Kaliyuga Ekalavya

NCERT Class X Chapter 7: Coordinate Geometry Example 2 Question: Show that the points (1, 7), (4, 2), (–1, –1) and (–4, 4) are the vertices of a square. Given: The points are A(1, 7), B(4, 2), C(–1, –1), and D(–4, 4). To Find: Show that these points are the vertices of a square. Formula: Distance between two points \( (x_1, y_1) \) and \( (x_2, y_2) \): $$ \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$ Solution: Step 1: Find the length of AB. $$ AB = \sqrt{(4 - 1)^2 + (2 - 7)^2} = \sqrt{3^2 + (-5)^2} = \sqrt{9 + 25} = \sqrt{34} $$ Step 2: Find the length of BC. $$ BC = \sqrt{(-1 - 4)^2 + (-1 - 2)^2} = \sqrt{(-5)^2 + (-3)^2} = \sqrt{25 + 9} = \sqrt{34} $$ Step 3: Find the length of CD. $$ CD = \sqrt{(-4 - (-1))^2 + (4 - (-1))^2} = \sqrt{(-3)^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34} $$ Step 4: Find the length of DA. $$ DA = \sqrt{(1 - (-4))^2 + (7 - 4)^2} = \sqrt{5^2 + 3^2} = \sqrt{25 + 9} = \sqrt{34} $$ Step 5: Fin...

NCERT Class X Chapter 12: Surface Areas and Volumes Question: The decorative block is made of two solids — a cube and a hemisphere. The base of the block is a cube with edge 5 cm, and the hemisphere fixed on the top has a diameter of 4.2 cm. Find the total surface area of the block. (Use \( \pi = \frac{22}{7} \)) Given: Edge of cube, \( a = 5 \) cm Diameter of hemisphere, \( d = 4.2 \) cm Radius of hemisphere, \( r = \frac{4.2}{2} = 2.1 \) cm To Find: Total surface area of the block Formula: Surface area of cube: \( 6a^2 \) Curved surface area of hemisphere: \( 2\pi r^2 \) Area of base of hemisphere: \( \pi r^2 \) Total surface area of block: \( \text{Surface area of cube} + \text{Curved surface area of hemisphere} - \text{Area of base of hemisphere} \) Solution: Step 1: Find the surface area of the cube. $$ \text{Surface area of cube} = 6a^2 = 6 \times (5)^2 = 6 \times 25 = 150 \ \text{cm}^2 $$ Step 2: Find the curved surface area ...

NCERT Class X Chapter 9: Some Applications of Trigonometry Example 2 Question: An electrician has to repair an electric fault on a pole of height 5 m. She needs to reach a point 1.3 m below the top of the pole to undertake the repair work. What should be the length of the ladder that she should use which, when inclined at an angle of 60° to the horizontal, would enable her to reach the required position? Also, how far from the foot of the pole should she place the foot of the ladder? (You may take \( \sqrt{3} = 1.73 \)) Given: Height of the pole = 5 m Point to be reached is 1.3 m below the top, so height to reach = \( 5 - 1.3 = 3.7 \) m Angle of inclination of ladder with horizontal = \( 60^\circ \) \( \sqrt{3} = 1.73 \) To Find: 1. The length of the ladder required. 2. The distance from the foot of the pole to the foot of the ladder. Formula: Trigonometric ratios in a right-angled triangle: \[ \sin \theta = \frac{\text{Opposite side}}{\text{Hypotenuse}} ...

NCERT Class X Chapter 1: Real Numbers Example 2 Question: Find the LCM and HCF of 6 and 20 by the prime factorisation method. Given: Numbers: 6 and 20 To Find: LCM and HCF of 6 and 20 Formula: Prime Factorisation Method: HCF: Product of the lowest powers of all common prime factors. LCM: Product of the highest powers of all prime factors present in any number. Solution: Step 1: Write the prime factorisation of each number. $$ 6 = 2 \times 3 $$ $$ 20 = 2 \times 2 \times 5 = 2^2 \times 5 $$ Step 2: List all prime factors with their powers. 6: \(2^1,\, 3^1\) 20: \(2^2,\, 5^1\) Step 3: Find the HCF by taking the product of the lowest powers of all common prime factors. Common prime factor: 2 Lowest power of 2: 1 $$ \text{HCF}(6,\,20) = 2^1 = 2 $$ Step 4: Find the LCM by taking the product of the highest powers of all prime factors present. 2: Highes...

NCERT Class X Chapter 11: Area Related To Circles Example 2 Question: Find the area of the segment AYB shown in Fig. 11.6, if radius of the circle is 21 cm and \( \angle AOB = 120^\circ \). (Use \( \pi = \frac{22}{7} \)) Given: Radius of the circle, \( r = 21 \) cm Central angle, \( \angle AOB = 120^\circ \) \( \pi = \frac{22}{7} \) To Find: Area of the segment AYB Formula: Area of segment = Area of sector − Area of triangle Area of sector = \( \frac{\theta}{360^\circ} \times \pi r^2 \) Area of triangle (when two sides and included angle are known): \( = \frac{1}{2} r^2 \sin\theta \) Solution: Step 1: Find the area of sector AOB. $$ \text{Area of sector} = \frac{120^\circ}{360^\circ} \times \pi \times (21)^2 $$ Step 2: Simplify the fraction and substitute the value of \( \pi \). $$ = \frac{1}{3} \times \frac{22}{7} \times 441 $$ Step 3: Calculate the area of the sector. $$ = \frac{1}{3} \times \frac{22}{7} \times 441 \\ = \frac{22}{7} \ti...

NCERT Class X Chapter 14: Probability Example 2 Question: A bag contains a red ball, a blue ball and a yellow ball, all the balls being of the same size. Kritika takes out a ball from the bag without looking into it. What is the probability that she takes out the following (i) yellow ball? (ii) red ball? (iii) blue ball? Given: A bag contains 3 balls: 1 red, 1 blue, and 1 yellow.  Number of red balls = 1 Number of blue balls = 1 Number of yellow balls = 1  Total number of balls = Number of red balls + Number of blue balls  +   Number of yellow balls = 3 Kritika picks one ball at random. To Find: The probability of picking ( i) a yellow ball,  (ii) a red ball, and  (iii) a blue ball. Formula: Probability = Number of favorable outcomes Total number of possible outcomes Solution: (i) Probability of picking a yellow ball: Number of yellow balls = 1 Total number of balls = 3 Pro...

NCERT Class X Chapter 5: Arithmetic Progression Example 2(iv) Question: Which of the following list of numbers form an AP? If they form an AP, write the next two terms 1, 1, 1, 2, 2, 2, 3, 3, 3, . . . Given: The sequence: 1, 1, 1, 2, 2, 2, 3, 3, 3, . . . To Find: Whether the given sequence forms an arithmetic progression (AP). If it does, find the next two terms. Formula: In an AP, the difference between consecutive terms is constant (common difference). Solution: Let the given list of numbers be a1, a2, a3, a4, ... Then, a1 = 1  a2 = 1  a3 = 1 a4 = 2 a5 = 2 a6 = 2 a7 = 3 a8 = 3 a9 = 3  Lets calculate the common difference, d = a2 - a1 = 1 - 1 = 0 d = a3 - a2 = 1 - 1 = 0 d = a4 - a3 = 2 - 1 = 1 The common difference changes and hence it is not a constant.  Hence, the sequence is not an AP because the difference between consecutive terms is not constant.  Result: The given sequence 1, 1, 1, 2, 2, 2, 3, 3, 3, . . . is not ...

NCERT Class X Chapter 5: Arithmetic Progression Example 2(iii) Question: Which of the following list of numbers form an AP? If they form an AP, write the next two terms -2,2,-2,2,-2, . . . Given: The sequence is -2, 2, -2, 2, -2, . . . To Find: Whether the given sequence forms an AP. If it does, find the next two terms. Formula: In an Arithmetic Progression (AP), the difference between consecutive terms is constant.  This constant difference is called the common difference (d). Solution: Let's examine the differences between consecutive terms: 2 - (-2) = 4 -2 - 2 = -4 Since the differences are not constant (4 and -4), the sequence does not have a common difference. Therefore, the given sequence is not an arithmetic progression. Result: The sequence -2, 2, -2, 2, -2, ... is not an arithmetic progression. Next question solution: NCERT Class X Chapter 5: Arithmetic Progression Example 2 (iv). Explore more in Arithmetic Progressions...

NCERT Class X Chapter 5: Arithmetic Progression Example 2(ii) Question: Which of the following list of numbers form an AP? If they form an AP, write the next two terms : 1, – 1, – 3, – 5, . . . Given: The list of numbers is 1, –1, –3, –5, . . . To Find: Whether the given list of numbers forms an arithmetic progression (AP). If it forms an AP, find the next two terms. Formula: In an arithmetic progression, the difference between consecutive terms is constant.  This constant difference is called the common difference (d). Solution: Let the given list of numbers be denoted by a 1 , a 2 , a 3 , a 4 , ... Then, a 1 = 1, a 2 = –1, a 3 = –3, a 4 = –5, ... Let's find the common difference (d): d = a 2 – a 1 = –1 – 1 = –2 d = a 3 – a 2 = –3 – (–1) = –2 d = a 4 – a 3 = –5 – (–3) = –2 Since the common difference is constant (d = –2), the given list of numbers forms an arithmetic progression. ...

NCERT Class X Chapter 5: Arithmetic Progression Example 2(i) Question: Which of the following list of numbers form an AP? If they form an AP, write the next two terms : (i) 4, 10, 16, 22, . . . Given: The list of numbers is 4, 10, 16, 22, . . . To Find: 1. Whether the given list of numbers forms an AP.  2. If they form an AP, find the next two terms. Formula: In an Arithmetic Progression (AP), the difference between consecutive terms is constant.  This constant difference is called the common difference (d). Solution: Let the given list of numbers be a 1 , a 2 , a 3 , a 4 , ... a 1 = 4 a 2 = 10 a 3 = 16 a 4 = 22 Let's find the common difference (d): d = a 2 - a 1 = 10 - 4 = 6 d = a 3 - a 2 = 16 - 10 = 6 d = a 4 - a 3 = 22 - 16 = 6 Since the common difference is constant (d = 6), the given list of numbers forms an AP. To find the next two terms, we add the co...

NCERT Class X Chapter 4: Quadratic Equation Example 2 (iii) Question: Check whether the following are quadratic equations:  x (2x + 3) = x 2 + 1 Given: Equation: x (2x + 3) = x 2 + 1 To Find: Whether the given equation is a quadratic equation. Formula: A quadratic equation is of the form ax 2 + bx + c = 0, where a ≠ 0. Solution: The given equation: x (2x + 3) = x 2 + 1 Expand the left side: ⇒ 2x 2 + 3x Substitute the expanded form back into the equation: ⇒ 2x 2 + 3x = x 2 + 1 Move all terms to one side to get the standard form ax 2 + bx + c = 0: ⇒ 2x 2 - x 2 + 3x - 1 = 0 ⇒ x 2 + 3x - 1 = 0 Compare this equation to the standard form ax 2 + bx + c = 0: Here, a = 1, b = 3, and c = -1. Since the a...

NCERT Class X Chapter 4: Quadratic Equation Example 2 (ii) Question: Check whether the following are quadratic equations:  x(x + 1) + 8 = (x + 2) (x – 2) Given: Equation: x(x + 1) + 8 = (x + 2) (x – 2) To Find: Whether the given equation is a quadratic equation . Formula: WKT, a quadratic equation is of the form ax 2 + bx + c = 0, where a ≠ 0. Solution: The given equation: x(x + 1) + 8 = (x + 2) (x – 2) Expand the left side: ⇒ x 2 + x + 8 WKT, the algebraic identity (a + b)(a - b) = a 2 - b 2 Expand the right side using the identity (a + b)(a - b) ⇒ x 2 - 2 2 = x 2 - 4 Substitute the expanded forms back into the equation: ⇒ x 2 + x + 8 = x 2 - 4 Move all terms to one side to get the standard form ax 2 + bx + c = 0: ⇒ x 2 - x 2 + x + 8 + 4 =...

NCERT Class X Chapter 4: Quadratic Equation Example 2 (i) Question: Check whether the following are quadratic equations:  (x – 2) 2 + 1 = 2x – 3 Given: Equation: (x – 2) 2 + 1 = 2x – 3 To Find: Whether the given equation is a quadratic equation . Formula: A quadratic equation is of the form ax 2 + bx + c = 0, where a ≠ 0. Solution: The given equation: (x – 2) 2 + 1 = 2x – 3 Expand (x – 2) 2 using (a - b) 2 formula (a - b) 2 = a 2 - 2ab + b 2 ⇒ x 2 - 2(x)(2) + 2 2 + 1 = 2x – 3 ⇒ x 2 - 4x + 4 + 1 = 2x – 3 Simplify the left side: ⇒ x 2 - 4x + 5 = 2x – 3 Move all terms to one side to get the standard form ax 2 + bx + c = 0: ⇒ x 2 - 4x - 2x + 5 + 3 = 0 ⇒  x 2 - 6x + 8 = 0 Compare this equation to the standard form ax 2 + bx + c = 0: Here, a = 1, b = -6, c = 8. Since a = 1 (which is not equal to 0), the equation...

NCERT Class X Chapter 4: Quadratic Equation Example 2 (iv) Question: Check whether the following are quadratic equations:  (x + 2) 3 = x 3 – 4 Given: Equation: (x + 2) 3 = x 3 – 4 To Find: Whether the given equation is a quadratic equation. Formula: A quadratic equation is of the form ax 2 + bx + c = 0, where a ≠ 0. Solution: The given equation:(x + 2) 3 = x 3 – 4 WKT,  (a + b) 3 = a 3 + b 3 + 3ab(a + b) Expand the left side using (a + b) 3 : Here, a = x and b = 2. ⇒ x 3 + 2 3 + 3(x)(2)(x + 2) ⇒ x 3 + 8 + 6x(x + 2) ⇒ x 3 + 8 + 6x 2 + 12x Substitute the expanded form back into the original equation: ⇒ x 3 + 6x 2 + 12x + 8 = x 3 – 4 Move all terms to one side to get the standard form ax 2 + bx + c = 0: ⇒ x 3 - x 3 + 6x 2 + 12x + 8 + 4 = 0 ⇒ 6x 2 + 12x + 12 = 0 The given equation (x + 2) 3 = x 3 – 4 simplifies to 6x 2 + 12x + 12 = 0. ...