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CBSE CLASS X – MATHEMATICS (STANDARD) SAMPLE QUESTION PAPER
SET – 5 | ACADEMIC SESSION 2026–2027
General Instructions:
- This question paper contains 38 questions divided into 5 Sections: A, B, C, D and E.
- All Questions are compulsory.
- Section A comprises 20 Multiple Choice Questions (MCQs) of 1 mark each (Q1 to Q18 are MCQs and Q19 & Q20 are Assertion–Reason based).
- Section B comprises 5 Very Short Answer (VSA) questions of 2 marks each (Q21 to Q25).
- Section C comprises 6 Short Answer (SA) questions of 3 marks each (Q26 to Q31).
- Section D comprises 4 Long Answer (LA) questions of 5 marks each (Q32 to Q35).
- Section E comprises 3 Case-Based Integrated Units of Assessment of 4 marks each (Q36 to Q38) with sub-parts of values 1, 1 and 2 marks respectively. Internal choice is provided in the 2-mark question.
- Use of calculators is strictly prohibited. Use $\pi = 22/7$ wherever required unless stated otherwise.
SECTION A – MULTIPLE CHOICE QUESTIONS (1 Mark Each)
Q1.
If $\text{HCF}(26, 169) = 13$, then $\text{LCM}(26, 169)$ is:
[1]
Q2.
If one zero of the quadratic polynomial $2x^2 + px + 4$ is 2, then the other zero is:
[1]
Q3.
The value of $k$ for which the system of linear equations $kx + 3y = k – 3$ and $12x + ky = k$ has infinitely many solutions is:
[1]
Q4.
If the quadratic equation $x^2 – kx + 9 = 0$ has real and equal roots, then the values of $k$ are:
[1]
Q5.
The first four terms of an AP whose first term is $-2$ and common difference is $-2$ are:
[1]
Q6.
The perpendicular distance of the point $P(2, 3)$ from the x-axis is:
[1]
Q7.
If $\triangle ABC \sim \triangle DEF$, $\angle A = 47^\circ$, and $\angle E = 83^\circ$, then the measure of $\angle C$ is:
[1]
Q8.
If $5\tan\theta = 4$, then the value of $\frac{5\sin\theta – 3\cos\theta}{5\sin\theta + 2\cos\theta}$ is:
[1]
Q9.
The value of $\frac{2\tan 30^\circ}{1 + \tan^2 30^\circ}$ is equal to:
[1]
Q10.
A ladder $15\text{ m}$ long just reaches the top of a vertical wall. If the ladder makes an angle of $60^\circ$ with the wall, then the height of the wall is:
[1]
Q11.
In a circle with centre $O$, $PQ$ is a tangent to the circle at point $P$. If $\triangle OPQ$ is an isosceles triangle with $OP = PQ$, then the measure of $\angle OQP$ is:
[1]
Q12.
The ratio of the areas of the in-circle and the circum-circle of a square is:
[1]
Q13.
If the radius of a solid sphere is doubled, then its total surface area becomes:
[1]
Q14.
The algebraic sum of the deviations of a set of values from their arithmetic mean is:
[1]
Q15.
An event is considered highly unlikely to happen. Its probability of occurrence is closest to:
[1]
Q16.
The perimeter of a triangle with vertices at $(0, 4)$, $(0, 0)$, and $(3, 0)$ is:
[1]
Q17.
If two fair coins are tossed simultaneously, the probability of getting at most one head is:
[1]
Q18.
The decimal expansion of the rational number $\frac{17}{8}$ will terminate after:
[1]
Q19.
Assertion (A): 2 is the only even prime number.
Reason (R): All other even numbers greater than 2 have more than two factors. [1]
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true.
Reason (R): All other even numbers greater than 2 have more than two factors. [1]
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true.
Q20.
Assertion (A): The point $(0, 2)$ is the point of intersection of the y-axis and the line $x – 2y + 4 = 0$.
Reason (R): The distance of the point $(0, 2)$ from the origin is 2 units. [1]
Reason (R): The distance of the point $(0, 2)$ from the origin is 2 units. [1]
SECTION B – VERY SHORT ANSWER QUESTIONS (2 Marks Each)
Q21.
Prove that $7\sqrt{5}$ is an irrational number, given that $\sqrt{5}$ is irrational.
[2]
Q22.
In $\triangle ABC$, $D$ and $E$ are points on sides $AB$ and $AC$ respectively such that $AD = 2\text{ cm}$, $DB = 3\text{ cm}$, $AE = 3.6\text{ cm}$, and $EC = 5.4\text{ cm}$. Show that $DE \parallel BC$.
[2]
OR
In a right-angled triangle $ABC$, right-angled at $B$, $BD \perp AC$. If $AC = 13\text{ cm}$ and $AB = 5\text{ cm}$, prove that $\triangle ADB \sim \triangle ABC$ and find the length of $AD$.
Q23.
Find the coordinates of a point on the y-axis which is equidistant from the points $A(6, 5)$ and $B(-4, 3)$.
[2]
Q24.
If $\tan(A + B) = 1$ and $\tan(A – B) = \frac{1}{\sqrt{3}}$, where $0^\circ < A + B \le 90^\circ$ and $A > B$, find the values of $A$ and $B$.
[2]
OR
Prove that:
$$\frac{\sin\theta}{1 – \cos\theta} = \csc\theta + \cot\theta$$
Q25.
From an external point $X$, two tangents $XP$ and $XQ$ are drawn to a circle with centre $O$. $R$ is a point on the minor arc $PQ$, and a tangent drawn through $R$ intersects $XP$ at $A$ and $XQ$ at $B$. Prove that $XP = \frac{1}{2}(\text{Perimeter of } \triangle XAB)$.
[2]
SECTION C – SHORT ANSWER QUESTIONS (3 Marks Each)
Q26.
If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $f(x) = ax^2 + bx + c$ ($a \neq 0, c \neq 0$), form a quadratic polynomial whose zeroes are $\frac{1}{\alpha}$ and $\frac{1}{\beta}$.
[3]
Q27.
Solve the following pair of linear equations for $x$ and $y$:
$$\frac{x}{a} + \frac{y}{b} = 2$$
$$ax – by = a^2 – b^2$$
[3]
OR
A taxi charges a fixed base fare together with a variable charge for the distance covered. For a distance of $10\text{ km}$, the charge paid is ₹105, and for a journey of $15\text{ km}$, the charge paid is ₹155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of $25\text{ km}$?
Q28.
How many terms of the AP: $9, 17, 25, \dots$ must be taken to give a total sum of 636?
[3]
Q29.
Prove the trigonometric identity:
$$\frac{\cos\theta – \sin\theta + 1}{\cos\theta + \sin\theta – 1} = \csc\theta + \cot\theta$$
[3]
Q30.
In a circle of radius $5\text{ cm}$, a chord $PQ$ of length $8\text{ cm}$ is drawn. The tangents at $P$ and $Q$ intersect at an external point $T$. Find the length of the tangent $TP$.
[3]
(a) The length of the arc
(b) The area of the sector formed by the arc
(c) The area of the minor segment formed by the corresponding chord
OR
In a circle of radius $21\text{ cm}$, an arc subtends an angle of $60^\circ$ at the centre. Find:
(a) The length of the arc
(b) The area of the sector formed by the arc
(c) The area of the minor segment formed by the corresponding chord
Q31.
A game consists of tossing a one-rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result (three heads or three tails) and loses otherwise. Calculate:
(a) The probability that Hanif will lose the game.
(b) The probability of getting at least two heads. [3]
(a) The probability that Hanif will lose the game.
(b) The probability of getting at least two heads. [3]
SECTION D – LONG ANSWER QUESTIONS (5 Marks Each)
Q32.
A fast train takes 3 hours less than a slow train for a journey of $600\text{ km}$. If the speed of the slow train is $10\text{ km/h}$ less than that of the fast train, find the speeds of both trains.
[5]
OR
Solve the following equation for $x$:
$$\frac{x – 1}{x – 2} + \frac{x – 3}{x – 4} = 3\frac{1}{3}, \quad x \neq 2, 4$$
Q33.
State and prove the Converse of Basic Proportionality Theorem.
[5]
Q34.
From a solid right circular cylinder of height $2.4\text{ cm}$ and diameter $1.4\text{ cm}$, a conical cavity of the same height and same base diameter is hollowed out. Find the total surface area and the volume of the remaining solid to the nearest $\text{cm}^3$. (Take $\pi = 22/7$).
[5]
OR
Water in a canal, $6\text{ m}$ wide and $1.5\text{ m}$ deep, is flowing with a speed of $10\text{ km/h}$. How much area will it irrigate in 30 minutes, if $8\text{ cm}$ of standing water is required for irrigation?
Q35.
The median of the following frequency distribution is $28.5$. If the total frequency is $60$, find the missing frequencies $x$ and $y$:
$$\begin{array}{|c|c||c|c|} \hline \text{Class Interval} & \text{Frequency} & \text{Class Interval} & \text{Frequency} \\ \hline 0-10 & 5 & 30-40 & 15 \\ 10-20 & x & 40-50 & y \\ 20-30 & 20 & 50-60 & 5 \\ \hline \end{array}$$ [5]
$$\begin{array}{|c|c||c|c|} \hline \text{Class Interval} & \text{Frequency} & \text{Class Interval} & \text{Frequency} \\ \hline 0-10 & 5 & 30-40 & 15 \\ 10-20 & x & 40-50 & y \\ 20-30 & 20 & 50-60 & 5 \\ \hline \end{array}$$ [5]
SECTION E – CASE-BASED INTEGRATED UNITS (4 Marks Each)
Q36. Case Study 1 (Arithmetic Progression):
[4]
(a) Find the total number of rungs in the ladder. [1]
(b) Find the common difference $d$ in the length of successive rungs from bottom to top. [1]
(c) Calculate the total length of wood required to manufacture all the rungs. [2]
A heavy-duty wooden ladder has rungs that decrease uniformly in length from bottom to top. The bottom rung is $45\text{ cm}$ long, and the top rung is $25\text{ cm}$ long. The top and bottom rungs are separated by a vertical distance of $2\frac{1}{2}\text{ m}$ ($250\text{ cm}$), and the rungs are spaced equally at a distance of $25\text{ cm}$ apart from each other.
Based on the above information, answer the following questions:
(a) Find the total number of rungs in the ladder. [1]
(b) Find the common difference $d$ in the length of successive rungs from bottom to top. [1]
(c) Calculate the total length of wood required to manufacture all the rungs. [2]
OR
Find the length of the 6th rung from the bottom. [2]
Q37. Case Study 2 (Coordinate Geometry):
[4]
(b) Find the straight-line distance between marker $B$ and marker $C$. [1]
(c) Show that quadrilateral $ABCD$ forms a square. [2]
During a sports conditioning drill, four sprint markers are placed on a field at coordinates $A(1, 7)$, $B(4, 2)$, $C(-1, -1)$, and $D(-4, 4)$, where 1 grid unit represents 10 metres.
(a) Find the straight-line distance between marker $A$ and marker $B$. [1]
(b) Find the straight-line distance between marker $B$ and marker $C$. [1]
(c) Show that quadrilateral $ABCD$ forms a square. [2]
OR
Find the coordinates of the central checkpoint $M$ located at the intersection of the diagonals $AC$ and $BD$. [2]
Q38. Case Study 3 (Some Applications of Trigonometry):
[4]
(b) Write the trigonometric relation connecting the height of the tower and the distance of the car when the angle of elevation is $60^\circ$. [1]
(c) Find the total time taken by the vehicle to reach the base of the tower from the $60^\circ$ observation point. [2]
A highway leads directly to the base of a $50\text{ m}$ tall traffic monitoring tower. An observer at the top of the tower spots a patrol vehicle moving towards the tower at a uniform speed. The angle of depression of the vehicle is initially observed to be $30^\circ$. Exactly 6 seconds later, the angle of depression of the vehicle increases to $60^\circ$.
(a) Draw a neat mathematical labelled schematic diagram representing this situation. [1]
(b) Write the trigonometric relation connecting the height of the tower and the distance of the car when the angle of elevation is $60^\circ$. [1]
(c) Find the total time taken by the vehicle to reach the base of the tower from the $60^\circ$ observation point. [2]
OR
Calculate the exact horizontal distance of the vehicle from the base of the tower when the angle of depression is $60^\circ$. (Take $\sqrt{3} \approx 1.732$). [2]
CBSE CLASS X MATHEMATICS (STANDARD) – SOLUTIONS & MARKING SCHEME (SET – 5)
SECTION A SOLUTIONS
Q1. (c) 338 [1 Mark]
Explanation: $\text{LCM}(a, b) = \frac{a \times b}{\text{HCF}(a, b)} = \frac{26 \times 169}{13} = 2 \times 169 = 338$.
Explanation: $\text{LCM}(a, b) = \frac{a \times b}{\text{HCF}(a, b)} = \frac{26 \times 169}{13} = 2 \times 169 = 338$.
Q2. (b) 1 [1 Mark]
Explanation: Product of zeroes $= \frac{c}{a} = \frac{4}{2} = 2$. If one zero is $\alpha = 2$, then $\alpha\beta = 2 \implies 2\beta = 2 \implies \beta = 1$.
Explanation: Product of zeroes $= \frac{c}{a} = \frac{4}{2} = 2$. If one zero is $\alpha = 2$, then $\alpha\beta = 2 \implies 2\beta = 2 \implies \beta = 1$.
Q3. (a) 6 [1 Mark]
Explanation: For infinitely many solutions: $\frac{k}{12} = \frac{3}{k} = \frac{k – 3}{k}$. From $\frac{k}{12} = \frac{3}{k} \implies k^2 = 36 \implies k = \pm 6$. For $k = 6$, $\frac{3}{6} = \frac{6-3}{6} = \frac{1}{2}$ (satisfied).
Explanation: For infinitely many solutions: $\frac{k}{12} = \frac{3}{k} = \frac{k – 3}{k}$. From $\frac{k}{12} = \frac{3}{k} \implies k^2 = 36 \implies k = \pm 6$. For $k = 6$, $\frac{3}{6} = \frac{6-3}{6} = \frac{1}{2}$ (satisfied).
Q4. (b) $\pm 6$ [1 Mark]
Explanation: Real and equal roots $\implies D = b^2 – 4ac = 0 \implies (-k)^2 – 4(1)(9) = 0 \implies k^2 = 36 \implies k = \pm 6$.
Explanation: Real and equal roots $\implies D = b^2 – 4ac = 0 \implies (-k)^2 – 4(1)(9) = 0 \implies k^2 = 36 \implies k = \pm 6$.
Q5. (b) $-2, -4, -6, -8$ [1 Mark]
Explanation: $a_1 = -2$, $a_2 = -2 + (-2) = -4$, $a_3 = -4 + (-2) = -6$, $a_4 = -6 + (-2) = -8$.
Explanation: $a_1 = -2$, $a_2 = -2 + (-2) = -4$, $a_3 = -4 + (-2) = -6$, $a_4 = -6 + (-2) = -8$.
Q6. (b) 3 units [1 Mark]
Explanation: The distance of any point $(x, y)$ from the x-axis is equal to its absolute y-coordinate, $|y| = |3| = 3\text{ units}$.
Explanation: The distance of any point $(x, y)$ from the x-axis is equal to its absolute y-coordinate, $|y| = |3| = 3\text{ units}$.
Q7. (a) 50° [1 Mark]
Explanation: $\triangle ABC \sim \triangle DEF \implies \angle B = \angle E = 83^\circ$. In $\triangle ABC$, $\angle C = 180^\circ – (47^\circ + 83^\circ) = 180^\circ – 130^\circ = 50^\circ$.
Explanation: $\triangle ABC \sim \triangle DEF \implies \angle B = \angle E = 83^\circ$. In $\triangle ABC$, $\angle C = 180^\circ – (47^\circ + 83^\circ) = 180^\circ – 130^\circ = 50^\circ$.
Q8. (a) 1/6 [1 Mark]
Explanation: $\tan\theta = \frac{4}{5}$. Dividing numerator and denominator by $\cos\theta$: $\frac{5\tan\theta – 3}{5\tan\theta + 2} = \frac{5(4/5) – 3}{5(4/5) + 2} = \frac{4 – 3}{4 + 2} = \frac{1}{6}$.
Explanation: $\tan\theta = \frac{4}{5}$. Dividing numerator and denominator by $\cos\theta$: $\frac{5\tan\theta – 3}{5\tan\theta + 2} = \frac{5(4/5) – 3}{5(4/5) + 2} = \frac{4 – 3}{4 + 2} = \frac{1}{6}$.
Q9. (a) $\sin 60^\circ$ [1 Mark]
Explanation: $\frac{2(1/\sqrt{3})}{1 + (1/\sqrt{3})^2} = \frac{2/\sqrt{3}}{1 + 1/3} = \frac{2/\sqrt{3}}{4/3} = \frac{2}{\sqrt{3}} \times \frac{3}{4} = \frac{\sqrt{3}}{2} = \sin 60^\circ$.
Explanation: $\frac{2(1/\sqrt{3})}{1 + (1/\sqrt{3})^2} = \frac{2/\sqrt{3}}{1 + 1/3} = \frac{2/\sqrt{3}}{4/3} = \frac{2}{\sqrt{3}} \times \frac{3}{4} = \frac{\sqrt{3}}{2} = \sin 60^\circ$.
Q10. (c) $7.5\text{ m}$ [1 Mark]
Explanation: The angle with the vertical wall is $60^\circ$. $\cos 60^\circ = \frac{\text{Height}}{\text{Hypotenuse}} \implies \frac{1}{2} = \frac{h}{15} \implies h = 7.5\text{ m}$.
Explanation: The angle with the vertical wall is $60^\circ$. $\cos 60^\circ = \frac{\text{Height}}{\text{Hypotenuse}} \implies \frac{1}{2} = \frac{h}{15} \implies h = 7.5\text{ m}$.
Q11. (b) 45° [1 Mark]
Explanation: Radius $\perp$ tangent $\implies \angle OPQ = 90^\circ$. Since $OP = PQ$, $\triangle OPQ$ is an isosceles right triangle $\implies \angle OQP = \frac{180^\circ – 90^\circ}{2} = 45^\circ$.
Explanation: Radius $\perp$ tangent $\implies \angle OPQ = 90^\circ$. Since $OP = PQ$, $\triangle OPQ$ is an isosceles right triangle $\implies \angle OQP = \frac{180^\circ – 90^\circ}{2} = 45^\circ$.
Q12. (a) 1 : 2 [1 Mark]
Explanation: For a square of side $a$, in-radius $r = \frac{a}{2}$ and circum-radius $R = \frac{a}{\sqrt{2}}$. Ratio of areas $= \frac{\pi r^2}{\pi R^2} = \frac{(a/2)^2}{(a/\sqrt{2})^2} = \frac{a^2/4}{a^2/2} = \frac{1}{2}$.
Explanation: For a square of side $a$, in-radius $r = \frac{a}{2}$ and circum-radius $R = \frac{a}{\sqrt{2}}$. Ratio of areas $= \frac{\pi r^2}{\pi R^2} = \frac{(a/2)^2}{(a/\sqrt{2})^2} = \frac{a^2/4}{a^2/2} = \frac{1}{2}$.
Q13. (c) 4 times [1 Mark]
Explanation: Initial TSA $= 4\pi r^2$. New radius $r’ = 2r \implies \text{New TSA} = 4\pi (2r)^2 = 16\pi r^2 = 4(4\pi r^2)$.
Explanation: Initial TSA $= 4\pi r^2$. New radius $r’ = 2r \implies \text{New TSA} = 4\pi (2r)^2 = 16\pi r^2 = 4(4\pi r^2)$.
Q14. (a) 0 [1 Mark]
Explanation: By property of arithmetic mean, $\sum f_i (x_i – \bar{x}) = \sum f_i x_i – \bar{x}\sum f_i = N\bar{x} – N\bar{x} = 0$.
Explanation: By property of arithmetic mean, $\sum f_i (x_i – \bar{x}) = \sum f_i x_i – \bar{x}\sum f_i = N\bar{x} – N\bar{x} = 0$.
Q15. (a) 0.0001 [1 Mark]
Explanation: An event that is very unlikely to occur has a probability value approaching 0.
Explanation: An event that is very unlikely to occur has a probability value approaching 0.
Q16. (b) 12 units [1 Mark]
Explanation: $AB = 4$, $BC = 3$, $AC = \sqrt{4^2 + 3^2} = 5$. $\text{Perimeter} = 4 + 3 + 5 = 12\text{ units}$.
Explanation: $AB = 4$, $BC = 3$, $AC = \sqrt{4^2 + 3^2} = 5$. $\text{Perimeter} = 4 + 3 + 5 = 12\text{ units}$.
Q17. (c) 3/4 [1 Mark]
Explanation: Sample space $S = \{HH, HT, TH, TT\}$. At most 1 head $= \{HT, TH, TT\}$ (3 outcomes). $P = \frac{3}{4}$.
Explanation: Sample space $S = \{HH, HT, TH, TT\}$. At most 1 head $= \{HT, TH, TT\}$ (3 outcomes). $P = \frac{3}{4}$.
Q18. (c) Three decimal places [1 Mark]
Explanation: $\frac{17}{8} = \frac{17}{2^3} = \frac{17 \times 5^3}{10^3} = \frac{2125}{1000} = 2.125$ (terminates after 3 places).
Explanation: $\frac{17}{8} = \frac{17}{2^3} = \frac{17 \times 5^3}{10^3} = \frac{2125}{1000} = 2.125$ (terminates after 3 places).
Q19. (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of (A). [1 Mark]
Explanation: Any other even number is divisible by 1, 2, and itself, so it cannot be prime.
Explanation: Any other even number is divisible by 1, 2, and itself, so it cannot be prime.
Q20. (b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of (A). [1 Mark]
Explanation: Setting $x = 0$ in $x – 2y + 4 = 0$ gives $y = 2$, so $(0, 2)$ is the y-intercept. The distance from origin is $\sqrt{0^2 + 2^2} = 2$.
Explanation: Setting $x = 0$ in $x – 2y + 4 = 0$ gives $y = 2$, so $(0, 2)$ is the y-intercept. The distance from origin is $\sqrt{0^2 + 2^2} = 2$.
SECTION B SOLUTIONS
Q21. Proof of Irrationality: [2 Marks]
- Let $7\sqrt{5}$ be rational $= \frac{a}{b}$, where $a, b$ are integers, $b \neq 0$, and $\text{HCF}(a, b) = 1$. [0.5 Mark]
- $\sqrt{5} = \frac{a}{7b}$. [0.5 Mark]
- Since $a$ and $b$ are integers, $\frac{a}{7b}$ is a rational number, which implies $\sqrt{5}$ is rational. [0.5 Mark]
- This contradicts the given fact that $\sqrt{5}$ is irrational. Hence, $7\sqrt{5}$ is irrational. [0.5 Mark]
Q22. Converse of BPT: [2 Marks]
$\frac{AD}{DB} = \frac{2}{3}$ [0.5 Mark]
$\frac{AE}{EC} = \frac{3.6}{5.4} = \frac{36}{54} = \frac{2}{3}$ [0.5 Mark]
Since $\frac{AD}{DB} = \frac{AE}{EC}$, by the Converse of Basic Proportionality Theorem, $\mathbf{DE \parallel BC}$. [1 Mark]
OR
In $\triangle ADB$ and $\triangle ABC$: $\angle ADB = \angle ABC = 90^\circ$ and $\angle A = \angle A$ (Common). By AA similarity, $\mathbf{\triangle ADB \sim \triangle ABC}$. [1 Mark]
$\frac{AD}{AB} = \frac{AB}{AC} \implies AD = \frac{AB^2}{AC} = \frac{5^2}{13} = \mathbf{\frac{25}{13}\text{ cm}} \approx 1.92\text{ cm}$. [1 Mark]
$\frac{AD}{DB} = \frac{2}{3}$ [0.5 Mark]
$\frac{AE}{EC} = \frac{3.6}{5.4} = \frac{36}{54} = \frac{2}{3}$ [0.5 Mark]
Since $\frac{AD}{DB} = \frac{AE}{EC}$, by the Converse of Basic Proportionality Theorem, $\mathbf{DE \parallel BC}$. [1 Mark]
OR
In $\triangle ADB$ and $\triangle ABC$: $\angle ADB = \angle ABC = 90^\circ$ and $\angle A = \angle A$ (Common). By AA similarity, $\mathbf{\triangle ADB \sim \triangle ABC}$. [1 Mark]
$\frac{AD}{AB} = \frac{AB}{AC} \implies AD = \frac{AB^2}{AC} = \frac{5^2}{13} = \mathbf{\frac{25}{13}\text{ cm}} \approx 1.92\text{ cm}$. [1 Mark]
Q23. Equidistant Point on y-axis: [2 Marks]
Let point on y-axis be $P(0, y)$. $PA^2 = PB^2$
$(0 – 6)^2 + (y – 5)^2 = (0 – (-4))^2 + (y – 3)^2 \implies 36 + y^2 – 10y + 25 = 16 + y^2 – 6y + 9$ [1 Mark]
$-10y + 61 = -6y + 25 \implies 4y = 36 \implies y = 9$.
The required point is $\mathbf{(0, 9)}$. [1 Mark]
Let point on y-axis be $P(0, y)$. $PA^2 = PB^2$
$(0 – 6)^2 + (y – 5)^2 = (0 – (-4))^2 + (y – 3)^2 \implies 36 + y^2 – 10y + 25 = 16 + y^2 – 6y + 9$ [1 Mark]
$-10y + 61 = -6y + 25 \implies 4y = 36 \implies y = 9$.
The required point is $\mathbf{(0, 9)}$. [1 Mark]
Q24. Trigonometric Values: [2 Marks]
$A + B = 45^\circ$ …(1); $A – B = 30^\circ$ …(2) [1 Mark]
Adding (1) and (2): $2A = 75^\circ \implies \mathbf{A = 37.5^\circ}$ (or $37\frac{1}{2}^\circ$).
Subtracting (2) from (1): $2B = 15^\circ \implies \mathbf{B = 7.5^\circ}$ (or $7\frac{1}{2}^\circ$). [1 Mark]
OR
$\text{LHS} = \frac{\sin\theta(1 + \cos\theta)}{(1 – \cos\theta)(1 + \cos\theta)} = \frac{\sin\theta(1 + \cos\theta)}{1 – \cos^2\theta} = \frac{\sin\theta(1 + \cos\theta)}{\sin^2\theta} = \frac{1 + \cos\theta}{\sin\theta}$ [1 Mark]
$= \frac{1}{\sin\theta} + \frac{\cos\theta}{\sin\theta} = \mathbf{\csc\theta + \cot\theta} = \text{RHS}$. Hence Proved. [1 Mark]
$A + B = 45^\circ$ …(1); $A – B = 30^\circ$ …(2) [1 Mark]
Adding (1) and (2): $2A = 75^\circ \implies \mathbf{A = 37.5^\circ}$ (or $37\frac{1}{2}^\circ$).
Subtracting (2) from (1): $2B = 15^\circ \implies \mathbf{B = 7.5^\circ}$ (or $7\frac{1}{2}^\circ$). [1 Mark]
OR
$\text{LHS} = \frac{\sin\theta(1 + \cos\theta)}{(1 – \cos\theta)(1 + \cos\theta)} = \frac{\sin\theta(1 + \cos\theta)}{1 – \cos^2\theta} = \frac{\sin\theta(1 + \cos\theta)}{\sin^2\theta} = \frac{1 + \cos\theta}{\sin\theta}$ [1 Mark]
$= \frac{1}{\sin\theta} + \frac{\cos\theta}{\sin\theta} = \mathbf{\csc\theta + \cot\theta} = \text{RHS}$. Hence Proved. [1 Mark]
Q25. Tangents and Perimeter: [2 Marks]
Tangents from external points are equal: $AP = AR$ and $BQ = BR$, and $XP = XQ$. [0.5 Mark]
$\text{Perimeter of } \triangle XAB = XA + AB + XB = XA + (AR + RB) + XB$
$= (XA + AP) + (XB + BQ) = XP + XQ = XP + XP = 2XP$. [1 Mark]
$\implies \mathbf{XP = \frac{1}{2}(\text{Perimeter of }\triangle XAB)}$. Hence Proved. [0.5 Mark]
Tangents from external points are equal: $AP = AR$ and $BQ = BR$, and $XP = XQ$. [0.5 Mark]
$\text{Perimeter of } \triangle XAB = XA + AB + XB = XA + (AR + RB) + XB$
$= (XA + AP) + (XB + BQ) = XP + XQ = XP + XP = 2XP$. [1 Mark]
$\implies \mathbf{XP = \frac{1}{2}(\text{Perimeter of }\triangle XAB)}$. Hence Proved. [0.5 Mark]
SECTION C SOLUTIONS
Q26. Reciprocal Zeroes Polynomial: [3 Marks]
$\alpha + \beta = -\frac{b}{a}$, $\alpha\beta = \frac{c}{a}$. [0.5 Mark]
$\text{Sum of new zeroes} = \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha+\beta}{\alpha\beta} = \frac{-b/a}{c/a} = -\frac{b}{c}$. [1 Mark]
$\text{Product of new zeroes} = \frac{1}{\alpha\beta} = \frac{1}{c/a} = \frac{a}{c}$. [0.5 Mark]
Required polynomial is $k\left(x^2 – \left(-\frac{b}{c}\right)x + \frac{a}{c}\right) \implies \mathbf{c x^2 + b x + a}$. [1 Mark]
$\alpha + \beta = -\frac{b}{a}$, $\alpha\beta = \frac{c}{a}$. [0.5 Mark]
$\text{Sum of new zeroes} = \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha+\beta}{\alpha\beta} = \frac{-b/a}{c/a} = -\frac{b}{c}$. [1 Mark]
$\text{Product of new zeroes} = \frac{1}{\alpha\beta} = \frac{1}{c/a} = \frac{a}{c}$. [0.5 Mark]
Required polynomial is $k\left(x^2 – \left(-\frac{b}{c}\right)x + \frac{a}{c}\right) \implies \mathbf{c x^2 + b x + a}$. [1 Mark]
Q27. Algebraic Solution: [3 Marks]
$bx + ay = 2ab$ …(1); $ax – by = a^2 – b^2$ …(2)
Multiplying (1) by $b$ and (2) by $a$: $b^2x + aby = 2ab^2$ and $a^2x – aby = a^3 – ab^2$.
Adding: $(a^2 + b^2)x = a^3 + ab^2 = a(a^2 + b^2) \implies \mathbf{x = a}$. [1.5 Marks]
From (1): $b(a) + ay = 2ab \implies ay = ab \implies \mathbf{y = b}$. [1.5 Marks]
OR
Let fixed charge be ₹$x$ and rate per km be ₹$y$.
$x + 10y = 105$ …(1); $x + 15y = 155$ …(2)
Subtracting (1) from (2): $5y = 50 \implies \mathbf{y = ₹10/\text{km}}$.
From (1): $x + 100 = 105 \implies \mathbf{x = ₹5}$ (Fixed charge). [2 Marks]
For $25\text{ km}$: $\text{Cost} = x + 25y = 5 + 25(10) = \mathbf{₹255}$. [1 Mark]
$bx + ay = 2ab$ …(1); $ax – by = a^2 – b^2$ …(2)
Multiplying (1) by $b$ and (2) by $a$: $b^2x + aby = 2ab^2$ and $a^2x – aby = a^3 – ab^2$.
Adding: $(a^2 + b^2)x = a^3 + ab^2 = a(a^2 + b^2) \implies \mathbf{x = a}$. [1.5 Marks]
From (1): $b(a) + ay = 2ab \implies ay = ab \implies \mathbf{y = b}$. [1.5 Marks]
OR
Let fixed charge be ₹$x$ and rate per km be ₹$y$.
$x + 10y = 105$ …(1); $x + 15y = 155$ …(2)
Subtracting (1) from (2): $5y = 50 \implies \mathbf{y = ₹10/\text{km}}$.
From (1): $x + 100 = 105 \implies \mathbf{x = ₹5}$ (Fixed charge). [2 Marks]
For $25\text{ km}$: $\text{Cost} = x + 25y = 5 + 25(10) = \mathbf{₹255}$. [1 Mark]
Q28. Number of Terms in AP: [3 Marks]
$a = 9, d = 8, S_n = 636$.
$S_n = \frac{n}{2}[2a + (n – 1)d] \implies 636 = \frac{n}{2}[18 + (n – 1)8] = n[9 + 4(n – 1)] = n(4n + 5)$ [1 Mark]
$4n^2 + 5n – 636 = 0 \implies 4n^2 + 53n – 48n – 636 = 0$ [1 Mark]
$n(4n + 53) – 12(4n + 53) = 0 \implies (n – 12)(4n + 53) = 0 \implies \mathbf{n = 12}$. [1 Mark]
$a = 9, d = 8, S_n = 636$.
$S_n = \frac{n}{2}[2a + (n – 1)d] \implies 636 = \frac{n}{2}[18 + (n – 1)8] = n[9 + 4(n – 1)] = n(4n + 5)$ [1 Mark]
$4n^2 + 5n – 636 = 0 \implies 4n^2 + 53n – 48n – 636 = 0$ [1 Mark]
$n(4n + 53) – 12(4n + 53) = 0 \implies (n – 12)(4n + 53) = 0 \implies \mathbf{n = 12}$. [1 Mark]
Q29. Trigonometric Identity: [3 Marks]
Dividing numerator and denominator by $\sin\theta$:
$\text{LHS} = \frac{\cot\theta – 1 + \csc\theta}{\cot\theta + 1 – \csc\theta} = \frac{(\cot\theta + \csc\theta) – (\csc^2\theta – \cot^2\theta)}{\cot\theta – \csc\theta + 1}$ [1 Mark]
$= \frac{(\csc\theta + \cot\theta)[1 – (\csc\theta – \cot\theta)]}{\cot\theta – \csc\theta + 1}$ [1 Mark]
$= \frac{(\csc\theta + \cot\theta)(1 – \csc\theta + \cot\theta)}{(\cot\theta – \csc\theta + 1)} = \mathbf{\csc\theta + \cot\theta} = \text{RHS}$. Hence Proved. [1 Mark]
Dividing numerator and denominator by $\sin\theta$:
$\text{LHS} = \frac{\cot\theta – 1 + \csc\theta}{\cot\theta + 1 – \csc\theta} = \frac{(\cot\theta + \csc\theta) – (\csc^2\theta – \cot^2\theta)}{\cot\theta – \csc\theta + 1}$ [1 Mark]
$= \frac{(\csc\theta + \cot\theta)[1 – (\csc\theta – \cot\theta)]}{\cot\theta – \csc\theta + 1}$ [1 Mark]
$= \frac{(\csc\theta + \cot\theta)(1 – \csc\theta + \cot\theta)}{(\cot\theta – \csc\theta + 1)} = \mathbf{\csc\theta + \cot\theta} = \text{RHS}$. Hence Proved. [1 Mark]
Q30. Tangent Length $TP$: [3 Marks]
$OT$ bisects chord $PQ$ at $R$ perpendicularly: $PR = 4\text{ cm}$.
In right $\triangle OPR$: $OR = \sqrt{OP^2 – PR^2} = \sqrt{5^2 – 4^2} = 3\text{ cm}$. [1 Mark]
$\triangle TPR \sim \triangle TPO \implies \frac{TP}{OP} = \frac{PR}{OR} \implies \frac{TP}{5} = \frac{4}{3} \implies \mathbf{TP = \frac{20}{3}\text{ cm}} \approx 6.67\text{ cm}$. [2 Marks]
OR
$r = 21\text{ cm}, \theta = 60^\circ$.
(a) $\text{Arc length} = \frac{60}{360} \times 2 \times \frac{22}{7} \times 21 = \frac{1}{6} \times 132 = \mathbf{22\text{ cm}}$. [1 Mark]
(b) $\text{Area of sector} = \frac{60}{360} \times \frac{22}{7} \times 21^2 = \frac{1}{6} \times 1386 = \mathbf{231\text{ cm}^2}$. [1 Mark]
(c) $\text{Area of segment} = 231 – \frac{\sqrt{3}}{4}(21)^2 = \mathbf{\left(231 – \frac{441\sqrt{3}}{4}\right)\text{ cm}^2} \approx 39.75\text{ cm}^2$. [1 Mark]
$OT$ bisects chord $PQ$ at $R$ perpendicularly: $PR = 4\text{ cm}$.
In right $\triangle OPR$: $OR = \sqrt{OP^2 – PR^2} = \sqrt{5^2 – 4^2} = 3\text{ cm}$. [1 Mark]
$\triangle TPR \sim \triangle TPO \implies \frac{TP}{OP} = \frac{PR}{OR} \implies \frac{TP}{5} = \frac{4}{3} \implies \mathbf{TP = \frac{20}{3}\text{ cm}} \approx 6.67\text{ cm}$. [2 Marks]
OR
$r = 21\text{ cm}, \theta = 60^\circ$.
(a) $\text{Arc length} = \frac{60}{360} \times 2 \times \frac{22}{7} \times 21 = \frac{1}{6} \times 132 = \mathbf{22\text{ cm}}$. [1 Mark]
(b) $\text{Area of sector} = \frac{60}{360} \times \frac{22}{7} \times 21^2 = \frac{1}{6} \times 1386 = \mathbf{231\text{ cm}^2}$. [1 Mark]
(c) $\text{Area of segment} = 231 – \frac{\sqrt{3}}{4}(21)^2 = \mathbf{\left(231 – \frac{441\sqrt{3}}{4}\right)\text{ cm}^2} \approx 39.75\text{ cm}^2$. [1 Mark]
Q31. Coin Toss Probability: [3 Marks]
Total outcomes $= 2^3 = 8$: $\{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$.
Total outcomes $= 2^3 = 8$: $\{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$.
- (a) Winning outcomes $= \{HHH, TTT\}$ (2 outcomes). Losing outcomes $= 8 – 2 = 6$.
$P(\text{Losing}) = \frac{6}{8} = \mathbf{\frac{3}{4}}$. [1.5 Marks] - (b) At least two heads $= \{HHT, HTH, THH, HHH\}$ (4 outcomes).
$P(\text{At least 2 heads}) = \frac{4}{8} = \mathbf{\frac{1}{2}}$. [1.5 Marks]
SECTION D SOLUTIONS
Q32. Speeds of Trains: [5 Marks]
Let speed of fast train be $x\text{ km/h} \implies$ slow train $= (x – 10)\text{ km/h}$.
$\frac{600}{x – 10} – \frac{600}{x} = 3 \implies 600\left[\frac{x – (x – 10)}{x(x – 10)}\right] = 3 \implies \frac{6000}{x^2 – 10x} = 3$ [1.5 Marks]
$x^2 – 10x – 2000 = 0 \implies (x – 50)(x + 40) = 0 \implies x = 50\text{ km/h}$. [2 Marks]
Speed of Fast train $= \mathbf{50\text{ km/h}}$ and Slow train $= \mathbf{40\text{ km/h}}$. [1.5 Marks]
OR
$\frac{(x – 1)(x – 4) + (x – 3)(x – 2)}{(x – 2)(x – 4)} = \frac{10}{3} \implies \frac{x^2 – 5x + 4 + x^2 – 5x + 6}{x^2 – 6x + 8} = \frac{10}{3}$ [2 Marks]
$\frac{2x^2 – 10x + 10}{x^2 – 6x + 8} = \frac{10}{3} \implies \frac{x^2 – 5x + 5}{x^2 – 6x + 8} = \frac{5}{3}$
$3x^2 – 15x + 15 = 5x^2 – 30x + 40 \implies 2x^2 – 15x + 25 = 0$ [1.5 Marks]
$(2x – 5)(x – 5) = 0 \implies \mathbf{x = 5\text{ or } x = \frac{5}{2}}$. [1.5 Marks]
Let speed of fast train be $x\text{ km/h} \implies$ slow train $= (x – 10)\text{ km/h}$.
$\frac{600}{x – 10} – \frac{600}{x} = 3 \implies 600\left[\frac{x – (x – 10)}{x(x – 10)}\right] = 3 \implies \frac{6000}{x^2 – 10x} = 3$ [1.5 Marks]
$x^2 – 10x – 2000 = 0 \implies (x – 50)(x + 40) = 0 \implies x = 50\text{ km/h}$. [2 Marks]
Speed of Fast train $= \mathbf{50\text{ km/h}}$ and Slow train $= \mathbf{40\text{ km/h}}$. [1.5 Marks]
OR
$\frac{(x – 1)(x – 4) + (x – 3)(x – 2)}{(x – 2)(x – 4)} = \frac{10}{3} \implies \frac{x^2 – 5x + 4 + x^2 – 5x + 6}{x^2 – 6x + 8} = \frac{10}{3}$ [2 Marks]
$\frac{2x^2 – 10x + 10}{x^2 – 6x + 8} = \frac{10}{3} \implies \frac{x^2 – 5x + 5}{x^2 – 6x + 8} = \frac{5}{3}$
$3x^2 – 15x + 15 = 5x^2 – 30x + 40 \implies 2x^2 – 15x + 25 = 0$ [1.5 Marks]
$(2x – 5)(x – 5) = 0 \implies \mathbf{x = 5\text{ or } x = \frac{5}{2}}$. [1.5 Marks]
Q33. Converse of BPT: [5 Marks]
- Statement: If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side. [1 Mark]
- Given, To Prove & Construction: Triangle $\triangle ABC$ with line $DE$ intersecting $AB$ and $AC$ such that $\frac{AD}{DB} = \frac{AE}{EC}$. Assume $DE$ is not parallel to $BC$, and draw $DE’ \parallel BC$. [1.5 Marks]
- Proof:
By BPT in $\triangle ABC$ with $DE’ \parallel BC$: $\frac{AD}{DB} = \frac{AE’}{E’C}$ …(1) [1 Mark]
Given: $\frac{AD}{DB} = \frac{AE}{EC}$ …(2)
From (1) and (2): $\frac{AE’}{E’C} = \frac{AE}{EC} \implies \frac{AE’ + E’C}{E’C} = \frac{AE + EC}{EC} \implies \frac{AC}{E’C} = \frac{AC}{EC} \implies E’C = EC$. [1 Mark]
This is possible only if $E’$ and $E$ coincide. Hence, $\mathbf{DE \parallel BC}$. [0.5 Mark]
Q34. Cylinder with Conical Cavity: [5 Marks]
Radius $r = 0.7\text{ cm}$, Height $h = 2.4\text{ cm}$.
Slant height of cone $l = \sqrt{r^2 + h^2} = \sqrt{(0.7)^2 + (2.4)^2} = \sqrt{0.49 + 5.76} = \sqrt{6.25} = 2.5\text{ cm}$. [1 Mark]
$\text{Total Surface Area} = 2\pi rh + \pi r^2 + \pi rl = \pi r(2h + r + l) = \frac{22}{7} \times 0.7 \times [2(2.4) + 0.7 + 2.5]$
$= 2.2 \times [4.8 + 3.2] = 2.2 \times 8 = \mathbf{17.6\text{ cm}^2} \approx \mathbf{18\text{ cm}^2}$. [2 Marks]
$\text{Remaining Volume} = \pi r^2 h – \frac{1}{3}\pi r^2 h = \frac{2}{3}\pi r^2 h = \frac{2}{3} \times \frac{22}{7} \times (0.7)^2 \times 2.4 = \mathbf{2.46\text{ cm}^3}$. [2 Marks]
OR
Water speed $= 10\text{ km/h} = 10,000\text{ m/h}$. In $30\text{ min} = \frac{1}{2}\text{ h}$, length of water flow $L = 5000\text{ m}$.
$\text{Volume of water} = 6 \times 1.5 \times 5000 = 45,000\text{ m}^3$. [2.5 Marks]
$\text{Area} \times \text{Height} = \text{Volume} \implies \text{Area} \times \frac{8}{100}\text{ m} = 45,000\text{ m}^3$
$\text{Area} = \frac{45000 \times 100}{8} = \mathbf{562,500\text{ m}^2}$ (or $56.25\text{ hectares}$). [2.5 Marks]
Radius $r = 0.7\text{ cm}$, Height $h = 2.4\text{ cm}$.
Slant height of cone $l = \sqrt{r^2 + h^2} = \sqrt{(0.7)^2 + (2.4)^2} = \sqrt{0.49 + 5.76} = \sqrt{6.25} = 2.5\text{ cm}$. [1 Mark]
$\text{Total Surface Area} = 2\pi rh + \pi r^2 + \pi rl = \pi r(2h + r + l) = \frac{22}{7} \times 0.7 \times [2(2.4) + 0.7 + 2.5]$
$= 2.2 \times [4.8 + 3.2] = 2.2 \times 8 = \mathbf{17.6\text{ cm}^2} \approx \mathbf{18\text{ cm}^2}$. [2 Marks]
$\text{Remaining Volume} = \pi r^2 h – \frac{1}{3}\pi r^2 h = \frac{2}{3}\pi r^2 h = \frac{2}{3} \times \frac{22}{7} \times (0.7)^2 \times 2.4 = \mathbf{2.46\text{ cm}^3}$. [2 Marks]
OR
Water speed $= 10\text{ km/h} = 10,000\text{ m/h}$. In $30\text{ min} = \frac{1}{2}\text{ h}$, length of water flow $L = 5000\text{ m}$.
$\text{Volume of water} = 6 \times 1.5 \times 5000 = 45,000\text{ m}^3$. [2.5 Marks]
$\text{Area} \times \text{Height} = \text{Volume} \implies \text{Area} \times \frac{8}{100}\text{ m} = 45,000\text{ m}^3$
$\text{Area} = \frac{45000 \times 100}{8} = \mathbf{562,500\text{ m}^2}$ (or $56.25\text{ hectares}$). [2.5 Marks]
Q35. Missing Frequencies: [5 Marks]
Total frequency $N = 60 \implies 45 + x + y = 60 \implies x + y = 15$ …(1) [1.5 Marks]
Median $= 28.5 \implies$ Median class is $20 – 30$ ($l = 20, cf = 5 + x, f = 20, h = 10$). [1 Mark]
$\text{Median} = l + \left(\frac{N/2 – cf}{f}\right) \times h \implies 28.5 = 20 + \left(\frac{30 – (5 + x)}{20}\right) \times 10$ [1 Mark]
$8.5 = \frac{25 – x}{2} \implies 17 = 25 – x \implies \mathbf{x = 8}$. [1 Mark]
From (1): $8 + y = 15 \implies \mathbf{y = 7}$. [0.5 Mark]
Total frequency $N = 60 \implies 45 + x + y = 60 \implies x + y = 15$ …(1) [1.5 Marks]
Median $= 28.5 \implies$ Median class is $20 – 30$ ($l = 20, cf = 5 + x, f = 20, h = 10$). [1 Mark]
$\text{Median} = l + \left(\frac{N/2 – cf}{f}\right) \times h \implies 28.5 = 20 + \left(\frac{30 – (5 + x)}{20}\right) \times 10$ [1 Mark]
$8.5 = \frac{25 – x}{2} \implies 17 = 25 – x \implies \mathbf{x = 8}$. [1 Mark]
From (1): $8 + y = 15 \implies \mathbf{y = 7}$. [0.5 Mark]
SECTION E SOLUTIONS
Q36. Case Study 1: [4 Marks]
- (a) Total number of rungs $n = \frac{250\text{ cm}}{25\text{ cm}} + 1 = 10 + 1 = \mathbf{11\text{ rungs}}$. [1 Mark]
- (b) $a_{11} = a + 10d \implies 25 = 45 + 10d \implies 10d = -20 \implies \mathbf{d = -2\text{ cm}}$. [1 Mark]
- (c) $S_{11} = \frac{11}{2}(a + l) = \frac{11}{2}(45 + 25) = \frac{11}{2}(70) = 11 \times 35 = \mathbf{385\text{ cm}}$ (or $3.85\text{ m}$). [2 Marks]
Q37. Case Study 2: [4 Marks]
- (a) $AB = \sqrt{(4 – 1)^2 + (2 – 7)^2} = \sqrt{9 + 25} = \mathbf{\sqrt{34}\text{ units}}$ ($10\sqrt{34}\text{ m}$). [1 Mark]
- (b) $BC = \sqrt{(-1 – 4)^2 + (-1 – 2)^2} = \sqrt{25 + 9} = \mathbf{\sqrt{34}\text{ units}}$ ($10\sqrt{34}\text{ m}$). [1 Mark]
- (c) $CD = \sqrt{(-4 – (-1))^2 + (4 – (-1))^2} = \sqrt{9 + 25} = \sqrt{34}$; $DA = \sqrt{(1 – (-4))^2 + (7 – 4)^2} = \sqrt{25 + 9} = \sqrt{34}$.
Diagonals: $AC = \sqrt{(-1 – 1)^2 + (-1 – 7)^2} = \sqrt{4 + 64} = \sqrt{68}$; $BD = \sqrt{(-4 – 4)^2 + (4 – 2)^2} = \sqrt{64 + 4} = \sqrt{68}$.
Since all 4 sides and both diagonals are equal, $ABCD$ is a square. [2 Marks]
Q38. Case Study 3: [4 Marks]
- (a) Labelled diagram showing tower $AB = 50\text{ m}$, car positions $C$ ($30^\circ$) and $D$ ($60^\circ$). [1 Mark]
- (b) $\tan 60^\circ = \frac{AB}{BD} \implies \sqrt{3} = \frac{50}{BD} \implies BD = \frac{50}{\sqrt{3}}$. [1 Mark]
- (c) In $\triangle ABC$, $BC = \frac{50}{\tan 30^\circ} = 50\sqrt{3}$.
Distance $CD = 50\sqrt{3} – \frac{50}{\sqrt{3}} = \frac{100}{\sqrt{3}}$. Distance $BD = \frac{50}{\sqrt{3}} = \frac{1}{2}CD$.
Since speed is uniform, time taken to cover $BD = \frac{1}{2} \times 6\text{ s} = \mathbf{3\text{ seconds}}$. [2 Marks]