Kaliyuga Ekalavya

NCERT Class X Chapter 7: Coordinate Geometry Example 4 Question: Find a relation between \( x \) and \( y \) such that the point \( (x, y) \) is equidistant from the points \( (7, 1) \) and \( (3, 5) \). Given: Let \( A = (7, 1) \), \( B = (3, 5) \), and \( P = (x, y) \) is a point equidistant from \( A \) and \( B \). To Find: A relation between \( x \) and \( y \) such that \( P \) is equidistant from \( A \) and \( B \). Formula: The distance between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by: $$ \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$ Solution: Step 1: Let the distances from \( P(x, y) \) to \( A(7, 1) \) and \( B(3, 5) \) be equal. $$ PA = PB $$ $$ \sqrt{(x - 7)^2 + (y - 1)^2} = \sqrt{(x - 3)^2 + (y - 5)^2} $$ Step 2: Square both sides to remove the square roots. $$ (x - 7)^2 + (y - 1)^2 = (x - 3)^2 + (y - 5)^2 $$ Step 3: Expand both sides. $$ (x^2 - 14x + 49) + (y^2 - 2y + 1) = (x^2 - 6x + 9) + ...

NCERT Class X Chapter 3: Pair of Linear Equations In Two Variables Example 4 Question: Solve the following pair of equations by substitution method: \(7x - 15y = 2\) and \(x + 2y = 3\) Given: The given pair of equations is: \(7x - 15y = 2 \)    ...(1) \(x + 2y = 3 \)    ...(2) To Find: Find the values of \(x\) and \(y\) using the substitution method. Formula: In the substitution method: Solve one equation for one variable in terms of the other. Substitute this value in the second equation to get the value of one variable. Substitute back to get the other variable. Solution: Step 1: Express \(x\) in terms of \(y\) using equation (2). $$ x + 2y = 3 \\ \Rightarrow x = 3 - 2y \quad ...(3) $$ Step 2: Substitute the value of \(x\) from (3) into equation (1). $$ 7x - 15y = 2 \\ 7(3 - 2y) - 15y = 2 \\ 21 - 14y - 15y = 2 \\ 21 - 29y = 2 $$ Step 3: Solve for \(y\). $$ 21 - 29y = 2 \\ -29y = 2 - 21 \\ -29y = -19 \\ y = ...

NCERT Class X Chapter 12: Surface Areas And Volumes Example 4 Question: Mayank made a bird-bath for his garden in the shape of a cylinder with a hemispherical depression at one end. The height of the cylinder is 1.45 m and its radius is 30 cm. Find the total surface area of the bird-bath. (Use \( \pi = \frac{22}{7} \)) Given: Height of the cylinder, \( h = 1.45\,\text{m} = 145\,\text{cm} \) Radius of the cylinder, \( r = 30\,\text{cm} \) \( \pi = \frac{22}{7} \) To Find: Total surface area of the bird-bath. Formula: Total surface area = Curved surface area of cylinder + Area of base + Curved surface area of hemisphere Curved surface area of cylinder: \( 2\pi r h \) Area of base: \( \pi r^2 \) Curved surface area of hemisphere: \( 2\pi r^2 \) Solution: Step 1: Write the expression for total surface area. $$ \text{Total surface area} = 2\pi r h + \pi r^2 + 2\pi r^2 = 2\pi r h + 3\pi r^2 $$ Step 2: Substitute the given val...

NCERT Class X Chapter 9: Some Applications of Trigonometry Example 4 Question: From a point P on the ground the angle of elevation of the top of a 10 m tall building is 30°. A flag is hoisted at the top of the building and the angle of elevation of the top of the flagstaff from P is 45°. Find the length of the flagstaff and the distance of the building from the point P. (You may take \( \sqrt{3} = 1.732 \)) Given: Height of the building = 10 m Angle of elevation of the top of the building from P = \( 30^\circ \) Angle of elevation of the top of the flagstaff from P = \( 45^\circ \) \( \sqrt{3} = 1.732 \) (to be used in calculation) To Find: Length of the flagstaff Distance of the building from point P Formula: Trigonometric ratio for tangent: $$ \tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}} $$ Solution: Step 1: Let the distance from point P to the base of the building be \( x \) meters. Let the length of the flagstaff be \( h \) meters...

NCERT Class X Chapter 1: Real Numbers Example 4 Question: Find the HCF and LCM of 6, 72, and 120 using the prime factorisation method. Given: The numbers are 6, 72, and 120. To Find: The Highest Common Factor (HCF) and Lowest Common Multiple (LCM) of 6, 72, and 120 using the prime factorisation method. Formula: HCF (by prime factorisation): Multiply the lowest powers of all common prime factors present in each number. LCM (by prime factorisation): Multiply the highest powers of all prime factors present in any of the numbers. Solution: Step 1: Write the prime factorisation of each number. $$6 = 2 \times 3$$ $$72 = 2 \times 2 \times 2 \times 3 \times 3$$ $$120 = 2 \times 2 \times 2 \times 3 \times 5$$ Step 2: Express each number in exponential form. $$6 = 2^1 \times 3^1$$ $$72 = 2^3 \times 3^2$$ $$120 = 2^3 \times 3^1 \times 5^1$$ Step 3: Find the HCF by taking the lowe...

NCERT Class X Chapter 14: Probability Example 4 Question: One card is drawn from a well-shuffled deck of 52 cards. Calculate the probability that the card will (i) be an ace, (ii) not be an ace. Given: A well-shuffled deck of 52 cards. To Find: (i) Probability of drawing an ace. (ii) Probability of not drawing an ace. Formula: Probability = Number of favorable outcomes Total number of possible outcomes Solution: (i) Probability of drawing an ace: Number of aces = 4 Total number of cards = 52 ⇒ Probability (Ace) = 4 52 = 1 13 (ii) Probability of not drawing an ace: Total cards - number of ace cards Number of non-aces = Total cards - number of ace cards  ⇒ Number of non-aces = 52 - 4  Number of non-aces = 48   Total number of cards = 52 ⇒ Probability (Not Ace) = 48 52 = 12 13 Result: (i) The probability of drawing an ace is 1 13 . (ii) The probability of not drawing an ace is 12 13...

NCERT Class X Chapter 5: Arithmetic Progression Example 4 Question: Which term of the AP : 21, 18, 15, . . . is – 81? Also, is any term 0? Give reason for your answer. Given: Arithmetic Progression (AP): 21, 18, 15, ... To Find: The term that is equal to -81 and whether any term is 0. Formula: n th term of an AP: a n = a + (n-1)d,  where,  a is the first term and  d is the common difference. Solution: Here,  the first term a = 21 and  the common difference d = 18 - 21 = -3. Let the n th term be -81.  Then, -81 = 21 + (n-1)(-3)  ⇒ -81 = 21 -3n + 3  ⇒ -81 = 24 -3n  ⇒ 3n = 105  ⇒ n = 105 3 = 35 Therefore, the 35 th term is -81. To check if any term is 0, let a n = 0.  Then, 0 = 21 + (n-1)(-3)  ⇒ 3(n-1) = 21  ⇒ n-1 = 7  ⇒ n = 8. Thus, the 8 th term is 0. Result: The 35 th term is -81. The 8 th term is 0. Next question solution: NCERT Class X Chapter 5: Ar...

NCERT Class X Chapter 4: Quadratic Equation Example 4 Question: Find the roots of the quadratic equation 6x 2 – x – 2 = 0. Given: The quadratic equation: 6x 2 – x – 2 = 0 To Find: The roots of the equation. Formula: For a quadratic equation ax² + bx + c = 0 , we can find two numbers m and n such that:                               ax² + bx + c = ax² + mx + nx + c where m × n = a × c m + n = b Cases: If (a × c > 0) and b < 0 , both m and n are negative . If (a × c > 0) and b > 0 , both m and n are positive . If (a × c < 0) , the numbers m and n have opposite signs . Solution: The given equation is 6x 2 – x – 2 = 0. Here, a = 6, b = -1, c = -2. Product (a × c) = 6 × (-2) = -12. Sum (b) = -1. We need to find two numbers whose ...