CBSE Class 10 Mathematics Standard Question Paper 2023 (March 21, Set 1- 30/1/1) Available: Download Solution PDF with Answer Key

The number of zeroes of a polynomial is equal to the number of times its graph intersects the x-axis. In this graph, the curve intersects the x-axis at one point. Therefore, the number of zeroes is 1.
Quick Tip: The zeroes of a polynomial are the x-values where the graph of the polynomial intersects the x-axis.

For a pair of linear equations in two variables \(a_1x + b_1y + c_1 = 0\) and \(a_2x + b_2y + c_2 = 0\) to have infinitely many solutions, the condition is: \[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \]

Rewriting the given equations in the standard form: \(kx - y - 2 = 0\) and \(6x - 2y - 3 = 0\)
Comparing with the standard form, we have: \(a_1 = k\), \(b_1 = -1\), \(c_1 = -2\) \(a_2 = 6\), \(b_2 = -2\), \(c_2 = -3\)

Applying the condition for infinitely many solutions: \[ \frac{k}{6} = \frac{-1}{-2} = \frac{-2}{-3} \] \[ \frac{k}{6} = \frac{1}{2} = \frac{2}{3} \]
From the first two ratios: \(\frac{k}{6} = \frac{1}{2} \implies k = 3\).
From the last two ratios: \(\frac{1}{2} \ne \frac{2}{3}\).
However, if \(k=3\), the ratios become \(\frac{3}{6} = \frac{1}{2}\) and \(\frac{1}{2}\) and \(\frac{2}{3}\) which does not satisfy the condition.
Let's rewrite the second equation as \(3x - y - \frac{3}{2}=0\). Now applying the condition: \(\frac{k}{3} = \frac{-1}{-1} = \frac{-2}{-3/2}\) \(\frac{k}{3} = 1 = \frac{4}{3}\).
This is not possible.
If we analyze the two equations:
kx = y+2. if k = 3, then 3x = y+2 or 6x = 2y+4.
The second equation is 6x = 2y+3. Since the two lines are parallel, there are no solutions. So, such a \(k\) does not exist.
Quick Tip: For infinite solutions, the ratios of corresponding coefficients must be equal. Make sure all three ratios are equal.

If three numbers \(a\), \(b\), and \(c\) are in arithmetic progression (A.P.), then the common difference between consecutive terms is constant. This means \(b - a = c - b\), or equivalently, \(2b = a + c\).

In this case, \(p-1\), \(p+3\), and \(3p-1\) are in A.P. So, \[ 2(p+3) = (p-1) + (3p-1) \] \[ 2p + 6 = 4p - 2 \] \[ 8 = 2p \] \[ p = 4 \]

Quick Tip: If \(a, b, c\) are in A.P., then \(2b = a + c\).

Let the x-axis divide the line segment AB in the ratio \(m:n\). The coordinates of the point dividing the line segment joining \((x_1, y_1)\) and \((x_2, y_2)\) in the ratio \(m:n\) are given by the section formula: \[ \left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n} \right) \]
Since the x-axis is doing the dividing, the y-coordinate of the dividing point will be 0. Here, \(x_1 = 3\), \(y_1 = 6\), \(x_2 = -12\), and \(y_2 = -3\). So, \[ \frac{m(-3) + n(6)}{m+n} = 0 \] \[ -3m + 6n = 0 \] \[ 3m = 6n \] \[ m = 2n \] \[ \frac{m}{n} = \frac{2}{1} \]
Therefore, the ratio is 2:1.
Quick Tip: Remember the section formula. For the x-axis division, the y-coordinate will be zero. For the y-axis division, the x-coordinate will be zero.

In the given figure, OA and OB are radii of the circle, and PQ is a tangent. \(\angle AOB = 95^\circ\) (Given).
Since OA = OB (radii), triangle AOB is isosceles. Therefore, \(\angle OAB = \angle OBA\).
In \(\triangle AOB\), the sum of angles is \(180^\circ\).
So, \( \angle OAB + \angle OBA + \angle AOB = 180^\circ \) \( 2\angle OBA + 95^\circ = 180^\circ \) \( 2\angle OBA = 85^\circ \) \( \angle OBA = 42.5^\circ \)

Since PQ is a tangent to the circle at point B, OB is perpendicular to PQ. Thus \(\angle OBQ = 90^\circ\). We have \(\angle OBA = 42.5^\circ\). Then, \(\angle ABQ = \angle OBQ - \angle OBA = 90^\circ - 42.5^\circ = 47.5^\circ\).
Quick Tip: The tangent to a circle is perpendicular to the radius at the point of contact. Also, the angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.

Given \(2 \tan A = 3\), so \(\tan A = \frac{3}{2}\).
We can write the given expression as: \[ \frac{4 \sin A + 3 \cos A}{4 \sin A - 3 \cos A} = \frac{4 \frac{\sin A}{\cos A} + 3}{4 \frac{\sin A}{\cos A} - 3} = \frac{4 \tan A + 3}{4 \tan A - 3} \]
Substituting \(\tan A = \frac{3}{2}\): \[ \frac{4(\frac{3}{2}) + 3}{4(\frac{3}{2}) - 3} = \frac{6 + 3}{6 - 3} = \frac{9}{3} = 3 \]
Another approach:
We are given \(2 \tan A = 3\) or \(\tan A = \frac{3}{2}\).
Then \(\sin A = \frac{3}{\sqrt{13}}\) and \(\cos A = \frac{2}{\sqrt{13}}\)
Substituting these values in the expression: \[ \frac{4(\frac{3}{\sqrt{13}}) + 3(\frac{2}{\sqrt{13}})}{4(\frac{3}{\sqrt{13}}) - 3(\frac{2}{\sqrt{13}})} = \frac{\frac{12+6}{\sqrt{13}}}{\frac{12-6}{\sqrt{13}}} = \frac{18}{6} = 3 \]
Quick Tip: Divide both numerator and denominator by \(\cos A\) to simplify the expression in terms of \(\tan A\).

If \(\alpha\) and \(\beta\) are the zeroes of the quadratic polynomial \(ax^2 + bx + c\), then
Sum of zeroes: \(\alpha + \beta = -\frac{b}{a}\)
Product of zeroes: \(\alpha\beta = \frac{c}{a}\)

In the given polynomial \(p(x) = x^2 + x - 1\), \(a = 1\), \(b = 1\), and \(c = -1\). Therefore, \(\alpha + \beta = -\frac{1}{1} = -1\) \(\alpha\beta = \frac{-1}{1} = -1\)

We are asked to find \(\frac{1}{\alpha} + \frac{1}{\beta}\). We can write this as: \[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} \]
Substituting the values we found: \[ \frac{-1}{-1} = 1 \]
Quick Tip: Remember the relationship between the zeroes and the coefficients of a quadratic polynomial.

A quadratic equation \(ax^2 + bx + c = 0\) has rational roots if its discriminant, \(D = b^2 - 4ac\), is a perfect square. In this case, \(a = 2\), \(b = k\), and \(c = -4\). \[ D = k^2 - 4(2)(-4) = k^2 + 32 \]
We want the least positive value of \(k\) for which \(k^2 + 32\) is a perfect square. When \(k = \pm 2\), we get \(k^2 + 32 = 4 + 32 = 36 = 6^2\), which is the smallest perfect square possible for \(D\). So the required minimum positive value is \(k=2\). It is also possible that \(k=-2\).
We need \(k^2+32 = m^2\) where \(m\) is an integer. \(k = \pm 2\) is the minimum positive integer for this to happen.
Then \(k^2+32 = 36 = 6^2\) which makes \(m\) an integer.
So, minimum value is \(\pm 2\).
Quick Tip: For rational roots, the discriminant (\(b^2 - 4ac\)) must be a perfect square.

We know the standard trigonometric values: \(\tan 30^\circ = \frac{1}{\sqrt{3}}\), so \(\tan^2 30^\circ = \frac{1}{3}\) \(\sec 45^\circ = \sqrt{2}\), so \(\sec^2 45^\circ = 2\) \(\sin 60^\circ = \frac{\sqrt{3}}{2}\), so \(\sin^2 60^\circ = \frac{3}{4}\)

Substituting these values into the given expression: \[ \frac{1}{3} - 2 + \frac{3}{4} = \frac{4 - 24 + 9}{12} = \frac{-11}{12} \] \[ \frac{1}{3} - 2 + \frac{3}{4} = \frac{4-24+9}{12} = \frac{-11}{12} \]
It seems there's a typo in the options provided. The correct answer based on the calculations is \(-\frac{11}{12}\).
Quick Tip: Memorize the values of trigonometric ratios for standard angles (\(0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ\)).

The curved surface area (CSA) of a cylinder is given by the formula:
CSA \(= 2\pi rh\), where \(r\) is the radius and \(h\) is the height.

Given CSA = 94.2 \(cm^2\) and \(h = 5\) cm. Substituting these values and \(\pi = 3.14\): \[ 94.2 = 2(3.14)r(5) \] \[ 94.2 = 31.4r \] \[ r = \frac{94.2}{31.4} = 3 cm \]
Quick Tip: Remember the formula for the curved surface area of a cylinder: \(2\pi rh\).

The modal class is the class interval with the highest frequency. To find this, we first convert the cumulative frequency table to a frequency distribution table:


The class interval 30-40 has the highest frequency (30). Therefore, the modal class is 30-40.
Quick Tip: In a cumulative frequency distribution, the frequency of a class interval is the difference between its cumulative frequency and the cumulative frequency of the previous interval. The modal class has the highest frequency.

The curved surface area (CSA) of a cone is given by \(\pi rl\), where \(r\) is the radius and \(l\) is the slant height. The slant height is given by \(l = \sqrt{r^2 + h^2}\), where \(h\) is the height.

Given \(r = 7\) cm and \(h = 24\) cm. \[ l = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25 cm \]
CSA \(= \pi rl = \frac{22}{7} \times 7 \times 25 = 22 \times 25 = 550 cm^2 \)
Quick Tip: The curved surface area of a cone is \(\pi rl\), where \(l\) is the slant height, given by \(l = \sqrt{r^2 + h^2}\).

The distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by the distance formula: \[ \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Let \((x_1, y_1) = (0, 2\sqrt{5})\) and \((x_2, y_2) = (-2\sqrt{5}, 0)\). Then the distance is
\begin{align* \label{eq:1 \sqrt{(-2\sqrt{5 - 0)^2 + (0 - 2\sqrt{5)^2 &= \sqrt{(-2\sqrt{5)^2 + (-2\sqrt{5)^2
&= \sqrt{20 + 20
&= \sqrt{40 = 2\sqrt{10 units\end{align*
Quick Tip: Remember the distance formula: \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2\).

A quadratic polynomial with zeroes \(\alpha\) and \(\beta\) can be written in the form \(k(x^2 - (\alpha+\beta)x + \alpha\beta)\), where \(k\) is a non-zero constant.

Here, \(\alpha = -\frac{2}{3}\) and \(\beta = \frac{2}{3}\).
Sum of zeroes: \(\alpha + \beta = -\frac{2}{3} + \frac{2}{3} = 0\)
Product of zeroes: \(\alpha\beta = -\frac{2}{3} \times \frac{2}{3} = -\frac{4}{9}\)

So, the quadratic polynomial is of the form \(k(x^2 - 0x - \frac{4}{9}) = k(x^2 - \frac{4}{9})\).
Multiplying by 9 to eliminate the fraction, we get \(9k(9x^2 - 4)\).
If we let \(k=\frac{5}{9}\) (a non-zero constant). then we have the polynomial:
\(\frac{5}{9} \times 9 (9x^2 - 4) = 5(9x^2 - 4)\), which matches option (d).
Quick Tip: A quadratic polynomial with zeroes \(\alpha\) and \(\beta\) can be written as \(k(x - \alpha)(x - \beta)\) or \(k(x^2 - (\alpha + \beta)x + \alpha\beta)\).

Let \(x_1, x_2, ..., x_n\) be the observations. The original mean is given by: \[ \bar{x} = \frac{x_1 + x_2 + ... + x_n}{n} \]
If each observation is increased by 3, the new observations are \(x_1 + 3, x_2 + 3, ..., x_n + 3\). The new mean is:
\begin{align* \bar{x' &= \frac{(x_1 + 3) + (x_2 + 3) + ... + (x_n + 3){n
&= \frac{(x_1 + x_2 + ... + x_n) + 3n{n
&= \frac{x_1 + x_2 + ... + x_n{n + \frac{3n{n
&= \bar{x + 3 \end{align*
So, the new mean is increased by 3.
Quick Tip: If each observation is increased or decreased by a constant value, the mean also increases or decreases by the same value.

The sum of the probabilities of an event happening and not happening is always 1. Therefore, \(p + q = 1\).
Quick Tip: The sum of probabilities of all possible outcomes of an event is always 1.

Let \(n\) be the number of tickets the girl bought.
The probability of winning is given by the ratio of the number of tickets she bought to the total number of tickets sold.
Probability = \(\frac{Number of tickets bought}{Total number of tickets sold}\)
Given that the probability of winning is 0.08 and the total tickets sold are 6000: \[ 0.08 = \frac{n}{6000} \] \[ n = 0.08 \times 6000 = 480 \]
So, the girl bought 480 tickets.
Quick Tip: Probability is calculated as the ratio of favorable outcomes to the total number of possible outcomes.

Total number of people = 20
Number of people who can't swim = 5
Number of people who can swim = \(20 - 5 = 15\)

Probability of selecting a person who can swim = \(\frac{Number of people who can swim}{Total number of people}\) \[ P(can swim) = \frac{15}{20} = \frac{3}{4} \]
Quick Tip: Subtract the number of unfavorable outcomes from the total outcomes to find the number of favorable outcomes.

Assertion (A): The point of intersection of a line with the y-axis occurs when \(x = 0\). Substituting \(x = 0\) in the equation \(3x + 2y = 4\), we get \(2y = 4\), which means \(y = 2\). Thus, the point of intersection with the y-axis is (0, 2), which is point P. So, Assertion (A) is true.

Reason (R): The distance of a point \((x, y)\) from the x-axis is given by \(|y|\). The distance of P(0, 2) from the x-axis is |2| = 2 units. So, Reason (R) is true.

However, the distance of P from the x-axis doesn't explain why P is the intersection point of the line with the y-axis. The intersection point is determined by setting \(x=0\) in the line's equation, not by the distance from the x-axis. Therefore, Reason (R) is not the correct explanation for Assertion (A).
Quick Tip: To find the intersection of a line with the y-axis, set \(x = 0\) and solve for \(y\). The distance of a point \((x, y)\) from the x-axis is \(|y|\).

Assertion (A):
In \(\triangle ABC\), AB = 2 cm and BC = 3 cm. Using the Pythagorean theorem, \(AC = \sqrt{AB^2 + BC^2} = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13}\) cm.
Perimeter = \(AB + BC + AC = 2 + 3 + \sqrt{13} = 5 + \sqrt{13}\). Since \(\sqrt{13}\) is irrational, the perimeter is irrational. So, Assertion (A) is false.

Reason (R): Let \(a\) and \(b\) be two rational numbers. Then \(a\) and \(b\) can be expressed as fractions \(\frac{p}{q}\) and \(\frac{r}{s}\), where \(p, q, r, s\) are integers and \(q, s \ne 0\). Then \(a^2 + b^2 = \frac{p^2}{q^2} + \frac{r^2}{s^2} = \frac{p^2s^2 + r^2q^2}{q^2s^2}\). Since integers are closed under multiplication and addition, \(p^2s^2 + r^2q^2\) and \(q^2s^2\) are integers, and \(q^2s^2 \ne 0\). Thus, \(a^2 + b^2\) can be expressed as a fraction of integers, making it a rational number. So, Reason (R) is true.
Quick Tip: The Pythagorean theorem states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Rational numbers can be expressed as \(\frac{p}{q}\), where \(p\) and \(q\) are integers, and \(q \ne 0\).

The equation \(x=3\) represents a vertical line passing through the point (3, 0) on the x-axis. The equation \(y=-4\) represents a horizontal line passing through the point (0, -4) on the y-axis.

To solve graphically, we plot these lines on a coordinate plane. The point where these two lines intersect is the solution to the pair of equations.

The lines intersect at the point (3, -4). Therefore, the solution is \(x=3\) and \(y=-4\).
Quick Tip: \(x = a\) represents a vertical line and \(y = b\) represents a horizontal line on the coordinate plane. Their intersection point is \((a, b)\).

The equation \(x = 0\) represents the y-axis. The equation \(y = -7\) represents a horizontal line passing through the point \((0, -7)\).

When we graph these two equations, they intersect at the point (0, -7). Since the lines intersect at a single point, the system of equations has a unique solution, and therefore, the system is consistent.
Quick Tip: A system of linear equations is consistent if it has at least one solution (either unique or infinitely many). If the lines are parallel and distinct, the system is inconsistent (no solution).

In \(\triangle ABC\), since \(XZ \parallel BC\), by the Basic Proportionality Theorem (Thales's Theorem), we have: \[ \frac{AX}{AB} = \frac{AZ}{AC} = \frac{XZ}{BC} \]
We are given \(AZ = 3\) cm and \(ZC = 2\) cm, so \(AC = AZ + ZC = 3 + 2 = 5\) cm. Thus, \(\frac{AZ}{AC} = \frac{3}{5}\).

Similarly, in \(\triangle ABM\), since \(XY \parallel BM\), we have: \[ \frac{AX}{AB} = \frac{AY}{AM} = \frac{XY}{BM} \]
Since \(\frac{AX}{AB} = \frac{AZ}{AC} = \frac{3}{5}\), we also have \(\frac{XY}{BM} = \frac{3}{5}\).
Given \(BM = 3\) cm, \[ \frac{XY}{3} = \frac{3}{5} \] \[ XY = \frac{3}{5} \times 3 = \frac{9}{5} = 1.8 cm \]
Quick Tip: When you have parallel lines intersecting triangles, use the Basic Proportionality Theorem (Thales's Theorem) to relate the lengths of the segments.

We are given \(\sin \theta + \cos \theta = \sqrt{3}\). Squaring both sides, we get: \[ (\sin \theta + \cos \theta)^2 = (\sqrt{3})^2 \] \[ \sin^2 \theta + 2\sin\theta\cos\theta + \cos^2 \theta = 3 \]
Since \(\sin^2 \theta + \cos^2 \theta = 1\), we have: \[ 1 + 2\sin\theta\cos\theta = 3 \] \[ 2\sin\theta\cos\theta = 2 \] \[ \sin\theta\cos\theta = 1 \]
Quick Tip: Use the trigonometric identity \(\sin^2 \theta + \cos^2 \theta = 1\). Squaring both sides of an equation can often help simplify trigonometric expressions.

Given \(\sin \alpha = \frac{1}{\sqrt{2}}\), we know that \(\csc \alpha = \frac{1}{\sin \alpha} = \frac{1}{\frac{1}{\sqrt{2}}} = \sqrt{2}\).

Given \(\cot \beta = \sqrt{3}\). Since \(\cot \beta = \frac{\cos \beta}{\sin \beta}\), we can consider a right-angled triangle where the adjacent side is \(\sqrt{3}\) and the opposite side is 1. The hypotenuse will be \(\sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3+1} = 2\). Thus, \(\sin \beta = \frac{1}{2}\) and \(\csc \beta = \frac{1}{\sin \beta} = 2\). \[ \csc \alpha + \csc \beta = \sqrt{2} + 2 \]
Quick Tip: Use the definitions \(\csc \theta = \frac{1}{\sin \theta}\) and \(\cot \theta = \frac{\cos \theta}{\sin \theta}\). Consider constructing a right triangle to find the values of \(\sin \beta\) and \(\cos \beta\) when \(\cot \beta\) is given.

Let the greatest number be \(d\).
When 85 is divided by \(d\), the remainder is 1. This means \(85 - 1 = 84\) is exactly divisible by \(d\).
When 72 is divided by \(d\), the remainder is 2. This means \(72 - 2 = 70\) is exactly divisible by \(d\).

We are looking for the greatest number \(d\) which divides both 84 and 70, which is the greatest common divisor (GCD) or highest common factor (HCF) of 84 and 70.

Using the prime factorization method: \(84 = 2^2 \times 3 \times 7\) \(70 = 2 \times 5 \times 7\)
The common factors are 2 and 7. The HCF is \(2 \times 7 = 14\).
Therefore, the greatest number is 14.
Quick Tip: When remainders are given, subtract the remainders from the respective numbers to find numbers that are exactly divisible. The GCD/HCF of these resulting numbers is the required divisor.

Total number of balls = \(4 (red) + 3 (blue) + 2 (yellow) = 9\)

(i) Number of red balls = 4
Probability of drawing a red ball = \(\frac{Number of red balls}{Total number of balls} = \frac{4}{9}\)

(ii) Number of yellow balls = 2
Probability of drawing a yellow ball = \(\frac{Number of yellow balls}{Total number of balls} = \frac{2}{9}\)
Quick Tip: Probability of an event = \(\frac{Number of favourable outcomes}{Total number of possible outcomes}\).

Let the two numbers be \(x\) and \(y\), where \(x > y\).
According to the first condition, half of the difference between the two numbers is 2: \[ \frac{x - y}{2} = 2 \] \[ x - y = 4 \quad (*)\label{eq:1} \]
According to the second condition, the sum of the greater number and twice the smaller number is 13: \[ x + 2y = 13 \quad (**)\label{eq:2} \]
Subtracting equation (*) from equation (**): \[ (x + 2y) - (x - y) = 13 - 4 \] \[ 3y = 9 \] \[ y = 3 \]
Substituting \(y = 3\) in equation (*): \[ x - 3 = 4 \] \[ x = 7 \]
Therefore, the two numbers are 7 and 3.
Quick Tip: Represent the unknowns with variables and form equations based on the given conditions. Solve the equations simultaneously using substitution, elimination, or any other suitable method.

Let's assume, for the sake of contradiction, that \(\sqrt{5}\) is a rational number. This means it can be expressed in the form \(\frac{a}{b}\), where \(a\) and \(b\) are coprime integers (i.e., their greatest common divisor is 1), and \(b \ne 0\). \[ \sqrt{5} = \frac{a}{b} \]
Squaring both sides: \[ 5 = \frac{a^2}{b^2} \] \[ 5b^2 = a^2 \]
This means that \(a^2\) is divisible by 5, which implies that \(a\) is also divisible by 5 (because if a prime number divides the square of an integer, it must also divide the integer itself).
So, we can write \(a = 5c\) for some integer \(c\). Substituting this into the equation above: \[ 5b^2 = (5c)^2 \] \[ 5b^2 = 25c^2 \] \[ b^2 = 5c^2 \]
This means \(b^2\) is divisible by 5, and thus, \(b\) is also divisible by 5.
We have shown that both \(a\) and \(b\) are divisible by 5. This contradicts our initial assumption that \(a\) and \(b\) are coprime. Therefore, our assumption that \(\sqrt{5}\) is rational must be false. Hence, \(\sqrt{5}\) is an irrational number.
Quick Tip: To prove a number is irrational, use proof by contradiction. Assume it is rational (\(\frac{a}{b}\) where a and b are coprime), and then show that this leads to a contradiction.

Let the vertices be A(-5, 3) and B(5, 3), and let the third vertex be C\((x, y)\).
Since it's an equilateral triangle, \(AB = BC = CA\). \[ AB = \sqrt{(5 - (-5))^2 + (3 - 3)^2} = \sqrt{10^2 + 0} = 10 \]

Now, we use the distance formula for BC and CA: \(BC^2 = (x - 5)^2 + (y - 3)^2 = 10^2 = 100\) \(CA^2 = (x + 5)^2 + (y - 3)^2 = 10^2 = 100\)

Since \(BC^2 = CA^2\), \((x - 5)^2 + (y - 3)^2 = (x + 5)^2 + (y - 3)^2\) \((x - 5)^2 = (x + 5)^2\) \(x^2 - 10x + 25 = x^2 + 10x + 25\) \(20x = 0 \implies x = 0\)

Now substitute \(x = 0\) in \(BC^2\): \((0 - 5)^2 + (y - 3)^2 = 100\) \(25 + (y - 3)^2 = 100\) \((y - 3)^2 = 75\) \(y - 3 = \pm\sqrt{75} = \pm 5\sqrt{3}\) \(y = 3 \pm 5\sqrt{3}\)
Since the origin lies inside the triangle, the third vertex must have a negative y-coordinate. Therefore, \(y = 3 - 5\sqrt{3} = 3 - 5(1.7) = 3 - 8.5 = -5.5\)
So, the third vertex is C(0, -5.5).
Quick Tip: In an equilateral triangle, all sides are equal. Use the distance formula to equate the lengths of the sides and solve for the unknown coordinates.

Let the circle have center O. TP and TQ are tangents drawn from an external point T.
We need to prove that \(\angle PTQ = 2\angle OPQ\).

Since TP and TQ are tangents from the same external point T to the circle with center O, we know that \(TP = TQ\). This means \(\triangle TPQ\) is an isosceles triangle.
Therefore, \(\angle TPQ = \angle TQP\).

In \(\triangle TPQ\), the sum of angles is \(180^\circ\). So, \[\angle PTQ + \angle TPQ + \angle TQP = 180^\circ\]
Since \(\angle TPQ = \angle TQP\), we have: \[\angle PTQ + 2\angle TPQ = 180^\circ \quad (*)\label{eq:1}\]

Also, OP is the radius and TP is the tangent at point P. Thus, \(OP \perp TP\), so \(\angle OPT = 90^\circ\).
In \(\triangle OPT\), \[\angle OPT + \angle OTP + \angle TOP = 180^\circ\] \[90^\circ + \angle OTP + \angle TOP = 180^\circ\] \[\angle OTP + \angle TOP = 90^\circ\] \[\angle TPQ + \angle OPQ = 90^\circ\] \[\angle TPQ = 90^\circ - \angle OPQ\]
Substituting this into equation (*): \[\angle PTQ + 2(90^\circ - \angle OPQ) = 180^\circ\] \[\angle PTQ + 180^\circ - 2\angle OPQ = 180^\circ\] \[\angle PTQ = 2\angle OPQ\]
Hence proved.
Quick Tip: Tangents drawn from an external point to a circle are equal in length. The tangent at any point of a circle is perpendicular to the radius through the point of contact.

Let the radius of the inscribed circle be \(r\).
Since the circle is inscribed in quadrilateral ABCD, the sides of the quadrilateral are tangents to the circle.
Let the circle touch the sides AB, BC, CD, and DA at P, Q, R, and S respectively.
Given \(\angle B = 90^\circ\), AB = 20 cm, AD = 17 cm, and DS = 3 cm.

Since tangents from an external point are equal, we have \(DS = DP = 3\) cm, \(AS = AP = AD - DS = 17 - 3 = 14\) cm, \(BP = BQ\), and \(CQ = CR\).
Also, \(AP = AB - BP = 20 - BP = 14\), so \(BP = BQ = 6\) cm.
Since \(\angle B = 90^\circ\), OPBQ is a square with side \(r\). Thus, \(r = BP = BQ = 6\) cm.
Therefore, the radius of the inscribed circle is 6 cm.
Quick Tip: In a quadrilateral with an inscribed circle, the tangents drawn from an external point to the circle are equal in length. If a quadrilateral has a right angle where the circle touches two adjacent sides, those tangent segments form a square with the radii.

We are asked to prove: \[\frac{\tan \theta + \sec \theta - 1}{\tan \theta - \sec \theta + 1} = \frac{1 + \sin \theta}{\cos \theta}\]
L.H.S. = \(\frac{\tan \theta + \sec \theta - 1}{\tan \theta - \sec \theta + 1}\)
We know that \(\sec^2 \theta - \tan^2 \theta = 1\).
\begin{align* \text{L.H.S. &= \frac{\tan\theta + \sec\theta - (\sec^2\theta - \tan^2\theta){\tan\theta - \sec\theta + 1
&= \frac{\tan\theta + \sec\theta - (\sec\theta + \tan\theta)(\sec\theta - \tan\theta){\tan\theta - \sec\theta + 1
&= \frac{(\tan\theta + \sec\theta)(1-\sec\theta+\tan\theta){\tan\theta-\sec\theta+1
&= \tan\theta + \sec\theta = \frac{\sin\theta{\cos\theta + \frac{1{\cos\theta
&= \frac{\sin\theta + 1{\cos\theta = \frac{1+\sin\theta{\cos\theta = \text{R.H.S.\end{align*
Hence Proved.
Quick Tip: Use trigonometric identities like \(\sec^2 \theta - \tan^2 \theta = 1\) to simplify expressions. Try expressing everything in terms of sine and cosine.

Let the radius of the hemispherical dome be \(r\) meters and the height of the cylindrical part be \(h\) meters.
According to the problem, the radius of the hemisphere is one-half the height of the cylinder, so \(r = \frac{h}{2}\).

The volume of the room is the sum of the volume of the cylinder and the volume of the hemisphere:
Volume of cylinder = \(\pi r^2 h\)
Volume of hemisphere = \(\frac{2}{3} \pi r^3\)
Total volume = \(\pi r^2 h + \frac{2}{3} \pi r^3 = \frac{1408}{21}\)

Substituting \(r = \frac{h}{2}\): \[ \pi \left(\frac{h}{2}\right)^2 h + \frac{2}{3} \pi \left(\frac{h}{2}\right)^3 = \frac{1408}{21} \] \[ \frac{\pi h^3}{4} + \frac{2\pi h^3}{24} = \frac{1408}{21} \] \[ \frac{\pi h^3}{4} + \frac{\pi h^3}{12} = \frac{1408}{21} \] \[ \frac{3\pi h^3 + \pi h^3}{12} = \frac{1408}{21} \] \[ \frac{4\pi h^3}{12} = \frac{\pi h^3}{3} = \frac{1408}{21} \]
Substituting \(\pi = \frac{22}{7}\): \[ \frac{22}{7} \times \frac{h^3}{3} = \frac{1408}{21} \] \[ h^3 = \frac{1408}{21} \times \frac{7}{22} \times 3 = \frac{1408 \times 21}{21 \times 22 \times 3} = 64 \] \[ h = \sqrt[3]{64} = 4 m \]
Since \(r = \frac{h}{2}\), \(r = \frac{4}{2} = 2\) m.
Total height of the room = height of cylinder + radius of hemisphere = \(h + r = 4 + 2 = 6\) m.
Quick Tip: The volume of a cylinder is \(\pi r^2 h\) and the volume of a hemisphere is \(\frac{2}{3}\pi r^3\).

Radius of the cone, \(r = 3\) cm
Height of the cone, \(h = 12\) cm

Volume of the cone = \(\frac{1}{3} \pi r^2 h = \frac{1}{3} \times 3.14 \times 3^2 \times 12 = 3.14 \times 36 = 113.04\) cm\(^3\)

Volume of the unfilled part of the cone = \(\frac{1}{6} \times Volume of cone = \frac{1}{6} \times 113.04 = 18.84\) cm\(^3\)

Volume of the filled part of the cone = Volume of cone - Volume of unfilled part = \(113.04 - 18.84 = 94.2\) cm\(^3\)

Radius of the hemisphere on top = Radius of the cone = 3 cm
Volume of the hemisphere = \(\frac{2}{3} \pi r^3 = \frac{2}{3} \times 3.14 \times 3^3 = 2 \times 3.14 \times 9 = 56.52\) cm\(^3\)

Total volume of the ice-cream = Volume of the filled cone + Volume of the hemisphere = \(94.2 + 56.52 = 150.72\) cm\(^3\)
Quick Tip: Volume of cone = \(\frac{1}{3}\pi r^2 h\). Volume of hemisphere = \(\frac{2}{3}\pi r^3\).

Theorem: Basic Proportionality Theorem (Thales's Theorem)


Given: In \(\triangle ABC\), a line \(DE\) is drawn parallel to side \(BC\), intersecting AB at D and AC at E.


To Prove: \(\frac{AD}{DB} = \frac{AE}{EC}\)


Construction:

Join B to E and C to D.
Draw \(DM \perp AC\) and \(EN \perp AB\).



Proof:
Consider \(\triangle ADE\) and \(\triangle BDE\). They share the same base DE and are between the same parallel lines DE and BC. Therefore, their areas are equal:
Area(\(\triangle ADE\)) = Area(\(\triangle BDE\))


Now, consider \(\triangle ADE\) and \(\triangle CDE\). They have the same base DE and are between the same parallel lines DE and BC. So,
Area(\(\triangle ADE\)) = Area(\(\triangle CDE\))


Therefore, Area(\(\triangle BDE\)) = Area(\(\triangle CDE\)).


The area of a triangle is given by \(\frac{1}{2} \times base \times height\).


Area(\(\triangle ADE\)) = \(\frac{1}{2} \times AD \times EN\)
Area(\(\triangle BDE\)) = \(\frac{1}{2} \times DB \times EN\)


Since Area(\(\triangle ADE\)) = Area(\(\triangle BDE\)), we have: \(\frac{1}{2} \times AD \times EN = \frac{1}{2} \times DB \times EN\)
\(\frac{AD}{DB} = 1\) (This seems to be an error in the source material or a very specific case. The general proof continues below.)



Generally, \(\frac{Area(\triangle ADE)}{Area(\triangle BDE)} = \frac{\frac{1}{2} AD \times EN}{\frac{1}{2} DB \times EN} = \frac{AD}{DB}\) \(\frac{Area(\triangle ADE)}{Area(\triangle CDE)} = \frac{\frac{1}{2} \times AE \times DM}{\frac{1}{2} \times EC \times DM} = \frac{AE}{EC}\)
Since Area(\(\triangle BDE\)) = Area(\(\triangle CDE\)), \(\frac{AD}{DB} = \frac{AE}{EC}\)
Hence Proved.
Quick Tip: The Basic Proportionality Theorem is a fundamental concept in geometry. Make sure you understand the proof and its applications.

Let the height of the first tower be \(h_1 = 24\) m and the height of the second tower be \(h_2\). Let the distance between the towers be \(d\).

From the foot of the second tower, the angle of elevation to the top of the first tower is \(30^\circ\). So, \(\tan 30^\circ = \frac{h_1}{d} \implies \frac{1}{\sqrt{3}} = \frac{24}{d} \implies d = 24\sqrt{3}\)


From the foot of the first tower, the angle of elevation to the top of the second tower is \(60^\circ\). So, \(\tan 60^\circ = \frac{h_2}{d} \implies \sqrt{3} = \frac{h_2}{24\sqrt{3}} \implies h_2 = 24\sqrt{3} \times \sqrt{3} = 72\) m


Distance between the towers, \(d = 24\sqrt{3} = 24(1.732) = 41.568\) m (approx.)
Height of the second tower, \(h_2 = 72\) m


Let \(l\) be the length of the wire connecting the tops of the towers. This forms a right triangle with the distance between the towers as one side and the difference in heights as the other side.
\(l = \sqrt{d^2 + (h_2 - h_1)^2} = \sqrt{(24\sqrt{3})^2 + (72 - 24)^2} = \sqrt{1728 + 2304} = \sqrt{4032} = 48\sqrt{2}\) m
Quick Tip: Draw a diagram to visualize the problem. Use trigonometric ratios (tangent) for angles of elevation and the Pythagorean theorem to find distances and heights.

Let O be the center of the balloon and let A be the eye of the observer. Let B be a point on the balloon such that AB is tangent to the balloon. Let h be the height of the center of the balloon from the ground, which is also the length of OA'. The radius of the balloon is \(r\), which is OB.

We are given that the angle subtended by the balloon at the observer's eye is \(60^\circ\), so \(\angle BAC = \frac{1}{2} \times 60^\circ = 30^\circ\), where C is the other point where AC touches the balloon.

The angle of elevation of the center of the balloon is \(45^\circ\), so \(\angle OAB = 45^\circ\).
We want to prove \(h = \sqrt{2}r\).




In right \(\triangle OAB\), \(\sin(\angle OAB) = \frac{OB}{OA}\), and \(\tan(\angle OAB) = \frac{OA'}{AB} = \frac{h}{OA'}\).


Since \(\angle OAB = 45^\circ\), \(\sin 45^\circ = \frac{r}{OA} = \frac{1}{\sqrt{2}} \implies OA = \sqrt{2}r\).

Also, \(\tan 45^\circ = \frac{h}{AB} = 1 \implies AB = OA'=h\).

Since OA\(=h\), \(OA=\sqrt{2}r\), thus \(h = r\sqrt{2}\).
Hence Proved.
Quick Tip: Draw a clear diagram. Use trigonometric ratios (sine, tangent) relating the angle of elevation, radius, and height of the balloon.

Radius of the circle, \(r = 14\) cm
Angle subtended by the chord at the center, \(\theta = 60^\circ\)

Area of the minor segment = Area of the sector - Area of the triangle
Area of the sector = \(\frac{\theta}{360^\circ} \pi r^2 = \frac{60^\circ}{360^\circ} \times \frac{22}{7} \times 14^2 = \frac{1}{6} \times \frac{22}{7} \times 196 = \frac{308}{3} = 102.67\) cm\(^2\) (approximately)

Area of the triangle formed by the chord and radii = \(\frac{1}{2}r^2 \sin\theta = \frac{1}{2} \times 14^2 \times \sin 60^\circ = \frac{1}{2} \times 196 \times \frac{\sqrt{3}}{2} = 49\sqrt{3} = 49(1.732) = 84.868\) cm\(^2\) (approximately)

Area of the minor segment = \(102.67 - 84.868 = 17.802\) cm\(^2\) (approximately)

Area of the circle = \(\pi r^2 = \frac{22}{7} \times 14^2 = 616\) cm\(^2\)

Area of the major segment = Area of the circle - Area of the minor segment = \(616 - 17.802 = 598.198\) cm\(^2\) (approximately)
Quick Tip: Area of minor segment = Area of sector - Area of triangle. Area of major segment = Area of circle - Area of minor segment.

Let the first term of the A.P. be \(a\) and the common difference be \(d\).
The \(n\)th term of an A.P. is given by \(a_n = a + (n-1)d\).

Given that \(a_{11} : a_{17} = 3 : 4\), \(\frac{a + 10d}{a + 16d} = \frac{3}{4}\) \(4(a + 10d) = 3(a + 16d)\) \(4a + 40d = 3a + 48d\) \(a = 8d\)

We need to find \(a_5 : a_{21}\): \(\frac{a_5}{a_{21}} = \frac{a + 4d}{a + 20d} = \frac{8d + 4d}{8d + 20d} = \frac{12d}{28d} = \frac{3}{7}\)


Sum of first \(n\) terms of an AP is \(S_n = \frac{n}{2}[2a+(n-1)d]\). \[S_5 = \frac{5}{2} [2a+4d] = \frac{5}{2} [16d+4d] = \frac{5}{2} \times 20d = 50d\] \[S_{21} = \frac{21}{2} [2a+20d] = \frac{21}{2} [16d+20d] = \frac{21}{2} \times 36d = 21 \times 18d = 378d\]
Then, \[\frac{S_5}{S_{21}} = \frac{50d}{378d} = \frac{25}{189}\]
Quick Tip: The \(n\)th term of an A.P. is \(a + (n-1)d\). The sum of the first \(n\) terms is \(\frac{n}{2}[2a + (n-1)d]\).

The number of logs in each row forms an arithmetic progression (A.P.) with the first term \(a = 22\) and common difference \(d = -1\). Let \(n\) be the number of rows. The sum of the logs in \(n\) rows is given by: \[ S_n = \frac{n}{2}[2a + (n-1)d] \]
We are given that \(S_n = 250\). Substituting the values of \(a\) and \(d\): \[ 250 = \frac{n}{2}[2(22) + (n-1)(-1)] \] \[ 500 = n[44 - n + 1] \] \[ 500 = n[45 - n] \] \[ 500 = 45n - n^2 \] \[ n^2 - 45n + 500 = 0 \]
We can solve this quadratic equation for \(n\) using factorization: \[ n^2 - 25n - 20n + 500 = 0 \] \[ n(n - 25) - 20(n - 25) = 0 \] \[ (n - 25)(n - 20) = 0 \]
So, \(n = 25\) or \(n = 20\).

If \(n = 20\), the number of logs in the top row is \(a + (n-1)d = 22 + (20-1)(-1) = 22 - 19 = 3\).
If \(n = 25\), the number of logs in the top row is \(22 + (25-1)(-1) = 22 - 24 = -2\), which is not possible.

Therefore, the number of rows is 20, and the number of logs in the top row is 3.
Quick Tip: When a sequence of numbers has a constant difference, it forms an arithmetic progression. The sum of an A.P. is given by \(S_n = \frac{n}{2}[2a + (n-1)d]\).

(I) Original length = 18 cm, Original width = 12 cm
Original area = \(18 \times 12 = 216\) cm\(^2\)
New length = \(18 + x\) cm, New width = \(12 + x\) cm
New area = \((18+x)(12+x)\) cm\(^2\)
The new area is double the original area: \((18+x)(12+x) = 2 \times 216 = 432\)

(II) Expanding the equation: \(216 + 18x + 12x + x^2 = 432\) \(x^2 + 30x + 216 = 432\) \(x^2 + 30x - 216 = 0\) (Standard form)

(III) Solving the quadratic equation using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a=1\), \(b=30\), and \(c=-216\). \[ x = \frac{-30 \pm \sqrt{30^2 - 4(1)(-216)}}{2(1)} = \frac{-30 \pm \sqrt{900 + 864}}{2} = \frac{-30 \pm \sqrt{1764}}{2} = \frac{-30 \pm 42}{2} \]
So, \(x = \frac{-30 + 42}{2} = \frac{12}{2} = 6\) or \(x = \frac{-30 - 42}{2} = \frac{-72}{2} = -36\). Since dimensions cannot be negative, \(x = 6\).
New length = \(18 + 6 = 24\) cm
New width = \(12 + 6 = 18\) cm
Quick Tip: Form a quadratic equation by relating the original and new areas. Use the quadratic formula to find the value of \(x\). Discard negative solutions as dimensions cannot be negative.

New area = \((18+x)(12+x) = 220\) \(216 + 30x + x^2 = 220\) \(x^2 + 30x - 4 = 0\)


Using the quadratic formula: \( x = \frac{-30 \pm \sqrt{30^2 - 4(1)(-4)}}{2(1)} = \frac{-30 \pm \sqrt{900 + 16}}{2} = \frac{-30 \pm \sqrt{916}}{2} = \frac{-30 \pm 2\sqrt{229}}{2} = -15 \pm \sqrt{229} \)


Since 229 is not a perfect square, \(\sqrt{229}\) is irrational. Therefore, \(x\) is irrational.

So, no rational value of \(x\) can make the new area equal to 220 cm\(^2\).
Quick Tip: The quadratic formula helps determine the nature of the roots. If the discriminant is not a perfect square, the roots are irrational.

(I) The modal class is the class interval with the highest frequency. In this case, the class interval 600-800 has the highest frequency (7). Therefore, the modal class is 600-800 mm.


(II) To find the median, we need the cumulative frequency:



Total number of subdivisions, \(N = 24\). \(\frac{N}{2} = \frac{24}{2} = 12\)
The cumulative frequency just greater than 12 is 13, which corresponds to the class interval 600-800.
So, the median class is 600-800.


Using the formula for median:
Median = \(l + \frac{\frac{N}{2} - cf}{f} \times h\)
where \(l\) is the lower limit of the median class, \(N\) is the total frequency, \(cf\) is the cumulative frequency of the class preceding the median class, \(f\) is the frequency of the median class, and \(h\) is the class width.


Here, \(l = 600\), \(N = 24\), \(cf = 6\), \(f = 7\), and \(h = 200\).

Median = \(600 + \frac{12 - 6}{7} \times 200 = 600 + \frac{6}{7} \times 200 = 600 + \frac{1200}{7} \approx 600 + 171.43 = 771.43\) mm.




(III) Subdivisions with at least 1000 mm rainfall are the ones in the intervals 1000-1200, 1200-1400, 1400-1600, and 1600-1800.
Number of such subdivisions = \(2 + 3 + 1 + 1 = 7\).
Quick Tip: The modal class is the class with the highest frequency. For the median, find the cumulative frequency and locate the median class. Then use the median formula.

To find the mean, we use the formula:
Mean = \(\frac{\sum f_i x_i}{\sum f_i}\)
where \(f_i\) is the frequency and \(x_i\) is the class mark (midpoint) of each class interval.

\begin{tabular{|c|c|c|c|
\hline
Rainfall (mm) & Number of Subdivisions (\(f_i\)) & Class Mark (\(x_i\)) & \(f_i x_i\)

\hline
200-400 & 2 & 300 & 600

400-600 & 4 & 500 & 2000

600-800 & 7 & 700 & 4900

800-1000 & 4 & 900 & 3600

1000-1200 & 2 & 1100 & 2200

1200-1400 & 3 & 1300 & 3900

1400-1600 & 1 & 1500 & 1500

1600-1800 & 1 & 1700 & 1700

\hline
Total & 24 & & 22400

\hline
\end{tabular

Mean = \(\frac{22400}{24} \approx 933.33\) mm.
Quick Tip: For mean calculation in grouped data, use the formula \(\frac{\sum f_i x_i}{\sum f_i}\), where \(x_i\) is the class mark.

(a) In \(\triangle OAB\), \(\angle ABO = 30^\circ\) and \(\angle OAB = 90^\circ\) (since AB is tangent to the circle at B).
We have \(OB = 75\) cm (radius).
Using trigonometric ratios, \(\tan(\angle ABO) = \frac{OA}{AB} \implies \tan 30^\circ = \frac{OA}{AB}\) \(\frac{1}{\sqrt{3}} = \frac{OA}{AB} \implies AB = OA\sqrt{3}\)

Also, \(\cos(\angle ABO) = \frac{OB}{AB} \implies \cos 30^\circ = \frac{75}{AB}\) \(\frac{\sqrt{3}}{2} = \frac{75}{AB} \implies AB = \frac{150}{\sqrt{3}} = 50\sqrt{3}\) cm.


(b) The length of OB is the radius of the circle, which is given as 75 cm.

(c) Since PQ is parallel to OA, \(\angle BPQ = \angle BOA = \theta\) (alternate interior angles), where \(\theta\) = \(\angle BOA\).
Also, \( \angle ABO = 30^{\circ}\). And \(\angle OAB = 90^{\circ}\). So \(\angle AOB = 60^{\circ}\).
In \(\triangle OAB\), \(OA = OB \tan(\angle ABO) = OB \tan(30^\circ) = 75 \times \frac{1}{\sqrt{3}} = 25\sqrt{3}\) cm.



Since \(\angle AOB + \angle ABO + \angle OAB = 180^{\circ}\), \(\angle AOB + 30^\circ + 90^\circ = 180^\circ\), or \(\angle AOB = 60^\circ\).

\(\angle BOA = 180^\circ - \angle ABO - \angle OAB = 180^\circ - 30^\circ - 90^\circ = 60^\circ\)


Since OB and OA are perpendicular to the tangents, and PQ || OA , APBO is a rectangle, thus \(AP = OB = 75\) cm.
Quick Tip: Use trigonometric ratios in the right-angled triangle formed by the radius, tangent, and the line connecting the external point to the center. Remember that the tangent to a circle is perpendicular to the radius at the point of contact.

Since PQ is parallel to OA, \(\angle BOA = \angle BPQ\) and OPBQ is a rectangle. Then \(PQ = OB = 75 cm\). Also, \(\angle AOB = 60^\circ\), so \(\angle BPQ = 60^\circ\).
Quick Tip: When parallel lines are involved, look for alternate interior angles and corresponding angles.