tower, trees, buildings, ships, mountains, etc. shall be considered as linear for mathematical convenience.
2. The angle of elevation or angle of depression is considered with reference to the horizontal line.
3. The height of the observer is neglected, if it is not given in the problem.
4. To find heights and distances, we need to draw figures and with the help of these figures we can solve the problems.

Students can go through AP SSC 10th Class Maths Notes Chapter 11 Trigonometry to understand and remember the concepts easily.

AP State Syllabus SSC 10th Class Maths Notes Chapter 11 Trigonometry

→ In our daily life, we can measure the heights, distances and slopes by using some mathematical techniques.

→ The mathematical techniques which come under a branch of mathematics is called ‘trigonometry’.

→ “Trigonometry” is the study of relationships between the sides and angles of a triangle.

→ Early astronomers used to find out the distances of the stars and planets from the Earth. Even today, most of the technologically advanced methods used in engineering and physical sciences are based on trigonometrical concepts.

→ Naming the sides in a right triangle:
Let’s take a right triangle ABC as shown in the figure.

Consider ∠CAB as A where angle ‘A’ is acute.
Since AC is the longest side, it is called “Hypotenuse”.

→ Now observe the position of side BC with respect to angle A. It is opposite to angle ‘A’ and we can call it as “Opposite side of angle A”.

→ And the remaining side AB can be called as “Adjacent side of angle A”.

→ Trigonometric Ratios:
The ratios of the sides of a right angled triangle with respect to its acute angles, are called Trigonometric ratios.

→ Consider a right angled triangle ABC having right angle at B as shown in the given figure.

Then, trigonometric ratios of the angle A in right angled triangle ABC are defined as follows:

→ There are three more ratios defined in trigonometry which are considered as multiplicative inverse of the above three ratios.

→ Multiplicative inverse of “sine A” is “cosecant A”.
i.e., cosec A = \(\frac{1}{\sin A}\) = \(\frac{\text { Hypotenuse }}{\text { Opposite side of the angle } A}\)

→ Multiplicative inverse of “cosine A” is “secant A”.
i.e., sec A = \(\frac{1}{\cos A}\) = \(\frac{\text { Hypotenuse }}{\text { Adjacent side of the angle } A}\)

→ Multiplicative inverse of “tangent A” is “cot A”.
i.e., cot A = \(\frac{1}{\tan A}\) = \(\frac{\text { Adjacent side of the angle } A}{\text { Opposite side of the angle } A}\)

Note:
i) The values of the trigonometric ratios of an angle do not vary with the lengths of the sides of the triangles, if the angle remains the same.
ii) Each trigonometric ratio is a real number and has no unit.
iii) “sin θ” is one symbol and sin, cos, tan etc., cannot be separated from θ.
iv) If one of the trigonometric ratios of an acute angle is known, the remaining trigonometric ratios of angle can be easily determined.
v)

→ Trigonometric ratios of some specific angles:
The values of various trigonometric ratios of 0°, 30°, 45°, 60° and 90°.

Note:
i) The values of “sin θ” and “cos θ” always lie between ‘0’ and ‘1’.
ii) In the case of tan θ, the values increase from 0 to ∞ (not determinate).
iii) In the case of cot θ, the values decrease from ∞ to 0.
iv) In the case of cosec θ, the values decrease from ∞ to 1.
v) In the case of sec θ, the values increase from 1 to ∞.

→ Trigonometric ratios of complementary angles:
Two angles are said to be complementary, if their sum is equal to 90°.
In a right angled triangle, if ∠B = 90°, then ∠A + ∠C = 90° i.e., ∠A and ∠C form a pair of complementary angles.
If ‘θ’ is an acute angle, then we can prove that
sin (90 – θ) = cos θ
cos (90 – θ) = sin θ
tan (90 – θ) = cot θ
cot (90 – θ) = tan θ
sec (90 – θ) = cosec θ
cosec (90 – θ) = sec θ

→ Trigonometric Identity: An identity equation having trigonometric ratios of an angle is called trigonometric identity. It is true for all the values of the angles involved in it.
We have three major trigonometric identities. They are
i) sin² A + cos² A = 1
ii) sec² A – tan² A = 1
iii) cosec² A – cot² A = 1
Note: sin² θ = (sin θ)² but sin θ² ≠ (sin θ)²

Students can go through AP SSC 10th Class Maths Notes Chapter 10 Mensuration to understand and remember the concepts easily.

AP State Syllabus SSC 10th Class Maths Notes Chapter 10 Mensuration

→ A solid is a geometrical shape with three dimensions namely length, breadth and height.
Eg:

→ A solid has two types of area namely,
a) Lateral Surface Area (L.S.A.)
b) Total Surface Area (T.S.A,)

→ In general, L.S.A. of a solid is the product of its base perimeter and height.
Eg : L.S.A. of a cuboid = 2h(l + b)
L.S.A. of a cylinder = 2πrh

→ The T.S.A. of a solid is the sum of L.S.A. and the areas of its top and base.
Eg : T.S.A. of a cylinder = 2πrh + 2πr²
= 2πr(r + h)

→ In general, the volume of a solid is the product of its base area and height.
V = A. h
Eg: Volume of a cube = a² . a = a³
Volume of a cylinder = πr² . h = πr²h

→ The volume of solid formed by joining two basic solids is the sum of volumes of the constituents.

→ Surface area of the combination of solids: In calculating the surface area of the solid which is a combination of two or more solids, we can’t add the surface areas of all its constituents, because some part of the surface area disappears in the process of joining them.

→ Surface areas and volume of different solid shapes:

→ Some solid figures and their combination shapes:

Students can go through AP SSC 10th Class Maths Notes Chapter 9 Tangents and Secants to a Circle to understand and remember the concepts easily.

AP State Syllabus SSC 10th Class Maths Notes Chapter 9 Tangents and Secants to a Circle

→ The locus of points which are joined by a curve and are equidistant from a fixed point is called a circle. The fixed point here is called the centre of the circle.
(or)
A simple closed curve consisting of all points in a plane which are equidistant from a fixed point is called a circle. The fixed point is its centre and the fixed distance is its radius.

→ The path followed by a circular object is a straight line.

→ The line segment joining any two points on a circle is called a ‘chord’. The longest of all chords of a circle passes through the centre and is called a diameter.

\(\overline{\mathrm{AB}}\) is a chord and \(\overline{\mathrm{PQ}}\) is a diameter.
\(\overline{\mathrm{OP}}\) is the radius of the circle,
diameter = 2 × radius d = 2r
r = \(\frac{d}{2}\)

→ There are three different possibilities for a given line and a circle.

Case (i): The line PQ and the circle have no point in common (or) they do not touch each other.
Case (ii): The line PQ and the circle have two common points (or)
The line PQ intersects the circle at two distinct points A and B. Here the line PQ is a “secant” of the circle.
Case (iii): The line PQ touches the circle at an unique point A (or) there is one and only one point common to both the line and circle.
Here \(\stackrel{\leftrightarrow}{\mathrm{PQ}}\) is called a tangent to the circle at ‘A’.

→ The word tangent is derived from the Latin word “TANGERE” which means “to touch” and was introduced by Danish mathematician“Thomas Fineke” in 1583.

→ There is only one tangent to the circle at one point.

→ The tangent at any point of a circle is perpendicular to the radius through the point of contact.

The radius OP is perpendicular to \(\stackrel{\leftrightarrow}{\mathrm{AB}}\) at P.
i.e, OP ⊥ AB.

→ Construction of a tangent to a circle:
Draw a circle with centre ‘O’.
Take a point ‘P’ on it. Join OP.
Draw a perpendicular line to OP through ‘P’.

Let it be \(\stackrel{\leftrightarrow}{\mathrm{XY}}\)
XY is the required tangent to the given circle passing through P.

→ Let ‘O’ be the centre of the given circle and \(\overline{\mathrm{AP}}\) is a tangent through A where OA is the radius, then the length of the tangent AP = \(\sqrt{\mathrm{OP}^{2}-\mathrm{OA}^{2}}\).

→ Two tangents can be drawn to a circle from an external point.

→ Let ‘O’ be the centre of the circle and P is an exterior point. There are exactly two tangents to the circle through P.
\(\overline{\mathrm{PA}}\) and \(\overline{\mathrm{PB}}\) are the tangents.

Here the lengths of the two tangents drawn from the external points are equal.
\(\overline{\mathrm{PA}}\) = \(\overline{\mathrm{PB}}\)

→ Construction of tangents to a circle from an external point:
Step – 1: Draw a circle with centre ‘O’ and with given radius.
Step – 2: Mark a point ‘P’ in the exterior of the circle and join ‘OP’.
Step – 3: Draw the perpendicular bisector \(\stackrel{\leftrightarrow}{\mathrm{XY}}\) to \(\overline{\mathrm{OP}}\), intersecting at M.
Step – 4: Taking M as centre, MP or OM as radius, draw a circle which intersects the given circle at A and B.

Step – 5: Join PA and PB. PA and PB are the required tangents.

→ Consider a circle with centre ‘O’. PA and PB are the tangents from an exterior point ‘P’. Then, the centre of the circle lies on the bisector of the angle between two tangents drawn from the exterior point P.
∠OPA = ∠OPB

→ Consider two concentric circles with centre ‘O’. Let the chord \(\overline{\mathrm{AB}}\) of the larger/ bigger circle just touches the smaller circle, then it is bisected at the point of contact with the smaller circle.
In the figure, \(\overline{\mathrm{AB}}\) is the chord of bigger circle touching the smaller circle at P then AP = PB.

→ If AP and AQ are two tangents to a circle with centre ‘O’, then ∠PAQ = 2∠OPQ = 2∠OQP.

→ If a circle touches the sides of a quadrilateral ABCD at points P, Q, R and S then AB + CD = BC + DA.
i.e., sum of the opposite sides are equal.

→ The region enclosed by a secant/chord and an arc is called a ‘segment of the circle’.

Case (i): If the arc is a minor arc then the segment is a minor segment.
Case (ii): If the arc is a semi arc then the segment is a semi circle.
Case (iii): If the arc is a major arc then the segment is a major segment.

→ Area of a segment between the chord AB and whose arc makes an angle ‘x’ at the centre = \(\frac{x}{360}\) × πr²

i.e., Area of the segment APB = (Area of the corresponding sector OAPB) – (Area of the corresponding triangle OAB).

Students can go through AP SSC 10th Class Maths Notes Chapter 8 Similar Triangles to understand and remember the concepts easily.

AP State Syllabus SSC 10th Class Maths Notes Chapter 8 Similar Triangles

→ The geometrical figures which have the same shape but are not necessarily of the same size are called similar figures.

→ The heights and distances of distant objects can be found using the principles of similar figures.

→ Two polygons with same number of sides are said to be similar if their corresponding angles are equal and their corresponding sides are in proportion.

→ A polygon in which all sides and all its angles are equal is called a regular polygon. Eg.:

→ The ratio of the corresponding sides is referred to as scale factor or representative factor.

→ All squares are similar.

→ All circles are similar.

→ All equilateral triangles are similar.

→ Two congruent figures are similar but two similar figures need not be congruent.

→ A square ABCD and a rectangle PQRS are of equal corresponding angles, but their corre¬sponding sides are not in proportion.

∴ The square ABCD and the rectangle PQRS are not similar.

→ The corresponding sides of a square ABCD and a rhombus PQRS are equal but their corresponding angles are not equal. So they are not similar.

→ If a line is drawn parallel to one side of a triangle intersecting the other two sides at two distinct points then the other two sides are divided in the same ratio.

In △ABC; DE // BC then \(\frac{AD}{DB}\) = \(\frac{AE}{EC}\).
This is called Basic proportionality theorem (or) Thale’s theorem.

→ If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.

In △ABC, a line ‘l’ intersecting AB in D and AC in E
such that \(\frac{AD}{DB}\) = \(\frac{AE}{EC}\) then l // BC.
This is converse of Thale’s theorem.

→ Two triangles are similar, if
i) their corresponding angles are equal.
ii) their corresponding sides are in the same ratio.

→ If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio or proportional and hence the two triangles are similar.

In △ABC, △DEF
∠A = ∠D
∠B = ∠E
∠C = ∠F
⇒ \(\frac{AB}{DE}\) = \(\frac{BC}{EF}\) = \(\frac{AE}{DF}\)
∴ △ABC ~ △DEF (A.A.A)

→ If in two triangles, sides of one triangle are proportional to the sides of other triangle, then their corresponding angles are equal and hence the two triangles are similar.

In △ABC, △DEF
if \(\frac{AB}{DE}\) = \(\frac{BC}{EF}\) = \(\frac{AE}{DF}\)
⇒ ∠A = ∠D
∠B = ∠E
∠C = ∠F
Hence, △ABC ~ △DEF (S.S.S)

→ If two angles of a triangle are equal to two corresponding angles of another triangle then the two triangles are similar.

In △ABC, △DEF
if ∠A = ∠D
∠B = ∠E
⇒ ∠C = ∠F (By Angle Sum property)
∴ △ABC ~ △DEF (A.A)

→ If one angle of a triangle is equal to one angle of other triangle and the sides including these angles are proportional, then the two triangles are similar.

In △ABC, △DEF if ∠A = ∠D, and
\(\frac{AB}{DE}\) = \(\frac{AC}{DF}\)
⇒ △ABC ~ △DEF (S.A.S)

→ The ratio of areas of two similar triangles is equal to the ratio corresponding sides.

→ If a perpendicular is drawn from the vertex, containing the right angle of a right angled – triangle to the hypotenuse, then the triangles on each side of perpendicular are similar to one another and to the original triangle. Also the square of the perpendicular is equal to the product of the lengths of the two parts of the hypotenuse.
In △ABC, ∠B = 90°
BD ⊥ AC

Then △ADB ~ △BDC ~ △ABC and
BD² = AD . DC

→ Pythagoras theorem: In a right angled triangle, the square of hypotenuse is equal to the sum of the squares of other two sides.

In △ABC; ∠A = 90°
AB² + AC² = BC²

→ In a triangle, if square of one side is equal to sum of squares of the other two sides, then the angle opposite to the first side is right angle.

In △ABC, if
AC² = AB² + BC² then ∠B = 90°
This is converse of Pythagoras theorem.

→ Baudhayan Theorem (about 800 BC):
The diagonal of a rectangle produces itself the same area as produced by its both sides (i.e., length and breadth).
In rectangle ABCD,

area produced by the diagonal AC = AC • AC
= AC²
area produced by the length = AB • BA = AB
area produced by the breadth = BC • CB = BC²
Hence, AC² = AB² + BC².

→ A sentence which is either true or false but not both is called a simple statement.

→ A statement formed by combining two or more simple statements is called a compound statement.

→ A compound statement of the form “If …… then ……” is called a Conditional or Implication.

→ A statement obtained by modifying the given statement by ‘NOT’ is called its negation.

Students can go through AP SSC 10th Class Maths Notes Chapter 7 Coordinate Geometry to understand and remember the concepts easily.

AP State Syllabus SSC 10th Class Maths Notes Chapter 7 Coordinate Geometry

→ The branch of mathematics which deals with points in a coordinate plane is called Coordinate/Analytical geometry.

→ There is a definite one-one correspondence between the position of a point in a plane and a pair of numbers called coordinates.

→ In a Cartesian plane, the position of a point P is determined by two coordinates x and y, where x is the distance of P from Y – axis and y is the distance of P from X – axis. So we denote it by P (x, y).

→ Signs of coordinates in different quadrants:

→ The point of intersection of the axes is called the- origin represented by (0, 0).

→ Any point lying on X – axis is denoted by (x, 0) as its y-coordinate is zero.

→ A point on Y – axis is denoted by (0, y) as its x-coordinate is zero.

→ Distance between any two points P(x₁, y₁) and Q(x₂, y₂) is given by
PQ = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)

→ Distance between any two points P(x₁, 0) and Q(x₂, 0) on the X – axis is | x₂ – x₁ |.
Distance between any two points P(0, y₁) and Q(0, y₂) on the Y – axis is | y₂ – y₁ |

→ The coordinates of the point P which divides the line joining of two points (x₁, y₁) and (x₂, y₂) internally in the ratio m₁ : m₂ is

→ The midpoint of a line joining any two points (x₁, y₁) and (x₂, y₂) is

→ If the ratio in which ‘P’ divides two points (x₁, y₁) and (x₂, y₂) is k : 1, then

→ The point of concurrence of medians of a triangle is called its centroid.

→ Centroid divides each median in the ratio 2 : 1.

→ If (x₁, y₁), (x₂, y₂) and (x₃, y₃) are the vertices of a triangle, then its centroid is

→ The points P, Q on a line segment \(\overline{\mathrm{AB}}\) are said to be the points of trisection, if P and Q divides \(\overline{\mathrm{AB}}\) into three equal points i.e., AP = PQ = QB.

→ If P, Q are points of trisection of \(\overline{\mathrm{AB}}\), then P divides AB in the ratio 1 : 2 and Q divides AB in the ratio 2 : 1.

→ Area of a triangle AOB formed by the points origin O(0, 0), a point on X – axis A (x, 0) and another point on Y – axis B(0, y) is \(\frac{1}{2}\)xy.

→ If (x₁, y₁), (x₂, y₂) and (x₃, y₃) are the vertices of a triangle, then its area is

→ If three points are collinear, then the area of the triangle formed by these points is zero.

→ Let A (x₁, y₁), B (x₂, y₂) and C (x₃, y₃) are the vertices of a triangle △ABC, then AB = c; BC = a and AC = b and

S = \(\frac{a+b+c}{2}\)(semi-perimeter)
Then area of △ABC = \(\sqrt{S(S-a)(S-b)(S-c)}\)
This is called Heron’s formula.

→ Let ax + by + c = 0 represents a straight line; then any pair of coordinates (x₁, y₁) satisfying the linear equation ax + by + c = 0 is called its solution, i.e., if (x₁, y₁) is a solution of ax + by + c = 0 then ax₁ + by₁ + c = 0.

→ The inclination or angle made by a straight line with positive X-axis is called the slope of the straight line. Slope of a line joining any two points (x₁, y₁) and (x₂, y₂) is m = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

→ If ‘θ’ is the angle made by a line with X-axis, then slope m = tan θ.

Students can go through AP SSC 10th Class Maths Notes Chapter 6 Progressions to understand and remember the concepts easily.

AP State Syllabus SSC 10th Class Maths Notes Chapter 6 Progressions

→ The array of numbers following some rule is called a number pattern.
E.g.: 4, 6, 4, 6, 4, 6,…….

→ There is a relationship between the numbers of a pattern.

→ Each number in a pattern is called a term.

→ The series or list of numbers formed by adding or subtracting a fixed number to / from the preceding terms is called an Arithmetic Progression, simply A.P.
E.g.: 3, 5, 7,9, 11, ……

→ In the above list, each term is obtained by adding ‘2’ to the preceding term except the first term.

→ Also, we find that the difference between any two successive terms is the same throughout the series. This is called “common difference”.

→ The general form of an A.P. is
a, a + d, a + 2d, a + 3d,………, a + (n – 1) d.
Where‘a’is the first term, d is common difference.
Here d = a₂ – a₁ = a₃ – a₂ = a₄ – a₃ = …….. = a_n – a_n-1

→ If the number of terms of an A.P. is finite, then it is a finite A.P.
E.g.: 10, 8, 6, 4, 2.

→ If the number of terms of an A.P. is infinite, then it is an infinite A.P.
E.g.: 4, 8, 12, 16, …….

→ If d > 0, then a_n > a_n-1 and if d < 0, then a_n < a_n-1

→ The general term or nth term of an A.P is a_n = a + (n – 1) d.
E.g.: The 10th term of 10, 6, 2, -2, ……. is
Here a = 10 ; d = a₂ – a₁ = 6 – 10 = -4
∴ a₁₀ = a + (n – 1) d = 10 + (10 – 1) × -4 = 10 – 40 + 4 = -26

→ Sum of first n-terms of an A.P. is S_n = \(\frac{n}{2}\)[a + l] where a is the first term and l is the last term.
E.g.: 1 + 2 + 3 + …… + 80 = \(\frac{80}{2}\)(1 + 80) = 40 × 81 = 3240.

→ Sum of the first n-terms of an A.P. is given by, S_n = \(\frac{n}{2}\)[2a + (n – 1)d]
Also, a_n = S_n – S_n-1.

→ In a series of numbers, if every number is obtained by multiplying the preceding number by a fixed number except for the first term, such arrangement is called geometric progression or G.P.
E.g.: 4, 8, 16, 32, 64,……
Here, starting from the second term, each term is obtained by multiplying the preceding term by 2.
The first term may be denoted by ‘a’, then we also see that
We call it “common ratio”, denoted by ‘r’

→ The general form of a G.P. is
a, ar, ar², ar³, ……. ar^n-1
i.e., a₁ = a, a₂ = ar, a₃ = ar², a_n = ar^n-1.

Students can go through AP SSC 10th Class Maths Notes Chapter 5 Quadratic Equations to understand and remember the concepts easily.

AP State Syllabus SSC 10th Class Maths Notes Chapter 5 Quadratic Equations

→ The general form of a linear equation in one variable is ax + b = c.

→ Any equation of the form p(x) = 0 where p(x) is a polynomial of degree 2, is a quadratic equation.

→ If p(x) = 0 whose degree is 2 is written in descending order of their degrees, then we say that the quadratic equation is written in the standard form.

→ The standard form of a quadratic equation is ax² + bx + c = 0 where a ≠ 0. We can write it as y = ax² + bx + c.

→ There are various occasions in which we make use of Q.E. in our day-to-day life.
Eg : The height of a rocket is defined by a Q.E.

→ Let ax² + bx + c = 0 be a quadratic equation. A real number α is called a root of the Q.E. if aα² + bα + c = 0. And x = α is called a solution of the Q.E.
(i.e.) the real value of x for which the Q.E ax² + bx + c = 0 is satisfied is called its solution.

→ Zeroes of the Q.E. ax² + bx + c = 0 and the roots of the Q.E. ax² + bx + c = 0 are the same.

→ To factorise a Q.E. ax² + bx + c = 0, we find p, q ∈ R such that p + q = b and pq = ac. This process is called Factorising a Q.E. by splitting its middle term.
Eg : 12x² + 13x + 3 = 0
here a = 12; b = 13; c = 3
a.c = 12 × 3 = 36 = 4 × 9 and
b = 4 + 9 here p = 9 and q = 4
Now 12x² + 13x + 3 = 0
⇒ 12x² + 9x + 4x + 3 = 0
⇒ 3x(4x + 3) + 1 (4x + 3) = 0
⇒ (4x + 3) (3x + 1) = 0
Here 4x + 3 = 0 or 3x + 1 = 0
⇒ 4x = -3 or 3x = -1
⇒ x = \(\frac{-3}{4}\) or \(\frac{-1}{3}\)
\(\frac{-3}{4}\) and \(\frac{-1}{3}\) are called the roots of the Q.E. 12x² + 13x + 3 = 0 and x = \(\frac{-3}{4}\) or \(\frac{-1}{3}\) is the solution of the Q.E. 12x² + 13x + 3 = 0.

→ In the above example, (4x + 3) and (3x +1) are called the linear factors of the Q.E. 12x² + 13x + 3 = 0.

→ We can factorise a Q.E. by adjusting its left side so that it becomes a perfect square.
Eg: x² + 6x + 8 = 0
⇒ x² + 2.x.3 + 8 = 0
⇒ x² + 2.x.3 = -8
The L.H.S. is of the form a² + 2ab
∴ By adding b² it becomes a perfect square
∴ x² + 2.x.3 + 3² = -8 + 3²
⇒ (x + 3)² = -8 + 9
⇒ (x + 3)² = 1
⇒ x + 3 = ± 1 Now we take x + 3 = 1 or x + 3 = -l
⇒ x = -2 or x = -4

→ Adjusting a Q.E. of the form ax² + bx + c = 0 so that it becomes a perfect square.
Step – 1 : ax² + bx + c = 0 ⇒ ax² + bx = -c ⇒ x² + \(\frac{b}{a}\)x = \(\frac{-c}{a}\)
Step – 2 :

Step – 3 :

Step – 4 : Solve the above.
E.g: 5x² – 6x + 2 = 0

→ Let ax² + bx + c = 0 (a ≠ 0) be a Q.E., then b² – 4ac is called the Discriminant of the Q.E.

→ If b² – 4ac > 0, then the roots of the Q.E. ax² + bx + c = 0 are given by
x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\). This is called quadratic formula to find the roots.

The nature of the roots of a Q.E. can be determined either by its discriminant or its graph.
Q.E.: y = ax² + bx + c.

Students can go through AP SSC 10th Class Maths Notes Chapter 4 Pair of Linear Equations in Two Variables to understand and remember the concepts easily.

AP State Syllabus SSC 10th Class Maths Notes Chapter 4 Pair of Linear Equations in Two Variables

→ An equation of the form ax + by + c = 0 where a, b, c are real numbers and where at least one of a or b is not zero (i.e, a² + b² ≠ 0) is called a linear equation in two variables x and y.

→ A pair of equations in the same two variables forms a pair of linear equations. The system of pair of equations in general a₁x + b₁y + c₁ = 0 ; a₂x + b₂y+ c₂ = 0
where a₁, a₂, b₁, b₂, c₁, c₂ ∈ R such that a₁² + b₁² ≠ 0 and a₂² + b₂² ≠ 0

→ A pair of linear equations in two variables can be represented and solved by
a) Graphical method
b) Model method
c) Algebraic methods
i) Substitution method
ii) Elimination method
iii) Cross-multiplication method.

→ Graphical method: The two linear equations in two variables are represented by two straight lines on a graph sheet.
a) If they intersect at a point, then the point gives the unique solution of the two equations. In this case the two equations are consistent.
b) If the lines coincide, then there are infinitely many solutions. Each point on the line represents a solution. In this case, we say that the pair of equations is dependent or consistent.
c) If the two lines are parallel to one another, then the pair of equations has no solution. In this case we say that the pair of equations is inconsistent. Substitution method : In this method, we make one of the variables x or y as the subject from the first equatiori. We substitute this value in the second equation and get the value of the variable involved, then by substituting this value in any of the equations we get the value of second variable.
Eg : 2x + 4y = 16 …….. (1)
3x – 8y = – 18 …….. (2)
From equation (1); 2x + 4y = 16 ⇒ 2x = 16 – 4y ⇒ x = \(\frac{16-4y}{2}\)
Substituting x = \(\frac{16-4y}{2}\) in the second equation
\(3\left(\frac{16-4 y}{2}\right)\) – 8y = -18
48 – 12y – 16y = – 18 × 2
-28y = -36 – 48
y = \(\frac{-84}{-28}\) = 3
Substituting y = 3 in (1), 2x + 4y = 16
2x + 4(3) = 16
2x = 16 – 12
x = 2

→ Elimination method: In this method we first try to eliminate one of the two variables and by reducing the system to equation in one variable. We then solve for the variable.
The following steps may be adopted:
Step – 1: Multiply both the equations by some suitable non-zero constants to make the co-efficients of one variable (either x or y) numerically equal.
Step – 2: Add or subtract one equation from other so that one variable gets eliminated. If we get an equation in one variable, proceed to step 3.
If in step – 2, we obtain a true statement involving no variable, then the original pair of equations has infinitely many solutions.
If in step – 2, we obtain a false statement involving no variable, then the original pair of equations has no solutions, i.e., it is inconsistent.
Step – 3: Solve the equation in one variable (x or y) so obtained to get its value.

→ Cross-multiplication method: Let the pair of equations be a₁x + b₁y + c₁ = 0 ; a₂x + b₂y+ c₂ = 0

→ In every case, the obtained solutions should always be verified with the original equations.
Let a₁x + b₁y + c₁ = 0 and a₂x + b₂y+ c₂ = 0, form a pair of linear equations. Then the following situations can arise,
Case – (i): \(\frac{a_{1}}{a_{2}}\) ≠ \(\frac{b_{1}}{b_{2}}\) – pair of linear equations is consistent.
Case – (ii): \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) ≠ \(\frac{c_{1}}{c_{2}}\) – pair of linear equations is inconsistent.
Case – (iii): \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) = \(\frac{c_{1}}{c_{2}}\) – pair of linear equations is dependent and consistent.

→ There are several situations which can be mathematically represented by two equa¬tions that are not linear to start with. But we alter them so that they are reduced to a pair of linear equations.

→ The pair of values of the variables x and y for which the pair of equations is satisfied is called the solution pair.

Students can go through AP SSC 10th Class Maths Notes Chapter 3 Polynomials to understand and remember the concepts easily.

AP State Syllabus SSC 10th Class Maths Notes Chapter 3 Polynomials

→ Polynomial: An algebraic expression in which the variables involved have only non-negative integer power is called a polynomial.
Ex: 2x + 5, 3x² + 5x + 6, -5y, x³, etc.

→ Polynomials are constructed using constants and variables.

→ Coefficients operate on variables, which can be raised to various powers of non negative integer exponents.
, etc. are not polynomials.

→ General form of a polynomial having n^th degree is p(x) = a₀xⁿ + a₁x^n-1 + a₂x^n-2 …….. + a_n-1x + a_n where a₀, a₁, a₂,…… a_n-1, a_n are real coefficients and a₀ ≠ 0.

→ Degree of a polynomial:
The exponent of the highest degree term in a polynomial is known as its degree.
In other words, the highest power of x in a polynomial f(x) is called the degree of a polynomial f(x).
Example:
i) f(x) = 5x + \(\frac{1}{3}\) is a polynomial in the variable x of degree 1.
ii) g(y) = 3y² – \(\frac{5}{2}\)y + 7 is a polynomial in the variable y of degree 2.

→ Zero polynomial: A polynomial of degree zero is called zero polynomial that are having only constants.
Ex: f(x) = 8, f(x) = –\(\frac{5}{2}\)

→ Linear polynomial: A polynomial of degree one is called linear polynomial.
Ex: f(x) = 3x + 5, g(y) = 7y – 1, p(z) = 5z – 3.
More generally, any linear polynomial in variable x with real coefficients is of the form f(x) = ax + b, where a and b are real numbers and a ≠ 0.
Note: A linear polynomial may be a monomial or a binomial because linear polynomial f(x) = \(\frac{7}{5}\)x – \(\frac{5}{2}\) is a binomial, whereas the linear polynomial g(x) = \(\frac{2}{5}\) x is a monomial.

→ Quadratic polynomial: A polynomial of degree two is called quadratic polynomial.
Ex: f(x) = 5x², f(x) = 7x² – 5x, f(x) = 8x² + 6x + 5.
More generally, any quadratic polynomial in variable x with real coefficients is of the form f(x) = ax² + bx + c, where a, b and c are real numbers and a ≠ 0.
Note: A quadratic polynomial may be a monomial or a binomial or a trinomial.
Ex: f(x) = \(\frac{1}{5}\)x² is a monomial, g(x) = 3x² – 5 is a binomial and
h(x) = 3x² – 2x + 5 is a trinomial.

→ Cubic polynomial: A polynomial of degree three is called cubic polynomial.
Ex: f(x) = 8x³, f(x) = 9x³ + 5x²
f(y) = 11y³ – 9y² + 7y,
f(z) = 13z³ – 12z² + 11z + 5.

→ Polynomial of degree ‘n’ in standard form: A polynomial in one variable x of degree n is an expression of the form f(x) = a_nxⁿ + a_n-1x^n-1 + …….. + a₁x + a₀ where a₀, a₁, a₂,…… a_n, a_n are constants and a_n ≠ 0.
In particular, if a₀ = a₁ = a₂ = …… = a_n = 0 (all the constants are zero; we get the constants zero), we get the zero polynomial which is not defined.

→ Value of a polynomial at a given point: If p(x) is a polynomial in x and α is a real number. Then the value obtained by putting x = a in p(x) is called the value of p(x) at x = α.
Ex : Let p(x) = 5x² – 4x + 2, then its value at x = 2 is given by
p(2) = 5(2)² – 4(2) + 2 = 5(4) – 8 + 2 = 20 – 8 + 2 = 14 Thus, the value of p(x) at x = 2 is 14.

→ Graph of a polynomial: In algebraic or in set theory language, the graph of a polynomial f(x) is the collection (or set) of all points (x, y) where y = f(x).
i) Graph of a linear polynomial ax + b is a straight line.
ii) The graph of a quadratic polynomial (ax² + bx + c) is U – shaped, called parabola.

→ If a > 0 in ax² + bx + c, the shape of parabola is opening upwards ‘∪’.

→ If a < 0 in ax² + bx + c, the shape of parabola is opening downwards ‘∩’

→ Relationship between the zeroes and the coefficient of a polynomial:

Note: Formation of a cubic polynomial : Let α, β, and γ be the zeroes of the polynomial.
∴ Required cubic polynomial = (x – α) (x – β) (x – γ).

→ How to make a quadratic polynomial with the given zeroes : Let the zeroes of a quadratic polynomial be α and β.
∴ x = α, x = β
Then, obviously the quadratic polynomial is (x – α) (x – β) i.e., x² – (α + β)x 4- ap.
i.e., x² – (sum of the zeroes) x + product of the zeroes.

→ Division Algorithm : If p(x) and g(x) are any two polynomials with g(x) ≠ 0, then we can find polynomials q(x) and r(x) such that, p(x) = g(x) × q(x) + r(x)
i.e., Dividend = Divisor × Quotient + Remainder
where, r(x) = 0 or degree of r(x) < degree of g(x). This result is known as the division algorithm for polynomials.

→ Some useful relations:
α² + β² = (α + β)² – 2αp
(α – β)² = (α + β)² – 4αβ
α² – β² = (α + β) (α – β) = (α + β)\(\sqrt{(\alpha+\beta)^{2}-4 \alpha \beta}\)
α³ + β³ = (α + β)³ – 3αβ(α + β)
α³ – β³ = (α – β)³ + 3αβ(α – β)