Mahesh

Students can go through AP Board 8th Class Maths Notes Chapter 5 Comparing Quantities Using Proportion to understand and remember the concepts easily.

AP State Board Syllabus 8th Class Maths Notes Chapter 5 Comparing Quantities Using Proportion

→ Two simple ratios are expressed like a single ratio as the ratio of product of antecedents to product of consequents and we call it compound ratio of the given two simple ratios.
a : b and c : d are any two ratios, then their compound ratio is \(\frac{a}{b}\) × \(\frac{c}{d}\) = \(\frac{ac}{bd}\) i.e. ac : bd.

→ A percentage(%) compares a number to 100. The word percent means “per every hundred” or “out of every hundred”. 100% = \(\frac{100}{100}\) it is also a fraction with denominator 100.

→ Discount is a decrease percent of marked price. Price reduction is called rebate or discount. It is calculated on marked price or list price.

→ Profit or loss is always calculated on cost price. Profit is an example of increase percent of cost price and loss is an example of decrease percent of cost price.

→ VAT will be charged on the selling price of an item and will be included in the bill.
VAT is an increase percent on selling price.

→ Simple interest is an increase percent on the principal.

→ Simple interest (I) = \(\frac{P \times T \times R}{100}\)
where P = Principal, T = Time inyears, R = Rate of interest.

→ Amount = Principal + Interest = P + \(\frac{P \times T \times R}{100}\) = P\(\left(1+\frac{T \times R}{100}\right)\)

→ Compound interest allows you to earn interest on interest.

→ Amount at the end of ‘n’ years using compound interest is A = P \(\left(1+\frac{R}{100}\right)^{n}\)

→ The time period after which interest is added to principal is called conversion period.
When interest is compounded halfyearly, there are two conversion periods in a year, each after 6 months. In such a case, ha If year rate will be half of the annual rate.

→ Note: 1.615 : 1 is called as golden ratio.
In ancient Greece, artists and architects believed there was a particular rectangular shape that looked very pleasing to the eye. For rectangles of this shape, the ratio of long side to the short side is roughly 1.615 : 1. This ratio is very close to what is known as golden ratio.

Students can go through AP Board 8th Class Maths Notes Chapter 4 Exponents and Powers to understand and remember the concepts easily.

AP State Board Syllabus 8th Class Maths Notes Chapter 4 Exponents and Powers

Laws of Exponents:

→ a × a × a …… m times = a^m

In a^m, a is called base; m is called exponent/ power.

→ a^m × aⁿ = a^m+n

→ \(\frac{\mathrm{a}^{\mathrm{m}}}{\mathrm{a}^{\mathrm{n}}}\) = a^m-n (m > n)
⇒ \(\frac{1}{a^{n-m}}\) (m < n)

→ (ab)^m = a^m . b^m

→ a⁰ = 1

→ a^-n = \(\frac{1}{a^{n}}\)

→ aⁿ = \(\frac{1}{a^{-n}}\)

→ \(\left(\frac{a}{b}\right)^{m}\) = \(\frac{a^{m}}{b^{m}}\)

→ \(\left(a^{m}\right)^{n}\) = a^mn

→

→ \(\sqrt[n]{a}\) = \((\mathrm{a})^{1 / \mathrm{n}}\)

Students can go through AP Board 8th Class Maths Notes Chapter 3 Construction of Quadrilaterals to understand and remember the concepts easily.

AP State Board Syllabus 8th Class Maths Notes Chapter 3 Construction of Quadrilaterals

→ A closed four sided polygon is called a quadrilateral.

→ A quadrilateral has 4 sides, 4 vertices, 4 angles and 2 diagonals.

→ The sum of the 4 angles of a quadrilateral is 360°.

→ Quadrilateral and their types:

→ Five independent measurements are required to draw a unique quadrilateral.

→ A quadrilateral can be constructed uniquely, if
a) The lengths of four sides and one angle are given
b) The lengths of four sides and one diagonal are given
c) The lengths of three sides and two diagonals are given
d) The lengths of two adjacent sides and three angles are given
e) The lengths of three sides and two included angles are given

→ The two special quadrilaterals, namely rhombus and square can be constructed when two diagonals are given.

Students can go through AP Board 8th Class Maths Notes Chapter 2 Linear Equations in One Variable to understand and remember the concepts easily.

AP State Board Syllabus 8th Class Maths Notes Chapter 2 Linear Equations in One Variable

→ An algebraic equation is equality of algebraic expressions involving variables and constants.

→ If the degree of an equation is one then it is called a linear equation.

→ If a linear equation has only one variable then it is called a linear equation in one variable or simple equation. ‘

→ The value which when substituted for the variable in the given equation makes L.H.S. = R.H.S. is called a solution or root of the given equation.

→ Just as numbers, variables can also be transposed from one side of the equation to the other side.
Note: When we transpose terms
‘+’ quantity becomes ’-‘ quantity,
‘-‘ quantity becomes ‘+’ quantity.
‘×’ quantity becomes ‘÷’ quantity.
‘÷’ quantity becomes ‘×’ quantity.
Also
Also,
(+) × (+) = +
(+) × (-) = –
(-) × (+) = –
(-) × (-) = +

Students can go through AP Board 8th Class Maths Notes Chapter 1 Rational Numbers to understand and remember the concepts easily.

AP State Board Syllabus 8th Class Maths Notes Chapter 1 Rational Numbers

→ The numbers which are expressed in the form of \(\frac{p}{q}\) where p and q are integers and q ≠ 0, are called “Rational Numbers” which are denoted by the letter ‘Q’.

→ Rational numbers are closed under the operations of addition, subtraction and multiplication.

→ Rational numbers are not closed on division.

→ Whole numbers:

→ Whole numbers:

→ The additive inverse of a is – a. (∵ a + (-a) = 0)

→ The multiplicative inverse of a is \(\frac{1}{a}\). (∵ a × \(\frac{1}{a}\) = 1)

→ The operations addition and multiplications are

1. Commutative for rational numbers.
2. Associative for rational numbers.

→ ‘0’ is the additive identity for rational number.

→ ‘1’ is the multiplicative identity for rational number.

→ A rational number and its additive inverse are opposite in their sign.

→ The multiplicative inverse of a rational number is its reciprocal.

→ Distributivity of rational numbers a, b and c is a(b + c) = ab + ac and a(b – c) = ab – ac.

→ Rational numbers can be represented on a number line.

→ There are infinite number of rational numbers between any two given rational numbers.

→ The concept of mean help us to find rational numbers between any two rational numbers.

→ The decimal representation of rational numbers is either in the form of terminating decimal or non-terminating recurring decimals.

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.