Maths Formula Sheet Class 10th PDF

FORMULA SHEET MATHS CLASS 10TH 2 CHAPTER 1 REAL NUMBERS 1. A number is prime if it has only two factors, 1 and itself. 2. Every composite number can be expressed as a product of prime factors....

FORMULA SHEET MATHS CLASS 10TH 2 CHAPTER 1 REAL NUMBERS 1. A number is prime if it has only two factors, 1 and itself. 2. Every composite number can be expressed as a product of prime factors. B 3. H C F of two numbers = Product of the smaller power of each common factor in the numbers. U H.C.F. of (30, 45) = 3 x 5 = 15. [30 = 2 x 3 x 5 ;45 = 32 x 5] H 4. L C M of two numbers = Product of the greatest power of each prime factor involved in the numbers. P L C M of ( 30, 45 ) = 2 x 32 x 5 = 90 EX 5. 𝐻 𝐶 𝐹 ( 𝑎, 𝑏 ) x 𝐿 𝐶 𝑀 ( 𝑎, 𝑏 ) = 𝑎 x 𝑏 6. 𝐻 𝐶 𝐹 ( 𝑎, 𝑏, 𝑐 ) x 𝐿 𝐶 𝑀 ( 𝑎, 𝑏, 𝑐 ) ≠ 𝑎 x 𝑏 x 𝑐 3 CHAPTER 2 POLYNOMIALS 1. x is a variable and 𝑎0 , 𝑎1 , … … ….. 𝑎𝑛 be real numbers, n is a positive integer then f(x) = 𝑎0 + 𝑎1 𝑥 + 𝑎2 𝑥 2 + − − − − 𝑎𝑛 𝑥 𝑛 is a polynomial in the variable of x. 2. The exponent of the highest degree term is called the degree of the polynomial. B 3. Constant Polynomial : 𝑓(𝑥) = 𝑎 , a is constant. U Linear Polynomial : 𝑓(𝑥) = 𝑎𝑥 + 𝑏 , 𝑎 ≠ 0 Quadratic Polynomial : 𝑓(𝑥)=𝑎𝑥 2 + 𝑏 𝑥 + 𝑐 , 𝑎 ≠ 0 H Cubic Polynomial : 𝑓(𝑥) = 𝑎𝑥 3 + 𝑏𝑥 2 + 𝑐 𝑥 + 𝑑 , 𝑎 ≠ 0 P 4. A real number ‘a’ is a zero of the polynomial 𝑓(𝑥) 𝑖𝑓 𝑓(𝑎) = 0 EX 5. A polynomial of degree ‘n’ can have at most ‘n’ real zeros. 6. Geometrically, the zeros of the polynomial 𝑓(𝑥) are the 𝑥 coordinates of the points where the graph 𝑦 = 𝑓 (𝑥 ) intersects the 𝑥 axis. 7. If 𝛼 𝑎𝑛𝑑 𝛽 are the zeroes of the quadratics polynomial 𝑓(𝑥) = 𝑎𝑥 2 + 𝑏 𝑥 + 𝑐 , 𝑡ℎ𝑒𝑛 𝑏 Sum of the Zeros = 𝜶 + 𝜷 = − 𝑎 𝑐 Product of the Zeros = 𝜶 𝜷 = 𝑎 4 8. Given the sum of the zeros and product of the zeros, the quadratic polynomial is k[ 𝑥 2 − 𝑥 (𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑧𝑒𝑟𝑜𝑒𝑠) + (𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑧𝑒𝑟𝑜𝑒𝑠)] ; 𝒌 [𝒙𝟐 − 𝒙 ( 𝜶 + 𝜷) + 𝜶 𝜷] 9. If 𝜶 + 𝜷 and 𝜶 𝜷 are given / known, then 𝛂𝟐 + 𝛃𝟐 = (𝛂 + 𝛃 ) 𝟐 − 𝟐 𝜶 𝜷 𝛂𝟑 + 𝛃𝟑 = (𝛂 + 𝛃 ) 𝟑 − 3 𝜶 𝜷(𝛂 + 𝛃) 𝛂𝟒 + 𝛃𝟒 = (𝛂𝟐 + 𝛃𝟐 ) 𝟐 − 𝟐( 𝛂 𝛃)𝟐 10. B If 𝛼, 𝛽, 𝛾 are the zeroes of the cubic polynomial U 𝑓(𝑥) = 𝑎𝑥 3 + 𝑏𝑥 2 + 𝑐 𝑥 + 𝑑 , then H 𝒃 𝜶+𝜷+𝜸=− 𝒂 P 𝒄 𝜶𝜷 + 𝜷𝜸 + 𝜸𝜶 = 𝒂 EX 𝒅 𝜶𝜷𝜸=− 𝒂 5 CHAPTER 3 PAIR OF LINEAR EQUATIONS IN TWO VARIABLES 1. 𝐸𝑎𝑐ℎ 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 (𝑥, 𝑦 ) 𝑜𝑓 𝑎 𝑙𝑖𝑛𝑒𝑎𝑟 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑎𝑥 + 𝑏𝑦 + 𝑐 = 0 𝑤ℎ𝑖𝑐ℎ 𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡𝑠 𝑎 𝑙𝑖𝑛𝑒, 𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑠 𝑡𝑜 𝑎 𝑝𝑜𝑖𝑛𝑡 𝑜𝑛 𝑡ℎ𝑒 𝑙𝑖𝑛𝑒. 2. A pair of linear equations in two variables 𝑥, 𝑦 is 𝑎1 𝑥 + 𝑏1 𝑦 + 𝑐1 = 0; 𝑎2 𝑥 + 𝑏2 𝑦 + 𝑐2 = 0 3. 𝑎1 𝑥 + 𝑏1 𝑦 + 𝑐1 = 0; 𝑎2 𝑥 + 𝑏2 𝑦 + 𝑐2 = 0 Graphically the two straight lines are, B a) Intersecting lines, if 𝑎1 U ≠ 𝑏1 𝑎2 𝑏2 H 𝑎1 𝑏1 𝑐1 Parallel lines, if = ≠ 𝑎2 𝑏2 𝑐2 P 𝑎1 𝑏1 𝑐1 b) Coincident lines, if = = (The equations are 𝑎2 𝑏2 𝑐2 EX dependent linear equations) 4. 𝑎1 𝑥 + 𝑏1 𝑦 + 𝑐1 = 0; 𝑎2 𝑥 + 𝑏2 𝑦 + 𝑐2 = 0 a) Consistent and have unique solution if 𝑎1 𝑏 ≠ 1 𝑎2 𝑏2 b) Consistent and have infinite number of solutions if 𝑎1 𝑏1 𝑐1 = = 𝑎2 𝑏2 𝑐2 c) Inconsistent and no unique solution if 6 𝑎1 𝑏1 𝑐1 = ≠ 𝑎2 𝑏2 𝑐2 5. Certain basic facts to know : (𝒊) 𝒙 = 𝟎 𝒊𝒔 𝒕𝒉𝒆 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝒐𝒇 𝒚 𝒂𝒙𝒊𝒔 𝒂𝒏𝒅 𝒚 = 𝟎 𝒊𝒔 𝒕𝒉𝒆 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝒐𝒇 𝒙 𝒂𝒙𝒊𝒔. (𝒊𝒊 ) 𝒙 = 𝒂 ( 𝒔𝒐𝒎𝒆 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 )𝒓𝒆𝒑𝒓𝒆𝒔𝒆𝒏𝒕𝒔 𝒂 𝒍𝒊𝒏𝒆 𝒑𝒂𝒓𝒂𝒍𝒍𝒆𝒍 𝒕𝒐 𝒚 𝒂𝒙𝒊𝒔. (𝒊𝒊𝒊 ) 𝒚 = 𝒃 ( 𝒔𝒐𝒎𝒆 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 )𝒓𝒆𝒑𝒓𝒆𝒔𝒆𝒏𝒕𝒔 𝒂 𝒍𝒊𝒏𝒆 𝒑𝒂𝒓𝒂𝒍𝒍𝒆𝒍 𝒕𝒐 𝒙 𝒂𝒙𝒊𝒔. B (𝒊𝒗 ) 𝑫𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝒕𝒓𝒂𝒗𝒆𝒍𝒍𝒆𝒅 = 𝒔𝒑𝒆𝒆𝒅 𝑿 𝒕𝒊𝒎𝒆 U (𝒗 ) 𝑰𝒏 𝒔𝒖𝒎𝒔 𝒄𝒐𝒏𝒔𝒊𝒔𝒕𝒊𝒏𝒈 𝒐𝒇 𝒔𝒖𝒎 𝒐𝒓 𝒅𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒄𝒆 𝒐𝒇 𝒕𝒉𝒆 𝒅𝒊𝒈𝒊𝒕𝒔 𝒂𝒏𝒅 𝒓𝒆𝒗𝒆𝒓𝒔𝒊𝒏𝒈 𝒕𝒉𝒆 𝒅𝒊𝒈𝒊𝒕𝒔, H 𝒊𝒇 𝒕𝒉𝒆 𝒖𝒏𝒊𝒕 𝒅𝒊𝒈𝒊𝒕 𝒊𝒔 𝒙 𝒂𝒏𝒅 𝒕𝒆𝒏𝒕𝒉 𝒅𝒊𝒈𝒊𝒕 𝒊𝒔 𝒚, 𝒕𝒉𝒆 𝒏𝒖𝒎𝒃𝒆𝒓 𝒔𝒉𝒐𝒖𝒍𝒅 𝒃𝒆 𝒕𝒂𝒌𝒆𝒏 𝒂𝒔 𝟏𝟎 𝒚 + 𝒙. P EX 6. Consider the two linear equations 49 𝑥 + 51 𝑦 = 499 𝑎𝑛𝑑 51 𝑥 + 49 𝑦 = 501. The coefficients of 𝑥 𝑎𝑛𝑑 𝑦 are interchanged in the two equations. In such cases add the two equations, simplify ; subtract the two equations , simplify. Very easy method to solve. 7 CHAPTER 4 QUADRATIC EQUATIONS 1.Standard form of a Quadratic Equation is 𝑎𝑥 2 +𝑏 𝑥 + 𝑐 =0,𝑎 ≠ 0 2. Examples of Quadratic Eq. : 𝑥 2 −6 𝑥 + 4 = 0,2𝑥 2 - 7 𝑥 = 0 3 3. Examples of Equations which are not Quadratic:𝑥 + = 𝑥 2 𝑥 2 + 𝑥 2√𝑥 − 3 = 0. 4. A real number 𝛼 is called a root of the quadratic Equation 𝑎𝑥 2 + 𝑏 𝑥 + 𝑐 if 𝛼 satisfies the quadratic equation i.e. 𝑖𝑓 𝒂 𝛼 2 + 𝒃 𝛼 + 𝒄 = 𝟎. 𝒙 = B 𝛼 is a solution of the quadratic equation. 5. Zeros of the quadratic polynomial is 𝑎𝑥 2 +𝑏 𝑥 + 𝑐 = 0 are the same as the roots of the quadratic Equation U H 𝒂𝒙𝟐 +𝑏 𝑥 + 𝑐 = 0 6. Solving a Quadratic Equation by Factorisation Method: If the P Quadratic Equation 𝒂𝒙𝟐 +𝑏 𝑥 + 𝑐 = 0 is factorizable into a product of EX two linear factors, then the roots of the quadratic equation can be found by equating each linear factor to ‘0’. 7. Solving a Quadratic Equation by Formula : The real roots of the −𝑏 ±√𝑏2 −4𝑎𝑐 Quadratic Equation 𝒂𝒙𝟐 +𝑏 𝑥 + 𝑐 = 0 are given by ; 𝑏2 − 2𝑎 4𝑎𝑐 ≥ 0. 8. Nature of the roots of the Quadratic Equation depends on 𝐷 = 𝑏 2 − 4𝑎𝑐, which is the Discriminant. 9. The Quadratic Equation 𝒂𝒙𝟐 +𝑏 𝑥 + 𝑐 = 0 has (𝑖) 𝑡𝑤𝑜 𝑑𝑖𝑠𝑡𝑖𝑛𝑐𝑡 𝑟𝑒𝑎𝑙 𝑟𝑜𝑜𝑡𝑠 , 𝑖𝑓 𝑏 2 − 4𝑎𝑐 > 0 (𝑖𝑖) 𝑡𝑤𝑜 𝑟𝑒𝑎𝑙 𝑎𝑛𝑑 𝑒𝑞𝑢𝑎𝑙 𝑟𝑜𝑜𝑡𝑠 ( 𝑐𝑜𝑖𝑛𝑐𝑖𝑑𝑒𝑛𝑡 𝑟𝑜𝑜𝑡𝑠 ), 𝑖𝑓 𝑏 2 − 4𝑎𝑐 =0 (𝑖𝑖𝑖)𝑛𝑜 𝑟𝑒𝑎𝑙 𝑟𝑜𝑜𝑡𝑠 , 𝑖𝑓𝑏 2 − 4𝑎𝑐 < 0 8 CHAPTER 5 ARITHMETIC PROGRESSION 1. A sequence 𝑎1 , 𝑎2 , … … ….. 𝑎𝑛 is an Arithmetic Progression ( A.P ) if the difference between any two consecutive terms is the same and that is the common difference ‘d’ which can be positive or negative. 2. General A.P is a, a + d, a + 2d, a + 3d________where a is the first term and d is the common difference. B 3. The 𝑛𝑡ℎ term of an A.P with a as the first term and d as the common difference is given by, U an = a + (n-1) d ( a10 = a + 9 d ; a16 = a + 15 d ) H 4. Easy way to find some term from the end of the sequence which is an A.P : P Find the 7𝑡ℎ term from the end of the sequence EX 17,14,11----- (− 40 ). Just write the sequence from the end.( - 40 ), (-37 )-----17. Now for this A.P a = - 40.d = 3 ; a7 = a + 6 d a7 = (- 40) + (6 x 3) = - 22. 5. If you want to find out whether a given number belongs to the A.P,you can find out ‘a’ the first term and ‘d’ the common difference of that A.P. Use the formula an = a + (n-1) d. In the place of an substitute the given number, and the values of ‘a’ and ‘d’. Find ‘n’. If n is a positive whole number, then the 9 given number belongs to the A.P. Otherwise it doesn’t belong to that A.P. 6. Sum to n terms of an A.P is denoted as Sn. Sn = n/2 [ 2 a + (n-1) d ] or n/2 [ a + 𝒍 ] where 𝑙 is the last term. 7. Sum of the first ‘n’ natural numbers = 1 + 2 +3 +-------+n = n ( n+1 ) /2. 8. Sum of the first ‘n’ odd natural numbers = 1 + 3 + 5 + ---- + ( 2n -1 ) = 𝑛2 9. Sum of the first ‘n’ even natural numbers = 2 + 4 + 6 + ----- + 2n = n ( n+1 ) B U 10. S1 = a1 ( first term ) ; S2 = a 1 + a 2 ; S3 = a 1 + a 2 + a 3 H and so on. S2 – S1 = a 2 ; S3 – S2 = a3 ; ------ In general an = Sn – Sn-1 P EX 10 CHAPTER 6 TRIANGLES 1. Two figures having the same shape and same size are congruent figures. 2. Two figures having the same shape but B not the same size are similar figures. U 3. All congruent figures are similar but all similar figures H need not be congruent. P 4. Two polygons having the same number of sides will be similar if the corresponding angles of the two polygons are EX equal and the corresponding sides are proportional. 5. Basic Proportionality Theorem : If a line is drawn parallel to one side of a triangle to intersect the other two sides in two distinct points, then the other two sides are divided in the same ratio. 𝐴𝑃 𝐴𝑄 PQ || BC; 𝑃𝐵 = 𝑄𝐶 6. Converse of a Basic Proportionality Theorem: If a line divides any two sides of a triangle in the same ratio, then the 11 line is parallel to the third side. 7. SIMILAR TRIANGLES : (A)A A A SIMILARITY : If the corresponding angles of two triangles are equal, then their corresponding sides are proportional. The two triangles are similar. (B)A A SIMILARITY : If two angles of one triangle are equal to the corresponding two angles of another triangle, the two triangles are similar. (C)S S S SIMILARITY : If the corresponding sides of two triangles are proportional ( the corresponding angles are equal ) then the two triangles are similar. B (D)S A S SIMILARITY : If one angle of a triangle is equal to one angle of U another triangle and the sides including these angles are in the same ratio, then the two triangles are similar. H (E)R H S SIMILARITY : In two right triangles, if the hypotenuse and one side are proportional then the two triangles are similar. P 8.PYTHAGORUS THEOREM : In a right triangle, the square on the EX hypotenuse is equal to the sum of the squares on the other two sides. 9.CONVERSE OF PYTHAGORUS THEOREM : If in a triangle, square of one side is equal to the sum of the squares of the other two sides, then the angle opposite to the first side is a right angle. 10. The line joining the mid points of two sides of a triangle is parallel to the third side and half the third side. 11. The diagonals of a trapezium divide each other proportionally. 12 CHAPTER 7 COORDINATE GEOMETRY 1.The distance of a point from y axis is called the x coordinate or abscissa. 2. The distance of a point from x axis is called the y coordinate or ordinate. 3.Any point on the x-axis will be of the form ( 𝒙, 𝟎 ) 4.Any point on the y-axis will be of the form ( 𝟎, 𝒚 ) B 5. A is ( 𝒙𝟏, 𝒚𝟏 ) and B is ( 𝒙𝟐, 𝒚𝟐 ) Distance AB = √(𝑥2 − 𝑥1 )2 + (𝑦2 − 𝑦1 )2 U 6. Distance of a point A (x,y) from the origin O = OA =√(𝑥)2 + (𝑦)2 H 7. Distance of a point 𝑷 ( 𝒙, 𝒚 ) 𝒇𝒓𝒐𝒎 𝒕𝒉𝒆 𝒙 − 𝒂𝒙𝒊𝒙 𝒊𝒔 | 𝒚 | 𝒖𝒏𝒊𝒕𝒔. P 8. Distance of a point 𝑷 ( 𝒙, 𝒚 ) 𝒇𝒓𝒐𝒎 𝒕𝒉𝒆 𝒚 − 𝒂𝒙𝒊𝒙 𝒊𝒔 | 𝒙 | 𝒖𝒏𝒊𝒕𝒔 EX 9.Three points A, B, C are collinear if the sum of the distances between two pairs of points = the distance between the third pair. 𝑷𝑸 + 𝑸𝑹 = 𝑷𝑹. 𝑷, 𝑸, 𝑹 𝒂𝒓𝒆 𝒄𝒐𝒍𝒍𝒊𝒏𝒆𝒂𝒓 𝒑𝒐𝒊𝒏𝒕𝒔. 𝑨𝑪 + 𝑪𝑩 ≠ 𝑨𝑩. 𝑨, 𝑩. 𝑪 𝒂𝒓𝒆 𝒏𝒐𝒏 𝒄𝒐𝒍𝒍𝒊𝒏𝒆𝒂𝒓 𝒑𝒐𝒊𝒏𝒕𝒔. 10. SECTION FORMULA : INTERNAL DIVISION : 𝑷(𝒙, 𝒚) 𝒅𝒊𝒗𝒊𝒅𝒆𝒔 𝒕𝒉𝒆 𝒍𝒊𝒏𝒆 𝒋𝒐𝒊𝒏𝒊𝒏𝒈 𝑨(𝒙𝟏, 𝒚𝟏) 𝒂𝒏𝒅 𝑩(𝒙𝟐, 𝒚𝟐) 13 𝒊𝒏 𝒕𝒉𝒆 𝒓𝒂𝒕𝒊𝒐 𝒎 ∶ 𝒏 𝒊𝒏𝒕𝒆𝒓𝒏𝒂𝒍𝒍𝒚. 𝑚𝑥2 +𝑛𝑥1 𝑚𝑦2 +𝑚𝑦1 P(x,y) = , 𝑚+𝑛 𝑚+𝑛 EXTERNAL DIVISION : 𝑷(𝒙, 𝒚) 𝒅𝒊𝒗𝒊𝒅𝒆𝒔 𝒕𝒉𝒆 𝒍𝒊𝒏𝒆 𝒋𝒐𝒊𝒏𝒊𝒏𝒈 𝑨(𝒙𝟏, 𝒚𝟏) 𝒂𝒏𝒅 𝑩(𝒙𝟐, 𝒚𝟐) 𝒊𝒏 𝒕𝒉𝒆 𝒓𝒂𝒕𝒊𝒐 𝒎 ∶ 𝒏 𝒆𝒙𝒕𝒆𝒓𝒏𝒂𝒍𝒍𝒚. 𝑚𝑥2 −𝑛𝑥1 𝑚𝑦2 −𝑚𝑦1 P(x,y) = , 𝑚−𝑛 𝑚−𝑛 11. MID – POINT FORMULA : B 𝑰𝒇 𝑷 ( 𝒙, 𝒚 ) 𝒊𝒔 𝒕𝒉𝒆 𝒎𝒊𝒅𝒑𝒐𝒊𝒏𝒕 𝒐𝒇 𝒕𝒉𝒆 𝒍𝒊𝒏𝒆 𝒋𝒐𝒊𝒏𝒊𝒏𝒈 𝑨(𝒙𝟏, 𝒚𝟏) 𝒂𝒏𝒅 𝑩(𝒙𝟐, 𝒚𝟐) 𝑥2 +𝑥1 𝑦2 +𝑦1 U 𝑷 ( 𝒙, 𝒚 ) = , 2 2 H 12. Points of Trisection : The points which divide the line joining A and B in the ratio 2 : 1 and 1 : 2 are called the points of Trisection. P 13.To prove that a quadrilateral is a parallelogram, ( given the vertices EX ) prove that the midpoints of both the diagonals are the same. Use the same concept to find the fourth vertex of the parallelogram if three vertices in order are given. 14.To prove that the quadrilateral is a rectangle, ( given the vertices ) prove that the opposite sides are equal and the two diagonals are equal. 15.To prove that the quadrilateral is a rhombus, ( given the vertices ) prove that all the four sides are equal. [ Diagonals are not equal. The diagonals of a rhombus bisect each other at right angles. Important property of a Rhombus ] 16.To prove that the quadrilateral is a square, ( given the vertices ) prove that the four sides are equal and the two diagonals are equal. 14 CHAPTER 8 and 9 INTRODUCTION TO TRIGONOMETRY AND APPLICATION OF TRIGONOMETRY TRIGONOMETRIC RATIOS: In Right triangle ABC, ∠B = 90 ° Side AC opposite to that is the hypotenuse. B U P H EX Let ∠A =𝜃 AB is the adjacent side and BC is the opposite side. The six trigonometric ratios are explained in the figure 1 1 1 𝑐𝑜𝑠 𝜃 cosec θ = ; sec θ = ; cot θ = ; cot θ = 𝑠𝑖𝑛 𝜃 𝑐𝑜𝑠 𝜃 𝑡𝑎𝑛 𝜃 𝑠𝑖𝑛 𝜃 15 VALUES OF TRIGONOMETRIC RATIOS OF STANDARD ANGLES: Values of Trigonometric ratios of an angle do not vary with the lengths of the sides of the triangle. PYTHOGORUS THEOREM : The square on the hypotenuse is equal to the sum of the squares on the other two sides containing the right angle. Examples of some Pythagorian Triplets : 3, 4, 5 ; 5, 12, 13 ; 8, 15, 17 ; 7, 24, 25 and any multiples of these numbers. B U P H TRIGONOMETRIC IDENTITIES: EX ANGLE OF ELEVATION AND DEPRESSION: 16 CHAPTER 11 CIRCLES RECAPITULATION OF CONCEPTS LEARNT IN IX STANDARD : 1. Equal Chords subtend equal angles at the centre of the circle. 2. Equal chords of congruent circles subtend equal angles at the centre of the circle. 3. Equal chords of a circle are equidistant from the centre of the circle. 4. Equal chords of congruent circles are equidistant from their corresponding centres. B 5. If you consider two chords of a circle, the one which is nearer to the centre is larger than the other. U 6. The perpendicular drawn from the centre of a circle bisects the chord. H 7. Angles in the same segment of a circle are equal. 8. Angle subtended by an arc at the centre of the circle is double the angle P subtended at any other point in the remaining part of a circle. 9. Angle in a semi-circle is a right angle. EX 10. If the four vertices of a quadrilateral lie on a circle, it is a cyclic quadrilateral. 11. The opposite angles of a cyclic quadrilateral are supplementary (Their sum is 𝟏𝟖𝟎°). 12. If one side of a cyclic quadrilateral is produced, then the exterior angle is equal to the interior opposite angle. CONCEPTS IN X STANDARD 1. Tangent is a line which intersects a circle at only one point. 2. Secant is a line which intersects a circle in two points. 3. From a point inside the circle, no tangent can be drawn to that circle. 4. From a point on the circle, only one tangent can be drawn to the circle. 5. From a point outside the circle two tangents can be drawn to the circle. 17 CHAPTER 12 AREAS RELATED TO CIRCLES 1. Perimeter ( Circumference ) of a circle 𝑷 = 𝟐 𝝅 𝒓 𝒐𝒓 𝝅 𝒅 𝒓 = 𝒓𝒂𝒅𝒊𝒖𝒔; 𝒅 = 𝒅𝒊𝒂𝒎𝒆𝒕𝒆𝒓 2. Area of a Circle 𝑨 = 𝝅 𝑟 2 3. Distance travelled by a wheel in one revolution is equal to its circumference. 𝜃 4. Length of an arc of a sector of angle 𝜃 𝑖𝑠 𝑙 = × 2𝜋𝑟 B 360 5. Perimeter of a sector P = 𝒍 + 𝟐 𝒓 U 𝜃 6. Area of a sector of angle 𝜃 𝑖𝑠 𝑨 = × 𝝅 𝑟2 360 H 𝑙𝑟 7. Area of a sector is 𝑨 = 2 P 8. Area of the major sector = 𝝅 𝑟 2 − 𝑨𝒓𝒆𝒂 𝒐𝒇 𝒎𝒊𝒏𝒐𝒓 𝒔𝒆𝒄𝒕𝒐𝒓 EX 9. Area of the minor segment ACB = Area of the minor sector OACBO – Area of the triangle AOB. 𝜃 𝜃 10. Area of triangle AOB if angle AOB is 𝜃 𝑖𝑠 𝝅 𝑟 2 𝐬𝐢𝐧 𝑐𝑜𝑠 𝜽 2 2 ( Can be used if the angle is 60° , 90°, 120° ) 11. Area of Major segment = 𝝅 𝑟 2 − 𝑨𝒓𝒆𝒂 𝒐𝒇 𝒎𝒊𝒏𝒐𝒓 𝒔𝒆𝒈𝒎𝒆𝒏𝒕 18 CHAPTER 13 SURFACE AREA AND VOLUMES CUBOID : 𝑙 = 𝑙𝑒𝑛𝑔𝑡ℎ 𝑏 = 𝑏𝑟𝑒𝑎𝑑𝑡ℎ ℎ = ℎ𝑒𝑖𝑔ℎ𝑡 1. TSA of the Cuboid = 𝟐 ( 𝒍𝒃 + 𝒃𝒉 + 𝒉𝒍 ) 𝒔𝒒 𝒖𝒏𝒊𝒕𝒔 2. CSA of the Cuboid = 𝟐 𝒉 ( 𝒍 + 𝒃 ) 𝒔𝒒. 𝒖𝒏𝒊𝒕𝒔 3. Volume of the Cuboid = 𝒍 𝒃 𝒉 𝒄𝒖𝒃𝒊𝒄 𝒖𝒏𝒊𝒕𝒔 4. Diagonal of the Cuboid = √𝑙 2 + 𝑏 2 + ℎ2 B CUBE : 𝒂 = 𝒔𝒊𝒅𝒆 𝒐𝒇 𝒕𝒉𝒆 𝑪𝒖𝒃𝒆 U 1. TSA of the Cube = 𝟔 𝑎2 𝒔𝒒. 𝒖𝒏𝒊𝒕𝒔 H 2. CSA of the Cube = 𝟒 𝑎2 𝒔𝒒. 𝒖𝒏𝒊𝒕𝒔 3. Volume of the Cube = 𝑎3 𝒄𝒖. 𝒖𝒏𝒊𝒕𝒔 P 4. Diagonal of the Cube =√3 a units EX CYLINDER : r = radius h = height 1. CSA of the Cylinder = 𝟐 𝝅 𝒓 𝒉 𝒔𝒒. 𝒖𝒏𝒊𝒕𝒔 2. TSA of the cylinder = 𝟐 𝝅 𝒓 (𝒉+r) sq units 3. Volume of the cylinder = 𝜋𝑟 2 ℎ cu units CONE : 𝒓 = 𝒓𝒂𝒅𝒊𝒖𝒔 𝒐𝒇 𝒕𝒉𝒆 𝒃𝒂𝒔𝒆 𝒉 = 𝒗𝒆𝒓𝒕𝒊𝒄𝒂𝒍 𝒉𝒆𝒊𝒈𝒉𝒕 𝒍 = 𝒔𝒍𝒂𝒏𝒕 𝒉𝒆𝒊𝒈𝒉𝒕 1. CSA of the Cone= 𝝅𝒓𝒍 𝒔𝒒. 𝒖𝒏𝒊𝒕𝒔 2. TSA of the Cone = 𝝅r(𝒍+r) 𝒔𝒒. 𝒖𝒏𝒊𝒕𝒔 19 1 3. Volume of the cone = 𝜋𝑟 2 ℎ cu units 3 4. Relation between r,h,l is 𝑙 2 = 𝑟 2 + ℎ2 SPHERE : 𝒓 = 𝒓𝒂𝒅𝒊𝒖𝒔 𝒐𝒇 𝒕𝒉𝒆 𝒔𝒑𝒉𝒆𝒓𝒆 1. Surface Area ( TSA = CSA ) of the Sphere = 𝟒 𝝅𝑟 2 𝒔𝒒. 𝒖𝒏𝒊𝒕𝒔 4 2. Volume of the Sphere = 𝜋𝑟 3 cu units 3 HEMISPHERE : B 𝒓 = 𝒓𝒂𝒅𝒊𝒖𝒔 𝒐𝒇 𝒕𝒉𝒆 𝒔𝒑𝒉𝒆𝒓𝒆 1. CSA of the Hemisphere = 𝟐 𝝅 𝒓𝟐 𝒔𝒒. 𝒖𝒏𝒊𝒕𝒔 U 2. TSA of the Hemisphere = 𝟑 𝝅 𝒓𝟐 𝒔𝒒. 𝒖𝒏𝒊𝒕𝒔 H 𝟐 3. Volume of the Hemisphere = 𝝅𝒓𝟑 c𝒖. 𝒖𝒏𝒊𝒕𝒔 𝟑 P HOLLOW CYLINDER : EX 𝑅 𝑖𝑠 𝑡ℎ𝑒 𝑟𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑜𝑢𝑡𝑒𝑟 𝑐𝑖𝑟𝑐𝑙𝑒 ; 𝑟 𝑖𝑠 𝑡ℎ𝑒 𝑟𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑖𝑛𝑛𝑒𝑟 𝑐𝑖𝑟𝑐𝑙𝑒 ; ℎ = ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 1. CSA of the Hollow Cylinder = 𝟐𝝅𝑹𝒉 + 𝟐𝝅𝒓𝒉 = 𝟐𝝅𝒉 ( 𝑹 + 𝒓 ) 𝒔𝒒. 𝒖𝒏𝒊𝒕𝒔 2. Area of the top and bottom of the Hollow Cylinder = 𝟐 ( 𝝅 𝑹𝟐 − 𝝅 𝒓𝟐 ) = 𝟐 𝝅 ( 𝑹𝟐 − 𝒓𝟐 ) 𝒔𝒒. 𝒖𝒏𝒊𝒕𝒔 3. TSA of the Hollow Cylinder = 𝟐 𝝅 𝒉 ( 𝑹 + 𝒓 ) + 𝟐 𝝅 ( 𝑹𝟐 − 𝒓𝟐 ) 𝒔𝒒. 𝒖𝒏𝒊𝒕𝒔 4. Volume of the material of the Hollow Cylinder = 𝝅𝑹𝟐 𝒉− 𝝅 𝒓𝟐 𝒉 =𝝅𝒉(𝑹𝟐 −𝒓𝟐 ) 𝒄𝒖. 𝒖𝒏𝒊𝒕𝒔 20 SPHERICAL SHELL : 𝑅 𝑖𝑠 𝑡ℎ𝑒 𝑟𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑜𝑢𝑡𝑒𝑟 𝑐𝑖𝑟𝑐𝑙𝑒 ; 𝑟 𝑖𝑠 𝑡ℎ𝑒 𝑟𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑖𝑛𝑛𝑒𝑟 𝑐𝑖𝑟𝑐𝑙𝑒 1. Surface area = outer surface area = 𝟒𝝅𝑹𝟐 𝒔𝒒. 𝒖𝒏𝒊𝒕𝒔 4 2. Volume of the material of the Spherical shell = 𝜋(𝑅3 − 𝑟 3 ) cu units 3 COMBINED SOLIDS : 1. WHEN THREE EQUAL CUBES OF SIDE ‘a ’ ARE JOINED END TO END A CUBOID IS FORMED. Length of the Cuboid = 3 a Breadth of the Cuboid = a B Height of the Cuboid = a U (Only the length will vary depending on the number of cubes joined) P H EX 2. A CONE SURMOUNTED BY A HEMISPHERE : Surface area of the solid = CSA of the Cone + CSA of the Hemisphere = 𝝅 𝒓 𝒍 + 𝟐 𝝅 𝒓𝟐 𝒔𝒒. 𝒖𝒏𝒊𝒕𝒔 3. A CUBE SURMOUNTED BY A HEMISPHERE : 21 Surface Area of the solid = TSA of the Cube – Area of the base circle of the Hemisphere + CSA of the Hemisphere. 4. CYLINDER SURMOUNTED BY A CONE : (CIRCUS TENT SUMS ) B Area of the canvas required for the Circus Tent = CSA of the Cylinder + CSA of the Cone U P H EX 5. A CONE OF MAXIMUM SIZE CARVED OUT FROM A CUBE OF SIDE ‘a’ The diameter of the base of the cone = 𝟐 𝒓 = 𝒂 = 𝒔𝒊𝒅𝒆 𝒐𝒇 𝒕𝒉𝒆 𝒄𝒖𝒃𝒆. The height ′𝒉′ 𝒐𝒇 𝒕𝒉𝒆 𝒄𝒐𝒏𝒆 = 𝒂 = 𝒔𝒊𝒅𝒆 𝒐𝒇 𝒕𝒉𝒆 𝒄𝒖𝒃𝒆. Volume of the remaining solid = Volume of the Cube – Volume of the Cone. TSA of the remaining solid = TSA of the Cube – Area of the base circle of the Cone + CSA of the Cone 22 6. A cylinder of diameter 7 cm is drilled out from a cubical block of dimensions 15 cm X 10 cm X 5 cm. Surface Area of the Remaining Solid = TSA of the Cuboid - Area of the top and bottom Circles of the Cylinder + CSA of the Cylinder. B U H 7. A Cone of same height and same base radius as a cylinder is hollowed out P from the Cylinder : T S A of the remaining solid = CSA of the cylinder + Area of the top Circle of the EX cylinder + CSA of the Cone 8. When lead shots or marbles are dropped in a cylindrical or conical vessel ( vessel of any shape ), the volume of the water raised or the volume of the water overflown = Volume of the lead shots or marbles dropped. ( Principle of Archimedes ) 23 CHAPTER 14 STATISTICS 1. Mean or average 𝑥̅ of ‘n’ number of observations = sum of all observations / Total no of observations. ∑ 𝑥𝑖 𝒙̅ = ( i = 1,2,----n) 𝑛 ∑ 𝑓𝑖 𝑥𝑖 2. Mean for ungrouped data = 𝒙̅ = ∑ 𝑓𝑖 ( i = 1,2,----n) 3. Median of ‘n’ observations : a. Arrange the given observations in ascending or descending order. B b. If ‘n’ is odd, median is the [(𝒏+𝟏)/𝟐] 𝒕𝒉 observation. U c. If ‘n’ is even, median=[𝒏/𝟐 𝒕𝒉 𝒐𝒃𝒔𝒆𝒓𝒗𝒂𝒕𝒊𝒐𝒏+(𝒏/𝟐+ 𝟏) 𝒕𝒉 𝒐𝒃𝒔𝒆𝒓𝒗𝒂𝒕𝒊𝒐𝒏÷ 𝟐 4. Mode : The most frequently repeated observation. H 5. Mean, Median and Mode are called the Measures of Central Tendency. P CONCEPTS IN X STANDARD : EX 1. Class Mark is the Mid-Point of the Class Interval. (Upper Class Limit + Lower Class Limit) /2 2. Mean of Grouped Data : Mean by Assumed Mean Method : ∑ 𝑓𝑖 𝑑𝑖 𝑥̅= 𝑎 + ∑ 𝑓𝑖 ; a = Assumed Mean ; 𝑑𝑖 = 𝑥𝑖 − 𝑎 ∑ 𝑓 𝑖 𝑢𝑖 3. Mean by Step Deviation Method : 𝑥̅= 𝑎 + [ ∑ 𝑓𝑖 (h)] 𝑢𝑖 = 𝑥𝑖−𝑎/ℎ 𝑓1 −𝑓0 4. Mode of Grouped Data : Mode = 𝑙 + [([( ) × ℎ] 2𝑓1 −𝑓0 −𝑓2 𝑙 = 𝑙𝑜𝑤𝑒𝑟 𝑙𝑖𝑚𝑖𝑡 𝑜𝑓 𝑚𝑜𝑑𝑎𝑙 𝑐𝑙𝑎𝑠𝑠. ( 𝑴𝒐𝒅𝒂𝒍 𝒄𝒍𝒂𝒔𝒔 𝒊𝒔 𝒕𝒉𝒆 𝒐𝒏𝒆 𝒘𝒊𝒕𝒉 𝒉𝒊𝒈𝒉𝒆𝒔𝒕 𝒇𝒓𝒆𝒒𝒖𝒆𝒏𝒄𝒚 ) 𝑓1 = 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑜𝑑𝑎𝑙 𝑐𝑙𝑎𝑠𝑠 24 𝑓0 = 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑟𝑒𝑐𝑒𝑒𝑑𝑖𝑛𝑔 𝑐𝑙𝑎𝑠𝑠 𝑓2 = 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑢𝑐𝑐𝑒𝑑𝑖𝑛𝑔 𝑐𝑙𝑎𝑠𝑠 ℎ = 𝑠𝑖𝑧𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑜𝑑𝑎𝑙 𝑐𝑙𝑎𝑠𝑠 𝑁 ( −𝐹) 2 5. Median for grouped Data : Median = 𝑙 + [ × ℎ] 𝑓 𝑙 = 𝑙𝑜𝑤𝑒𝑟 𝑙𝑖𝑚𝑖𝑡 𝑜𝑓 𝑚𝑒𝑑𝑖𝑎𝑛 𝑐𝑙𝑎𝑠𝑠 Median class is the corresponding class in which the cumulative frequency is just greater than N/2 𝑓 = 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑒𝑑𝑖𝑎𝑛 𝑐𝑙𝑎𝑠𝑠 𝐹 = 𝑐𝑢𝑚𝑢𝑙𝑎𝑡𝑖𝑣𝑒 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑟𝑒𝑐𝑒𝑒𝑑𝑖𝑛𝑔 𝑐𝑙𝑎𝑠𝑠 B 𝑁 = ∑ 𝑓𝑖 ℎ = 𝑠𝑖𝑧𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑒𝑑𝑖𝑎𝑛 𝑐𝑙𝑎𝑠𝑠. U 6. For finding median, the class interval should be continuous. If not, make it continuous. H Ex : Consider 3 Class intervals which are not continuous. 110 – 119 , 120 – 129, 130 – 139. P The difference between the upper limit of one class interval and the lower limit EX of the consecutive class interval is 1. ( 120 – 119 = 1 ) Divide the difference 1 by 2. You get 0.5. Subtract 0.5 from lower limits, add 0.5 to upper limits. The new continuous Class Intervals are 109.5 – 119.5 , 119.5 – 129.5 , 129.5 – 139.5. Now h = 10. 7. If instead of Class intervals, given less than 20, less than 25 -----, then cumulative frequency is given. A proper tabular column with Class interval and frequency to be prepared. 8. Empirical Relationship between Mean, Median and Mode : 3 Median = Mode + 2 Mean 25 CHAPTER 15 PROBABILITY 1. Probability is the likelihood of occurrence of an event. 2. Probability of an event A is denoted by P ( A ). 𝑁𝑜.𝑜𝑓 𝑓𝑎𝑣𝑜𝑢𝑟𝑎𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 𝑛(𝐴) 𝑃(𝐴) = = S is called the sample space. 𝑁𝑜.𝑜𝑓 𝑎𝑙𝑙 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 𝑛(𝑆) 3. TOSSING A COIN : A ) When a coin is tossed once, 𝑆 = {𝐻 , 𝑇} 𝑛( 𝑆 ) = 2 B ) When one coin is tossed twice or 2 coins are tossed once, B 𝑆 = {( 𝐻, 𝐻 ), ( 𝐻, 𝑇 ), ( 𝑇, 𝐻 ), (𝑇, 𝑇)} 𝑛 ( 𝑆 ) = 22 = 4 U C ) When one coin is tossed thrice or 3 coins are tossed once, 𝑆 = {( 𝐻,𝐻,𝐻 ), (𝐻,𝐻,𝑇), (𝐻,T,𝐻 ), (𝐻,𝑇,𝑇), ( 𝑇,𝐻,𝐻), (𝑇,𝐻,𝑇), (𝑇,𝑇,𝐻), 𝑇, 𝑇, 𝑇)} H 𝑛(𝑆) = 23 = 8. In general if one coin is tossed ‘n’ times or ‘n’ coins are tossed once , 𝑛(𝑆) = 2𝑛 P 4. THROWING A DIE : EX A ) When a die is thrown once, 𝑆 = {1, 2, 3, 4, 5, 6 } 𝑛 (𝑆) = 6 B ) When a pair of dice are thrown once or a die is thrown twice (1,1), (1,2), − − −(1,6) (2,1), (2,2), − − −(2,6) (3,1), (3,2), − − −(3,6) 𝑆= (4,1), (4,2) − − − (4,6) (5,1), (5,2), − − −(5,6) { (6,1), (6,2), − − −(6,6) 𝑛(𝑆) = 62 = 36 C ) When three dice are thrown once or one die is thrown thrice 𝑛(𝑆) = 63 =216. 5. Playing Cards : 26 Total number of cards in a deck = 52. No. of red cards = No. of black cards = 26 No. of cards in each suit = 13 No. of face cards = 12 6. 𝑃 ( 𝐴̅) is the probability of event A not happening. 7. 𝑃 ( 𝐴 ) + 𝑃 ( 𝐴̅) = 1 ; 𝑃 ( 𝐴̅) = 1 − 𝑃 ( 𝐴 ) 𝐴 𝑎𝑛𝑑 𝐴̅𝑎𝑟𝑒 𝑐𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝑒𝑣𝑒𝑛𝑡𝑠. 8. 𝑃𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑎 𝑠𝑢𝑟𝑒 𝑒𝑣𝑒𝑛𝑡 𝑖𝑠 1. 9. 𝑃𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑎𝑛 𝑖𝑚𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑒𝑣𝑒𝑛𝑡 𝑖𝑠 0. 10. 𝟎 ≤ 𝑷 ≤ 𝟏 11. The sum of the probabilities of all the outcomes ( elementary events ) of an B experiment is 1. 12. 𝑨𝒕 𝒍𝒆𝒂𝒔𝒕 𝒎𝒆𝒂𝒏𝒔 ≥ ( 𝒈𝒓𝒆𝒂𝒕𝒆𝒓 𝒕𝒉𝒂𝒏 𝒐𝒓 𝒆𝒒𝒖𝒂𝒍 𝒕𝒐 ) U 13. 𝑨𝒕 𝒎𝒐𝒔𝒕 𝒎𝒆𝒂𝒏𝒔 ≤ ( 𝒍𝒆𝒔𝒔𝒆𝒓 𝒕𝒉𝒂𝒏 𝒐𝒓 𝒆𝒒𝒖𝒂𝒍 𝒕𝒐 ) H 14. Difference between ‘or’ and ‘and ‘ in probability sums. Ex : Consider numbers from 1 to 20. Find the probability of numbers divisible by P (i)2 or 3(ii) 2 and 3. EX For the ( i ) subdivision, consider numbers divisible by 2 and numbers divisible by 3 for favourable outcomes subtracting numbers divisible by both 2 and 3. ( since repeated in both ). For the ( ii ) subdivision, consider numbers divisible by both 2 and 3 ( numbers divisible by 6 ) as favourable outcome