ADEPT YOURSELF
www.adeptyourself.com β’ Academic Excellence Series
CBSE CLASS IX β MATHEMATICS (CODE NO. 041)
SAMPLE QUESTION PAPER β SET 3 | 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.
The decimal representation of the rational number 8⁄125 is:
[1]
Q2.
If p(x) = 5x − 4x2 + 3, then the value of p(−1) is:
[1]
Q3.
One of the factors of (1 + 7x)2 + (49x2 − 1) is:
[1]
Q4.
The ordinate of any point on the x-axis is always:
[1]
Q5.
The graph of the linear equation 2x + 3y = 6 cuts the y-axis at the point:
[1]
Q6.
Boundaries of solids are:
[1]
Q7.
If two interior angles on the same side of a transversal intersecting two parallel lines are in the ratio 2 : 3, then the smaller angle is:
[1]
Q8.
Which of the following is not a valid criterion for congruence of triangles?
[1]
Q9.
Three angles of a quadrilateral are 75°, 90°, and 75°. The measure of the fourth angle is:
[1]
Q10.
In a circle with radius 5 cm, the length of the longest chord is:
[1]
Q11.
The area of a triangle with base 12 cm and corresponding altitude 8 cm is:
[1]
Q12.
The total surface area of a solid right circular cone of base radius r and slant height l is given by:
[1]
Q13.
If the radius of the base of a cone is 3 cm and its perpendicular height is 4 cm, then its curved surface area is:
[1]
Q14.
The mean of the first five prime numbers (2, 3, 5, 7, 11) is:
[1]
Q15.
The value of 20 + 7050 is:
[1]
Q16.
Points (2, −2), (3, −3), (4, −5), and (−3, −4):
[1]
Q17.
In the given circle with centre O, if ∠AOB = 100°, then the angle ∠APB subtended at the major arc is:
[1]
Q18.
The curved surface area of a sphere of diameter 14 cm is:
[1]
Q19.
Assertion (A): The value of 272/3 is 9.
Reason (R): For any positive real number a and rational exponent m⁄n, am/n = (a1/n)m. [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): For any positive real number a and rational exponent m⁄n, am/n = (a1/n)m. [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): If two lines intersect each other, then the vertically opposite angles are equal.
Reason (R): The sum of all angles around a point on one side of a line is 180°. [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 sum of all angles around a point on one side of a line is 180°. [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.
Visualise 3.765 on the number line using successive magnification.
[2]
Q22.
Find two different solutions of the linear equation 3x − 2y = 12 in two variables.
[2]
Q23.
In ΔABC, side BC is produced to point D. If the exterior ∠ACD = 120° and ∠B = 2∠A, find the measures of ∠A and ∠B.
[2]
OR
In the given figure, if AB ∥ CD, ∠APQ = 50° and ∠PRD = 127°, find the values of x and y.
Q24.
Two opposite angles of a parallelogram are (3x − 2)° and (50 − x)°. Find the measure of each angle of the parallelogram.
[2]
Q25.
A hemispherical bowl has an inner radius of 3.5 cm. Find the volume of water it can hold in litres.
[2]
OR
The circumference of the base of a conical tent is 44 m and its vertical height is 9 m. Find the volume of air inside the tent.
SECTION C β SHORT ANSWER QUESTIONS TYPE-II (18 Marks)
Q26.
If x = 2 + √3, find the value of x2 + 1x2.
[3]
Q27.
Factorise using suitable identities:
(i) 4x2 + 9y2 + 16z2 + 12xy − 24yz − 16xz
(ii) 27p3 − 1216 − 92p2 + 14p [3]
(i) 4x2 + 9y2 + 16z2 + 12xy − 24yz − 16xz
(ii) 27p3 − 1216 − 92p2 + 14p [3]
Q28.
In which quadrant or on which axis do each of the following points lie?
(i) (−2, 4) (ii) (3, −1) (iii) (−1, 0) (iv) (1, 2) (v) (−3, −5) (vi) (0, −4.5) [3]
(i) (−2, 4) (ii) (3, −1) (iii) (−1, 0) (iv) (1, 2) (v) (−3, −5) (vi) (0, −4.5) [3]
Q29.
In ΔABC, AD is the perpendicular bisector of BC. Show that ΔABC is an isosceles triangle in which AB = AC.
[3]
OR
E and F are respectively the mid-points of equal sides AB and AC of ΔABC. Show that BF = CE.
Q30.
Prove that the line drawn through the centre of a circle to bisect a chord is perpendicular to the chord.
[3]
Q31.
A company selected 2400 families at random and surveyed them to determine a relationship between income level and the number of vehicles in a family:
Find the probability that a family chosen at random:
(i) Earns βΉ10000 − βΉ13000 per month and owns exactly 2 vehicles.
(ii) Earns βΉ16000 or more per month and owns exactly 1 vehicle.
(iii) Owns not more than 1 vehicle. [3]
| Monthly Income (in βΉ) | 0 Vehicles | 1 Vehicle | 2 Vehicles | Above 2 |
|---|---|---|---|---|
| Less than 7000 | 10 | 160 | 25 | 0 |
| 7000β10000 | 0 | 305 | 27 | 2 |
| 10000β13000 | 1 | 535 | 29 | 1 |
| 13000β16000 | 2 | 469 | 59 | 25 |
| 16000 or more | 1 | 579 | 82 | 88 |
(i) Earns βΉ10000 − βΉ13000 per month and owns exactly 2 vehicles.
(ii) Earns βΉ16000 or more per month and owns exactly 1 vehicle.
(iii) Owns not more than 1 vehicle. [3]
OR
Represent the following data on the distribution of marks scored by 60 students in a mathematics test using a histogram:| Marks | 0β10 | 10β20 | 20β30 | 30β40 | 40β50 | 50β60 |
|---|---|---|---|---|---|---|
| No. of Students | 4 | 8 | 15 | 20 | 9 | 4 |
SECTION D β LONG ANSWER QUESTIONS (20 Marks)
Q32.
Factorise the cubic polynomial completely using the Factor Theorem:
p(x) = 2x3 + x2 − 2x − 1
[5]OR
Verify the algebraic identity:
x3 + y3 + z3 − 3xyz = 1⁄2 (x + y + z) [ (x − y)2 + (y − z)2 + (z − x)2 ]
Q33.
Prove that angles in the same segment of a circle are equal.
In a circle with centre O, chords AB and CD intersect at point E inside the circle. If ∠BEC = 130° and ∠ECD = 20°, find ∠BAC. [5]
In a circle with centre O, chords AB and CD intersect at point E inside the circle. If ∠BEC = 130° and ∠ECD = 20°, find ∠BAC. [5]
Q34.
In ΔABC, D, E, and F are respectively the mid-points of sides AB, BC, and CA. Show that ΔABC is divided into four congruent triangles by joining D, E, and F.
[5]
(i) ∠A = ∠B
(ii) ∠C = ∠D
(iii) ΔABC ≅ ΔBAD
(iv) Diagonal AC = Diagonal BD.
OR
ABCD is a trapezium in which AB ∥ CD and AD = BC. Show that:
(i) ∠A = ∠B
(ii) ∠C = ∠D
(iii) ΔABC ≅ ΔBAD
(iv) Diagonal AC = Diagonal BD.
Q35.
A cloth material merchant has a storage godown in the shape of a cuboid measuring 60 m × 40 m × 30 m.
(i) Find the total capacity (volume) of the godown in m3.
(ii) How many standard wooden crates each measuring 1.5 m × 1.25 m × 0.5 m can be stored in the godown?
(iii) Find the total surface area of one wooden crate. [5]
(i) Find the total capacity (volume) of the godown in m3.
(ii) How many standard wooden crates each measuring 1.5 m × 1.25 m × 0.5 m can be stored in the godown?
(iii) Find the total surface area of one wooden crate. [5]
SECTION E β CASE-BASED INTEGRATED UNITS (12 Marks)
Q36. Read the following text and answer the questions that follow:
[4](a) Name the type of quadrilateral formed by joining the points A, B, C, and D in order. [1]
(b) Find the length of the sides AB and BC. [1]
(c) Find the length of the diagonal track running from corner A to corner C. [2]
To promote sports in a school, a quadrilateral sports ground ABCD was marked on a field with corner points having coordinates:
β’ A(0, 0)
β’ B(8, 0)
β’ C(8, 6)
β’ D(0, 6)
(b) Find the length of the sides AB and BC. [1]
(c) Find the length of the diagonal track running from corner A to corner C. [2]
OR
Find the total area enclosed by the sports ground ABCD in square units. [2]
Q37. Read the following text and answer the questions that follow:
[4](a) Calculate the semi-perimeter s of the triangular face. [1]
(b) Write Heron’s formula for the area of a triangle with sides a, b, and c. [1]
(c) Calculate the area of the triangular wall face using Heron’s formula. [2]
A builder designed an ornamental residential boundary wall. The front triangular portion has sides measuring 13 m, 14 m, and 15 m. The wall requires decorative tiling on one face.
(b) Write Heron’s formula for the area of a triangle with sides a, b, and c. [1]
(c) Calculate the area of the triangular wall face using Heron’s formula. [2]
OR
If the cost of decorative tiling is βΉ50 per m2, find the total cost of tiling the wall face. [2]
Q38. Read the following text and answer the questions that follow:
[4](a) What is the base radius r of the cone? [1]
(b) Write the formula for the volume of a right circular cone. [1]
(c) Calculate the capacity (volume) of ice cream that can be filled inside one cone. [2]
An ice-cream parlor serves ice cream in conical waffle cones. Each cone has a base diameter of 7 cm and a vertical height of 12 cm.
(b) Write the formula for the volume of a right circular cone. [1]
(c) Calculate the capacity (volume) of ice cream that can be filled inside one cone. [2]
OR
If the cost of ice cream is βΉ0.15 per cm3, find the cost of filling one ice cream cone completely. [2]
CBSE CLASS IX MATHEMATICS (041) β SOLUTIONS & MARKING SCHEME β SET 3
SECTION A β ANSWERS
Q1. (a) 0.064 [8⁄125 = (8 × 8)⁄(125 × 8) = 64⁄1000 = 0.064] [1]
Q2. (a) −6 [p(−1) = 5(−1) − 4(−1)2 + 3 = −5 − 4 + 3 = −6] [1]
Q3. (c) 7x + 1 [(1 + 7x)2 + (7x − 1)(7x + 1) = (7x + 1)[(1 + 7x) + (7x − 1)] = (7x + 1)(14x)] [1]
Q4. (a) 0 [Any point on x-axis is of the form (x, 0)] [1]
Q5. (c) (0, 2) [Put x = 0 ⇒ 3y = 6 ⇒ y = 2] [1]
Q6. (b) Surfaces [Euclidean geometry definition] [1]
Q7. (b) 72° [2x + 3x = 180° ⇒ 5x = 180° ⇒ x = 36°; Smaller angle = 2(36°) = 72°] [1]
Q8. (c) SSA [Side-Side-Angle is not a sufficient congruence criterion] [1]
Q9. (d) 120° [Fourth angle = 360° − (75° + 90° + 75°) = 360° − 240° = 120°] [1]
Q10. (b) 10 cm [Longest chord of a circle is its diameter = 2r = 2 × 5 = 10 cm] [1]
Q11. (b) 48 cm2 [Area = 1⁄2 × Base × Height = 1⁄2 × 12 × 8 = 48 cm2] [1]
Q12. (b) πr(l + r) [1]
Q13. (b) 15π cm2 [Slant height l = √32 + 42 = 5 cm; CSA = πrl = π(3)(5) = 15π cm2] [1]
Q14. (b) 5.6 [Mean = (2 + 3 + 5 + 7 + 11)⁄5 = 28⁄5 = 5.6] [1]
Q15. (c) 2 [(1 + 1)⁄1 = 2] [1]
Q16. (d) Do not all lie in the same quadrant [Point (−3, −4) lies in Q-III while others lie in Q-IV] [1]
Q17. (a) 50° [Angle at circumference = 1⁄2 × Angle at centre = 100°⁄2 = 50°] [1]
Q18. (c) 616 cm2 [Radius r = 7 cm; Surface area = 4πr2 = 4 × 22⁄7 × 7 × 7 = 616 cm2] [1]
Q19. (a) Both A and R are true and R is the correct explanation of A. [272/3 = (33)2/3 = 32 = 9] [1]
Q20. (b) Both A and R are true but R is not the correct explanation of A. [1]
SECTION B β ANSWERS
Q21.
3.765 lies between 3 and 4; divide into 10 parts to locate interval (3.7, 3.8); magnify to locate (3.76, 3.77); divide again into 10 parts to locate exactly 3.765 on the magnified number line. [2]
Q22.
Given: 3x − 2y = 12 ⇒ 2y = 3x − 12 ⇒ y = (3x − 12)⁄2.
β’ Put x = 0 ⇒ y = −6 ⇒ (0, −6)
β’ Put x = 4 ⇒ y = 0 ⇒ (4, 0). [2]
β’ Put x = 0 ⇒ y = −6 ⇒ (0, −6)
β’ Put x = 4 ⇒ y = 0 ⇒ (4, 0). [2]
Q23.
Exterior angle = ∠A + ∠B ⇒ 120° = ∠A + 2∠A ⇒ 3∠A = 120° ⇒ ∠A = 40°.
∠B = 2(40°) = 80°. [2]
OR x = ∠APQ = 50° (Alternate interior angles, AB ∥ CD).
∠APR = ∠PRD = 127° ⇒ 50° + y = 127° ⇒ y = 77°. [2]
∠B = 2(40°) = 80°. [2]
OR x = ∠APQ = 50° (Alternate interior angles, AB ∥ CD).
∠APR = ∠PRD = 127° ⇒ 50° + y = 127° ⇒ y = 77°. [2]
Q24.
Opposite angles of a parallelogram are equal:
3x − 2 = 50 − x ⇒ 4x = 52 ⇒ x = 13°.
First angle = 3(13°) − 2° = 37°.
Adjacent angle = 180° − 37° = 143°.
The four angles are 37°, 143°, 37°, and 143°. [2]
3x − 2 = 50 − x ⇒ 4x = 52 ⇒ x = 13°.
First angle = 3(13°) − 2° = 37°.
Adjacent angle = 180° − 37° = 143°.
The four angles are 37°, 143°, 37°, and 143°. [2]
Q25.
Volume V = 2⁄3 π r3 = 2⁄3 × 22⁄7 × (3.5)3 = 2⁄3 × 22⁄7 × 42.875 = 89.83 cm3.
Capacity in litres = 89.83⁄1000 = 0.08983 litres. [2]
OR Base circumference 2πr = 44 m ⇒ r = (44 × 7)⁄44 = 7 m. Height h = 9 m.
Volume = 1⁄3 π r2 h = 1⁄3 × 22⁄7 × 7 × 7 × 9 = 22 × 7 × 3 = 462 m3. [2]
Capacity in litres = 89.83⁄1000 = 0.08983 litres. [2]
OR Base circumference 2πr = 44 m ⇒ r = (44 × 7)⁄44 = 7 m. Height h = 9 m.
Volume = 1⁄3 π r2 h = 1⁄3 × 22⁄7 × 7 × 7 × 9 = 22 × 7 × 3 = 462 m3. [2]
SECTION C β ANSWERS
Q26.
x = 2 + √3 ⇒ 1⁄x = 1⁄(2 + √3) = (2 − √3)⁄(4 − 3) = 2 − √3.
x + 1⁄x = (2 + √3) + (2 − √3) = 4.
x2 + 1⁄x2 = (x + 1⁄x)2 − 2 = (4)2 − 2 = 16 − 2 = 14. [3]
x + 1⁄x = (2 + √3) + (2 − √3) = 4.
x2 + 1⁄x2 = (x + 1⁄x)2 − 2 = (4)2 − 2 = 16 − 2 = 14. [3]
Q27.
(i) (2x)2 + (3y)2 + (−4z)2 + 2(2x)(3y) + 2(3y)(−4z) + 2(−4z)(2x) = (2x + 3y − 4z)2.
(ii) (3p)3 − (1⁄6)3 − 3(3p)2(1⁄6) + 3(3p)(1⁄6)2 = (3p − 1⁄6)3. [3]
(ii) (3p)3 − (1⁄6)3 − 3(3p)2(1⁄6) + 3(3p)(1⁄6)2 = (3p − 1⁄6)3. [3]
Q28.
(i) (−2, 4) → Quadrant II
(ii) (3, −1) → Quadrant IV
(iii) (−1, 0) → Negative x-axis
(iv) (1, 2) → Quadrant I
(v) (−3, −5) → Quadrant III
(vi) (0, −4.5) → Negative y-axis. [3]
(ii) (3, −1) → Quadrant IV
(iii) (−1, 0) → Negative x-axis
(iv) (1, 2) → Quadrant I
(v) (−3, −5) → Quadrant III
(vi) (0, −4.5) → Negative y-axis. [3]
Q29.
In ΔABD and ΔACD:
1. BD = CD (AD bisects BC)
2. ∠ADB = ∠ADC = 90° (AD ⊥ BC)
3. AD = AD (Common side)
⇒ ΔABD ≅ ΔACD (By SAS congruence rule).
⇒ AB = AC (CPCT). Hence ΔABC is isosceles. [3]
OR In ΔABF and ΔACE:
1. AB = AC (Given)
2. ∠A = ∠A (Common angle)
3. AF = AE (Halves of equal sides)
⇒ ΔABF ≅ ΔACE (SAS rule) ⇒ BF = CE (CPCT). [3]
1. BD = CD (AD bisects BC)
2. ∠ADB = ∠ADC = 90° (AD ⊥ BC)
3. AD = AD (Common side)
⇒ ΔABD ≅ ΔACD (By SAS congruence rule).
⇒ AB = AC (CPCT). Hence ΔABC is isosceles. [3]
OR In ΔABF and ΔACE:
1. AB = AC (Given)
2. ∠A = ∠A (Common angle)
3. AF = AE (Halves of equal sides)
⇒ ΔABF ≅ ΔACE (SAS rule) ⇒ BF = CE (CPCT). [3]
Q30.
Let circle have centre O and chord AB with mid-point M (AM = BM).
In ΔOAM and ΔOBM:
1. OA = OB (Radii)
2. AM = BM (Given)
3. OM = OM (Common)
⇒ ΔOAM ≅ ΔOBM (SSS rule) ⇒ ∠OMA = ∠OMB.
Since ∠OMA + ∠OMB = 180° ⇒ ∠OMA = 90° ⇒ OM ⊥ AB. [3]
In ΔOAM and ΔOBM:
1. OA = OB (Radii)
2. AM = BM (Given)
3. OM = OM (Common)
⇒ ΔOAM ≅ ΔOBM (SSS rule) ⇒ ∠OMA = ∠OMB.
Since ∠OMA + ∠OMB = 180° ⇒ ∠OMA = 90° ⇒ OM ⊥ AB. [3]
Q31.
Total surveyed families = 2400.
(i) Families earning βΉ10000ββΉ13000 owning 2 vehicles = 29 ⇒ P = 29⁄2400.
(ii) Families earning βΉ16000+ owning 1 vehicle = 579 ⇒ P = 579⁄2400 = 193⁄800.
(iii) Families owning not more than 1 vehicle = (10+0+1+2+1) + (160+305+535+469+579) = 14 + 2048 = 2062 ⇒ P = 2062⁄2400 = 1031⁄1200. [3]
OR Histogram plotted with class boundaries along horizontal axis and student frequency along vertical axis. [3]
(i) Families earning βΉ10000ββΉ13000 owning 2 vehicles = 29 ⇒ P = 29⁄2400.
(ii) Families earning βΉ16000+ owning 1 vehicle = 579 ⇒ P = 579⁄2400 = 193⁄800.
(iii) Families owning not more than 1 vehicle = (10+0+1+2+1) + (160+305+535+469+579) = 14 + 2048 = 2062 ⇒ P = 2062⁄2400 = 1031⁄1200. [3]
OR Histogram plotted with class boundaries along horizontal axis and student frequency along vertical axis. [3]
SECTION D β ANSWERS
Q32.
Let p(x) = 2x3 + x2 − 2x − 1.
For x = 1: p(1) = 2(1) + 1 − 2 − 1 = 0 ⇒ (x − 1) is a factor.
Dividing p(x) by (x − 1): p(x) = (x − 1)(2x2 + 3x + 1).
Factoring quadratic: 2x2 + 2x + x + 1 = 2x(x + 1) + 1(x + 1) = (x + 1)(2x + 1).
Complete Factorisation: (x − 1)(x + 1)(2x + 1). [5]
OR RHS = 1⁄2(x + y + z)[x2 + y2 − 2xy + y2 + z2 − 2yz + z2 + x2 − 2zx]
= 1⁄2(x + y + z)[2x2 + 2y2 + 2z2 − 2xy − 2yz − 2zx]
= (x + y + z)(x2 + y2 + z2 − xy − yz − zx) = x3 + y3 + z3 − 3xyz = LHS. Hence verified. [5]
For x = 1: p(1) = 2(1) + 1 − 2 − 1 = 0 ⇒ (x − 1) is a factor.
Dividing p(x) by (x − 1): p(x) = (x − 1)(2x2 + 3x + 1).
Factoring quadratic: 2x2 + 2x + x + 1 = 2x(x + 1) + 1(x + 1) = (x + 1)(2x + 1).
Complete Factorisation: (x − 1)(x + 1)(2x + 1). [5]
OR RHS = 1⁄2(x + y + z)[x2 + y2 − 2xy + y2 + z2 − 2yz + z2 + x2 − 2zx]
= 1⁄2(x + y + z)[2x2 + 2y2 + 2z2 − 2xy − 2yz − 2zx]
= (x + y + z)(x2 + y2 + z2 − xy − yz − zx) = x3 + y3 + z3 − 3xyz = LHS. Hence verified. [5]
Q33.
Theorem: Let chord AB subtend ∠ACB and ∠ADB in the same segment.
∠AOB = 2∠ACB and ∠AOB = 2∠ADB ⇒ 2∠ACB = 2∠ADB ⇒ ∠ACB = ∠ADB.
Calculation: In ΔDEC, exterior ∠BEC = ∠EDC + ∠ECD ⇒ 130° = ∠EDC + 20° ⇒ ∠EDC = 110°.
Since angles in the same segment are equal: ∠BAC = ∠BDC = 110°. [5]
∠AOB = 2∠ACB and ∠AOB = 2∠ADB ⇒ 2∠ACB = 2∠ADB ⇒ ∠ACB = ∠ADB.
Calculation: In ΔDEC, exterior ∠BEC = ∠EDC + ∠ECD ⇒ 130° = ∠EDC + 20° ⇒ ∠EDC = 110°.
Since angles in the same segment are equal: ∠BAC = ∠BDC = 110°. [5]
Q34.
By Mid-point Theorem:
DE ∥ AC and DE = 1⁄2AC = AF.
EF ∥ AB and EF = 1⁄2AB = AD.
DF ∥ BC and DF = 1⁄2BC = BE.
Thus, ADEF, BDFE, and DFCE are parallelograms.
Diagonal divides each parallelogram into two congruent triangles ⇒ ΔADF ≅ ΔDBE ≅ ΔEFC ≅ ΔFED. [5]
OR Extend AB and draw line through C parallel to AD meeting extended AB at E.
ADCE is a parallelogram ⇒ AD = CE. Given AD = BC ⇒ BC = CE ⇒ ∠CBE = ∠CEB.
(i) ∠A + ∠CEB = 180° and ∠B + ∠CBE = 180° ⇒ ∠A = ∠B.
(ii) ∠A + ∠D = 180° and ∠B + ∠C = 180° ⇒ ∠C = ∠D.
(iii) In ΔABC and ΔBAD: AB = BA, BC = AD, ∠B = ∠A ⇒ ΔABC ≅ ΔBAD.
(iv) AC = BD (CPCT). [5]
DE ∥ AC and DE = 1⁄2AC = AF.
EF ∥ AB and EF = 1⁄2AB = AD.
DF ∥ BC and DF = 1⁄2BC = BE.
Thus, ADEF, BDFE, and DFCE are parallelograms.
Diagonal divides each parallelogram into two congruent triangles ⇒ ΔADF ≅ ΔDBE ≅ ΔEFC ≅ ΔFED. [5]
OR Extend AB and draw line through C parallel to AD meeting extended AB at E.
ADCE is a parallelogram ⇒ AD = CE. Given AD = BC ⇒ BC = CE ⇒ ∠CBE = ∠CEB.
(i) ∠A + ∠CEB = 180° and ∠B + ∠CBE = 180° ⇒ ∠A = ∠B.
(ii) ∠A + ∠D = 180° and ∠B + ∠C = 180° ⇒ ∠C = ∠D.
(iii) In ΔABC and ΔBAD: AB = BA, BC = AD, ∠B = ∠A ⇒ ΔABC ≅ ΔBAD.
(iv) AC = BD (CPCT). [5]
Q35.
(i) Volume of godown = 60 × 40 × 30 = 72,000 m3.
(ii) Volume of one crate = 1.5 × 1.25 × 0.5 = 0.9375 m3 (or 15⁄16 m3).
Number of crates = 72000⁄0.9375 = 76,800 crates.
(iii) TSA of crate = 2(lb + bh + hl) = 2(1.875 + 0.625 + 0.75) = 2(3.25) = 6.5 m2. [5]
(ii) Volume of one crate = 1.5 × 1.25 × 0.5 = 0.9375 m3 (or 15⁄16 m3).
Number of crates = 72000⁄0.9375 = 76,800 crates.
(iii) TSA of crate = 2(lb + bh + hl) = 2(1.875 + 0.625 + 0.75) = 2(3.25) = 6.5 m2. [5]
SECTION E β ANSWERS
Q36.
- (a) Rectangle (opposite sides are parallel, adjacent sides perpendicular). [1]
- (b) Side AB = 8 − 0 = 8 units; Side BC = 6 − 0 = 6 units. [1]
- (c) Diagonal AC = √82 + 62 = √64 + 36 = √100 = 10 units. [2]
Q37.
- (a) Semi-perimeter s = (13 + 14 + 15)⁄2 = 42⁄2 = 21 m. [1]
- (b) Area = √s(s − a)(s − b)(s − c). [1]
- (c) Area = √21(21 − 13)(21 − 14)(21 − 15) = √21 × 8 × 7 × 6 = √7056 = 84 m2. [2]
Q38.
- (a) Radius r = 7⁄2 = 3.5 cm. [1]
- (b) Volume of cone = 1⁄3 π r2 h. [1]
- (c) Volume = 1⁄3 × 22⁄7 × 7⁄2 × 7⁄2 × 12 = 154 cm3. [2]