CBSE Class 10 Mathematics Standard Question Paper 2024 (Set 1- 30/5/1) with Answer Key

The given sequence \( \sqrt{18}, \ \sqrt{50}, \ \sqrt{98}, \ \ldots \) is in arithmetic progression (A.P.) since the difference between consecutive terms is constant.

Step 1: Calculate the common difference (\( d \)):
\[ d = \sqrt{50} - \sqrt{18}. \]
Simplify each term: \[ \sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2}, \quad \sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt{2}. \]
Thus: \[ d = 5\sqrt{2} - 3\sqrt{2} = 2\sqrt{2}. \]

Step 2: Find the 4^th term:

The general term of an A.P. is given by: \[ a_n = a + (n - 1)d, \]
where \( a = \sqrt{18} = 3\sqrt{2} \), \( n = 4 \), and \( d = 2\sqrt{2} \). Substituting: \[ a_4 = 3\sqrt{2} + (4 - 1)(2\sqrt{2}) = 3\sqrt{2} + 6\sqrt{2} = 9\sqrt{2}. \]
Simplify: \[ a_4 = \sqrt{81 \cdot 2} = \sqrt{162}. \]

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Conclusion:

The 4^th term is \( \sqrt{162} \). Quick Tip: To find the next term in an A.P., calculate the common difference and apply the formula for the \( n \)-th term.

We are given: \[ \frac{x}{3} = 2\sin A \quad and \quad \frac{y}{3} = 2\cos A. \]

Multiplying through by 3: \[ x = 6\sin A \quad and \quad y = 6\cos A. \]

The sum of squares is: \[ x^2 + y^2 = (6\sin A)^2 + (6\cos A)^2 = 36(\sin^2 A + \cos^2 A). \]

Using \(\sin^2 A + \cos^2 A = 1\): \[ x^2 + y^2 = 36. \]

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Conclusion:

The value of \(x^2 + y^2\) is 36. Quick Tip: Remember the Pythagorean identity: \(\sin^2\theta + \cos^2\theta = 1\), which simplifies problems involving sums of squares.

Given: \[ 4\sec\theta - 5 = 0 \quad \implies \quad \sec\theta = \frac{5}{4}. \]

We know: \[ \sec^2\theta = 1 + \tan^2\theta. \]

Substitute \(\sec\theta = \frac{5}{4}\): \[ \left(\frac{5}{4}\right)^2 = 1 + \tan^2\theta. \]

Simplify: \[ \frac{25}{16} = 1 + \tan^2\theta. \]
\[ \tan^2\theta = \frac{25}{16} - 1 = \frac{9}{16}. \]
\[ \tan\theta = \frac{3}{4}. \]

The reciprocal of \(\tan\theta\) gives \(\cot\theta\): \[ \cot\theta = \frac{1}{\tan\theta} = \frac{4}{3}. \]

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Conclusion:

The value of \(\cot\theta\) is \(\frac{4}{3}\). Quick Tip: For problems involving trigonometric functions, use known identities like \(\sec^2\theta = 1 + \tan^2\theta\) to relate and calculate values.

The given equations are: \[ 1. \ 3x + 4y = 5 \quad and \quad 2. \ 6x + 8y = 7 \]

Rewrite Equation 2 to check if it is a multiple of Equation 1: \[ 6x + 8y = 7 \quad \implies \quad 2(3x + 4y) = 7 \]

Since the constant terms (\(5\) and \(7\)) are not in the same ratio as the coefficients, the lines are not coincident. The coefficients of \(x\) and \(y\) are in the same ratio: \[ \frac{3}{6} = \frac{4}{8}. \]

This implies the lines are parallel.

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Conclusion:

The lines represented by the equations are parallel. Quick Tip: For a system of linear equations, lines are parallel if the ratios of the coefficients of \(x\) and \(y\) are equal but differ in constant terms.

For a quadratic equation of the form: \[ ax^2 + bx + c = 0, \]
the sum of the roots is: \[ -\frac{b}{a}. \]

The product of the roots is: \[ \frac{c}{a}. \]

Substitute the values \(a = 5\), \(b = -6\), and \(c = 21\): \[ Sum of the roots: -\frac{-6}{5} = \frac{6}{5}. \] \[ Product of the roots: \frac{21}{5}. \]

Find the ratio of the sum to the product: \[ Ratio = \frac{\frac{6}{5}}{\frac{21}{5}} = \frac{6}{21} = \frac{2}{7}. \]

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Conclusion:

The ratio of the sum and product of the roots is \(2 : 7\). Quick Tip: For quadratic equations, use the relationships between coefficients and roots: \(Sum of roots = -\frac{b}{a}\), \(Product of roots = \frac{c}{a}\).

The mean of a data set is given by: \[ Mean = \frac{Sum of observations}{Number of observations}. \]

Here, the sum of observations is: \[ 2 + 9 + (x+6) + (2x+3) + 5 + 10 + 5 = 40 + 3x. \]

The number of observations is 7. Substituting into the formula for mean: \[ 7 = \frac{40 + 3x}{7}. \]

Multiply through by 7: \[ 40 + 3x = 49. \]

Simplify: \[ 3x = 9 \quad \implies \quad x = 3. \]

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Conclusion:

The value of \(x\) is 3. Quick Tip: For mean calculations, multiply the given mean by the number of terms to equate it to the sum of the terms.

The multiples of 7 between 1 and 40 are: \[ 7, 14, 21, 28, 35. \]

There are 5 multiples of 7, and the total number of tickets is 40. The probability is: \[ P = \frac{Favorable outcomes}{Total outcomes} = \frac{5}{40} = \frac{1}{8}. \]

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Conclusion:

The probability is \(\frac{1}{8}\). Quick Tip: For probability problems, identify the total possible outcomes and favorable outcomes clearly before simplifying the ratio.

The perimeter of the sector is given by: \[ Perimeter = 2r + Arc length. \]

The arc length is: \[ Arc length = \frac{\theta}{360^\circ} \cdot 2\pi r. \]

Substitute \(\theta = 60^\circ\), \(r = 21\), and \(\pi = \frac{22}{7}\): \[ Arc length = \frac{60}{360} \cdot 2 \cdot \frac{22}{7} \cdot 21 = \frac{1}{6} \cdot 2 \cdot \frac{22}{7} \cdot 21 = 22 \, cm. \]

The perimeter is: \[ Perimeter = 2(21) + 22 = 42 + 22 = 64 \, cm. \]

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Conclusion:

The perimeter is 64 cm. Quick Tip: For sector problems, always add the arc length and twice the radius to find the total perimeter.

The length of an arc is given by: \[ Arc length = \frac{\theta}{360^\circ} \cdot 2\pi r. \]

Substitute \(Arc length = 10\pi\), \(r = 12\), and solve for \(\theta\): \[ 10\pi = \frac{\theta}{360} \cdot 2\pi \cdot 12. \]

Simplify: \[ 10 = \frac{\theta}{360} \cdot 24 \quad \implies \quad \frac{\theta}{360} = \frac{10}{24}. \]
\[ \theta = \frac{10}{24} \cdot 360 = 150^\circ. \]

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Conclusion:

The angle subtended by the arc is \(150^\circ\). Quick Tip: For arc length problems, substitute known values into the formula and solve for the unknown step-by-step.

If a number \(N\) divides 281 leaving a remainder of 5 and 1249 leaving a remainder of 7, then: \[ 281 - 5 = 276 \quad and \quad 1249 - 7 = 1242 \]
must be divisible by \(N\).

Thus, \(N\) is the greatest common divisor (GCD) of 276 and 1242.

Step 1: Find the GCD of 276 and 1242.
Using the Euclidean algorithm: \[ 1242 = 276 \cdot 4 + 138 \] \[ 276 = 138 \cdot 2 + 0 \]

The GCD is \(138\).

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Conclusion:

The greatest number which divides 281 and 1249, leaving the specified remainders, is \(138\). Quick Tip: For such problems, subtract the remainders from the given numbers and find the GCD of the resulting values.

The general formula for the \(n\)-th term of an A.P. is: \[ a_n = a + (n-1)d, \]
where \(a\) is the first term, \(d\) is the common difference, and \(n\) is the number of terms.

Here: \[ a = 3, \ d = 3, \ a_n = 111. \]

Substitute into the formula: \[ 111 = 3 + (n-1) \cdot 3. \]

Simplify: \[ 111 - 3 = 3(n-1). \] \[ 108 = 3(n-1). \] \[ n-1 = 36 \quad \implies \quad n = 37. \]

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Conclusion:

The number of terms in the A.P. is \(37\). Quick Tip: For A.P. problems, rearrange the formula for the \(n\)-th term to find the total number of terms.

Given:
- Radius \(r = 10 \, cm\).
- The chord subtends a \(90^\circ\) angle at the center.

In \(\triangle OAB\), \(O\) is the center, and \(\angle AOB = 90^\circ\). Using the Pythagoras theorem: \[ Chord length (AB) = \sqrt{OA^2 + OB^2}. \]

Substitute \(OA = OB = r = 10\): \[ Chord length (AB) = \sqrt{10^2 + 10^2} = \sqrt{200} = 10\sqrt{2}. \]

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Conclusion:

The length of the chord is \(10\sqrt{2}\). Quick Tip: For chords subtending \(90^\circ\) at the center, use the Pythagoras theorem to calculate the chord length.

To find the LCM of \(28, 44, 132\):
1. Perform prime factorization: \[ 28 = 2^2 \cdot 7, \quad 44 = 2^2 \cdot 11, \quad 132 = 2^2 \cdot 3 \cdot 11. \]

2. Take the highest powers of all prime factors: \[ LCM = 2^2 \cdot 3 \cdot 7 \cdot 11 = 924. \]

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Conclusion:

The LCM of \(28, 44, 132\) is \(924\). Quick Tip: For LCM, take the highest powers of all prime factors common or unique to the given numbers.

Co-prime numbers are numbers that have no common factors other than 1.

By definition, the HCF of any two co-prime numbers is always \(1\).

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Conclusion:

The HCF of the two co-prime numbers is \(1\). Quick Tip: Co-prime numbers always have an HCF of 1, as they share no common factors other than 1.

The polynomial is: \[ p(x) = kx^2 - 30x + 45k. \]

The sum and product of the roots (\(\alpha + \beta\) and \(\alpha \beta\)) are given by: \[ \alpha + \beta = -\frac{Coefficient of x}{Coefficient of x^2} = -\frac{-30}{k} = \frac{30}{k}. \] \[ \alpha \beta = \frac{Constant term}{Coefficient of x^2} = \frac{45k}{k} = 45. \]

Given: \[ \alpha + \beta = \alpha \beta. \]

Substitute the values: \[ \frac{30}{k} = 45. \]

Solve for \(k\): \[ 30 = 45k \quad \implies \quad k = \frac{30}{45} = \frac{2}{3}. \]

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Conclusion:

The value of \(k\) is \(\frac{2}{3}\). Quick Tip: For problems involving zeroes of polynomials, use the relationships \(\alpha + \beta = -\frac{Coefficient of x}{Coefficient of x^2}\) and \(\alpha \beta = \frac{Constant term}{Coefficient of x^2}\).

Given: \[ \angle RJL = 42^\circ, \quad RJ and RL are tangents. \]

In a circle, the angle formed by the tangents at the external point (\(\angle RJL\)) is half the angle subtended by the chord at the center (\(\angle JOL\)): \[ \angle JOL = 2 \cdot \angle RJL. \]

Substitute the given value: \[ \angle JOL = 2 \cdot 42^\circ = 84^\circ. \]

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Conclusion:

The measure of \(\angle JOL\) is \(84^\circ\). Quick Tip: For tangents and angles in circles, the angle subtended at the center is twice the angle between the tangents.

In the given figure, \(DE \parallel BC\), so by the Basic Proportionality Theorem (Thales’ Theorem): \[ \frac{AD}{DB} = \frac{AE}{EC}. \]

Step 1: Use the property of proportions.

Given: \[ AD = 2.4 \, cm, \quad DB = 4 \, cm, \quad AE = 2 \, cm. \]

Substitute the values: \[ \frac{AD}{DB} = \frac{AE}{EC}. \]
\[ \frac{2.4}{4} = \frac{2}{EC}. \]

Simplify: \[ EC = \frac{4 \cdot 2}{2.4} = \frac{8}{2.4} = 3.33 \, cm. \]

Step 2: Find the total length of \(AC\).

The total length of \(AC\) is: \[ AC = AE + EC. \]

Substitute the values: \[ AC = 2 + 3.33 = 5.33 \, cm. \]

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Conclusion:

The length of \(AC\) is \(16 \frac{2}{3} \, cm\). Quick Tip: For parallel lines in triangles, use the proportionality rule: \(\frac{AD}{DB} = \frac{AE}{EC}\).

The problem involves similar triangles formed by the pole and its shadow, and the tower and its shadow. The ratios of their heights to their respective shadow lengths are equal.

Let the height of the tower be \(H\). Using the property of similar triangles: \[ \frac{Height of the pole}{Length of the pole's shadow} = \frac{Height of the tower}{Length of the tower's shadow}. \]

Substitute the given values: \[ \frac{7.5}{5} = \frac{H}{24}. \]

Simplify: \[ H = \frac{7.5}{5} \cdot 24 = 36 \, m. \]

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Conclusion:

The height of the tower is \(36 \, m\). Quick Tip: For shadow problems, use the proportionality property of similar triangles: \(\frac{Height}{Shadow length}\) is constant for objects under the same light source.

Both Assertion (A) and Reason (R) are true. The line \(EF\), being parallel to the parallel sides of the trapezium (\(AB\) and \(DC\)), divides the non-parallel sides proportionally. The Reason (R) provides the correct explanation for Assertion (A).

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% Correct Answer
Correct Answer: (A) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A). Quick Tip: Use the property of proportionality in trapeziums: A line parallel to the parallel sides divides the non-parallel sides proportionally.

Both Assertion (A) and Reason (R) are true. However, Reason (R) is not the correct explanation for Assertion (A). The degree of a zero polynomial is undefined because it does not have any terms, while the degree of a non-zero constant polynomial is defined as \(0\).

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% Correct Answer
Correct Answer: (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A). Quick Tip: Remember: The degree of a zero polynomial is undefined, while the degree of any non-zero constant polynomial is \(0\).

Given:
- Radius of the circle: \(r = 3 \, cm\),
- Angle between the tangents: \(\angle APB = 60^\circ\).


\begin{tikzpicture[scale=1]
% Draw the circle
\draw[thick] (0,0) circle (2); % Circle radius is 2 units (scaled for clarity)

% Draw the center
\filldraw[black] (0,0) circle (1pt) node[below right] {\(O\);

% Draw the tangents
\draw[thick] (-3.5,2) -- (0,0) -- (3.5,2);
\draw[thick] (-3.5,2) -- (3.5,2);

% Draw the radius
\draw[thick] (0,0) -- (-3.5,2) node[midway,left] {\(r=3\,cm\);
\draw[thick] (0,0) -- (3.5,2);

% Mark angle
\draw[thick] (0.5,0) arc[start angle=0,end angle=30,radius=0.5];
\node at (0.7,0.1) {\(30^\circ\);

% Label points
\node at (-3.5,2.3) {\(A\);
\node at (3.5,2.3) {\(B\);
\node at (0,-2.5) {\(P\);

% Draw tangent lengths
\node at (-2,2.5) {Tangent \(AP\);
\node at (2,2.5) {Tangent \(BP\);
\end{tikzpicture


Step 1: Analyze the geometry.

The tangents form two right triangles with the center of the circle. In \(\triangle APO\), where \(O\) is the center of the circle: \[ \angle APO = \frac{\angle APB}{2} = \frac{60^\circ}{2} = 30^\circ. \]

Step 2: Use trigonometric ratios.

From \(\triangle APO\), using \(\tan 30^\circ\): \[ \tan 30^\circ = \frac{Opposite side (radius)}{Adjacent side (tangent length)}. \]

Substitute the values: \[ \tan 30^\circ = \frac{1}{\sqrt{3}}, \quad \frac{1}{\sqrt{3}} = \frac{3}{AP}. \]

Solve for \(AP\): \[ AP = 3\sqrt{3} \, cm. \]

Step 3: Finalize the result.

Since the tangents are symmetrical, the length of each tangent is: \[ AP = 3\sqrt{3} \, cm. \]

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Conclusion:

The length of each tangent is \(3\sqrt{3} \, cm\). Quick Tip: When two tangents are drawn to a circle, use trigonometric ratios in the formed right triangles to find tangent lengths.

Let \(AB\) be the diameter of a circle with center \(O\), and let \(P\) and \(Q\) be the tangents at \(A\) and \(B\), respectively.

Step 1: Analyze the geometry.

The tangent to a circle is perpendicular to the radius at the point of tangency. Therefore: \[ \angle OAY = 90^\circ \quad and \quad \angle OBP = 90^\circ. \]

Step 2: Prove parallelism.

The angles \(\angle OAY\) and \(\angle OBP\) are equal and form alternate interior angles between the lines \(PQ\) and \(XY\). Hence, by the property of alternate interior angles: \[ PQ \parallel XY. \]

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Conclusion:

The tangents drawn at the ends of a diameter of a circle are parallel.


Diagram:


\begin{tikzpicture[scale=1.2]
% Draw the circle
\draw[thick] (0,0) circle (2); % Circle radius is 2 units

% Draw the diameter
\draw[thick] (-2,0) -- (2,0) node[midway, below] {\(A\) node[midway, above] {\(B\);

% Draw the center
\filldraw[black] (0,0) circle (1pt) node[below left] {\(O\);

% Draw tangents
\draw[thick,->] (-2,0) -- (-2,-1) node[below] {\(X\);
\draw[thick,->] (-2,0) -- (-2,1) node[above] {\(P\);
\draw[thick,->] (2,0) -- (2,-1) node[below] {\(Y\);
\draw[thick,->] (2,0) -- (2,1) node[above] {\(Q\);

% Mark right angles
\draw[thick] (-2.2,0) -- (-1.8,0) -- (-1.8,0.2);
\draw[thick] (2.2,0) -- (1.8,0) -- (1.8,0.2);
\end{tikzpicture Quick Tip: The tangents at the ends of the diameter are always parallel because they form equal alternate interior angles with the line joining the ends of the diameter.

Substitute the trigonometric values: \[ \tan 30^\circ = \frac{1}{\sqrt{3}}, \quad \sec 60^\circ = 2, \quad \tan 45^\circ = 1, \quad \sin^2 60^\circ = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}. \]

Simplify the numerator: \[ 2 \cdot \frac{1}{\sqrt{3}} \cdot 2 \cdot 1 = \frac{4}{\sqrt{3}}. \]

Simplify the denominator: \[ 1 - \sin^2 60^\circ = 1 - \frac{3}{4} = \frac{1}{4}. \]

The entire expression becomes: \[ \frac{\frac{4}{\sqrt{3}}}{\frac{1}{4}} = \frac{4}{\sqrt{3}} \cdot 4 = \frac{16}{\sqrt{3}}. \]

Rationalize the denominator: \[ \frac{16}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{16\sqrt{3}}{3}. \]

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Conclusion:

The value of the expression is: \[ \frac{16\sqrt{3}}{3}. \] Quick Tip: Always simplify trigonometric functions step-by-step and rationalize the denominator when necessary.

For a polynomial \(ax^2 + bx + c\), the sum and product of the roots are: \[ \alpha + \beta = -\frac{Coefficient of x}{Coefficient of x^2} = -\frac{-6}{5} = \frac{6}{5}. \]
\[ \alpha \beta = \frac{Constant term}{Coefficient of x^2} = \frac{1}{5}. \]

Add \(\alpha + \beta\) and \(\alpha \beta\): \[ \alpha + \beta + \alpha \beta = \frac{6}{5} + \frac{1}{5} = \frac{7}{5}. \]

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Conclusion:

The value of \(\alpha + \beta + \alpha \beta\) is: \[ \frac{7}{5}. \] Quick Tip: For zeroes of a polynomial, use the relationships: \[ \alpha + \beta = -\frac{b}{a}, \quad \alpha \beta = \frac{c}{a}. \]

Let the ratio be \(k:1\). The coordinates of the point dividing a line segment are given by the section formula: \[ Coordinates of P = \left(\frac{kx_2 + x_1}{k+1}, \frac{ky_2 + y_1}{k+1}\right). \]

Here: \[ P(-4, 6), \quad A(-6, 10), \quad B(3, -8). \]

Equating the \(x\)-coordinate of \(P\): \[ -4 = \frac{3k - 6}{k+1}. \]

Simplify: \[ -4(k+1) = 3k - 6. \] \[ -4k - 4 = 3k - 6. \] \[ -4k - 3k = -6 + 4. \] \[ -7k = -2 \quad \implies \quad k = \frac{2}{7}. \]

Thus, the required ratio is: \[ k:1 = 2:7. \]

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Conclusion:

The point \(P(-4, 6)\) divides the line segment joining \(A(-6, 10)\) and \(B(3, -8)\) in the ratio \(2:7\).


Diagram:


\begin{tikzpicture[scale=1.5]
% Draw the line segment
\draw[thick] (-3,2.5) -- (1,-1.5);

% Draw points
\filldraw[black] (-3,2.5) circle (1pt) node[above left] {\(A(-6, 10)\);
\filldraw[black] (1,-1.5) circle (1pt) node[below right] {\(B(3, -8)\);
\filldraw[black] (-1,1) circle (1pt) node[below right] {\(P(-4, 6)\);

% Label the ratio
\node at (-2,1.5) {\(k\);
\node at (0,-0.5) {\(1\);
\end{tikzpicture Quick Tip: Use the section formula: \[ (x, y) = \left(\frac{kx_2 + x_1}{k+1}, \frac{ky_2 + y_1}{k+1}\right), \] and equate the coordinates to find the ratio.

To prove that the given points form an isosceles triangle, we calculate the lengths of the sides using the distance formula: \[ Distance between two points (x_1, y_1) and (x_2, y_2) = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}. \]

Step 1: Calculate \(AB\).
\[ AB = \sqrt{(6 - 3)^2 + (4 - 0)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5. \]

Step 2: Calculate \(BC\).
\[ BC = \sqrt{(-1 - 6)^2 + (3 - 4)^2} = \sqrt{(-7)^2 + (-1)^2} = \sqrt{49 + 1} = \sqrt{50}. \]

Step 3: Calculate \(CA\).
\[ CA = \sqrt{(3 - (-1))^2 + (0 - 3)^2} = \sqrt{(3 + 1)^2 + (-3)^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5. \]

Step 4: Compare the side lengths.

Here: \[ AB = CA = 5, \quad BC = \sqrt{50}. \]

Since two sides (\(AB\) and \(CA\)) are equal, the triangle formed by the points \(A(3, 0)\), \(B(6, 4)\), and \(C(-1, 3)\) is an isosceles triangle.

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Conclusion:
\(\triangle ABC\) is an isosceles triangle because \(AB = CA\). Quick Tip: To prove a triangle is isosceles, calculate all three side lengths using the distance formula and check if two of them are equal.

For the shirt to be acceptable to Anmol, it must either be a good shirt or a shirt with minor defects. Therefore, we exclude only the shirts with major defects.

Step 1: Calculate the number of shirts without major defects.

Total shirts without major defects: \[ 48 \, (good shirts) + 4 \, (minor defects) = 52. \]

Step 2: Calculate the probability.

Total number of shirts in the carton: \[ 60. \]

Probability that the shirt is acceptable to Anmol: \[ P(Acceptable to Anmol) = \frac{Number of shirts without major defects}{Total number of shirts} = \frac{52}{60}. \]

Simplify the fraction: \[ P(Acceptable to Anmol) = \frac{13}{15}. \]

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Conclusion:

The probability that the shirt is acceptable to Anmol is: \[ \frac{13}{15}. \] Quick Tip: When calculating probabilities, carefully exclude only the items explicitly stated as unacceptable.

Let us assume, for the sake of contradiction, that \(\sqrt{3}\) is a rational number. Then it can be expressed as: \[ \sqrt{3} = \frac{p}{q}, \]
where \(p\) and \(q\) are integers, \(q \neq 0\), and \(p\) and \(q\) are coprime (have no common factors other than 1).

Step 1: Square both sides.
\[ 3 = \frac{p^2}{q^2} \quad \implies \quad p^2 = 3q^2. \]

Step 2: Analyze divisibility of \(p\).

Since \(p^2\) is divisible by 3, it follows that \(p\) must also be divisible by 3 (property of prime numbers). Let: \[ p = 3a, \quad where a is an integer. \]

Step 3: Substitute \(p = 3a\) into the equation.
\[ p^2 = 3q^2 \quad \implies \quad (3a)^2 = 3q^2 \quad \implies \quad 9a^2 = 3q^2 \quad \implies \quad q^2 = 3a^2. \]

Step 4: Analyze divisibility of \(q\).

Since \(q^2\) is divisible by 3, it follows that \(q\) must also be divisible by 3.

Step 5: Contradiction.

If both \(p\) and \(q\) are divisible by 3, then \(p\) and \(q\) are not coprime, which contradicts our assumption that \(\frac{p}{q}\) is in its simplest form.

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Conclusion:

The assumption that \(\sqrt{3}\) is a rational number leads to a contradiction.

Therefore, \(\sqrt{3}\) is an irrational number. Quick Tip: To prove a number is irrational, assume it is rational and derive a contradiction using properties of divisibility.

Expand \((\sqrt{2} + \sqrt{3})^2\): \[ (\sqrt{2} + \sqrt{3})^2 = 2 + 3 + 2\sqrt{6} = 5 + 2\sqrt{6}. \]

Step 1: Assume, for contradiction, that \(5 + 2\sqrt{6}\) is a rational number.

Let: \[ 5 + 2\sqrt{6} = \frac{a}{b}, \]
where \(a, b\) are integers, and \(b \neq 0\).

Step 2: Rearrange to isolate \(\sqrt{6}\).
\[ 2\sqrt{6} = \frac{a}{b} - 5 \quad \implies \quad \sqrt{6} = \frac{a - 5b}{2b}. \]

Step 3: Analyze rationality.

Since \(a\) and \(b\) are integers, \(\frac{a - 5b}{2b}\) is a rational number. However, it is given that \(\sqrt{6}\) is an irrational number. This leads to a contradiction.

Step 4: Conclude irrationality.

The assumption that \(5 + 2\sqrt{6}\) is rational is incorrect. Therefore: \[ 5 + 2\sqrt{6} = (\sqrt{2} + \sqrt{3})^2 \]
is an irrational number.

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Conclusion:
\((\sqrt{2} + \sqrt{3})^2\) is an irrational number. Quick Tip: When proving irrationality, assume rationality, isolate the square root term, and demonstrate that it contradicts the given property of irrationality.

The sum of the first \(n\) terms of an A.P. is given by: \[ S_n = \frac{n}{2} \left[2a + (n-1)d\right]. \]

Here: \[ S_{14} = 1050, \quad a = 10, \quad n = 14. \]

Substitute the values: \[ \frac{14}{2} \left[2(10) + 13d\right] = 1050. \]

Simplify: \[ 7 \left[20 + 13d\right] = 1050 \quad \implies \quad 20 + 13d = 150 \quad \implies \quad d = 10. \]

Find the 20th term (\(a_{20}\)):

The \(n\)-th term of an A.P. is given by: \[ a_n = a + (n-1)d. \]

For \(n = 20\): \[ a_{20} = 10 + (20-1)10 = 10 + 190 = 200. \]

Find the general \(n\)-th term (\(a_n\)):

Substitute \(a = 10\) and \(d = 10\): \[ a_n = 10 + (n-1)10 = 10n. \]

---

Conclusion:

The 20th term is \(200\) and the \(n\)-th term is \(10n\). Quick Tip: Use the formula \(S_n = \frac{n}{2}[2a + (n-1)d]\) to calculate the sum of \(n\) terms and \(a_n = a + (n-1)d\) for specific terms.

The sum of the first \(n\) terms is given by: \[ S_n = \frac{n}{2}(a + l), \]
where \(a = 5\), \(l = 45\), and \(S_n = 400\).

Substitute the values: \[ \frac{n}{2}(5 + 45) = 400. \]

Simplify: \[ \frac{n}{2}(50) = 400 \quad \implies \quad 25n = 400 \quad \implies \quad n = 16. \]

Find the common difference (\(d\)):

The last term of an A.P. is given by: \[ a_n = a + (n-1)d. \]

Substitute \(a_n = 45\), \(a = 5\), and \(n = 16\): \[ 45 = 5 + (16-1)d \quad \implies \quad 45 = 5 + 15d \quad \implies \quad 15d = 40 \quad \implies \quad d = \frac{40}{15} = \frac{8}{3}. \]

---

Conclusion:

The number of terms is \(16\) and the common difference is \(\frac{8}{3}\). Quick Tip: For problems involving the sum of an A.P., use \(S_n = \frac{n}{2}(a + l)\) when the first and last terms are given.

Let the parallelogram \(ABCD\) circumscribe a circle, touching the sides \(AB\), \(BC\), \(CD\), and \(DA\) at points \(P\), \(Q\), \(R\), and \(S\), respectively.

Step 1: Use the property of tangents.

The lengths of tangents drawn from an external point to a circle are equal. Therefore: \[ AP = AS, \quad BP = BQ, \quad CR = CQ, \quad DR = DS. \]

Step 2: Add the tangent pairs.

Adding all the equal tangents: \[ (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ). \]

Simplify: \[ AB + CD = AD + BC. \]

Step 3: Use the properties of a parallelogram.

In a parallelogram, opposite sides are equal: \[ AB = CD \quad and \quad AD = BC. \]

Substitute these values: \[ AB + AB = AB + AB \quad \implies \quad 2AB = 2BC. \]

Divide by 2: \[ AB = BC. \]

Step 4: Conclude that the parallelogram is a rhombus.

Since all sides of the parallelogram are equal (\(AB = BC = CD = DA\)), \(ABCD\) is a rhombus.

---

Conclusion:

The parallelogram circumscribing a circle is a rhombus.


Diagram:


\begin{tikzpicture[scale=1.5]
% Draw the parallelogram
\draw[thick] (0,0) -- (4,0) -- (5,2) -- (1,2) -- cycle;

% Draw the circle
\draw[thick] (2.5,1) circle (1);

% Tangent points
\node at (0,0) [below left] {\(A\);
\node at (4,0) [below right] {\(B\);
\node at (5,2) [above right] {\(C\);
\node at (1,2) [above left] {\(D\);
\node at (2.5,1) {\(O\);
\node at (0.5,1.5) [left] {\(S\);
\node at (4.5,1.5) [right] {\(R\);
\node at (3.5,0) [below] {\(Q\);
\node at (0.5,0) [below] {\(P\);

% Tangent lines
\draw[thick] (0,0) -- (0.5,0);
\draw[thick] (0,0) -- (0.5,1.5);
\draw[thick] (4,0) -- (3.5,0);
\draw[thick] (4,0) -- (4.5,1.5);
\draw[thick] (5,2) -- (4.5,1.5);
\draw[thick] (5,2) -- (3.5,0);
\draw[thick] (1,2) -- (0.5,1.5);
\draw[thick] (1,2) -- (0.5,0);
\end{tikzpicture Quick Tip: The key to solving this problem is using the property that the sum of tangents from opposite sides of a circumscribed circle is equal.

Step 1: Simplify the Left-Hand Side (LHS).

We start with: \[ LHS = \frac{\tan A}{1 - \cot A} + \frac{\cot A}{1 - \tan A}. \]

Substitute \(\tan A = \frac{\sin A}{\cos A}\) and \(\cot A = \frac{\cos A}{\sin A}\): \[ LHS = \frac{\frac{\sin A}{\cos A}}{1 - \frac{\cos A}{\sin A}} + \frac{\frac{\cos A}{\sin A}}{1 - \frac{\sin A}{\cos A}}. \]

Simplify the denominators: \[ 1 - \frac{\cos A}{\sin A} = \frac{\sin A - \cos A}{\sin A}, \quad 1 - \frac{\sin A}{\cos A} = \frac{\cos A - \sin A}{\cos A}. \]

Rewrite the fractions: \[ LHS = \frac{\frac{\sin A}{\cos A}}{\frac{\sin A - \cos A}{\sin A}} + \frac{\frac{\cos A}{\sin A}}{\frac{\cos A - \sin A}{\cos A}}. \]

Simplify each term: \[ \frac{\frac{\sin A}{\cos A}}{\frac{\sin A - \cos A}{\sin A}} = \frac{\sin^2 A}{\cos A (\sin A - \cos A)}, \] \[ \frac{\frac{\cos A}{\sin A}}{\frac{\cos A - \sin A}{\cos A}} = \frac{\cos^2 A}{\sin A (\cos A - \sin A)}. \]

Since \((\cos A - \sin A) = -(\sin A - \cos A)\), we combine the terms: \[ LHS = \frac{\sin^2 A}{\cos A (\sin A - \cos A)} - \frac{\cos^2 A}{\sin A (\sin A - \cos A)}. \]

Step 2: Combine into a single fraction.
\[ LHS = \frac{\sin^3 A - \cos^3 A}{\sin A \cos A (\sin A - \cos A)}. \]

Factorize \(\sin^3 A - \cos^3 A\) using the difference of cubes: \[ \sin^3 A - \cos^3 A = (\sin A - \cos A)(\sin^2 A + \sin A \cos A + \cos^2 A). \]

Simplify: \[ \sin^2 A + \cos^2 A = 1, \quad so: \sin^2 A + \sin A \cos A + \cos^2 A = 1 + \sin A \cos A. \]

Substitute back: \[ LHS = \frac{(\sin A - \cos A)(1 + \sin A \cos A)}{\sin A \cos A (\sin A - \cos A)}. \]

Cancel \((\sin A - \cos A)\) in numerator and denominator: \[ LHS = \frac{1 + \sin A \cos A}{\sin A \cos A}. \]

Split the fraction: \[ LHS = \frac{1}{\sin A \cos A} + 1. \]

Use identities: \[ \frac{1}{\sin A \cos A} = \sec A \csc A. \]

Thus: \[ LHS = 1 + \sec A \csc A = RHS. \]

---

Conclusion:
\[ \frac{\tan A}{1 - \cot A} + \frac{\cot A}{1 - \tan A} = 1 + \sec A \csc A. \] Quick Tip: When dealing with trigonometric proofs, substitute \(\tan A\) and \(\cot A\) in terms of \(\sin A\) and \(\cos A\) and simplify step-by-step.

Step 1: Total number of outcomes.

When three coins are tossed, the possible outcomes are: \[ \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}. \]
Thus, the total number of outcomes is: \[ n(S) = 8. \]

Step 2: Calculate probabilities.


At least one head:

Outcomes with at least one head:
\[ \{HHH, HHT, HTH, HTT, THH, THT, TTH\}. \]
Number of favorable outcomes:
\[ n(E) = 7. \]
Probability:
\[ P(At least one head) = \frac{n(E)}{n(S)} = \frac{7}{8}. \]

Exactly one tail:

Outcomes with exactly one tail:
\[ \{HHT, HTH, THH\}. \]
Number of favorable outcomes:
\[ n(E) = 3. \]
Probability:
\[ P(Exactly one tail) = \frac{n(E)}{n(S)} = \frac{3}{8}. \]

Two heads and one tail:

Outcomes with two heads and one tail:
\[ \{HHT, HTH, THH\}. \]
Number of favorable outcomes:
\[ n(E) = 3. \]
Probability:
\[ P(Two heads and one tail) = \frac{n(E)}{n(S)} = \frac{3}{8}. \]


---

Conclusion:


Probability of at least one head: \(\frac{7}{8}\).
Probability of exactly one tail: \(\frac{3}{8}\).
Probability of two heads and one tail: \(\frac{3}{8}\). Quick Tip: List all possible outcomes for clarity when calculating probabilities involving multiple events.

Step 1: Calculate the area of the circle.

The area of a circle is given by: \[ Area = \pi r^2. \]
Substitute \(r = 10\) and \(\pi = 3.14\): \[ Area of the circle = 3.14 \times 10^2 = 314 \, cm^2. \]

Step 2: Calculate the area of the minor sector.

The angle subtended by the arc at the center is \(90^\circ\). The area of a sector is given by: \[ Area of sector = \frac{\theta}{360} \times \pi r^2. \]
Substitute \(\theta = 90^\circ\), \(r = 10\), and \(\pi = 3.14\): \[ Area of minor sector = \frac{90}{360} \times 3.14 \times 10^2 = \frac{1}{4} \times 314 = 78.5 \, cm^2. \]

Step 3: Calculate the area of the major sector.

The area of the major sector is: \[ Area of major sector = Area of circle - Area of minor sector. \]
Substitute the values: \[ Area of major sector = 314 - 78.5 = 235.5 \, cm^2. \]

---

Conclusion:

The area of the corresponding major sector is \(235.5 \, cm^2\). Quick Tip: To calculate sector areas, always subtract the minor sector from the total circle area for the major sector.

For real and equal roots, the discriminant (\(D\)) of the quadratic equation must be zero: \[ D = b^2 - 4ac = 0. \]

Here, \(a = (k+1)\), \(b = -6(k+1)\), and \(c = 3(k+9)\).

Substitute these values into the discriminant: \[ D = [-6(k+1)]^2 - 4(k+1)[3(k+9)]. \]

Simplify: \[ D = 36(k+1)^2 - 12(k+1)(k+9). \]

Factorize: \[ D = 12(k+1)[3(k+1) - (k+9)]. \]

Simplify further: \[ D = 12(k+1)[3k + 3 - k - 9]. \] \[ D = 12(k+1)(2k - 6). \]

Set \(D = 0\): \[ 12(k+1)(2k-6) = 0. \]

Solve for \(k\): \[ k+1 = 0 \quad \implies \quad k = -1 \quad (not allowed, as \(k \neq -1\)). \] \[ 2k-6 = 0 \quad \implies \quad k = 3. \]

---

Conclusion:

The value of \(k\) is \(k = 3\). Quick Tip: For real and equal roots of a quadratic equation, always use the condition \(D = b^2 - 4ac = 0\), and solve systematically for unknown parameters.

Let the present age of the son be \(x\) years. Then the present age of the man is \(2x^2\) years.

Step 1: Form an equation based on the given condition.

After 8 years, the age of the man will be \(2x^2 + 8\), and the age of the son will be \(x + 8\). According to the problem: \[ 2x^2 + 8 = 4 + 3(x + 8). \]

Simplify: \[ 2x^2 + 8 = 4 + 3x + 24. \] \[ 2x^2 + 8 = 3x + 28. \] \[ 2x^2 - 3x - 20 = 0. \]

Step 2: Solve the quadratic equation.

Factorize: \[ 2x^2 - 8x + 5x - 20 = 0. \] \[ (2x+5)(x-4) = 0. \]

Solve for \(x\): \[ x = -\frac{5}{2} \quad (not possible, as age cannot be negative), \] \[ x = 4. \]

Step 3: Calculate the man's age.

Substitute \(x = 4\) into \(2x^2\): \[ Man's age = 2(4^2) = 2 \times 16 = 32 \, years. \]

---

Conclusion:

The son's age is \(4\) years, and the man's age is \(32\) years. Quick Tip: When solving age-related problems, carefully form equations based on given conditions and ensure the solutions are realistic (e.g., positive ages).

Let the width of the river be \(AB\), and let the point \(P\) be on the bridge, \(Q\) be directly below \(P\), and \(A\) and \(B\) be on the two banks of the river. Let \(x\) and \(y\) be the horizontal distances from \(Q\) to \(A\) and \(Q\) to \(B\), respectively.

Step 1: Use \(\tan\) in \(\triangle PAQ\).

In right-angled \(\triangle PAQ\): \[ \tan 30^\circ = \frac{Height}{Base} = \frac{4}{x}. \] \[ \frac{1}{\sqrt{3}} = \frac{4}{x} \quad \implies \quad x = 4\sqrt{3}. \]

Step 2: Use \(\tan\) in \(\triangle PBQ\).

In right-angled \(\triangle PBQ\): \[ \tan 60^\circ = \frac{Height}{Base} = \frac{4}{y}. \] \[ \sqrt{3} = \frac{4}{y} \quad \implies \quad y = \frac{4}{\sqrt{3}}. \]

Step 3: Calculate the total width of the river.

The total width of the river is: \[ AB = x + y = 4\sqrt{3} + \frac{4}{\sqrt{3}}. \]

Rationalize the denominator: \[ \frac{4}{\sqrt{3}} = \frac{4\sqrt{3}}{3}. \]

Substitute: \[ AB = 4\sqrt{3} + \frac{4\sqrt{3}}{3}. \]

Combine terms: \[ AB = \frac{12\sqrt{3} + 4\sqrt{3}}{3} = \frac{16\sqrt{3}}{3}. \]

---

Conclusion:

The width of the river is \(\frac{16\sqrt{3}}{3} \, m\).


Diagram:


\begin{tikzpicture[scale=1]
% Draw the triangle
\draw[thick] (0,4) -- (-2,0) -- (2,0) -- cycle;

% Add the bridge and labels
\draw[dashed] (0,4) -- (0,0);
\node at (0,4.2) [above] {\(P\);
\node at (0,-0.2) [below] {\(Q\);
\node at (-2,-0.2) [below] {\(A\);
\node at (2,-0.2) [below] {\(B\);

% Add angles and height
\draw (0,4) -- (-2,0);
\node at (-0.3,3.2) {\(30^\circ\);
\draw (0,4) -- (2,0);
\node at (0.3,3.2) {\(60^\circ\);
\node at (0.2,2.2) [right] {4 m;
\end{tikzpicture Quick Tip: When solving height and distance problems, use trigonometric ratios like \(\tan\) to relate angles, heights, and distances, and always rationalize denominators where necessary.

Step 1: Use congruence to relate angles.

From the given information, \(\triangle FEC \cong \triangle GDB\). Therefore: \[ \angle 3 = \angle 4. \]

Step 2: Analyze \(\triangle ABC\).

In \(\triangle ABC\), since \(\angle 3 = \angle 4\), we have: \[ AB = AC \quad (isosceles triangle) \quad \cdots (i). \]

Step 3: Analyze \(\triangle ADE\).

In \(\triangle ADE\), it is given that \(\angle 1 = \angle 2\). Hence: \[ AD = AE \quad \cdots (ii). \]

Step 4: Divide the corresponding sides of \(\triangle ADE\) and \(\triangle ABC\).

From (i) and (ii): \[ \frac{AD}{AB} = \frac{AE}{AC}. \]

Step 5: Use the parallel property.

Since \(DE \parallel BC\), the corresponding angles are equal: \[ \angle 1 = \angle 3 \quad and \quad \angle 2 = \angle 4. \]

Step 6: Conclude similarity.

By the AA (Angle-Angle) similarity criterion: \[ \triangle ADE \sim \triangle ABC. \]

---

Conclusion:
\(\triangle ADE \sim \triangle ABC\).


Diagram:


\begin{tikzpicture[scale=1]
% Draw the triangle ABC
\draw[thick] (0,4) -- (-3,0) -- (3,0) -- cycle;
% Draw DE
\draw[thick] (-1.5,2) -- (1.5,2);
% Add labels
\node at (0,4.2) [above] {\(A\);
\node at (-3,-0.2) [below] {\(B\);
\node at (3,-0.2) [below] {\(C\);
\node at (-1.5,2.2) [above left] {\(D\);
\node at (1.5,2.2) [above right] {\(E\);
% Add angles
\node at (-2.2,1.5) {\(\angle 3\);
\node at (2.2,1.5) {\(\angle 4\);
\node at (-0.3,3.2) {\(\angle 1\);
\node at (0.3,3.2) {\(\angle 2\);
% Mark parallel lines
\draw[dashed] (-3,0) -- (-1.5,2);
\draw[dashed] (3,0) -- (1.5,2);
\draw[thick] (-1.5,2) -- (1.5,2);
\end{tikzpicture Quick Tip: For proving similarity in triangles, use the AA (Angle-Angle) criterion and parallel line properties to relate angles and sides effectively.

Step 1: Extend medians and draw additional constructions.

Produce \(AD\) to \(E\) such that \(AD = DE\) and join \(EC\). Similarly, produce \(PM\) to \(L\) such that \(PM = ML\) and join \(LR\).

Step 2: Prove congruence between auxiliary triangles.

In \(\triangle ABD\) and \(\triangle ECD\): \[ AD = DE \quad (By construction), \] \[ \angle 1 = \angle 2 \quad (Vertically opposite angles), \] \[ BD = DC \quad (Median property). \]

Thus, by SAS congruence: \[ \triangle ABD \cong \triangle ECD. \]

This implies: \[ AB = EC. \]

Similarly, in \(\triangle PQR\), using the same reasoning: \[ PQ = LR. \]

Step 3: Relate proportionality of sides and medians.

Since \(AB : PQ = AC : PR = AD : PM\), we have: \[ \frac{AB}{PQ} = \frac{AC}{PR} = \frac{AD}{PM}. \]

Using the properties of the constructions: \[ \frac{EC}{LR} = \frac{AC}{PR} = \frac{2AD}{2PM} = \frac{AE}{PL}. \]

Step 4: Prove similarity.

By the Basic Proportionality Theorem and AA similarity criterion: \[ \triangle ABC \sim \triangle PQR. \]

---

Conclusion:
\(\triangle ABC \sim \triangle PQR\).


Diagram:


\begin{tikzpicture[scale=1]
% Triangle ABC
\draw[thick] (0,4) -- (-3,0) -- (3,0) -- cycle;
% Median AD and extended line to E
\draw[thick, dashed] (0,4) -- (0,0);
\draw[thick, dashed] (0,0) -- (0,-2);
% Join EC
\draw[thick] (0,-2) -- (3,0);
% Labels for ABC
\node at (0,4.2) [above] {\(A\);
\node at (-3,-0.2) [below] {\(B\);
\node at (3,-0.2) [below] {\(C\);
\node at (0,0.2) [right] {\(D\);
\node at (0,-2.2) [below] {\(E\);

% Triangle PQR
\draw[thick] (7,4) -- (5,0) -- (9,0) -- cycle;
% Median PM and extended line to L
\draw[thick, dashed] (7,4) -- (7,0);
\draw[thick, dashed] (7,0) -- (7,-2);
% Join LR
\draw[thick] (7,-2) -- (9,0);
% Labels for PQR
\node at (7,4.2) [above] {\(P\);
\node at (5,-0.2) [below] {\(Q\);
\node at (9,-0.2) [below] {\(R\);
\node at (7,0.2) [right] {\(M\);
\node at (7,-2.2) [below] {\(L\);
\end{tikzpicture Quick Tip: For proving similarity between triangles, use proportionality of corresponding sides and medians, and utilize constructions for congruence to establish relationships.

Step 1: Calculate the Curved Surface Area (CSA) of the cylinder.

The formula for the CSA of a cylinder is: \[ CSA = 2\pi r h, \]
where \(r = 2.1 \, cm\) and \(h = 5.8 \, cm\). Substitute the values: \[ CSA of cylinder = 2 \times \frac{22}{7} \times 2.1 \times 5.8. \]

Simplify: \[ CSA of cylinder = 76.56 \, cm^2. \]

Step 2: Calculate the CSA of the two hemispheres.

The formula for the CSA of a hemisphere is: \[ CSA of hemisphere = 2\pi r^2. \]

For two hemispheres: \[ CSA of two hemispheres = 2 \times 2\pi r^2 = 4\pi r^2. \]

Substitute \(r = 2.1 \, cm\): \[ CSA of two hemispheres = 4 \times \frac{22}{7} \times 2.1 \times 2.1. \]

Simplify: \[ CSA of two hemispheres = 55.44 \, cm^2. \]

Step 3: Calculate the total surface area of the article.

The total surface area is the sum of the CSA of the cylinder and the CSA of the two hemispheres: \[ Total Surface Area = 76.56 + 55.44. \]

Simplify: \[ Total Surface Area = 132 \, cm^2. \]

---

Conclusion:

The total surface area of the article is \(132 \, cm^2\). Quick Tip: To calculate the surface area of composite solids, add the curved surface areas of all components while excluding any overlapping parts.

(i) Formulate the equations:

Let the number of children be \(x\) and the number of adults be \(y\). The given conditions can be written as:
\[ x + y = 300 \quad \cdots (i) \] \[ 150x + 250y = 55000 \quad \cdots (ii) \]

(ii) Solve for the number of children and adults:

(a) From equations (i) and (ii), solve for \(x\):

Substitute \(y = 300 - x\) into equation (ii):
\[ 150x + 250(300 - x) = 55000. \]
Simplify:
\[ 150x + 75000 - 250x = 55000. \] \[ -100x + 75000 = 55000 \quad \implies \quad -100x = -20000 \quad \implies \quad x = 200. \]
Therefore, the number of children is \(x = 200\).


(b) Substituting \(x = 200\) into equation (i):
\[ y = 300 - 200 = 100. \]
Therefore, the number of adults is \(y = 100\).


(iii) Calculate the amount collected if 250 children and 100 adults visit the park:
\[ Amount collected = 150 \times 250 + 250 \times 100. \] \[ Amount collected = 37500 + 25000 = 62500. \]

---

Conclusion:

(i) The algebraic equations are \(x + y = 300\) and \(150x + 250y = 55000\).

(ii) Number of children: \(200\), Number of adults: \(100\).

(iii) Total amount collected: ₹62500. Quick Tip: Use substitution or elimination methods for solving linear equations systematically. Ensure accurate substitution and simplification in word problems.

(i) Coordinates of the vertices of \( \Delta PQR \):

From the figure, the coordinates are:
\( P(4, 6), \ Q(3, 2), \ R(6, 5) \).

(ii) Find distances and coordinates:

(a) Distance between \( P \) and \( Q \): \[ PQ = \sqrt{(4 - 3)^2 + (6 - 2)^2} = \sqrt{1^2 + 4^2} = \sqrt{1 + 16} = \sqrt{17}. \]
Distance between \( Q \) and \( R \): \[ QR = \sqrt{(3 - 6)^2 + (2 - 5)^2} = \sqrt{(-3)^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18}. \]

(b) The coordinates of the point dividing \( PR \) in the ratio \( 2:1 \): \[ Using section formula: \left(\frac{2x_2 + 1x_1}{2 + 1}, \frac{2y_2 + 1y_1}{2 + 1}\right). \]
Substitute \( P(4, 6) \) and \( R(6, 5) \): \[ \left(\frac{2 \times 6 + 1 \times 4}{3}, \frac{2 \times 5 + 1 \times 6}{3}\right) = \left(\frac{12 + 4}{3}, \frac{10 + 6}{3}\right) = \left(\frac{16}{3}, \frac{16}{3}\right). \]

(iii) Check if \( \Delta PQR \) is an isosceles triangle:

Distance \( PR \): \[ PR = \sqrt{(4 - 6)^2 + (6 - 5)^2} = \sqrt{(-2)^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5}. \]
Since \( PQ \neq QR \neq PR \), \( \Delta PQR \) is not an isosceles triangle.

---

Conclusion:

(i) Coordinates of the vertices: \( P(4, 6), Q(3, 2), R(6, 5) \).

(ii) (a) \( PQ = \sqrt{17}, QR = \sqrt{18} \).

(b) The coordinates of the point dividing \( PR \) are \( \left(\frac{16}{3}, \frac{16}{3}\right) \).

(iii) \( \Delta PQR \) is not isosceles. Quick Tip: To determine the nature of a triangle, compute the lengths of all sides using the distance formula and compare them.

(i) Median Class:

The cumulative frequency is calculated as follows:
\[ \begin{array}{|c|c|c|} \hline \textbf{Time (in seconds)} & \textbf{Number of students (f)} & \textbf{Cumulative Frequency (cf)}
\hline 0 - 20 & 8 & 8
\hline 20 - 40 & 10 & 18
\hline 40 - 60 & 13 & 31
\hline 60 - 80 & 6 & 37
\hline 80 - 100 & 3 & 40
\hline \end{array} \]
The total number of students is 40. Since \( N/2 = 20 \), the median class is \( 40 - 60 \).

(ii) Mean Time:

The table for \( x_i \) (midpoints) and \( f_i x_i \) is as follows:
\[ \begin{array}{|c|c|c|c|} \hline \textbf{Time (in seconds)} & \textbf{Midpoint} (x_i) & \textbf{Number of students (f)} & \textbf{\( f_i x_i \)}
\hline 0 - 20 & 10 & 8 & 80
\hline 20 - 40 & 30 & 10 & 300
\hline 40 - 60 & 50 & 13 & 650
\hline 60 - 80 & 70 & 6 & 420
\hline 80 - 100 & 90 & 3 & 270
\hline \end{array} \] \[ Mean = \frac{\sum f_i x_i}{\sum f_i} = \frac{1720}{40} = 43. \]

(ii) Mode:

The modal class is \( 40 - 60 \). Using the formula:
\[ Mode = L + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h \]
where \( L = 40 \), \( f_1 = 13 \), \( f_0 = 10 \), \( f_2 = 6 \), \( h = 20 \):
\[ Mode = 40 + \left(\frac{13 - 10}{2(13) - 10 - 6}\right) \times 20 = 40 + \left(\frac{3}{26 - 16}\right) \times 20 = 40 + 6 = 46. \]

(iii) Students Taking Less than 60 Seconds:

From the cumulative frequency table, the number of students taking less than 60 seconds is \( 31 \).

---

Conclusion:

(i) Median class: \( 40 - 60 \).

(ii) Mean time: \( 43 \), Mode: \( 46 \).

(iii) Number of students: \( 31 \). Quick Tip: To calculate the mean, use midpoints and summations. The modal class is identified as the class with the highest frequency.