Kaliyuga Ekalavya

NCERT Class X Chapter 7: Coordinate Geometry Example 6 Question: Find the coordinates of the point which divides the line segment joining the points (4, –3) and (8, 5) in the ratio 3 : 1 internally. Given: Point A: (4, –3) Point B: (8, 5) The point divides AB in the ratio 3:1 internally. To Find: The coordinates of the point that divides the line segment AB in the ratio 3:1 internally. Formula: If a point \( P(x, y) \) divides the line segment joining \( A(x_1, y_1) \) and \( B(x_2, y_2) \) in the ratio \( m:n \) internally, then: $$ x = \frac{mx_2 + nx_1}{m + n}, \qquad y = \frac{my_2 + ny_1}{m + n} $$ Solution: Step 1: Assign values to the variables. Let \( A(x_1, y_1) = (4, -3) \), \( B(x_2, y_2) = (8, 5) \), and the ratio \( m:n = 3:1 \). Step 2: Write the section formula for \( x \)-coordinate. $$ x = \frac{m x_2 + n x_1}{m + n} $$ Substitute the values: $$ x = \frac{3 \times 8 + 1 \times 4}{3 + 1} $$ ...

NCERT Class X Chapter 3: Pair of Linear Equations In Two Variables Example 6 Question: In a shop the cost of 2 pencils and 3 erasers is ₹9 and the cost of 4 pencils and 6 erasers is ₹18. Find the cost of each pencil and each eraser. Given: The cost of 2 pencils and 3 erasers is ₹9. The cost of 4 pencils and 6 erasers is ₹18. To Find: The cost of one pencil. The cost of one eraser. Formula: Let the cost of one pencil be ₹\( x \). Let the cost of one eraser be ₹\( y \). Form equations using the given information: $$ \begin{align*} 2x + 3y &= 9 \\ 4x + 6y &= 18 \end{align*} $$ Solve the pair of linear equations to find \( x \) and \( y \). Solution: Step 1: Let the cost of one pencil be ₹\( x \) and the cost of one eraser be ₹\( y \). Write the equations from the given statements. $$ \begin{align*} 2x + 3y &= 9 \quad \cdots (1) \\ 4x + 6y &= 18 \quad \cdots (2) \end{align*} $$ Step 2: Observe that eq...

NCERT Class X Chapter 9: Some Applications of Trigonometry Example 6 Question: The angles of depression of the top and the bottom of an 8 m tall building from the top of a multi-storeyed building are 30° and 45°, respectively. Find the height of the multi-storeyed building and the distance between the two buildings. Given: Angle of depression of the top of the building = \(30^\circ\) Angle of depression of the bottom of the building = \(45^\circ\) Height of the smaller building = \(8\,\text{m}\) To Find: Height of the multi-storeyed building Distance between the two buildings Formula: \[ \tan \theta = \frac{\text{Perpendicular}}{\text{Base}} \] Solution: Step 1: Let the height of the multi-storeyed building be \(h\) meters and the distance between the two buildings be \(x\) meters. Step 2: Consider the angle of depression to the bottom of the 8 m building (\(45^\circ\)). From the top of the multi-storeyed building, the vertica...

NCERT Class X Chapter 1: Real Numbers Example 6 Question: Show that \(5 - \sqrt{3}\) is irrational. Given: Assume, for contradiction, that \(5 - \sqrt{3}\) is rational. To Find: Prove that \(5 - \sqrt{3}\) is irrational. Formula: Proof by contradiction using properties of rational and irrational numbers. Definitions: Rational number: A number that can be expressed as \(\frac{p}{q}\), where \(p\) and \(q\) are integers and \(q \ne 0\). Irrational number: A number that cannot be expressed as \(\frac{p}{q}\), where \(p\) and \(q\) are integers and \(q \ne 0\). Solution: Step 1: Assume that \(5 - \sqrt{3}\) is rational. Let \(5 - \sqrt{3} = \frac{a}{b}\), where \(a\) and \(b\) are integers, \(b \ne 0\). Step 2: Rearranging the equation to isolate \(\sqrt{3}\). $$ 5 - \sqrt{3} = \frac{a}{b} $$ Add \(\sqrt{3}\) to both sides: $$ 5 = \frac{a}{b} + \sqrt{3} $$ ...

NCERT Class X Chapter 14: Probability Example 6 Question: Savita and Hamida are friends. What is the probability that both will have (i) different birthdays? (ii) the same birthday? (Ignoring a leap year). Given: Two friends, Savita and Hamida. To Find: (i) Probability that both have different birthdays. (ii) Probability that both have the same birthday. Formula: Probability = Favorable Outcomes Total Outcomes Solution: (i) Probability of different birthdays: Total number of days in a year = 365 Let Savita's birthday be on any day.  The probability that Hamida's birthday is different from Savita's is 365-1 365 The probability that Hamida's birthday is different from Savita's is \( \frac{364}{365}\) (ii) Probability of same birthdays: Total number of days in a year = 365 The probability that Hamida's birthday is the same as Savita's is 1 365 Result: (i) Probability that both have different ...

NCERT Class X Chapter 4: Quadratic Equation Example 6 Question: Charity trust decides to build a prayer hall having a carpet area of 300 square metres with its length one metre more than twice its breadth. What should be the length and breadth of the hall?. Given: Carpet area of the prayer hall = 300 m 2 Length = 1 metre more than twice its breadth. To Find: The length of the hall. The breadth of the hall Formula: Area of a rectangle = Length × Breadth. For a quadratic equation ax 2 + bx + c = 0, the roots are given by the quadratic formula: x = -b ± √(Δ) 2a Δ = b 2 - 4ac Solution: Let the breadth of the hall (in metres) = x According to the given condition, the length is one metre more than twice its breadth. ⇒ Length (in metres) = ( 1 + 2x ) = ( 2x + 1 ) Area of the hall = Length × Breadth. ⇒ 300 = (2x + 1) × x ⇒ 300 = 2x 2 + x Group the terms on one side of t...