CBSE Class 10 Maths (Basic) Question Paper 2025 Solved Code 430-1-3

SECTION A

1. The ratio of the area of a quadrant of a circle to the area of the same circle is : [1 Mark]
(A) 1 : 2
(B) 2 : 1
(C) 1 : 4
(D) 4 : 1

Answer: (C) 1 : 4

Teacher's Note:
a) A quadrant is one-fourth of the circle, so its area is always \( \frac{1}{4} \) of the circle's area.
b) Do not confuse quadrant (quarter) with semicircle (half).

2. For which of the following solids is the lateral/curved surface area and total surface area the same ? [1 Mark]
(A) Cube
(B) Cuboid
(C) Hemisphere
(D) Sphere

Answer: (D) Sphere

Teacher's Note:
a) A sphere has no flat base, so its curved surface area itself is the total surface area.
b) A hemisphere has a flat circular base, so its TSA includes an extra term.

3. The class mark of the median class of the following data is : [1 Mark]
Class Interval: 10 - 25 | 25 - 40 | 40 - 55 | 55 - 70 | 70 - 85 | 85 - 100
Frequency: 2 | 3 | 7 | 6 | 6 | 6
(A) 40
(B) 55
(C) 47·5
(D) 62·5

Answer: (D) 62·5

Teacher's Note:
a) \( N = 30 \), so \( \frac{N}{2} = 15 \); cumulative frequency first exceeds 15 in class 55-70.
b) Class mark of 55-70 is \( \frac{55+70}{2} = 62.5 \).

4. The following distribution shows the number of runs scored by some batsmen in test matches :
Runs Scored: 3000 - 4000 | 4000 - 5000 | 5000 - 6000 | 6000 - 7000
Number of Batsmen: 5 | 10 | 9 | 8
The lower limit of the modal class is : [1 Mark]
(A) 3000
(B) 4000
(C) 5000
(D) 6000

Answer: (B) 4000

Teacher's Note:
a) The modal class has the highest frequency, here 10 in class 4000-5000.
b) The lower limit of that class is the required answer, 4000.

5. In an experiment of throwing a pair of dice, the probability of not getting a doublet is : [1 Mark]
(A) \( \frac{1}{6} \)
(B) \( \frac{5}{6} \)
(C) \( \frac{1}{5} \)
(D) \( \frac{1}{30} \)

Answer: (B) \( \frac{5}{6} \)

Teacher's Note:
a) There are 6 doublets out of 36 outcomes, so \( P(\text{doublet}) = \frac{6}{36} = \frac{1}{6} \).
b) \( P(\text{not doublet}) = 1 - \frac{1}{6} = \frac{5}{6} \).

6. If the HCF of two positive integers a and b is 1, then their LCM is : [1 Mark]
(A) a + b
(B) a
(C) b
(D) ab

Answer: (D) ab

Teacher's Note:
a) Product of two numbers = HCF \( \times \) LCM.
b) Since HCF = 1, LCM = ab directly.

7. \( \left(2+\sqrt{2}\right)^2 \) is : [1 Mark]
(A) a rational number
(B) an irrational number
(C) an integer
(D) a natural number

Answer: (B) an irrational number

Teacher's Note:
a) Expand: \( 4 + 4\sqrt{2} + 2 = 6 + 4\sqrt{2} \), which has an irrational term.
b) A rational number plus an irrational number is always irrational.

8. The discriminant of the quadratic equation \( 2x^2 - 3x - 5 = 0 \) is : [1 Mark]
(A) -31
(B) 49
(C) 7
(D) \( \sqrt{-31} \)

Answer: (B) 49

Teacher's Note:
a) Discriminant \( = b^2 - 4ac = (-3)^2 - 4(2)(-5) = 9+40 = 49 \).
b) Always substitute values carefully with correct signs.

9. The equation \( x + \frac{1}{x} = 3 \) (\( x \neq 0 \)) is expressed as a quadratic equation in the form of \( ax^2 + bx + c = 0 \). The value of \( a - b + c \) is : [1 Mark]
(A) 5
(B) 2
(C) 1
(D) -1

Answer: (A) 5

Teacher's Note:
a) Multiplying by x gives \( x^2 - 3x + 1 = 0 \), so \( a=1, b=-3, c=1 \).
b) \( a - b + c = 1-(-3)+1 = 5 \).

10. For a point X(a, b) where (b > a > 0), the value of its [distance from x-axis - distance from y-axis] is : [1 Mark]
(A) a - b
(B) b - a
(C) \( a^2 - b^2 \)
(D) \( b^2 - a^2 \)

Answer: (B) b - a

Teacher's Note:
a) Distance from x-axis = b, distance from y-axis = a.
b) Since \( b \gt a \), \( b - a \) is positive, matching the given condition.

11. The mid-point of a line segment divides the line segment in the ratio : [1 Mark]
(A) 1 : 2
(B) 2 : 1
(C) 1 : 1
(D) \( \frac{1}{2} \) : 2

Answer: (C) 1 : 1

Teacher's Note:
a) A mid-point divides the segment into two equal parts.
b) This is a direct definition-based question, no calculation needed.

12. Which of the following is not the criterion for similarity of triangles ? [1 Mark]
(A) AAA
(B) SSS
(C) SAS
(D) RHS

Answer: (D) RHS (the marking scheme states that none of the given options is the exact intended answer, and awards 1 mark to all candidates who attempted this question)

Teacher's Note:
a) AAA, SSS and SAS are standard similarity criteria for triangles.
b) RHS is generally a congruence criterion, not listed among the standard similarity criteria.

13. From the figures given below, which of the following is true about the measure of \( \angle P \) ? [1 Mark]
(A) \( \angle P = 60^{\circ} \)
(B) \( \angle P = 80^{\circ} \)
(C) \( \angle P = 40^{\circ} \)
(D) The measure of \( \angle P \) cannot be determined

[Figure: Triangle ABC with angle A = \( 80^{\circ} \), AB = 3.8 cm, AC = \( 3\sqrt{3} \) cm, angle B = \( 60^{\circ} \), BC = 6 cm. Triangle PQR with PR = \( 6\sqrt{3} \) cm, RQ = 7.6 cm, PQ = 12 cm.]

Answer: (C) \( \angle P = 40^{\circ} \)

Teacher's Note:
a) Check ratios: \( \frac{AC}{PR} = \frac{AB}{RQ} = \frac{BC}{PQ} = \frac{1}{2} \), so triangle ABC ~ triangle RQP.
b) Since angle C corresponds to angle P, \( \angle C = 180 - 80 - 60 = 40^{\circ} = \angle P \).

14. In the given figure, if AB is a tangent to the circle with centre O such that OB = 6 cm and \( \angle AOB = 60^{\circ} \), then the length of OA is : [1 Mark]
(A) 3 cm
(B) \( 3\sqrt{3} \) cm
(C) \( 4\sqrt{3} \) cm
(D) 12 cm

[Figure: Circle with centre O, radius OB = 6 cm, AB is a tangent at B, angle AOB = \( 60^{\circ} \) at centre O, A is an external point.]

Answer: (D) 12 cm

Teacher's Note:
a) Since AB is tangent, \( OB \perp AB \), so triangle OBA is right-angled at B.
b) \( \cos 60^{\circ} = \frac{OB}{OA} \Rightarrow OA = \frac{6}{\cos 60^{\circ}} = 12 \) cm.

15. Which of the following statements is false ? [1 Mark]
(A) \( \tan 45^{\circ} = \cot 45^{\circ} \)
(B) \( \sin 90^{\circ} = \tan 45^{\circ} \)
(C) \( \sin 30^{\circ} = \cos 30^{\circ} \)
(D) \( \sin 45^{\circ} = \cos 45^{\circ} \)

Answer: (C) \( \sin 30^{\circ} = \cos 30^{\circ} \)

Teacher's Note:
a) \( \sin 30^{\circ} = \frac{1}{2} \) but \( \cos 30^{\circ} = \frac{\sqrt{3}}{2} \), so they are not equal.
b) Remember the standard trigonometric ratio table for quick checking.

16. The value of \( \left(\frac{1}{\sec^2 A} + \frac{1}{\text{cosec}^2 A}\right) \) is : [1 Mark]
(A) more than 1
(B) 1
(C) 0
(D) -1

Answer: (B) 1

Teacher's Note:
a) \( \frac{1}{\sec^2 A} = \cos^2 A \) and \( \frac{1}{\text{cosec}^2 A} = \sin^2 A \).
b) \( \cos^2 A + \sin^2 A = 1 \) by the fundamental identity.

17. In the given figure, which of the following angles represents the angle of depression ? [1 Mark]
(A) x
(B) y
(C) z
(D) a

[Figure: Observer at top-left with a dashed horizontal line. Line of sight goes down to the Object at bottom-right. At the observer's point, angle z is between the horizontal line and the line of sight, and angle y is below z. At the bottom, angle a is a right angle and angle x is at the Object between the ground and the line of sight.]

Answer: (C) z

Teacher's Note:
a) The angle of depression is measured from the horizontal line at the observer's eye level down to the line of sight.
b) Do not confuse it with the angle of elevation, measured at the object's position.

18. The perimeter of the shaded region in the given figure is : [1 Mark]
(A) l
(B) l + a
(C) l + 2r
(D) l + 2r + a

[Figure: A circle with centre O and radius r. Two radii, each of length r, are drawn to two points on the circle, and the chord joining these points is labelled a. The minor arc between the two points is labelled l. The shaded (dotted) region is the sector bounded by the two radii and the arc l.]

Answer: (C) l + 2r

Teacher's Note:
a) The perimeter of a sector is the sum of the two radii and the arc length, not the chord.
b) The chord "a" is not part of the sector's boundary, so it is excluded.

19. Assertion (A) : For any two natural numbers a and b, the HCF of a and b is a factor of the LCM of a and b.
Reason (R) : HCF of any two natural numbers divides both the numbers. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

Answer: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

Teacher's Note:
a) Since LCM \( \times \) HCF = product of numbers, HCF always divides the LCM.
b) Both statements are individually true, and R directly explains A.

20. Assertion (A) : The value of p for which the system of equations \( 4x + py + 8 = 0 \) and \( 2x + 2y + 2 = 0 \) is consistent is 4.
Reason (R) : The system of equations \( a_1x + b_1y = c_1 \) and \( a_2x + b_2y = c_2 \) is consistent with infinitely many solutions, if \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \). [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

Answer: (D) Assertion (A) is false, but Reason (R) is true.

Teacher's Note:
a) Using \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \): \( \frac{4}{2} = \frac{p}{2} = \frac{8}{2} \) gives \( p = 4 \), but this makes the lines identical (infinitely many solutions), so the value of p makes them consistent for a different reason than stated numerically; correct value check shows Assertion's claim is false.
b) The Reason statement itself is a correct standard fact about consistency.

SECTION B

21. From a circular sheet of radius 70 cm, a quadrant is cut. Find the area of the remaining sheet. [2 Marks]

Answer:
1. Area of remaining sheet \( = \pi r^2 - \frac{1}{4}\pi r^2 = \frac{3}{4}\pi r^2 \).
2. \( = \frac{3}{4} \times \frac{22}{7} \times 70 \times 70 = 11550 \) sq. cm.

Teacher's Note:
a) Remaining area is always \( \frac{3}{4} \) of the full circle when a quadrant is removed.
b) Keep units in sq. cm throughout the calculation.

22. Solve for x and y :
3x + 5y = 8
5x - 3y = 2 [2 Marks]

Answer:
1. Multiply first equation by 3 and second by 5, then subtract to eliminate y: \( 9x+15y=24 \), \( 25x-15y=10 \), adding gives \( 34x=34 \Rightarrow x=1 \).
2. Substituting back: \( 3(1)+5y=8 \Rightarrow y=1 \). So \( x=1, y=1 \).

Teacher's Note:
a) Elimination method works well when coefficients can be matched by multiplication.
b) Always verify the solution by substituting into both original equations.

23. (a) In the given figure, if PQ || RS, then prove that \( \triangle POQ \sim \triangle SOR \). [2 Marks]

[Figure: Two line segments PQ and RS cross at point O, with P at top-left, Q at bottom-left, R at top-right, S at bottom-right, forming an X shape; arrows on PQ and RS indicate they are parallel.]

Answer:
1. Since PQ \( \parallel \) RS, \( \angle P = \angle S \) and \( \angle Q = \angle R \) (alternate interior angles).
2. Therefore \( \triangle POQ \sim \triangle SOR \) by AA similarity criterion.

Teacher's Note:
a) Vertically opposite angles at O are also equal, giving a third pair if needed.
b) AA similarity only needs two pairs of equal angles.

OR

(b) In the given figure, \( \triangle OSR \sim \triangle OQP \), \( \angle ROQ = 125^{\circ} \) and \( \angle ORS = 70^{\circ} \). Find the measures of \( \angle OSR \) and \( \angle OQP \). [2 Marks]

[Figure: Two parallel lines SR (top) and PQ (bottom) with transversal lines SP and RQ crossing at O; angle ORS = \( 70^{\circ} \) at R, angle ROQ = \( 125^{\circ} \) at O.]

Answer:
1. \( \angle OSR = 125^{\circ} - 70^{\circ} = 55^{\circ} \) (exterior angle property of triangle ORS).
2. Since \( \triangle OSR \sim \triangle OQP \), \( \angle OQP = \angle OSR = 55^{\circ} \) (corresponding angles of similar triangles).

Teacher's Note:
a) The exterior angle of a triangle equals the sum of the two opposite interior angles.
b) Corresponding angles of similar triangles are always equal.

24. Two concentric circles are of radii 6 cm and 10 cm. Find the length of the chord of the larger circle which touches the smaller circle. [2 Marks]

[Figure: Two concentric circles with centre O, radii 6 cm and 10 cm; a chord AB of the larger circle touches the smaller circle at point C, with OC perpendicular to AB.]

Answer:
1. \( BC^2 = OB^2 - OC^2 = 10^2 - 6^2 = 64 \Rightarrow BC = 8 \) cm.
2. Since OC bisects the chord, \( AB = 2 \times 8 = 16 \) cm.

Teacher's Note:
a) The perpendicular from the centre to a chord always bisects the chord.
b) Use the Pythagoras theorem in the right triangle OCB formed by the radius and the tangent point.

25. (a) Find the values of A and B (\( 0 \leq A \lt 90^{\circ} \), \( 0 \leq B \lt 90^{\circ} \)), if tan (A + B) = 1 and tan (A - B) = \( \frac{1}{\sqrt{3}} \). [2 Marks]

Answer:
1. \( \tan(A+B)=1 \Rightarrow A+B=45^{\circ} \); \( \tan(A-B)=\frac{1}{\sqrt{3}} \Rightarrow A-B=30^{\circ} \).
2. Solving: \( A = 37.5^{\circ} \), \( B = 7.5^{\circ} \).

Teacher's Note:
a) Convert the tangent values into standard angles using known tables.
b) Solve the two linear equations in A and B simultaneously.

OR

(b) Prove that tan \( 45^{\circ} \) = 1 geometrically. [2 Marks]

Answer:
1. Consider an isosceles right triangle ABC, right-angled at B, with \( \angle A = \angle C = 45^{\circ} \) by the angle sum property.
2. Since AB = BC = x (isosceles legs), \( \tan 45^{\circ} = \frac{AB}{BC} = \frac{x}{x} = 1 \).

Teacher's Note:
a) In an isosceles right triangle, the two legs opposite the equal angles are equal in length.
b) Draw the triangle clearly, labelling the right angle and the two 45° angles.

SECTION C

26. Prove the following trigonometric identity :
(sin A - cosec A) (cos A - sec A) = \( \frac{1}{\tan A + \cot A} \) [3 Marks]

Answer:
1. LHS \( = \left(\sin A - \frac{1}{\sin A}\right)\left(\cos A - \frac{1}{\cos A}\right) = \frac{\sin^2 A - 1}{\sin A} \times \frac{\cos^2 A - 1}{\cos A} \).
2. This simplifies to \( \sin A \cos A = \frac{\sin A \cos A}{\sin^2 A + \cos^2 A} \) (since \( \sin^2A+\cos^2A=1 \)).
3. This equals \( \frac{1}{\tan A + \cot A} \) = RHS.

Teacher's Note:
a) Use \( \sin^2A - 1 = -\cos^2A \) and \( \cos^2A - 1 = -\sin^2A \) to simplify the numerators.
b) Converting tan and cot in terms of sin and cos helps compare both sides easily.

27. A lot consists of 200 pens of which 180 are good and the rest are defective. A customer will buy a pen if it is not defective. The shopkeeper draws a pen at random and gives it to the customer. What is the probability that the customer will not buy it ? Another lot of 100 pens containing 80 good pens is mixed with the previous lot of 200 pens. The shopkeeper now draws one pen at random from the entire lot and gives it to the customer. What is the probability that the customer will buy the pen ? [3 Marks]

Answer:
1. Defective pens = 200 - 180 = 20, so P(customer will not buy) \( = \frac{20}{200} = \frac{1}{10} \).
2. After mixing, total pens = 300, good pens = 180 + 80 = 260.
3. P(customer will buy) \( = \frac{260}{300} = \frac{13}{15} \).

Teacher's Note:
a) The customer buys only non-defective pens, so identify the correct favourable outcomes for each part.
b) Always add total pens and total good pens carefully after mixing two lots.

28. (a) Prove that \( \sqrt{3} \) is an irrational number. [3 Marks]

Answer:
1. Assume \( \sqrt{3} \) is rational, so \( \sqrt{3} = \frac{p}{q} \), where p, q are co-prime integers, \( q \neq 0 \).
2. Then \( 3q^2 = p^2 \), so 3 divides \( p^2 \), hence 3 divides p. Let \( p = 3m \).
3. Substituting gives \( q^2 = 3m^2 \), so 3 divides q as well, contradicting that p and q are co-prime. Hence \( \sqrt{3} \) is irrational.

Teacher's Note:
a) This is a proof by contradiction; state the assumption clearly at the start.
b) The key step is showing 3 divides both p and q, which contradicts co-primality.

OR

(b) The factor tree of a number x is shown below :
Find the values of x, y, a and b. Hence, write the product of the prime factors of the number x so obtained. [3 Marks]

[Figure: Factor tree: x branches into 2 and y; y branches into 2 and 210; 210 branches into a and 70; 70 branches into 2 and 35; 35 branches into 5 and b.]

Answer:
1. From 35 = 5 \( \times \) b, b = 7.
2. From 70 = 2 \( \times \) 35, and 210 = a \( \times \) 70, so a = 3.
3. y = 2 \( \times \) 210 = 420, and x = 2 \( \times \) y = 840; so \( x = 2^3 \times 3 \times 5 \times 7 \).

Teacher's Note:
a) Work from the bottom of the tree upwards to find each missing value.
b) Express the final number as a product of prime factors in ascending order of primes.

29. Determine a quadratic polynomial, sum and product of whose zeroes are -10 and 24, respectively. Also, determine the zeroes of the polynomial so obtained. [3 Marks]

Answer:
1. Required polynomial: \( x^2 - (\text{sum})x + (\text{product}) = x^2 + 10x + 24 \).
2. Factorising: \( x^2 + 10x + 24 = (x+6)(x+4) \).
3. Zeroes are \( -6 \) and \( -4 \).

Teacher's Note:
a) Use the standard form \( x^2 - (\alpha+\beta)x + \alpha\beta \) to build the polynomial.
b) Factorise carefully, checking that the factors multiply to give the correct product term.

30. (a) Solve the following system of equations graphically :
x + 3y = 6; 2x - 3y = 12 [3 Marks]

Answer:
1. Plot points for x + 3y = 6: (0, 2) and (6, 0).
2. Plot points for 2x - 3y = 12: (0, -4) and (6, 0).
3. The two lines intersect at (6, 0), so the solution is x = 6, y = 0.

Teacher's Note:
a) Use at least two points per line, plotted accurately on graph paper.
b) The point of intersection of the two lines is the solution of the system.

OR

(b) x and y are complementary angles such that x : y = 1 : 2. Express the given information as a system of linear equations in two variables and hence solve it. [3 Marks]

Answer:
1. Since x and y are complementary, \( x + y = 90^{\circ} \).
2. Since \( x:y=1:2 \), \( 2x = y \).
3. Substituting: \( x + 2x = 90 \Rightarrow x = 30^{\circ} \), \( y = 60^{\circ} \).

Teacher's Note:
a) Complementary angles always add up to \( 90^{\circ} \).
b) Convert the given ratio into a linear equation before solving simultaneously.

31. Prove that a rectangle circumscribing a circle is a square. [3 Marks]

[Figure: A rectangle ABCD with an inscribed circle touching all four sides at points P, Q, R, S.]

Answer:
1. Since tangents from an external point to a circle are equal in length: AP = AS, BP = BQ, DR = DS, CR = CQ.
2. Adding these: AB + CD = BC + AD.
3. Since AB = CD and BC = AD (opposite sides of a rectangle), we get AB = AD, so all sides are equal, hence ABCD is a square.

Teacher's Note:
a) The tangent-length property from an external point is the key idea used here.
b) Showing all four sides are equal is enough to conclude the rectangle is a square.

SECTION D

32. A life insurance agent found the following data for the distribution of 100 policy holders on the basis of their ages.
Age (in years): 15 - 20 | 20 - 25 | 25 - 30 | 30 - 35 | 35 - 40 | 40 - 45 | 45 - 50 | 50 - 55 | 55 - 60
Number of policy holders: 2 | 4 | 18 | 21 | 33 | 11 | 3 | 6 | 2
Find the median age of the policy holders. [5 Marks]

Answer:
1. Cumulative frequencies: 2, 6, 24, 45, 78, 89, 92, 98, 100.
2. \( N/2 = 50 \), which lies in the class 35-40, so this is the median class; here \( l=35, cf=45, f=33, h=5 \).
3. Median \( = l + \frac{\frac{N}{2}-cf}{f} \times h = 35 + \frac{50-45}{33} \times 5 \).
4. Median \( = 35 + 0.76 = 35.76 \).
5. Thus, the median age of the policy holders is 35.76 years.

Teacher's Note:
a) Build the cumulative frequency column carefully before locating the median class.
b) Use the correct median formula with lower boundary, class width, and cumulative frequency of the class before the median class.

33. (a) The difference of the squares of two positive numbers is 180. The square of the smaller number is 8 times the greater number. Find the two numbers. [5 Marks]

Answer:
1. Let the greater number be x and the smaller number be y.
2. \( x^2 - y^2 = 180 \) and \( y^2 = 8x \).
3. Substituting: \( x^2 - 8x = 180 \Rightarrow x^2 - 8x - 180 = 0 \).
4. Factorising: \( (x-18)(x+10) = 0 \Rightarrow x = 18 \) (rejecting x = -10 as it is negative).
5. Then \( y^2 = 8 \times 18 = 144 \Rightarrow y = 12 \). The numbers are 18 and 12.

Teacher's Note:
a) Form a single-variable quadratic equation by substitution.
b) Reject negative or invalid roots since both numbers are stated to be positive.

OR

(b) Find the value(s) of k for which the equation \( 2x^2 + kx + 3 = 0 \) has real and equal roots. Hence, find the roots of the equations so obtained. [5 Marks]

Answer:
1. For equal roots, discriminant = 0: \( k^2 - 4(2)(3) = 0 \Rightarrow k^2 = 24 \Rightarrow k = \pm 2\sqrt{6} \).
2. For \( k = 2\sqrt{6} \): equation is \( 2x^2 + 2\sqrt{6}x + 3 = 0 \), roots are \( x = -\sqrt{\frac{3}{2}} \), \( -\sqrt{\frac{3}{2}} \).
3. For \( k = -2\sqrt{6} \): equation is \( 2x^2 - 2\sqrt{6}x + 3 = 0 \), roots are \( x = \sqrt{\frac{3}{2}} \), \( \sqrt{\frac{3}{2}} \).

Teacher's Note:
a) Real and equal roots always occur when the discriminant equals zero.
b) Both positive and negative values of k should be considered and their roots found separately.

34. State the converse of "Basic Proportionality Theorem" and use it to prove the following :
Line segment joining mid-points of any two sides of a triangle is parallel to the third side. [5 Marks]

Answer:
1. Converse of BPT: If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
2. Given: In \( \triangle ABC \), D and E are mid-points of AB and AC respectively. To prove: DE \( \parallel \) BC.
3. Since D is the mid-point of AB, AD = DB, so \( \frac{AD}{DB} = 1 \).
4. Since E is the mid-point of AC, AE = EC, so \( \frac{AE}{EC} = 1 \); hence \( \frac{AD}{DB} = \frac{AE}{EC} \).
5. By the converse of BPT, DE \( \parallel \) BC.

Teacher's Note:
a) State the converse theorem exactly as it is a bookwork statement carrying separate marks.
b) Showing equal ratios on both sides of the triangle is the key proof step.

35. (a) A toy is in the form of a cone surmounted on a hemisphere. The cone and hemisphere have the same radii. The height of the conical part of the toy is equal to the diameter of its base. If the radius of the conical part is 5 cm, find the volume of the toy. [5 Marks]

Answer:
1. Radius r = 5 cm; height of cone h = diameter = 10 cm.
2. Volume of toy = volume of hemisphere + volume of cone \( = \frac{2}{3}\pi r^3 + \frac{1}{3}\pi r^2 h \).
3. \( = \frac{2}{3} \times \frac{22}{7} \times 5 \times 5 \times 5 + \frac{1}{3} \times \frac{22}{7} \times 5 \times 5 \times 10 \).
4. \( = \frac{5500}{21} + \frac{5500}{21} = \frac{11000}{21} \) cu. cm.
5. Thus, the volume of the toy is \( \frac{11000}{21} \) cu. cm or approximately 523.81 cu. cm.

Teacher's Note:
a) The total volume is the sum of the two separate solid volumes, not their difference.
b) Use the same radius value in both formulas since cone and hemisphere share the same base.

OR

(b) A cubical block is surmounted by a hemisphere of radius 3·5 cm. What is the smallest possible length of the edge of the cube so that the hemisphere can totally lie on the cube ? Find the total surface area of the solid so formed. [5 Marks]

Answer:
1. The smallest edge of the cube must equal the diameter of the hemisphere: \( a = 3.5 \times 2 = 7 \) cm.
2. Total surface area of solid = surface area of cube + curved surface area of hemisphere - area of circle covered \( = 6a^2 + 2\pi r^2 - \pi r^2 = 6a^2 + \pi r^2 \).
3. \( = 6 \times 7 \times 7 + \frac{22}{7} \times 3.5 \times 3.5 = 294 + 38.5 = 332.5 \) sq. cm.

Teacher's Note:
a) The edge of the cube must equal the diameter, not the radius, of the hemisphere.
b) Subtract the base circle area once since it is covered by the hemisphere and not exposed.

SECTION E

Case Study - 1

36. In a society, there is a circular park having two gates. The gates are placed at points A(10, 20) and B(50, 50), as shown in the figure below. Two fountains are installed at points P and Q on AB such that AP = PQ = QB.

[Figure: A picture of a circular park with fountains and gates; a geometric diagram showing a circle with points A and B on the circumference, centre C, and points P and Q on line segment AB with AP = PQ = QB.]

(i) Find the coordinates of the centre C. [1 Mark]

Answer: C is the mid-point of AB, so \( C = \left(\frac{10+50}{2}, \frac{20+50}{2}\right) = (30, 35) \).

Teacher's Note:
a) AB is a diameter, so its mid-point gives the centre.
b) Use the mid-point formula directly for a quick answer.

(ii) Find the radius of the circular park. [1 Mark]

Answer: Radius \( = \sqrt{(30-10)^2 + (35-20)^2} = \sqrt{400+225} = \sqrt{625} = 25 \).

Teacher's Note:
a) The radius is the distance from the centre C to any point on the circle, such as A.
b) Use the distance formula between two coordinate points.

(iii) (a) Find the coordinates of the point P. [2 Marks]

Answer:
1. Since AP = PQ = QB, P divides AB in the ratio 1 : 2.
2. Using the section formula: \( P = \left(\frac{1(50)+2(10)}{3}, \frac{1(50)+2(20)}{3}\right) = \left(\frac{70}{3}, 30\right) \).

Teacher's Note:
a) Identify the correct ratio in which P divides AB before applying the section formula.
b) Keep the coordinates as fractions unless a decimal approximation is asked for.

OR

(b) Find the distance of the fountain at Q from gate A. [2 Marks]

Answer:
1. AB = diameter = 2 \( \times \) 25 = 50.
2. Since AQ = \( \frac{2}{3} \) of AB, \( AQ = \frac{2}{3} \times 50 = \frac{100}{3} \).

Teacher's Note:
a) Since AP = PQ = QB, AQ covers two of the three equal parts of AB.
b) Use the already found diameter value instead of recalculating with coordinates.

Case Study - 2

37. An injured bird was found on the roof of a building. The building is 15 m high. A fireman was called to rescue the bird. The fireman used an adjustable ladder to reach the roof. He placed the ladder in such a way that the ladder makes an angle of 60° with the ground in order to reach the roof.

[Figure: A picture of a fireman with a ladder on a building roof; a right triangle diagram showing the building height 15 m, the ladder as the hypotenuse, and an angle of 60° with the ground.]

(i) Find the length of the ladder used by the fireman to reach the roof. [1 Mark]

Answer: Using \( \sin 60^{\circ} = \frac{15}{a} \), \( a = \frac{30}{\sqrt{3}} = 10\sqrt{3} \) m.

Teacher's Note:
a) The building height is the side opposite to the given angle, so use sine.
b) Rationalise the denominator to simplify the final answer.

(ii) Find the distance of the point on the ground at which the ladder was fixed from the bottom of the building. [1 Mark]

Answer: Using \( \tan 60^{\circ} = \frac{15}{x} \), \( x = \frac{15}{\sqrt{3}} = 5\sqrt{3} \) m.

Teacher's Note:
a) The distance on the ground is the side adjacent to the given angle, so use tangent.
b) Keep the building height as the opposite side in this right triangle.

(iii) In order to avoid skidding, the fireman placed the ladder in such a way that the bottom of the ladder touches the base of the wall which is opposite to the building, making an angle of 30° with the ground.
(a) Draw a neat diagram to represent the above situation and hence find the width of the road between the building and the wall. [2 Marks]

[Figure: A right triangle with the building of height 15 m on one side, the ladder as the hypotenuse making an angle of 30° with the ground at the base of the opposite wall, and the road width as the base.]

Answer:
1. Using \( \tan 30^{\circ} = \frac{15}{y} \), \( y = 15\sqrt{3} \).
2. Thus, the width of the road is \( 15\sqrt{3} \) m.

Teacher's Note:
a) The diagram should clearly show the 15 m height and the 30° angle at the base of the opposite wall.
b) Use tangent since the height (opposite) and road width (adjacent) are both known/unknown sides.

OR

(b) Find the length of the ladder used by the fireman in this case. [2 Marks]

Answer:
1. Using \( \sin 30^{\circ} = \frac{15}{l} \), \( l = \frac{15}{0.5} = 30 \).
2. Thus, the length of the ladder is 30 m.

Teacher's Note:
a) The building height remains the side opposite to the given angle in this triangle too.
b) \( \sin 30^{\circ} = \frac{1}{2} \) is a standard value worth memorising.

Case Study - 3

38. In a garden, saplings of rose flowers were planted at equal intervals to form a spiral pattern. The spiral is made up of successive semicircles, with centres alternatively at A and B, starting with centre at A, of radii 50 cm, 100 cm, 150 cm, ....... as shown in the figure given below. Spiral 1 has 10 flowers, Spiral 2 has 20 flowers, Spiral 3 has 30 flowers and so on.

[Figure: A picture of a rose flower spiral pattern; a geometric diagram showing semicircles with alternating centres A and B, radii increasing as \( l_1, l_2, l_3, l_4 \), forming an outward spiral.]

(i) What is the radius of the 13^th spiral ? [1 Mark]

Answer: Since radii form an AP with first term 50 and common difference 50, \( a_{13} = 50 \times 13 = 650 \) cm.

Teacher's Note:
a) The radii of successive spirals form an arithmetic progression with a = 50, d = 50.
b) \( a_n = 50n \) is a quick shortcut here since a = d.

(ii) If the radius of the n^th spiral is 500 cm, find the value of n. [1 Mark]

Answer: \( 50 + (n-1)50 = 500 \Rightarrow n = 10 \).

Teacher's Note:
a) Use the general AP term formula \( a_n = a+(n-1)d \).
b) Solve the resulting linear equation for n carefully.

(iii) (a) Find the total number of saplings till the 11^th spiral. [2 Marks]

Answer:
1. Number of flowers per spiral forms an AP with a = 10, d = 10.
2. \( S_{11} = \frac{11}{2}[2(10) + (11-1)(10)] = \frac{11}{2}[20+100] = 660 \).

Teacher's Note:
a) Use the sum of AP formula \( S_n = \frac{n}{2}[2a+(n-1)d] \).
b) Note that this AP for flower counts is separate from the AP of radii.

OR

(b) Till which spiral, will there be a total of 450 saplings ? [2 Marks]

Answer:
1. \( 450 = \frac{n}{2}[2(10)+(n-1)(10)] \Rightarrow n^2+n-90=0 \).
2. Solving the quadratic: \( n = 9 \) (rejecting the negative root).

Teacher's Note:
a) Form a quadratic equation in n from the sum formula and solve by factorisation.
b) Reject the negative value of n since the number of spirals must be a positive integer.

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