ADEPT YOURSELF
www.adeptyourself.com β’ Academic Excellence Series
CBSE CLASS IX β MATHEMATICS (CODE NO. 041)
SAMPLE QUESTION PAPER β SET 2 | ACADEMIC SESSION 2026β2027
General Instructions:
- This question paper contains 38 questions divided into Five Sections: A, B, C, D and E.
- All questions are compulsory. However, an internal choice in 2 questions of 2 marks, 2 questions of 3 marks, and 2 questions of 5 marks has been provided. An internal choice is also provided in the 2 marks sub-question of Section E.
- Section A comprises 20 Multiple Choice Questions (Q1 to Q20) of 1 mark each. (Q1 to Q18 are MCQs, Q19 and Q20 are Assertion-Reason based).
- Section B comprises 5 Short Answer Type-I questions (Q21 to Q25) of 2 marks each.
- Section C comprises 6 Short Answer Type-II questions (Q26 to Q31) of 3 marks each.
- Section D comprises 4 Long Answer questions (Q32 to Q35) of 5 marks each.
- Section E comprises 3 Case-Based integrated units of assessment (Q36 to Q38) of 4 marks each with sub-parts.
- Use of calculators is not permitted. Take π = 22⁄7 wherever required unless stated otherwise.
SECTION A β MULTIPLE CHOICE QUESTIONS (20 Marks)
Q1.
Between any two rational numbers, there are:
[1]
Q2.
If p(x) = x2 − 2√2x + 1, then p(2√2) is equal to:
[1]
Q3.
The value of (249)2 − (248)2 is:
[1]
Q4.
If the coordinates of a point are (−2, 0), then this point lies:
[1]
Q5.
Any point on the line y = x is of the form:
[1]
Q6.
The number of straight lines that can pass through two distinct points is:
[1]
Q7.
Two complementary angles are in the ratio 4 : 5. The measure of the larger angle is:
[1]
Q8.
In ΔPQR, if ∠R > ∠Q, then:
[1]
Q9.
In a parallelogram ABCD, if ∠A = (3x − 20)° and ∠C = (x + 40)°, then the value of x is:
[1]
Q10.
In a circle, chords equidistant from the centre of the circle are:
[1]
Q11.
The perimeter of an equilateral triangle is 60 cm. Its area is:
[1]
Q12.
If the volume of a sphere is numerically equal to its surface area, then the radius of the sphere is:
[1]
Q13.
A cone and a cylinder have equal base radii and equal heights. The ratio of the volume of the cone to the volume of the cylinder is:
[1]
Q14.
In a frequency distribution, the mid-value of a class is 10 and the width of each class is 6. The lower limit of the class is:
[1]
Q15.
The value of (64)1/2 × (64)1/3 × (64)1/6 is:
[1]
Q16.
The point of intersection of the coordinate axes is called:
[1]
Q17.
ABCD is a cyclic quadrilateral such that ∠A = 70°. The measure of its opposite angle ∠C is:
[1]
Q18.
The total surface area of a solid hemisphere of radius r is:
[1]
Q19.
Assertion (A): If the side of a rhombus is 10 cm and one diagonal is 12 cm, then the length of the other diagonal is 16 cm.
Reason (R): The diagonals of a rhombus bisect each other perpendicularly at right angles. [1]
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Reason (R): The diagonals of a rhombus bisect each other perpendicularly at right angles. [1]
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Q20.
Assertion (A): (x − 2) is a factor of the polynomial p(x) = x3 − 4x.
Reason (R): By Factor Theorem, (x − a) is a factor of p(x) if p(a) = 0. [1]
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Reason (R): By Factor Theorem, (x − a) is a factor of p(x) if p(a) = 0. [1]
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
SECTION B β SHORT ANSWER QUESTIONS TYPE-I (10 Marks)
Q21.
Find two rational numbers and two irrational numbers between √2 and √3.
[2]
Q22.
Check whether the graph of the linear equation 3x + 4y = 12 passes through the points (0, 3) and (4, 0).
[2]
Q23.
In the given figure, lines PQ and RS intersect each other at point O. If ∠POR : ∠ROQ = 5 : 7, find all the four angles formed at O.
[2]
OR
If a transversal intersects two parallel lines, prove that the interior angles on the same side of the transversal are supplementary (sum = 180°).
Q24.
Prove that the diagonals of a rectangle are equal in length.
[2]
Q25.
The diameter of a spherical moon is approximately one-fourth (1⁄4) of the diameter of the earth. Find the ratio of their surface areas.
[2]
OR
Find the curved surface area and total surface area of a right circular cone of base radius 5 cm and height 12 cm.
SECTION C β SHORT ANSWER QUESTIONS TYPE-II (18 Marks)
Q26.
Find the values of a and b if:
√3 − 1√3 + 1 = a + b√3
[3]
Q27.
Evaluate the following products without multiplying directly, using standard algebraic identities:
(i) 103 × 107
(ii) 95 × 96
(iii) (104)3 [3]
(i) 103 × 107
(ii) 95 × 96
(iii) (104)3 [3]
Q28.
Find the coordinates of the vertices of a rectangle placed in the first quadrant whose length is 6 units along the x-axis and breadth is 4 units along the y-axis, with one vertex at the origin. Also, find its area.
[3]
Q29.
In ΔABC, the bisector AD of ∠A is perpendicular to side BC. Show that AB = AC and hence ΔABC is an isosceles triangle.
[3]
(i) ΔAOB ≅ ΔDOC
(ii) O is also the mid-point of BC.
OR
Line segment AB is parallel to another line segment CD. O is the mid-point of AD. Show that:
(i) ΔAOB ≅ ΔDOC
(ii) O is also the mid-point of BC.
Q30.
Prove that the perpendicular from the centre of a circle to a chord bisects the chord.
[3]
Q31.
The length of 40 leaves of a plant are measured correct to one millimetre, and the obtained data is represented in the following table:
(i) Convert the discontinuous class intervals into continuous class intervals.
(ii) Draw a histogram to represent the continuous data. [3]
Draw a histogram to represent the data when class intervals are of unequal widths.
| Length (in mm) | 118β126 | 127β135 | 136β144 | 145β153 | 154β162 | 163β171 |
|---|---|---|---|---|---|---|
| Number of Leaves | 3 | 5 | 9 | 12 | 5 | 6 |
(ii) Draw a histogram to represent the continuous data. [3]
OR
A random survey of the number of children of various age groups playing in a park was found as follows:| Age (in years) | 1β2 | 2β3 | 3β5 | 5β7 | 7β10 | 10β15 |
|---|---|---|---|---|---|---|
| Number of Children | 5 | 3 | 6 | 12 | 9 | 10 |
SECTION D β LONG ANSWER QUESTIONS (20 Marks)
Q32.
If x + y + z = 0, prove that:
x3 + y3 + z3 = 3xyz
Hence, without actually calculating the cubes, find the value of:
(28)3 + (−15)3 + (−13)3
[5]OR
Using Factor Theorem, show that (x + 2) is a factor of p(x) = x3 + 13x2 + 32x + 20. Hence, factorise p(x) completely into linear factors.
Q33.
Prove that the sum of either pair of opposite angles of a cyclic quadrilateral is 180°.
In a cyclic quadrilateral ABCD, if ∠B = (5x − 10)° and ∠D = (3x + 30)°, find the measures of ∠B and ∠D. [5]
In a cyclic quadrilateral ABCD, if ∠B = (5x − 10)° and ∠D = (3x + 30)°, find the measures of ∠B and ∠D. [5]
Q34.
Prove that a diagonal of a parallelogram divides it into two congruent triangles.
Show that the bisectors of angles of a parallelogram form a rectangle. [5]
Show that the bisectors of angles of a parallelogram form a rectangle. [5]
OR
ABCD is a rhombus and P, Q, R, and S are the mid-points of the sides AB, BC, CD, and DA respectively. Prove that the quadrilateral PQRS is a rectangle.
Q35.
A conical tent of radius 7 m and height 24 m is to be made from canvas cloth.
(i) Find the slant height of the tent.
(ii) Find the area of the canvas cloth required to make the tent.
(iii) If the canvas cloth is 2 m wide, find the length of the cloth used.
(iv) Calculate the volume of air contained inside the tent. [5]
(i) Find the slant height of the tent.
(ii) Find the area of the canvas cloth required to make the tent.
(iii) If the canvas cloth is 2 m wide, find the length of the cloth used.
(iv) Calculate the volume of air contained inside the tent. [5]
SECTION E β CASE-BASED INTEGRATED UNITS (12 Marks)
Q36. Read the following text and answer the questions that follow:
[4](a) Find the coordinates of Terminal Stations A and B. [1]
(b) Does the Metro Red Line pass through the origin O(0, 0)? Justify your answer. [1]
(c) Find the coordinates of the junction station P where the Blue Line and Red Line intersect. [2]
In an urban smart-city plan, two major transit railway lines cross each other at the central control hub O(0, 0).
β’ The Metro Blue Line is represented by the linear equation x + y = 6.
β’ The Metro Red Line is represented by the linear equation 2x − y = 0.
β’ Terminal Station A is located where the Blue Line crosses the x-axis.
β’ Terminal Station B is located where the Blue Line crosses the y-axis.
(b) Does the Metro Red Line pass through the origin O(0, 0)? Justify your answer. [1]
(c) Find the coordinates of the junction station P where the Blue Line and Red Line intersect. [2]
OR
Find the area of the triangular region ΔOAB formed by the Blue Line and the coordinate axes. [2]
Q37. Read the following text and answer the questions that follow:
[4](a) What is the side length a of the equilateral triangle? [1]
(b) Find the semi-perimeter s of the signboard. [1]
(c) Using Heron’s formula, calculate the area of the traffic signboard. [2]
A traffic awareness committee designed an equilateral triangular traffic signboard with the message “SCHOOL AHEAD”. The perimeter of the triangular signboard is 180 cm.
(b) Find the semi-perimeter s of the signboard. [1]
(c) Using Heron’s formula, calculate the area of the traffic signboard. [2]
OR
If the cost of painting the signboard is βΉ0.50 per cm2, find the total cost of painting both sides of the signboard. [Take √3 = 1.732] [2]
Q38. Read the following text and answer the questions that follow:
[4](a) Write the mathematical formula for the volume of a sphere of radius R. [1]
(b) Find the radius of the original sphere and the radius of each small sphere. [1]
(c) Calculate the total number of small spherical lead shots obtained from the melted sphere. [2]
A solid lead sphere of radius 6 cm is melted and recast into small identical spherical lead shots, each of radius 0.6 cm, with no loss of material during the process.
(b) Find the radius of the original sphere and the radius of each small sphere. [1]
(c) Calculate the total number of small spherical lead shots obtained from the melted sphere. [2]
OR
Find the ratio of the surface area of the original large sphere to the total surface area of all the small spherical lead shots combined. [2]
CBSE CLASS IX MATHEMATICS (041) β SOLUTIONS & MARKING SCHEME β SET 2
SECTION A β ANSWERS
Q1. (b) Infinitely many rational numbers [Dense property of rational numbers] [1]
Q2. (b) 1 [p(2√2) = (2√2)2 − 2√2(2√2) + 1 = 8 − 8 + 1 = 1] [1]
Q3. (d) 497 [a2 − b2 = (a + b)(a − b) = (249 + 248)(249 − 248) = 497 × 1 = 497] [1]
Q4. (c) On the negative x-axis [Since y-coordinate is 0 and x < 0] [1]
Q5. (a) (a, a) [Since x and y coordinates are equal] [1]
Q6. (b) Only one [Euclid’s Postulate 1: A unique line passes through two points] [1]
Q7. (c) 50° [4x + 5x = 90° ⇒ 9x = 90° ⇒ x = 10°; Larger angle = 5(10°) = 50°] [1]
Q8. (b) PQ > PR [Side opposite to greater angle is longer] [1]
Q9. (a) 30° [Opposite angles of parallelogram are equal ⇒ 3x − 20 = x + 40 ⇒ 2x = 60 ⇒ x = 30°] [1]
Q10. (b) Equal in length [Chords equidistant from centre are equal] [1]
Q11. (a) 100√3 cm2 [Side a = 60⁄3 = 20 cm; Area = √34(20)2 = 100√3 cm2] [1]
Q12. (c) 3 units [4⁄3 π r3 = 4πr2 ⇒ r = 3] [1]
Q13. (a) 1 : 3 [(1⁄3 π r2 h)⁄(π r2 h) = 1⁄3] [1]
Q14. (b) 7 [Lower limit = Mid-value − Width⁄2 = 10 − 3 = 7] [1]
Q15. (c) 64 [(64)(1/2 + 1/3 + 1/6) = (64)(3+2+1)/6 = (64)1 = 64] [1]
Q16. (c) Origin [1]
Q17. (b) 110° [Opposite angles of cyclic quadrilateral are supplementary: ∠C = 180° − 70° = 110°] [1]
Q18. (b) 3πr2 [Curved area (2πr2) + Base circular area (πr2)] [1]
Q19. (a) Both A and R are true and R is the correct explanation of A. [Side = √(d1/2)2 + (d2/2)2 ⇒ 10 = √62 + 82 = 10; d2 = 16 cm] [1]
Q20. (a) Both A and R are true and R is the correct explanation of A. [p(2) = (2)3 − 4(2) = 8 − 8 = 0] [1]
SECTION B β ANSWERS
Q21.
√2 ≈ 1.4142… and √3 ≈ 1.7320…
β’ Two rational numbers between them: 1.5 (or 3⁄2) and 1.6 (or 8⁄5).
β’ Two irrational numbers between them: 1.5010010001… and 1.6010010001… [2]
β’ Two rational numbers between them: 1.5 (or 3⁄2) and 1.6 (or 8⁄5).
β’ Two irrational numbers between them: 1.5010010001… and 1.6010010001… [2]
Q22.
Given equation: 3x + 4y = 12.
β’ For point (0, 3): LHS = 3(0) + 4(3) = 12 = RHS. (Passes through (0, 3)).
β’ For point (4, 0): LHS = 3(4) + 4(0) = 12 = RHS. (Passes through (4, 0)).
Yes, the graph passes through both given points. [2]
β’ For point (0, 3): LHS = 3(0) + 4(3) = 12 = RHS. (Passes through (0, 3)).
β’ For point (4, 0): LHS = 3(4) + 4(0) = 12 = RHS. (Passes through (4, 0)).
Yes, the graph passes through both given points. [2]
Q23.
∠POR and ∠ROQ form a linear pair ⇒ ∠POR + ∠ROQ = 180°.
Let ∠POR = 5x and ∠ROQ = 7x ⇒ 12x = 180° ⇒ x = 15°.
β’ ∠POR = 5(15°) = 75°
β’ ∠ROQ = 7(15°) = 105°
β’ ∠SOQ = ∠POR = 75° (Vertically opposite angles)
β’ ∠POS = ∠ROQ = 105° (Vertically opposite angles). [2]
OR Let parallel lines be AB ∥ CD intersected by transversal line EF.
∠1 + ∠2 = 180° (Linear pair). Since alternate interior angles are equal (∠1 = ∠3):
∠3 + ∠2 = 180°. Hence, consecutive interior angles are supplementary. [2]
Let ∠POR = 5x and ∠ROQ = 7x ⇒ 12x = 180° ⇒ x = 15°.
β’ ∠POR = 5(15°) = 75°
β’ ∠ROQ = 7(15°) = 105°
β’ ∠SOQ = ∠POR = 75° (Vertically opposite angles)
β’ ∠POS = ∠ROQ = 105° (Vertically opposite angles). [2]
OR Let parallel lines be AB ∥ CD intersected by transversal line EF.
∠1 + ∠2 = 180° (Linear pair). Since alternate interior angles are equal (∠1 = ∠3):
∠3 + ∠2 = 180°. Hence, consecutive interior angles are supplementary. [2]
Q24.
Let ABCD be a rectangle.
In ΔABC and ΔBAD:
1. BC = AD (Opposite sides of rectangle)
2. ∠ABC = ∠BAD = 90°
3. AB = BA (Common side)
⇒ ΔABC ≅ ΔBAD (By SAS congruence rule).
⇒ AC = BD (By CPCT). Diagonals of a rectangle are equal. [2]
In ΔABC and ΔBAD:
1. BC = AD (Opposite sides of rectangle)
2. ∠ABC = ∠BAD = 90°
3. AB = BA (Common side)
⇒ ΔABC ≅ ΔBAD (By SAS congruence rule).
⇒ AC = BD (By CPCT). Diagonals of a rectangle are equal. [2]
Q25.
Let diameter of earth = D ⇒ Radius of earth R = D⁄2.
Diameter of moon = D⁄4 ⇒ Radius of moon r = D⁄8.
Ratio of surface areas = 4πr2⁄4πR2 = (r⁄R)2 = ((D/8)⁄(D/2))2 = (2⁄8)2 = (1⁄4)2 = 1 : 16. [2]
OR Radius r = 5 cm, Height h = 12 cm.
Slant height l = √r2 + h2 = √52 + 122 = √169 = 13 cm.
β’ CSA = πrl = 22⁄7 × 5 × 13 = 1430⁄7 cm2 ≈ 204.28 cm2.
β’ TSA = πr(l + r) = 22⁄7 × 5 × (13 + 5) = 22⁄7 × 5 × 18 = 1980⁄7 cm2 ≈ 282.86 cm2. [2]
Diameter of moon = D⁄4 ⇒ Radius of moon r = D⁄8.
Ratio of surface areas = 4πr2⁄4πR2 = (r⁄R)2 = ((D/8)⁄(D/2))2 = (2⁄8)2 = (1⁄4)2 = 1 : 16. [2]
OR Radius r = 5 cm, Height h = 12 cm.
Slant height l = √r2 + h2 = √52 + 122 = √169 = 13 cm.
β’ CSA = πrl = 22⁄7 × 5 × 13 = 1430⁄7 cm2 ≈ 204.28 cm2.
β’ TSA = πr(l + r) = 22⁄7 × 5 × (13 + 5) = 22⁄7 × 5 × 18 = 1980⁄7 cm2 ≈ 282.86 cm2. [2]
SECTION C β ANSWERS
Q26.
LHS = √3 − 1√3 + 1 × √3 − 1√3 − 1
= (√3 − 1)2(√3)2 − (1)2
= 3 + 1 − 2√33 − 1
= 4 − 2√32 = 2 − √3.
Comparing with a + b√3 ⇒ a = 2 and b = −1. [3]
Comparing with a + b√3 ⇒ a = 2 and b = −1. [3]
Q27.
(i) 103 × 107 = (100 + 3)(100 + 7) = (100)2 + (3 + 7)(100) + (3)(7) = 10000 + 1000 + 21 = 11021.
(ii) 95 × 96 = (100 − 5)(100 − 4) = (100)2 + (−5 − 4)(100) + (−5)(−4) = 10000 − 900 + 20 = 9120.
(iii) (104)3 = (100 + 4)3 = (100)3 + (4)3 + 3(100)(4)(100 + 4) = 1000000 + 64 + 1200(104) = 1000064 + 124800 = 1124864. [3]
(ii) 95 × 96 = (100 − 5)(100 − 4) = (100)2 + (−5 − 4)(100) + (−5)(−4) = 10000 − 900 + 20 = 9120.
(iii) (104)3 = (100 + 4)3 = (100)3 + (4)3 + 3(100)(4)(100 + 4) = 1000000 + 64 + 1200(104) = 1000064 + 124800 = 1124864. [3]
Q28.
Vertices of rectangle:
β’ O(0, 0)
β’ A(6, 0)
β’ B(6, 4)
β’ C(0, 4)
Area of rectangle = Length × Breadth = 6 × 4 = 24 square units. [3]
β’ O(0, 0)
β’ A(6, 0)
β’ B(6, 4)
β’ C(0, 4)
Area of rectangle = Length × Breadth = 6 × 4 = 24 square units. [3]
Q29.
In ΔABD and ΔACD:
1. ∠BAD = ∠CAD (AD bisects ∠A)
2. AD = AD (Common side)
3. ∠ADB = ∠ADC = 90° (AD ⊥ BC)
⇒ ΔABD ≅ ΔACD (By ASA congruence rule).
⇒ AB = AC (By CPCT). Hence ΔABC is isosceles. [3]
OR In ΔAOB and ΔDOC:
1. ∠OAB = ∠ODC (Alternate interior angles, AB ∥ CD)
2. OA = OD (O is mid-point of AD)
3. ∠AOB = ∠DOC (Vertically opposite angles)
⇒ ΔAOB ≅ ΔDOC (ASA rule).
⇒ OB = OC (CPCT) ⇒ O is the mid-point of BC. [3]
1. ∠BAD = ∠CAD (AD bisects ∠A)
2. AD = AD (Common side)
3. ∠ADB = ∠ADC = 90° (AD ⊥ BC)
⇒ ΔABD ≅ ΔACD (By ASA congruence rule).
⇒ AB = AC (By CPCT). Hence ΔABC is isosceles. [3]
OR In ΔAOB and ΔDOC:
1. ∠OAB = ∠ODC (Alternate interior angles, AB ∥ CD)
2. OA = OD (O is mid-point of AD)
3. ∠AOB = ∠DOC (Vertically opposite angles)
⇒ ΔAOB ≅ ΔDOC (ASA rule).
⇒ OB = OC (CPCT) ⇒ O is the mid-point of BC. [3]
Q30.
Let circle have centre O and chord AB. Let OL ⊥ AB.
In right ΔOLA and right ΔOLB:
1. OA = OB (Radii of same circle, Hypotenuse)
2. OL = OL (Common side)
3. ∠OLA = ∠OLB = 90°
⇒ ΔOLA ≅ ΔOLB (By RHS congruence rule).
⇒ AL = BL (By CPCT). Perpendicular bisects the chord. [3]
In right ΔOLA and right ΔOLB:
1. OA = OB (Radii of same circle, Hypotenuse)
2. OL = OL (Common side)
3. ∠OLA = ∠OLB = 90°
⇒ ΔOLA ≅ ΔOLB (By RHS congruence rule).
⇒ AL = BL (By CPCT). Perpendicular bisects the chord. [3]
Q31.
(i) Adjustment factor = (127 − 126)⁄2 = 0.5.
Continuous classes: 117.5β126.5 (3), 126.5β135.5 (5), 135.5β144.5 (9), 144.5β153.5 (12), 153.5β162.5 (5), 162.5β171.5 (6).
(ii) Histogram drawn using continuous limits along horizontal axis with a kink at the origin. [3]
OR Adjusted frequency formula = (Minimum class width × Frequency)⁄Current class width.
Minimum class width = 1.
Adjusted frequencies:
β’ Age 1β2 (width 1): 1⁄1 × 5 = 5
β’ Age 2β3 (width 1): 1⁄1 × 3 = 3
β’ Age 3β5 (width 2): 1⁄2 × 6 = 3
β’ Age 5β7 (width 2): 1⁄2 × 12 = 6
β’ Age 7β10 (width 3): 1⁄3 × 9 = 3
β’ Age 10β15 (width 5): 1⁄5 × 10 = 2.
Histogram drawn with adjusted frequency densities as bar heights. [3]
Continuous classes: 117.5β126.5 (3), 126.5β135.5 (5), 135.5β144.5 (9), 144.5β153.5 (12), 153.5β162.5 (5), 162.5β171.5 (6).
(ii) Histogram drawn using continuous limits along horizontal axis with a kink at the origin. [3]
OR Adjusted frequency formula = (Minimum class width × Frequency)⁄Current class width.
Minimum class width = 1.
Adjusted frequencies:
β’ Age 1β2 (width 1): 1⁄1 × 5 = 5
β’ Age 2β3 (width 1): 1⁄1 × 3 = 3
β’ Age 3β5 (width 2): 1⁄2 × 6 = 3
β’ Age 5β7 (width 2): 1⁄2 × 12 = 6
β’ Age 7β10 (width 3): 1⁄3 × 9 = 3
β’ Age 10β15 (width 5): 1⁄5 × 10 = 2.
Histogram drawn with adjusted frequency densities as bar heights. [3]
SECTION D β ANSWERS
Q32.
Identity: x3 + y3 + z3 − 3xyz = (x + y + z)(x2 + y2 + z2 − xy − yz − zx).
If x + y + z = 0:
x3 + y3 + z3 − 3xyz = (0)(…) = 0 ⇒ x3 + y3 + z3 = 3xyz.
Let x = 28, y = −15, z = −13.
x + y + z = 28 + (−15) + (−13) = 28 − 28 = 0.
Value = 3(28)(−15)(−13) = 84 × 195 = 16380. [5]
OR p(−2) = (−2)3 + 13(−2)2 + 32(−2) + 20 = −8 + 52 − 64 + 20 = 0 ⇒ (x + 2) is a factor.
Dividing p(x) by (x + 2):
p(x) = (x + 2)(x2 + 11x + 10).
Factoring quadratic: x2 + 10x + x + 10 = (x + 10)(x + 1).
Complete Factorisation: (x + 1)(x + 2)(x + 10). [5]
If x + y + z = 0:
x3 + y3 + z3 − 3xyz = (0)(…) = 0 ⇒ x3 + y3 + z3 = 3xyz.
Let x = 28, y = −15, z = −13.
x + y + z = 28 + (−15) + (−13) = 28 − 28 = 0.
Value = 3(28)(−15)(−13) = 84 × 195 = 16380. [5]
OR p(−2) = (−2)3 + 13(−2)2 + 32(−2) + 20 = −8 + 52 − 64 + 20 = 0 ⇒ (x + 2) is a factor.
Dividing p(x) by (x + 2):
p(x) = (x + 2)(x2 + 11x + 10).
Factoring quadratic: x2 + 10x + x + 10 = (x + 10)(x + 1).
Complete Factorisation: (x + 1)(x + 2)(x + 10). [5]
Q33.
Theorem: Let cyclic quadrilateral be ABCD inscribed in circle with centre O.
Arc BCD subtends ∠BOD at centre and ∠BAD at circumference ⇒ ∠BOD = 2∠BAD.
Reflex ∠BOD = 2∠BCD.
∠BOD + Reflex ∠BOD = 360° ⇒ 2(∠BAD + ∠BCD) = 360° ⇒ ∠BAD + ∠BCD = 180°.
Application: ∠B + ∠D = 180° ⇒ (5x − 10) + (3x + 30) = 180 ⇒ 8x + 20 = 180 ⇒ 8x = 160 ⇒ x = 20°.
β’ ∠B = 5(20°) − 10° = 90°
β’ ∠D = 3(20°) + 30° = 90°. [5]
Arc BCD subtends ∠BOD at centre and ∠BAD at circumference ⇒ ∠BOD = 2∠BAD.
Reflex ∠BOD = 2∠BCD.
∠BOD + Reflex ∠BOD = 360° ⇒ 2(∠BAD + ∠BCD) = 360° ⇒ ∠BAD + ∠BCD = 180°.
Application: ∠B + ∠D = 180° ⇒ (5x − 10) + (3x + 30) = 180 ⇒ 8x + 20 = 180 ⇒ 8x = 160 ⇒ x = 20°.
β’ ∠B = 5(20°) − 10° = 90°
β’ ∠D = 3(20°) + 30° = 90°. [5]
Q34.
1. Diagonal divides into congruent triangles: In ΔABC and ΔCDA: ∠CAB = ∠ACD (alternate angles), AC = CA, ∠BCA = ∠DAC ⇒ ΔABC ≅ ΔCDA (ASA).
2. Angle bisectors form a rectangle: In parallelogram ABCD, ∠A + ∠D = 180° (co-interior) ⇒ 1⁄2∠A + 1⁄2∠D = 90°.
In ΔAPD: ∠APD = 180° − 90° = 90°.
Similarly all four interior angles are 90° ⇒ Enclosed figure is a rectangle. [5]
OR Join diagonals AC and BD of rhombus ABCD (AC ⊥ BD).
In ΔABC: PQ ∥ AC and PQ = 1⁄2AC.
In ΔADC: SR ∥ AC and SR = 1⁄2AC ⇒ PQ ∥ SR and PQ = SR.
Similarly PS ∥ BD and QR ∥ BD.
Since AC ⊥ BD, PQ ⊥ PS ⇒ ∠P = 90° ⇒ PQRS is a rectangle. [5]
2. Angle bisectors form a rectangle: In parallelogram ABCD, ∠A + ∠D = 180° (co-interior) ⇒ 1⁄2∠A + 1⁄2∠D = 90°.
In ΔAPD: ∠APD = 180° − 90° = 90°.
Similarly all four interior angles are 90° ⇒ Enclosed figure is a rectangle. [5]
OR Join diagonals AC and BD of rhombus ABCD (AC ⊥ BD).
In ΔABC: PQ ∥ AC and PQ = 1⁄2AC.
In ΔADC: SR ∥ AC and SR = 1⁄2AC ⇒ PQ ∥ SR and PQ = SR.
Similarly PS ∥ BD and QR ∥ BD.
Since AC ⊥ BD, PQ ⊥ PS ⇒ ∠P = 90° ⇒ PQRS is a rectangle. [5]
Q35.
Given: Radius r = 7 m, Height h = 24 m.
(i) Slant height l = √r2 + h2 = √72 + 242 = √49 + 576 = √625 = 25 m.
(ii) Canvas area = CSA = πrl = 22⁄7 × 7 × 25 = 550 m2.
(iii) Length of canvas cloth = Area⁄Width = 550⁄2 = 275 m.
(iv) Volume of air = 1⁄3 π r2 h = 1⁄3 × 22⁄7 × 7 × 7 × 24 = 22 × 7 × 8 = 1232 m3. [5]
(i) Slant height l = √r2 + h2 = √72 + 242 = √49 + 576 = √625 = 25 m.
(ii) Canvas area = CSA = πrl = 22⁄7 × 7 × 25 = 550 m2.
(iii) Length of canvas cloth = Area⁄Width = 550⁄2 = 275 m.
(iv) Volume of air = 1⁄3 π r2 h = 1⁄3 × 22⁄7 × 7 × 7 × 24 = 22 × 7 × 8 = 1232 m3. [5]
SECTION E β ANSWERS
Q36.
Area = 1⁄2 × 6 × 6 = 18 square units. [2]
- (a) Blue Line: x + y = 6.
Crossing x-axis (y=0) ⇒ A(6, 0); Crossing y-axis (x=0) ⇒ B(0, 6). [1] - (b) Yes, 2(0) − (0) = 0 ⇒ Line 2x − y = 0 satisfies (0, 0) and passes through origin. [1]
- (c) Substitute y = 2x into x + y = 6 ⇒ x + 2x = 6 ⇒ 3x = 6 ⇒ x = 2, y = 4.
Junction Station P = (2, 4). [2]
Area = 1⁄2 × 6 × 6 = 18 square units. [2]
Q37.
Cost of painting = 3117.6 × βΉ0.50 = βΉ1558.80. [2]
- (a) Equilateral triangle side a = Perimeter⁄3 = 180⁄3 = 60 cm. [1]
- (b) Semi-perimeter s = Perimeter⁄2 = 180⁄2 = 90 cm. [1]
- (c) Area = √90(90 − 60)3 = √90 × 30 × 30 × 30 = 30 × 30√3 = 900√3 cm2. [2]
Cost of painting = 3117.6 × βΉ0.50 = βΉ1558.80. [2]
Q38.
Total surface area of 1000 small spheres = 1000 × 4πr2.
Ratio = 4πR2⁄1000 × 4πr2 = (6)2⁄1000 × (0.6)2 = 36⁄1000 × 0.36 = 36⁄360 = 1 : 10. [2]
- (a) Volume of sphere = 4⁄3 π R3. [1]
- (b) Radius of large sphere R = 6 cm; Radius of small sphere r = 0.6 cm. [1]
- (c) Number of lead shots n = Volume of large sphere⁄Volume of one small sphere = R3⁄r3 = (6⁄0.6)3 = (10)3 = 1000 lead shots. [2]
Total surface area of 1000 small spheres = 1000 × 4πr2.
Ratio = 4πR2⁄1000 × 4πr2 = (6)2⁄1000 × (0.6)2 = 36⁄1000 × 0.36 = 36⁄360 = 1 : 10. [2]