CBSE Class 10 Mathematics Standard Compartment Question Paper 2024 with Answer Key (Set 1- 30/S/1)

Step 1: Substituting \( x = 5 \) in the given quadratic equation: \[ 2(5)^2 + (k-1)(5) + 10 = 0 \] \[ 2(25) + 5(k-1) + 10 = 0 \] \[ 50 + 5k - 5 + 10 = 0 \]

Step 2: Simplifying the equation: \[ 55 + 5k = 0 \] \[ 5k = -55 \] \[ k = -11 \] Quick Tip: When a value is given as a solution to a quadratic equation, substitute it into the equation and solve for the unknown variable.

Step 1: The least common multiple (LCM) of two numbers is found by taking the highest power of each prime factor appearing in their prime factorizations.

Given: \[ m = p^5 q^2, \quad n = p^3 q^4 \]

Step 2: Determine the LCM by selecting the highest power of each prime factor:

- The highest power of \( p \) is \( p^5 \).
- The highest power of \( q \) is \( q^4 \).

Thus, \[ LCM(m, n) = p^5 q^4. \] Quick Tip: To find the LCM of two numbers, take the highest power of each prime factor present in their prime factorization.

Step 1: Understanding the given equations: \[ x = 2a \quad and \quad y = 3b \]

Step 2: The equation \( x = 2a \) represents a vertical line passing through \( x = 2a \).
The equation \( y = 3b \) represents a horizontal line passing through \( y = 3b \).

Step 3:
A vertical and a horizontal line always intersect at a single point, which is determined by their given values of \( x \) and \( y \).
Thus, the lines intersect at: \[ (2a, 3b) \] Quick Tip: When given equations of the form \( x = c \) and \( y = d \), they represent vertical and horizontal lines, respectively. Their intersection point is always \( (c, d) \).

Step 1: In an arithmetic progression, the difference between consecutive terms is constant.
Thus, the common difference should be the same for both: \[ (2k - 2) - (k + 7) = (2k + 6) - (2k - 2) \]

Step 2: Solve for \( k \):
\[ 2k - 2 - k - 7 = 2k + 6 - 2k + 2 \]
\[ k - 9 = 8 \]
\[ k = 17 \] Quick Tip: For three consecutive terms in an arithmetic progression, use the property that the middle term is the average of the other two: \[ 2 \times middle term = sum of first and third terms \]

Step 1: Given that \( PA \) and \( PB \) are tangents to the circle from external point \( P \), they are equal in length.

Step 2: The quadrilateral \( OAPB \) is a kite, where \( OA = OB = 5 \) cm (radius of the circle) and \( \angle APB = 60^\circ \).

Step 3: The perpendiculars \( OA \) and \( OB \) bisect \( \angle APB \) into two right-angled triangles \( \triangle OAP \) and \( \triangle OBP \) with \[ \angle OAP = 30^\circ. \]
Using the right-angled triangle properties, the tangent length \( PA \) is given by: \[ PA = OA \tan 30^\circ = 5 \times \frac{1}{\sqrt{3}} = \frac{5}{\sqrt{3}}. \] Quick Tip: In problems involving tangents to a circle, use symmetry and right-angled triangle properties to find unknown lengths. The tangent-secant theorem and trigonometry often simplify the calculations.

Step 1: Since \( MN \) is parallel to \( PS \), the triangles \( \triangle OMN \) and \( \triangle OPS \) are similar by the Basic Proportionality Theorem (Thales Theorem).

Step 2: Using the properties of similar triangles: \[ \frac{OM}{OP} = \frac{ON}{OS} = \frac{MN}{PS} \]

Step 3: Given the proportional values, we solve for \( OP \). Using the given data and proportion calculations, we determine: \[ OP = 15.3 \] Quick Tip: In geometry problems involving parallel lines and proportional segments, use the Basic Proportionality Theorem (Thales Theorem) to set up the correct proportion equations.

Step 1: A standard deck has 52 cards. After removing queens, jacks, and aces, the remaining cards are: \[ 52 - (4+4+4) = 40 \]

Step 2: There are 4 kings in a standard deck, and all kings remain in the modified deck. Thus, the probability of drawing a king is: \[ \frac{4}{40} = \frac{1}{10} \] Quick Tip: When calculating probability, determine the total number of favorable outcomes and divide by the total number of possible outcomes.

Step 1: The center \( O(x_c, y_c) \) of the circle is the midpoint of the diameter \( P(-4,5) \) and \( Q(x,y) \). Using the midpoint formula: \[ \left(\frac{x + (-4)}{2}, \frac{y + 5}{2}\right) = (2,-4) \]

Step 2: Solving for \( x \) and \( y \):
\[ \frac{x - 4}{2} = 2 \quad \Rightarrow \quad x - 4 = 4 \quad \Rightarrow \quad x = 8 \]
\[ \frac{y + 5}{2} = -4 \quad \Rightarrow \quad y + 5 = -8 \quad \Rightarrow \quad y = -13 \]

Thus, the coordinates of \( Q \) are \( (8, -13) \). Quick Tip: The midpoint of a line segment joining two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by: \[ \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \]

Step 1: Using the trigonometric identity: \[ 1 + \tan^2\theta = \sec^2\theta \]
we can rewrite the given expression as: \[ \sin^2\theta + \frac{1}{\sec^2\theta} \]

Step 2: Since \( \frac{1}{\sec^2\theta} = \cos^2\theta \), the expression simplifies to: \[ \sin^2\theta + \cos^2\theta \]

Step 3: Using the fundamental identity of trigonometry: \[ \sin^2\theta + \cos^2\theta = 1 \]

Thus, the value of the given expression is \( 1 \). Quick Tip: The fundamental trigonometric identity states that: \[ \sin^2\theta + \cos^2\theta = 1 \] This identity is useful in simplifying many trigonometric expressions.

Step 1: The volume of a cylinder is given by: \[ V_{cylinder} = \pi r^2 h_1 \]

Step 2: The volume of a cone is given by: \[ V_{cone} = \frac{1}{3} \pi r^2 h_2 \]

Step 3: Since the volumes are equal: \[ \pi r^2 h_1 = \frac{1}{3} \pi r^2 h_2 \]

Canceling \( \pi r^2 \) from both sides: \[ h_1 = \frac{1}{3} h_2 \]

Step 4: The ratio of heights is: \[ h_1 : h_2 = 1:3 \] Quick Tip: For problems involving volumes, always write down the volume formulas before solving for the required ratio or value.

Step 1: Identify the first term (\(a\)) and common difference (\(d\)) of the arithmetic progression (AP).
Given AP: -8, -3, 2, ..., 49 \(a = -8\) \(d = -3 - (-8) = 5\)

Step 2: Find the total number of terms (\(n\)) in the AP.
The \(n\)-th term of an AP is given by:
\[ 47 = -8 + (n-1) \times 5 \] \[ 47 + 8 = (n-1) \times 5 \] \[ 55 = (n-1) \times 5 \] \[ n-1 = 11 \Rightarrow n = 12 \]

Step 3: Find the 7th term from the end.
The \(k\)-th term from the end of an AP is given by: \[ a_{n-k+1} \]
Here, \(k = 7\) and \(n = 12\): \[ a_{12-7+1} = a_6 \]
Calculate \(a_6\): \[ a_6 = a + (6-1)d = -8 + 5 \times 5 = -8 + 25 = 17 \] Quick Tip: To find the \(k\)-th term from the end of an AP, use: \[ a_{n-k+1} \] where \(n\) is the total number of terms.

Step 1: The area of a sector is given by the formula: \[ A = \frac{\theta}{360^\circ} \times \pi r^2 \]

Step 2: The diagonals of a rhombus bisect each other at right angles, so the given angle \( \angle AOD \) is \( 90^\circ \). The radius of the arc is given as \( r = 6 \) cm.

Step 3: Substituting the values: \[ A = \frac{90^\circ}{360^\circ} \times \pi \times (6)^2 \]
\[ A = \frac{1}{4} \times \pi \times 36 \]
\[ A = 9\pi cm^2 \]

Thus, the area of sector \( OEA \) is \( 9\pi \) cm\(^2\). Quick Tip: The area of a sector of a circle is calculated using: \[ A = \frac{\theta}{360^\circ} \times \pi r^2 \] where \( \theta \) is the central angle and \( r \) is the radius.

Step 1: The probability formula is: \[ P = \frac{Number of favorable outcomes}{Total outcomes} \]

Step 2: Rearranging for the number of chocolate-flavored ice creams: \[ Number of chocolate ice creams = P \times Total ice creams \]
\[ = 0.055 \times 600 \]
\[ = 33 \]

Thus, the number of chocolate-flavored ice creams in the lot is \( 33 \). Quick Tip: To find the expected number of outcomes, multiply the total sample size by the probability of a single occurrence.

Step 1: Using the identity: \[ \tan^2\alpha + \cot^2\alpha = 2 \]
Since, \[ \cot^2\alpha = \frac{1}{\tan^2\alpha} \]
we substitute: \[ \tan^2\alpha + \frac{1}{\tan^2\alpha} = 2 \]

Step 2: Let \( x = \tan^2\alpha \), then: \[ x + \frac{1}{x} = 2 \]

Step 3: Solving the equation: \[ x^2 - 2x + 1 = 0 \]
\[ (x-1)^2 = 0 \Rightarrow x = 1 \]

Step 4: Since \( x = \tan^2\alpha = 1 \), we get: \[ \tan\alpha = 1 \Rightarrow \alpha = 45^\circ \] Quick Tip: For solving trigonometric equations, use standard identities and simplify algebraically before finding the required values.

Step 1: The point \( (x,0) \) on the x-axis is equidistant from \( (5,-3) \) and \( (4,2) \), meaning: \[ Distance from (x,0) to (5,-3) = Distance from (x,0) to (4,2) \]

Using the distance formula: \[ \sqrt{(x-5)^2 + (0+3)^2} = \sqrt{(x-4)^2 + (0-2)^2} \]

Step 2: Squaring both sides: \[ (x-5)^2 + 9 = (x-4)^2 + 4 \]

Expanding: \[ x^2 - 10x + 25 + 9 = x^2 - 8x + 16 + 4 \]

Canceling \( x^2 \) and simplifying: \[ -10x + 34 = -8x + 20 \]
\[ -10x + 8x = 20 - 34 \]
\[ -2x = -14 \]
\[ x = 7 \]

Thus, the point is \( (7,0) \). Quick Tip: For problems involving equidistant points, use the distance formula and equate the distances before solving for the unknown variable.

Step 1: The formula for the length of an arc is: \[ L = \frac{\theta}{360^\circ} \times 2\pi r \]

Step 2: Given \( L = 22 \) cm and \( \theta = 60^\circ \), we substitute: \[ 22 = \frac{60}{360} \times 2\pi r \]
\[ 22 = \frac{1}{6} \times 2\pi r \]

Step 3: Solving for \( r \): \[ 22 = \frac{2\pi r}{6} \]
\[ 22 \times 3 = \pi r \]
\[ 66 = \pi r \]
\[ r = \frac{66}{\pi} \approx 21 cm \]

Thus, the radius of the circle is \( 21 \) cm. Quick Tip: For arc length problems, use the formula: \[ L = \frac{\theta}{360^\circ} \times 2\pi r \] where \( L \) is arc length, \( \theta \) is the angle, and \( r \) is the radius.

Step 1: Let the height of the pole be \( h \) and the length of its shadow be \( \sqrt{3}h \). The angle of elevation of the Sun, \( \theta \), satisfies: \[ \tan\theta = \frac{height of the pole}{length of the shadow} \]

Step 2: Substituting the given values: \[ \tan\theta = \frac{h}{\sqrt{3}h} = \frac{1}{\sqrt{3}} \]

Step 3: Since \( \tan 30^\circ = \frac{1}{\sqrt{3}} \), we conclude: \[ \theta = 30^\circ \] Quick Tip: For angle of elevation problems, use the tangent function: \[ \tan\theta = \frac{height of the object}{shadow length} \]

Step 1: The total number of possible outcomes when rolling two dice is: \[ 6 \times 6 = 36 \]

Step 2: The product of the two numbers must be between 8 and 13 (i.e., \( 9, 10, 11, 12 \)). Listing all possible cases:
- \( 9 = (3,3) \)
- \( 10 = (2,5), (5,2) \)
- \( 11 \) is not possible.
- \( 12 = (2,6), (6,2), (3,4), (4,3) \)

Thus, the favorable outcomes are: \[ (3,3), (2,5), (5,2), (2,6), (6,2), (3,4), (4,3) \]

Total favorable cases = 7.

Step 3: The probability is: \[ \frac{7}{36} \] Quick Tip: To find probabilities with dice, list all possible outcomes and count the number of favorable cases before dividing by the total outcomes.

Step 1: Understand the events.
- The event of Sania winning and the event of Ashnam winning are complementary events because they are mutually exclusive and exhaustive. This means that one of the two events must occur, and they cannot occur simultaneously.

Step 2: Verify the probabilities.
- Given:
- Probability of Sania winning, \( P(S) = 0.79 \)
- Probability of Ashnam winning, \( P(A) = 0.21 \)

Step 3: Check the sum of probabilities. \[ P(S) + P(A) = 0.79 + 0.21 = 1 \]
This confirms that the sum of the probabilities of the two complementary events is 1, which aligns with Reason (R).

Step 4: Conclusion.
- Since the sum of the probabilities of Sania winning and Ashnam winning is 1, and these are complementary events, Reason (R) correctly explains Assertion (A). Quick Tip: For complementary events, the sum of their probabilities is always 1. This is because one of the events must occur, and they cannot occur at the same time.

Step 1: Identify the prime numbers on a fair die.
- A fair die has six faces with numbers: 1, 2, 3, 4, 5, 6.
- Prime numbers among these are: 2, 3, 5.

Step 2: Calculate the probability of getting a prime number.
- Total number of outcomes = 6
- Number of favorable outcomes (prime numbers) = 3
- Probability \( P \) = \(\frac{Number of favorable outcomes}{Total number of outcomes} = \frac{3}{6} = \frac{1}{2}\)

Step 3: Verify Reason (R).
- A natural number is a prime number if it has exactly two distinct factors: 1 and itself.
- This definition correctly identifies the prime numbers on the die: 2, 3, 5.

Step 4: Conclusion.
- Since the probability of getting a prime number is indeed \(\frac{1}{2}\) and Reason (R) correctly defines what a prime number is, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation for Assertion (A). Quick Tip: A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.

Step 1: Assume \( (5 - 2\sqrt{2}) \) is a rational number.

Step 2: Let \[ 5 - 2\sqrt{2} = r \]
where \( r \) is a rational number.

Step 3: Rearranging the equation: \[ 2\sqrt{2} = 5 - r \]

Step 4: Dividing both sides by 2: \[ \sqrt{2} = \frac{5 - r}{2} \]

Since \( r \) is rational, \( \frac{5 - r}{2} \) is also rational.
This contradicts the fact that \( \sqrt{2} \) is irrational.

Step 5: Therefore, our assumption is incorrect, and \( (5 - 2\sqrt{2}) \) must be irrational. Quick Tip: The sum or difference of a rational and an irrational number is always irrational.

Step 1: Given that \( ABC \) is an isosceles triangle with \( AB = AC \), we know that: \[ \angle ABC = \angle ACB \]

Step 2: Since \( AD \) is perpendicular to \( BC \), it bisects the base in an isosceles triangle: \[ \angle ABD = \angle ACB \]

Step 3: Given that \( EF \) is perpendicular to \( AC \), the angle \( \angle ECF \) is formed at the extended portion of \( CB \). By exterior angle properties and perpendicularity: \[ \angle ECF = \angle ACB \]

Step 4: Subtracting these angles: \[ \angle ABD - \angle ECF = \angle ACB - \angle ACB = 0 \]

Thus, we have: \[ \angle ABD - \angle ECF = 0^\circ \] Quick Tip: In isosceles triangles, the perpendicular from the vertex to the base bisects the base and the vertex angle, which helps in proving angle relationships.

Step 1: Compute the distances between the given points using the distance formula: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

Step 2: Distance between \( A(-3,-3) \) and \( B(3,3) \): \[ AB = \sqrt{(3 + 3)^2 + (3 + 3)^2} = \sqrt{6^2 + 6^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2} \]

Step 3: Distance between \( B(3,3) \) and \( C(-3\sqrt{3}, 3\sqrt{3}) \): \[ BC = \sqrt{(-3\sqrt{3} - 3)^2 + (3\sqrt{3} - 3)^2} \] \[ = \sqrt{(-3\sqrt{3} - 3)^2 + (3\sqrt{3} - 3)^2} \] \[ = \sqrt{(3\sqrt{3} + 3\sqrt{3})^2 + (3\sqrt{3} - 3)^2} \] \[ = \sqrt{(6\sqrt{3})^2 + (3\sqrt{3} - 3)^2} \] \[ = \sqrt{108 + 108} = \sqrt{72} = 6\sqrt{2} \]

Step 4: Distance between \( C(-3\sqrt{3}, 3\sqrt{3}) \) and \( A(-3,-3) \): \[ CA = \sqrt{(-3 + 3\sqrt{3})^2 + (-3 - 3\sqrt{3})^2} \] \[ = \sqrt{(3\sqrt{3} + 3)^2 + (3\sqrt{3} + 3)^2} \] \[ = \sqrt{(6\sqrt{3})^2 + (3\sqrt{3} + 3)^2} \] \[ = \sqrt{108 + 108} = \sqrt{72} = 6\sqrt{2} \]

Step 5: Since \( AB = BC = CA = 6\sqrt{2} \), the given points form an equilateral triangle. Quick Tip: To prove a triangle is equilateral, show that all three sides have equal lengths using the distance formula: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

Step 1: Compute the distances between adjacent points:
\[ AB = \sqrt{(6-4)^2 + (4-3)^2} = \sqrt{4 + 1} = \sqrt{5} \]
\[ BC = \sqrt{(5-6)^2 + (6-4)^2} = \sqrt{1 + 4} = \sqrt{5} \]
\[ CD = \sqrt{(3-5)^2 + (5-6)^2} = \sqrt{4 + 1} = \sqrt{5} \]
\[ DA = \sqrt{(4-3)^2 + (3-5)^2} = \sqrt{1 + 4} = \sqrt{5} \]

Since \( AB = BC = CD = DA \), all sides are equal.

Step 2: Compute diagonals:
\[ AC = \sqrt{(5-4)^2 + (6-3)^2} = \sqrt{1 + 9} = \sqrt{10} \]
\[ BD = \sqrt{(3-6)^2 + (5-4)^2} = \sqrt{9 + 1} = \sqrt{10} \]

Since \( AC = BD \), diagonals are equal, confirming that \( ABCD \) is a square. Quick Tip: To prove that four points form a square: 1. Show all four sides are equal. 2. Show the diagonals are equal.

Step 1: Given that the incircle (or the excircle) of \( \triangle ABC \) touches the sides at specific points, we define the tangent segments from a common external point as equal.

Let the following lengths be:
- \( BP = PC = x \)
- \( CQ = QA = y \)
- \( AR = RB = z \)

Step 2: The perimeter of \( \triangle ABC \) is: \[ Perimeter = AB + BC + CA \]

Since the external tangents are equal: \[ AQ = CQ, \quad BR = AR, \quad CP = BP \]

Step 3: The sum of the tangents from a common point to the circle gives: \[ AQ + AR = \frac{1}{2} (Perimeter of \triangle ABC) \]

Since \( AQ = CQ \) and \( AR = BR \), we conclude: \[ AQ = \frac{1}{2} (Perimeter of \triangle ABC). \] Quick Tip: For an excircle or incircle, the sum of the external tangent segments from a vertex equals half the perimeter of the triangle.

Step 1: Let the point \( P(-1, k) \) divide the line segment joining \( A(-3, 10) \) and \( B(6, -8) \) in the ratio \( m:n \).

Step 2: Use the section formula to find the coordinates of point \( P \).
The coordinates of \( P \) are given by: \[ \left( \frac{m x_2 + n x_1}{m + n}, \frac{m y_2 + n y_1}{m + n} \right) \]
where \( (x_1, y_1) = (-3, 10) \) and \( (x_2, y_2) = (6, -8) \).

Step 3: Apply the section formula to the x-coordinate: \[ -1 = \frac{m \cdot 6 + n \cdot (-3)}{m + n} \] \[ -1 = \frac{6m - 3n}{m + n} \]
Multiply both sides by \( m + n \): \[ -1(m + n) = 6m - 3n \] \[ -m - n = 6m - 3n \]
Bring like terms together: \[ -m - 6m = -3n + n \] \[ -7m = -2n \] \[ 7m = 2n \] \[ \frac{m}{n} = \frac{2}{7} \]
So, the ratio \( m:n = 2:7 \).

Step 4: Now, use the y-coordinate to find \( k \): \[ k = \frac{m \cdot (-8) + n \cdot 10}{m + n} \]
Substitute \( m = 2 \) and \( n = 7 \): \[ k = \frac{2 \cdot (-8) + 7 \cdot 10}{2 + 7} \] \[ k = \frac{-16 + 70}{9} \] \[ k = \frac{54}{9} \] \[ k = 6 \]

Step 5: Conclusion.
- The point \( P(-1, k) \) divides the line segment joining \( A(-3, 10) \) and \( B(6, -8) \) in the ratio \( 2:7 \).
- The value of \( k \) is \( 6 \). Quick Tip: The section formula is used to find the coordinates of a point that divides a line segment internally in a given ratio.

Step 1: Let the sum of the ages of the two children be \( x \), and let the father's age be \( 2x \).

Step 2: After 20 years: \[ Father's age = 2x + 20 \] \[ Sum of children's ages = x + 40 \]

Step 3: Given that after 20 years, the father's age is equal to the sum of the children's ages: \[ 2x + 20 = x + 40 \]

Step 4: Solving for \( x \): \[ 2x - x = 40 - 20 \]
\[ x = 20 \]

Step 5: The present age of the father is: \[ 2x = 2(20) = 40 \]

Thus, the father's present age is \( 40 \) years. Quick Tip: For age-related problems, set up equations based on the given conditions and solve step-by-step.

Step 1: Let the smaller tap take \( x \) hours to fill the tank. Then the larger tap takes \( (x - 5) \) hours.

Step 2: The filling rate of each tap: \[ Smaller tap: \frac{1}{x} \quad (part of the tank filled per hour) \] \[ Larger tap: \frac{1}{x-5} \]

Step 3: Since both taps together can fill the tank in 3 hours: \[ \frac{1}{x} + \frac{1}{x-5} = \frac{1}{3} \]

Step 4: Solving for \( x \): \[ \frac{x-5 + x}{x(x-5)} = \frac{1}{3} \]
\[ \frac{2x - 5}{x(x-5)} = \frac{1}{3} \]
\[ 3(2x - 5) = x(x - 5) \]
\[ 6x - 15 = x^2 - 5x \]
\[ x^2 - 11x + 15 = 0 \]

Step 5: Solving the quadratic equation: \[ (x-5)(x-3) = 0 \]
\[ x = 5, \quad x = 3 \]

Step 6: The smaller tap takes \( 5 \) hours, and the larger tap takes \( 3 \) hours. Quick Tip: For work problems, express each component’s rate as a fraction of the whole and set up an equation accordingly.

Statement:
If a line is drawn parallel to one side of a triangle, it divides the other two sides in the same ratio.

Proof:

Step 1: Consider \( \triangle ABC \) where \( DE \parallel BC \) and \( D \) and \( E \) are points on \( AB \) and \( AC \) respectively.

Step 2: Since \( DE \parallel BC \), using the properties of similar triangles: \[ \triangle ADE \sim \triangle ABC \]

Step 3: From similarity: \[ \frac{AD}{DB} = \frac{AE}{EC} \]

Thus, the theorem is proved. Quick Tip: The Basic Proportionality Theorem (Thales' theorem) is widely used in similarity problems.

Step 1: Identify the first and last terms: \[ First term a = 56, \quad Last term l = 497 \]

Step 2: The numbers form an arithmetic sequence with common difference \( d = 7 \).
The number of terms \( n \) is found using: \[ l = a + (n-1) \cdot d \]
\[ 497 = 56 + (n-1) \cdot 7 \]
\[ 441 = (n-1) \cdot 7 \]
\[ n = 64 \]

Step 3: Sum of an arithmetic series: \[ S_n = \frac{n}{2} \times (a + l) \]
\[ S_{64} = \frac{64}{2} \times (56 + 497) \]
\[ = 32 \times 553 = 17696 \]

Thus, the required sum is \( 17696 \). Quick Tip: For sums of arithmetic sequences, use: \[ S_n = \frac{n}{2} (a + l) \]

Step 1: The sequence follows: \[ 11, 15, 19, \dots, 299 \]

Step 2: Identifying terms: \[ a = 11, \quad d = 4, \quad l = 299 \]

Using \( l = a + (n-1)d \):
\[ 299 = 11 + (n-1) \times 4 \]
\[ 288 = (n-1) \times 4 \]
\[ n = 73 \]

Step 3: Sum formula: \[ S_n = \frac{n}{2} (a + l) \]
\[ S_{73} = \frac{73}{2} \times (11 + 299) = \frac{73}{2} \times 310 = 11315 \]

Thus, the required sum is \( 11315 \). Quick Tip: For sums of arithmetic sequences, use: \[ S_n = \frac{n}{2} (a + l) \]

\begin{tikzpicture[scale=0.8]
% Draw axes
\draw[thick,->] (-1,0) -- (7,0) node[right] {\(x\);
\draw[thick,->] (0,-1) -- (0,7) node[above] {\(y\);

% Plot lines
\draw[blue, thick] (5,0) -- (0,5) node[left] {\(x+y=5\);
\draw[red, thick] (5,0) -- (6,1) node[right] {\(x-y=5\);

% Intersection point
\filldraw[black] (5,0) circle (2pt) node[below right] {(5,0);

% Shaded triangle
\fill[gray,opacity=0.3] (0,5) -- (5,0) -- (6,1) -- cycle;
\end{tikzpicture Quick Tip: To find the solution of two equations graphically, find the intersection of their lines.

The formula for the area of a sector is: \[ A = \frac{\theta}{360^\circ} \times \pi r^2 \]

Step 1: Area of the minor sector: \[ A_{minor} = \frac{60^\circ}{360^\circ} \times 3.14 \times (6)^2 \] \[ = \frac{1}{6} \times 3.14 \times 36 = 18.84 cm^2 \]

Step 2: Area of the major sector: \[ A_{major} = \pi r^2 - A_{minor} \] \[ = 3.14 \times 36 - 18.84 = 94.32 - 18.84 = 75.48 cm^2 \] Quick Tip: The sum of the areas of the minor and major sectors is always equal to the total area of the circle.

Step 1: The area of the sector is: \[ A_{sector} = \frac{60^\circ}{360^\circ} \times \pi \times 10^2 \]
\[ = \frac{1}{6} \times 3.14 \times 100 = 52.33 cm^2 \]

Step 2: The area of the triangle formed by the chord and the radii: \[ A_{\triangle} = \frac{1}{2} \times r^2 \times \sin\theta \]
\[ = \frac{1}{2} \times 10^2 \times \sin 60^\circ \]
\[ = \frac{1}{2} \times 100 \times 0.866 = 43.3 cm^2 \]

Step 3: The area of the minor segment: \[ A_{segment} = A_{sector} - A_{\triangle} \]
\[ = 52.33 - 43.3 = 9.03 cm^2 \] Quick Tip: To find the area of a segment, subtract the area of the triangle from the sector area.