ADEPT YOURSELF
www.adeptyourself.com β’ Academic Excellence Series
CBSE CLASS IX β MATHEMATICS (CODE NO. 041)
SAMPLE QUESTION PAPER β SET 1 | 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.
Which of the following is an irrational number?
[1]
Q2.
The degree of the zero polynomial is:
[1]
Q3.
If (x + 1) is a factor of the polynomial p(x) = 2x2 + kx, then the value of k is:
[1]
Q4.
The point (−3, −5) lies in which quadrant of the Cartesian coordinate plane?
[1]
Q5.
The linear equation 2x − 5y = 7 in two variables has:
[1]
Q6.
Euclid’s Axiom 1 states: “Things which are equal to the same thing are…”
[1]
Q7.
An exterior angle of a triangle is 105° and its two interior opposite angles are equal. Each of these equal angles is:
[1]
Q8.
In ΔABC and ΔPQR, AB = AC, ∠C = ∠P and ∠B = ∠Q. The two triangles are:
[1]
Q9.
The diagonals of a rhombus:
[1]
Q10.
In a circle with centre O and radius 13 cm, a chord AB is at a perpendicular distance of 5 cm from O. The length of the chord AB is:
[1]
Q11.
The sides of a triangle are 56 cm, 60 cm, and 52 cm. The area of the triangle is:
[1]
Q12.
If the radius of a sphere is doubled, its surface area will increase by:
[1]
Q13.
The curved surface area of a right circular cone of base radius 7 cm and slant height 10 cm is:
[1]
Q14.
The class mark of the class interval 130 − 150 is:
[1]
Q15.
The simplified value of (125)−1/3 is:
[1]
Q16.
If the coordinates of two points are P(−2, 3) and Q(−3, 5), then (Abscissa of P) − (Abscissa of Q) is:
[1]
Q17.
The angle subtended by a semicircle at any point on the circumference of a circle is:
[1]
Q18.
The volume of a solid hemisphere of radius 3 cm is:
[1]
Q19.
Assertion (A): The polynomial p(x) = x3 − 3x2 + 2x has at most 3 real zeroes.
Reason (R): A polynomial of degree n can have at most n real zeroes. [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): A polynomial of degree n can have at most n real zeroes. [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): The perimeter of an equilateral triangle of side 6 cm is 18 cm and its area is 9√3 cm2.
Reason (R): The area of an equilateral triangle with side a is given by √34a2. [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 area of an equilateral triangle with side a is given by √34a2. [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.
Express 0.47 in the rational form pq, where p and q are integers and q ≠ 0.
[2]
Q22.
Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k. Hence, write one more solution of this equation.
[2]
Q23.
In the given figure, lines AB and CD intersect at point O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.
[2]
OR
In ΔABC, ∠A = 50°. If the angle bisectors of ∠B and ∠C meet at point O inside the triangle, find the measure of ∠BOC.
Q24.
Prove that the diagonals of a square are equal in length and bisect each other at right angles (90°).
[2]
Q25.
A solid right circular cone of base radius 6 cm has a curved surface area of 60π cm2. Find its slant height and vertical height.
[2]
OR
Find the total surface area of a solid hemisphere of radius 10 cm. [Take π = 3.14]
SECTION C β SHORT ANSWER QUESTIONS TYPE-II (18 Marks)
Q26.
Rationalise the denominator and simplify:
7 + 3√57 − 3√5
[3]
Q27.
Factorise completely using an appropriate algebraic identity:
8x3 + y3 + 27z3 − 18xyz
[3]
Q28.
Plot the points A(1, 3), B(1, −1), C(7, −1), and D(7, 3) on a Cartesian plane. Name the geometrical figure ABCD formed by joining the points in order and calculate its area in square units.
[3]
Q29.
AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB. Prove that:
(i) ΔDAP ≅ ΔEBP
(ii) AD = BE [3]
(i) ΔDAP ≅ ΔEBP
(ii) AD = BE [3]
OR
In an isosceles triangle ABC with AB = AC, D and E are points on side BC such that BE = CD. Prove that AD = AE.
Q30.
Prove that equal chords of a circle subtend equal angles at the centre of the circle.
[3]
Q31.
The following frequency table shows the daily wages of 50 workers in a workshop:
Construct a histogram and a frequency polygon for the given distribution.
[3]
A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O, A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O.
(i) Represent this data in the form of a frequency distribution table.
(ii) Which is the most common and which is the rarest blood group among these students?
| Daily Wages (in βΉ) | 100β120 | 120β140 | 140β160 | 160β180 | 180β200 |
|---|---|---|---|---|---|
| Number of Workers | 12 | 14 | 8 | 6 | 10 |
OR
The blood groups of 30 students of Class IX are recorded as follows:
A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O, A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O.
(i) Represent this data in the form of a frequency distribution table.
(ii) Which is the most common and which is the rarest blood group among these students?
SECTION D β LONG ANSWER QUESTIONS (20 Marks)
Q32.
If x = √3 + √2√3 − √2 and y = √3 − √2√3 + √2, find the value of x2 + y2 + xy.
[5]
OR
Using the Factor Theorem, factorise the cubic polynomial completely:
p(x) = x3 − 23x2 + 142x − 120
Q33.
Prove that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
[5]
Q34.
State and prove the Mid-point Theorem for a triangle.
Using this theorem, prove that the quadrilateral formed by joining the mid-points of the sides of any quadrilateral ABCD, taken in order, is a parallelogram. [5]
(i) D is the mid-point of AC.
(ii) MD ⊥ AC.
(iii) CM = MA = 1⁄2 AB.
Using this theorem, prove that the quadrilateral formed by joining the mid-points of the sides of any quadrilateral ABCD, taken in order, is a parallelogram. [5]
OR
ABC is a triangle right-angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects side AC at D. Show that:
(i) D is the mid-point of AC.
(ii) MD ⊥ AC.
(iii) CM = MA = 1⁄2 AB.
Q35.
A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the total cost of βΉ498.96. If the rate of white-washing is βΉ2.00 per square metre, find:
(i) The inside surface area of the dome.
(ii) The volume of the air inside the dome. [5]
(i) The inside surface area of the dome.
(ii) The volume of the air inside the dome. [5]
SECTION E β CASE-BASED INTEGRATED UNITS (12 Marks)
Q36. Read the following text and answer the questions that follow:
[4](a) Write the abscissa of point A and ordinate of point C. [1]
(b) In which quadrants do the points B and D lie? [1]
(c) Find the perimeter and area of the rectangular boundary formed by joining the points A, B, C, and D in order. [2]
National Green Corps organized an environmental plantation project in a municipal park. The layout plan was plotted on a Cartesian coordinate plane where the main entrance gate is taken as the origin O(0, 0).
β’ Ashoka trees were planted at point A(3, 4).
β’ Neem saplings were planted at point B(−3, 4).
β’ Gulmohar trees were planted at point C(−3, −4).
β’ Peepal trees were planted at point D(3, −4).
(b) In which quadrants do the points B and D lie? [1]
(c) Find the perimeter and area of the rectangular boundary formed by joining the points A, B, C, and D in order. [2]
OR
Find the length of the diagonal pathway AC connecting point A to point C. [2]
Q37. Read the following text and answer the questions that follow:
[4](a) Find the semi-perimeter s of the triangular ground. [1]
(b) Calculate the actual length of each of the three sides of the playground. [1]
(c) Using Heron’s formula, calculate the total area of the triangular playground. [2]
A school playground is in the shape of a triangle whose perimeter is 300 m. Its sides are in the ratio 3 : 5 : 7. The school administration decides to lay natural grass turf across the entire field and install a protective boundary fence around it, leaving an entrance space of 3 m wide for a gate.
(b) Calculate the actual length of each of the three sides of the playground. [1]
(c) Using Heron’s formula, calculate the total area of the triangular playground. [2]
OR
Find the total cost of fencing the playground with wire at the rate of βΉ25 per metre, leaving a 3 m wide space for the gate. [2]
Q38. Read the following text and answer the questions that follow:
[4]
(a) What is the class width (class size) of the given class intervals? [1]
(b) Find the class mark of the class interval 60 − 70. [1]
(c) How many persons have body weight less than 70 kg? What percentage of the total persons does this represent? [2]
A health awareness camp recorded the body weights (in kg) of 50 senior citizens in a residential locality. The recorded data is summarized in the frequency distribution table below:
| Weight (in kg) | 40β50 | 50β60 | 60β70 | 70β80 | 80β90 |
|---|---|---|---|---|---|
| Number of Persons | 6 | 12 | 18 | 10 | 4 |
(b) Find the class mark of the class interval 60 − 70. [1]
(c) How many persons have body weight less than 70 kg? What percentage of the total persons does this represent? [2]
OR
Construct a cumulative frequency distribution table (less than type) for the given dataset. [2]
CBSE CLASS IX MATHEMATICS (041) β SOLUTIONS & MARKING SCHEME β SET 1
SECTION A β ANSWERS
Q1. (c) 0.101001000100001… [A non-terminating, non-repeating decimal represents an irrational number] [1]
Q2. (d) Not defined [Degree of zero polynomial is not defined] [1]
Q3. (b) 2 [p(−1) = 0 ⇒ 2(−1)2 + k(−1) = 0 ⇒ 2 − k = 0 ⇒ k = 2] [1]
Q4. (c) Quadrant III [Both x and y coordinates are negative] [1]
Q5. (c) Infinitely many solutions [A linear equation in two variables represents a straight line with infinite points] [1]
Q6. (a) Equal to one another [Euclid’s Axiom 1] [1]
Q7. (b) 52.5° [Exterior angle = Sum of opposite interior angles ⇒ 2x = 105° ⇒ x = 52.5°] [1]
Q8. (b) Isosceles but not necessarily congruent [Corresponding sides may not be equal] [1]
Q9. (b) Bisect each other at right angles (90°) [1]
Q10. (b) 24 cm [Half chord = √132 − 52 = √169 − 25 = 12 cm ⇒ AB = 2 × 12 = 24 cm] [1]
Q11. (a) 1344 cm2 [s = (56+60+52)⁄2 = 84 cm; Area = √84 × 28 × 24 × 32 = 1344 cm2] [1]
Q12. (c) 300% [S1 = 4πr2; S2 = 4π(2r)2 = 4S1; Increase = 3S1 ⇒ 3S1⁄S1 × 100 = 300%] [1]
Q13. (a) 220 cm2 [CSA = πrl = 22⁄7 × 7 × 10 = 220 cm2] [1]
Q14. (c) 140 [Class mark = (130+150)⁄2 = 140] [1]
Q15. (c) 1⁄5 [(125)−1/3 = (53)−1/3 = 5−1 = 1⁄5] [1]
Q16. (b) 1 [Abscissa of P = −2, Abscissa of Q = −3; (−2) − (−3) = −2 + 3 = 1] [1]
Q17. (c) 90° [Angle subtended by a semicircle is a right angle] [1]
Q18. (a) 18π cm3 [V = 2⁄3 π r3 = 2⁄3 π (3)3 = 18π cm3] [1]
Q19. (a) Both A and R are true and R is the correct explanation of A. [1]
Q20. (a) Both A and R are true and R is the correct explanation of A. [1]
SECTION B β ANSWERS
Q21.
Let x = 0.47 = 0.4777… —— (1)
Multiplying (1) by 10: 10x = 4.777… —— (2)
Multiplying (1) by 100: 100x = 47.777… —— (3)
Subtracting (2) from (3): 90x = 43 ⇒ x = 43⁄90. [2]
Multiplying (1) by 10: 10x = 4.777… —— (2)
Multiplying (1) by 100: 100x = 47.777… —— (3)
Subtracting (2) from (3): 90x = 43 ⇒ x = 43⁄90. [2]
Q22.
Substitute x = 2, y = 1 in 2x + 3y = k:
2(2) + 3(1) = k ⇒ 4 + 3 = k ⇒ k = 7.
The equation is 2x + 3y = 7.
Put x = 5: 2(5) + 3y = 7 ⇒ 3y = −3 ⇒ y = −1.
Hence, another solution is (5, −1). [2]
2(2) + 3(1) = k ⇒ 4 + 3 = k ⇒ k = 7.
The equation is 2x + 3y = 7.
Put x = 5: 2(5) + 3y = 7 ⇒ 3y = −3 ⇒ y = −1.
Hence, another solution is (5, −1). [2]
Q23.
∠AOC = ∠BOD = 40° (Vertically opposite angles).
Given: ∠AOC + ∠BOE = 70° ⇒ 40° + ∠BOE = 70° ⇒ ∠BOE = 30°.
Since AB is a straight line: ∠AOC + ∠COE + ∠BOE = 180° ⇒ 70° + ∠COE = 180° ⇒ ∠COE = 110°.
Reflex ∠COE = 360° − 110° = 250°. [2]
OR In ΔABC: ∠A + ∠B + ∠C = 180° ⇒ ∠B + ∠C = 180° − 50° = 130°.
1⁄2∠B + 1⁄2∠C = 65°.
In ΔBOC: ∠BOC = 180° − (1⁄2∠B + 1⁄2∠C) = 180° − 65° = 115°. [2]
Given: ∠AOC + ∠BOE = 70° ⇒ 40° + ∠BOE = 70° ⇒ ∠BOE = 30°.
Since AB is a straight line: ∠AOC + ∠COE + ∠BOE = 180° ⇒ 70° + ∠COE = 180° ⇒ ∠COE = 110°.
Reflex ∠COE = 360° − 110° = 250°. [2]
OR In ΔABC: ∠A + ∠B + ∠C = 180° ⇒ ∠B + ∠C = 180° − 50° = 130°.
1⁄2∠B + 1⁄2∠C = 65°.
In ΔBOC: ∠BOC = 180° − (1⁄2∠B + 1⁄2∠C) = 180° − 65° = 115°. [2]
Q24.
Let ABCD be a square.
1. Equality: In ΔABC and ΔBAD, AB = BA, BC = AD, ∠B = ∠A = 90° ⇒ ΔABC ≅ ΔBAD (SAS) ⇒ AC = BD.
2. Perpendicular Bisector: Since a square is a parallelogram, diagonals bisect each other (OA = OC, OB = OD).
In ΔAOB and ΔCOB: AB = CB, OA = OC, OB = OB ⇒ ΔAOB ≅ ΔCOB (SSS) ⇒ ∠AOB = ∠COB = 90°. [2]
1. Equality: In ΔABC and ΔBAD, AB = BA, BC = AD, ∠B = ∠A = 90° ⇒ ΔABC ≅ ΔBAD (SAS) ⇒ AC = BD.
2. Perpendicular Bisector: Since a square is a parallelogram, diagonals bisect each other (OA = OC, OB = OD).
In ΔAOB and ΔCOB: AB = CB, OA = OC, OB = OB ⇒ ΔAOB ≅ ΔCOB (SSS) ⇒ ∠AOB = ∠COB = 90°. [2]
Q25.
CSA = πrl ⇒ 60π = π(6)l ⇒ l = 60⁄6 = 10 cm.
Vertical height h = √l2 − r2 = √102 − 62 = √64 = 8 cm. [2]
OR Total surface area of solid hemisphere = 3πr2 = 3 × 3.14 × (10)2 = 3 × 3.14 × 100 = 942 cm2. [2]
Vertical height h = √l2 − r2 = √102 − 62 = √64 = 8 cm. [2]
OR Total surface area of solid hemisphere = 3πr2 = 3 × 3.14 × (10)2 = 3 × 3.14 × 100 = 942 cm2. [2]
SECTION C β ANSWERS
Q26.
7 + 3√57 − 3√5 × 7 + 3√57 + 3√5
= (7 + 3√5)2(7)2 − (3√5)2
= 49 + 45 + 42√549 − 45
= 94 + 42√54
= 47 + 21√52. [3]
Q27.
Rewrite expression as: (2x)3 + (y)3 + (3z)3 − 3(2x)(y)(3z).
Using identity a3 + b3 + c3 − 3abc = (a + b + c)(a2 + b2 + c2 − ab − bc − ca):
= (2x + y + 3z)(4x2 + y2 + 9z2 − 2xy − 3yz − 6zx). [3]
Using identity a3 + b3 + c3 − 3abc = (a + b + c)(a2 + b2 + c2 − ab − bc − ca):
= (2x + y + 3z)(4x2 + y2 + 9z2 − 2xy − 3yz − 6zx). [3]
Q28.
Length of side AB = 3 − (−1) = 4 units; Length of side BC = 7 − 1 = 6 units.
Opposite sides are equal and parallel with right angles at vertices ⇒ Figure ABCD is a Rectangle.
Area of rectangle = Length × Breadth = 6 × 4 = 24 square units. [3]
Opposite sides are equal and parallel with right angles at vertices ⇒ Figure ABCD is a Rectangle.
Area of rectangle = Length × Breadth = 6 × 4 = 24 square units. [3]
Q29.
Given: ∠EPA = ∠DPB. Adding ∠EPD to both sides:
∠EPA + ∠EPD = ∠DPB + ∠EPD ⇒ ∠APD = ∠BPE.
In ΔDAP and ΔEBP:
1. ∠PAD = ∠PBE (Given ∠BAD = ∠ABE)
2. AP = BP (P is the mid-point of AB)
3. ∠APD = ∠BPE (Proved above)
⇒ ΔDAP ≅ ΔEBP (By ASA congruence criterion).
⇒ AD = BE (By CPCT). [3]
OR In ΔABC, AB = AC ⇒ ∠B = ∠C.
Given BE = CD. Subtracting DE from both sides: BE − DE = CD − DE ⇒ BD = CE.
In ΔABD and ΔACE: AB = AC, ∠B = ∠C, BD = CE ⇒ ΔABD ≅ ΔACE (SAS rule) ⇒ AD = AE (CPCT). [3]
∠EPA + ∠EPD = ∠DPB + ∠EPD ⇒ ∠APD = ∠BPE.
In ΔDAP and ΔEBP:
1. ∠PAD = ∠PBE (Given ∠BAD = ∠ABE)
2. AP = BP (P is the mid-point of AB)
3. ∠APD = ∠BPE (Proved above)
⇒ ΔDAP ≅ ΔEBP (By ASA congruence criterion).
⇒ AD = BE (By CPCT). [3]
OR In ΔABC, AB = AC ⇒ ∠B = ∠C.
Given BE = CD. Subtracting DE from both sides: BE − DE = CD − DE ⇒ BD = CE.
In ΔABD and ΔACE: AB = AC, ∠B = ∠C, BD = CE ⇒ ΔABD ≅ ΔACE (SAS rule) ⇒ AD = AE (CPCT). [3]
Q30.
Let AB and CD be two equal chords of a circle with centre O.
In ΔAOB and ΔCOD:
1. OA = OC (Radii of the same circle)
2. OB = OD (Radii of the same circle)
3. AB = CD (Given equal chords)
⇒ ΔAOB ≅ ΔCOD (By SSS congruence rule).
⇒ ∠AOB = ∠COD (By CPCT). Hence proved. [3]
In ΔAOB and ΔCOD:
1. OA = OC (Radii of the same circle)
2. OB = OD (Radii of the same circle)
3. AB = CD (Given equal chords)
⇒ ΔAOB ≅ ΔCOD (By SSS congruence rule).
⇒ ∠AOB = ∠COD (By CPCT). Hence proved. [3]
Q31.
Histogram plotted with continuous class intervals along the horizontal axis and frequencies as rectangular column heights; polygon formed by joining mid-points of column tops. [3]
OR Frequency Distribution Table:
β’ Blood group A: 9 | Blood group B: 6 | Blood group O: 12 | Blood group AB: 3 (Total = 30).
β’ Most common blood group: O (Frequency = 12).
β’ Rarest blood group: AB (Frequency = 3). [3]
OR Frequency Distribution Table:
β’ Blood group A: 9 | Blood group B: 6 | Blood group O: 12 | Blood group AB: 3 (Total = 30).
β’ Most common blood group: O (Frequency = 12).
β’ Rarest blood group: AB (Frequency = 3). [3]
SECTION D β ANSWERS
Q32.
x = √3 + √2√3 − √2 = (√3 + √2)2 = 3 + 2 + 2√6 = 5 + 2√6.
y = √3 − √2√3 + √2 = (√3 − √2)2 = 3 + 2 − 2√6 = 5 − 2√6.
x + y = (5 + 2√6) + (5 − 2√6) = 10.
xy = (5 + 2√6)(5 − 2√6) = 25 − 24 = 1.
x2 + y2 + xy = (x + y)2 − xy = (10)2 − 1 = 100 − 1 = 99. [5]
OR Let p(x) = x3 − 23x2 + 142x − 120.
For x = 1: p(1) = 1 − 23 + 142 − 120 = 0 ⇒ (x − 1) is a factor.
Dividing p(x) by (x − 1): p(x) = (x − 1)(x2 − 22x + 120).
Factoring quadratic: x2 − 12x − 10x + 120 = (x − 12)(x − 10).
Complete Factorisation: (x − 1)(x − 10)(x − 12). [5]
y = √3 − √2√3 + √2 = (√3 − √2)2 = 3 + 2 − 2√6 = 5 − 2√6.
x + y = (5 + 2√6) + (5 − 2√6) = 10.
xy = (5 + 2√6)(5 − 2√6) = 25 − 24 = 1.
x2 + y2 + xy = (x + y)2 − xy = (10)2 − 1 = 100 − 1 = 99. [5]
OR Let p(x) = x3 − 23x2 + 142x − 120.
For x = 1: p(1) = 1 − 23 + 142 − 120 = 0 ⇒ (x − 1) is a factor.
Dividing p(x) by (x − 1): p(x) = (x − 1)(x2 − 22x + 120).
Factoring quadratic: x2 − 12x − 10x + 120 = (x − 12)(x − 10).
Complete Factorisation: (x − 1)(x − 10)(x − 12). [5]
Q33.
Theorem: Consider arc AB subtending ∠AOB at centre O and ∠APB at point P on the remaining circle.
Construction: Join PO and produce it to point Q.
In ΔAPO: OA = OP (Radii) ⇒ ∠OPA = ∠OAP.
Exterior angle ∠AOQ = ∠OPA + ∠OAP = 2∠OPA —— (1).
Similarly in ΔBPO: ∠BOQ = 2∠OPB —— (2).
Adding (1) and (2): ∠AOQ + ∠BOQ = 2(∠OPA + ∠OPB) ⇒ ∠AOB = 2∠APB. [5]
Construction: Join PO and produce it to point Q.
In ΔAPO: OA = OP (Radii) ⇒ ∠OPA = ∠OAP.
Exterior angle ∠AOQ = ∠OPA + ∠OAP = 2∠OPA —— (1).
Similarly in ΔBPO: ∠BOQ = 2∠OPB —— (2).
Adding (1) and (2): ∠AOQ + ∠BOQ = 2(∠OPA + ∠OPB) ⇒ ∠AOB = 2∠APB. [5]
Q34.
Mid-Point Theorem: In ΔABC, let E and F be mid-points of AB and AC.
Proof: Extend EF to D such that EF = FD and join CD.
ΔAEF ≅ ΔCDF (SAS) ⇒ AE = CD and AB ∥ CD.
Since BE = AE = CD and BE ∥ CD, BCDE is a parallelogram ⇒ EF ∥ BC and EF = 1⁄2 BC.
Quadrilateral Mid-point Proof: Join diagonal AC. In ΔABC, PQ ∥ AC and PQ = 1⁄2AC. In ΔADC, SR ∥ AC and SR = 1⁄2AC. Thus PQ = SR and PQ ∥ SR ⇒ PQRS is a parallelogram. [5]
OR (i) In ΔABC, M is mid-point of AB and MD ∥ BC. By converse of Mid-point theorem, D is the mid-point of AC.
(ii) ∠ADM = ∠C = 90° (Corresponding angles) ⇒ MD ⊥ AC.
(iii) ΔADM ≅ ΔCDM (SAS) ⇒ CM = MA = 1⁄2 AB. [5]
Proof: Extend EF to D such that EF = FD and join CD.
ΔAEF ≅ ΔCDF (SAS) ⇒ AE = CD and AB ∥ CD.
Since BE = AE = CD and BE ∥ CD, BCDE is a parallelogram ⇒ EF ∥ BC and EF = 1⁄2 BC.
Quadrilateral Mid-point Proof: Join diagonal AC. In ΔABC, PQ ∥ AC and PQ = 1⁄2AC. In ΔADC, SR ∥ AC and SR = 1⁄2AC. Thus PQ = SR and PQ ∥ SR ⇒ PQRS is a parallelogram. [5]
OR (i) In ΔABC, M is mid-point of AB and MD ∥ BC. By converse of Mid-point theorem, D is the mid-point of AC.
(ii) ∠ADM = ∠C = 90° (Corresponding angles) ⇒ MD ⊥ AC.
(iii) ΔADM ≅ ΔCDM (SAS) ⇒ CM = MA = 1⁄2 AB. [5]
Q35.
(i) Inside surface area of dome = Total cost⁄Rate = 498.96⁄2.00 = 249.48 m2.
(ii) 2πr2 = 249.48 ⇒ 2 × 22⁄7 × r2 = 249.48 ⇒ r2 = (249.48 × 7)⁄44 = 39.69 ⇒ r = 6.3 m.
Volume of dome = 2⁄3 π r3 = 2⁄3 × 22⁄7 × (6.3)3 = 523.908 m3. [5]
(ii) 2πr2 = 249.48 ⇒ 2 × 22⁄7 × r2 = 249.48 ⇒ r2 = (249.48 × 7)⁄44 = 39.69 ⇒ r = 6.3 m.
Volume of dome = 2⁄3 π r3 = 2⁄3 × 22⁄7 × (6.3)3 = 523.908 m3. [5]
SECTION E β ANSWERS
Q36.
- (a) Abscissa of A(3, 4) = 3; Ordinate of C(−3, −4) = −4. [1]
- (b) Point B(−3, 4) lies in Quadrant II; Point D(3, −4) lies in Quadrant IV. [1]
- (c) Length AB = 3 − (−3) = 6 units; Breadth AD = 4 − (−4) = 8 units.
Perimeter = 2(6 + 8) = 28 units; Area = 6 × 8 = 48 sq. units. [2]
Q37.
Cost of fencing = 297 × βΉ25 = βΉ7,425. [2]
- (a) Semi-perimeter s = 300⁄2 = 150 m. [1]
- (b) Ratio sum = 3 + 5 + 7 = 15.
a = 3⁄15 × 300 = 60 m; b = 5⁄15 × 300 = 100 m; c = 7⁄15 × 300 = 140 m. [1] - (c) Area = √150(150 − 60)(150 − 100)(150 − 140) = √150 × 90 × 50 × 10 = 1500√3 m2. [2]
Cost of fencing = 297 × βΉ25 = βΉ7,425. [2]
Q38.
β’ Less than 50 kg: 6 | Less than 60 kg: 18 | Less than 70 kg: 36 | Less than 80 kg: 46 | Less than 90 kg: 50. [2]
- (a) Class width = 50 − 40 = 10. [1]
- (b) Class mark = (60 + 70)⁄2 = 65. [1]
- (c) Number of persons with weight < 70 kg = 6 + 12 + 18 = 36 persons.
Percentage = 36⁄50 × 100 = 72%. [2]
β’ Less than 50 kg: 6 | Less than 60 kg: 18 | Less than 70 kg: 36 | Less than 80 kg: 46 | Less than 90 kg: 50. [2]