RBSE Class 9th 2017 Maths-APS-910-E-2017 Previous Year Papers

Annual Exam, 2016 – 2017

Class – 9

Subject – Maths

Time: 3 Hours

Marks: 100

Note: Draw a graph paper in Q. No. 26.

Section A (1 mark each)

1. If sin θ = 4/5, then find the value of cos θ.

   Solution: Using identity sin²θ + cos²θ = 1, cos²θ = 1 - (16/25) = 9/25, so cos θ = 3/5.

2. If three angles of a quadrilateral are 75°, 90° and 75°, then find the fourth angle.

   Solution: Sum of angles in a quadrilateral = 360°. Fourth angle = 360° - (75° + 90° + 75°) = 360° - 240° = 120°.

3. Write the value of each angle in an equilateral triangle.

   Solution: Each angle in an equilateral triangle is 60°.

4. What are the coordinates of the origin?

   Solution: The coordinates of the origin are (0, 0).

5. What is the definition of a trapezium?

   Solution: A trapezium is a quadrilateral with at least one pair of parallel sides.

6. In right-angled triangle ABC, ∠B = 90°, which side is the longest?

   Solution: The side opposite the right angle (hypotenuse) is the longest. Here, side AC is the longest.

7. If y = 2x + 5 and x = 5, then find the value of y.

   Solution: y = 2(5) + 5 = 10 + 5 = 15.

8. In ΔABC, AB = 3 cm, BC = 4 cm, and CA = 5 cm. Write the perimeter of ΔABC.

   Solution: Perimeter = 3 + 4 + 5 = 12 cm.

9. In a two-digit number, the unit digit is y and the tens digit is x. Write the number.

   Solution: The number is 10x + y.

Section B (2 marks each)

10. Multiply by sutra Urdhva Tiryagbhyam: 362 × 43.

    Solution: Using Urdhva Tiryagbhyam (vertically and crosswise):
    362 × 43 = (3×4) | (3×3 + 6×4) | (6×3 + 2×4) | (2×3) = 12 | (9+24) | (18+8) | 6 = 12 | 33 | 26 | 6 = 15566.

11. Factorise x³ - 64 using identities.

    Solution: Using identity a³ - b³ = (a - b)(a² + ab + b²): x³ - 64 = x³ - 4³ = (x - 4)(x² + 4x + 16).

12. In the given figure, ∠1 and ∠2 are linear. If ∠2 - ∠1 = 67°, then find the values of ∠1 and ∠2.

    Solution: Linear pair means ∠1 + ∠2 = 180°. Given ∠2 - ∠1 = 67°. Adding: 2∠2 = 247°, so ∠2 = 123.5°. Then ∠1 = 180° - 123.5° = 56.5°.

13. Prove that 1 unit = 90°.

    Solution: In angle measurement, 1 right angle = 90°. So 1 unit (right angle) = 90°.

14. How much time will be needed for the minute arm of a clock to make a 3π/2 radian angle?

    Solution: 3π/2 radians = 270°. Minute hand moves 360° in 60 minutes, so 270° takes (270/360) × 60 = 45 minutes.

15. Locate √7 on the number line.

    Solution: Draw a number line. Mark 0 and 1. Construct a right triangle with base 2 and height √3 (since 2² + (√3)² = 7). The hypotenuse is √7. Use compass to mark √7 on the number line.

Section C (3 marks each)

16. The ratio of two numbers is 3:4. If 5 is subtracted from each, the ratio becomes 5:7. Find the numbers.

    Solution: Let numbers be 3x and 4x. After subtracting 5: (3x - 5)/(4x - 5) = 5/7. Cross-multiply: 7(3x - 5) = 5(4x - 5) → 21x - 35 = 20x - 25 → x = 10. Numbers are 30 and 40.

17. In the given figure, AB = AC and BE = CD. Prove that ΔABE ≅ ΔACD.

    Solution: In ΔABE and ΔACD: AB = AC (given), BE = CD (given), and ∠ABE = ∠ACD (angles opposite equal sides in isosceles triangle). So by SAS congruence, ΔABE ≅ ΔACD.

18. The line drawn through the midpoint of one side of a triangle, parallel to another side, bisects the third side.

    Solution: In ΔABC, let D be midpoint of AB. Draw DE ∥ BC meeting AC at E. By basic proportionality theorem, AD/DB = AE/EC. Since AD = DB, AE = EC, so E is midpoint of AC. Hence proved.

19. ABC and BDE are two equilateral triangles such that D is the midpoint of side BC. Show that ar(BDE) = 1/4 ar(ABC).

    Solution: Let side of ΔABC = a. Then side of ΔBDE = a/2 (since D is midpoint). Area of equilateral triangle = (√3/4) × side². So ar(ABC) = (√3/4)a², ar(BDE) = (√3/4)(a/2)² = (√3/4)(a²/4) = (1/4) × (√3/4)a² = (1/4) ar(ABC).

20. Find the area of quadrilateral ABCD, given AC = 5 cm, AB = 7 cm, BC = 12 cm, CD = 12 cm, and AD = 9 cm.

    Solution: Divide quadrilateral into two triangles by diagonal AC. For ΔABC: sides 7, 12, 5. Using Heron's formula: s = (7+12+5)/2 = 12, area = √[12(12-7)(12-12)(12-5)] = √[12×5×0×7] = 0 (invalid triangle). Check: 7+5=12, so ΔABC is degenerate. For ΔADC: sides 9, 12, 5. s = (9+12+5)/2 = 13, area = √[13(13-9)(13-12)(13-5)] = √[13×4×1×8] = √416 ≈ 20.4 cm². Total area ≈ 20.4 cm².

21. A box 1 m long, 60 cm wide, and 40 cm deep is to be made. Find the cost of painting at the rate of Rs. 20 per m².

    Solution: Dimensions in meters: length = 1 m, width = 0.6 m, depth = 0.4 m. Surface area = 2(lb + bh + hl) = 2(1×0.6 + 0.6×0.4 + 0.4×1) = 2(0.6 + 0.24 + 0.4) = 2×1.24 = 2.48 m². Cost = 2.48 × 20 = Rs. 49.60.

22. Construct ΔABC given side AB = 6 cm, ∠ABC = 60°, and ∠ACB = 30°.

    Solution: Steps: (1) Draw AB = 6 cm. (2) At B, draw angle 60°. (3) At C (on line from B), angle sum gives ∠BAC = 90°. (4) At A, draw angle 90° to meet line from B at C. (5) Triangle ABC is formed.

23. Construct a grouped frequency distribution table with class size 5 for the data: 3, 1, 8, 8, 0, 44, 27, 9, 8, 35, 33, 27, 30, 4, 43, 94, 99, 43, 77, 7.

    Solution: Range: 0 to 99. Class intervals (size 5): 0-4, 5-9, 10-14, ..., 95-99. Frequency: 0-4: 3 (0,3,1,4), 5-9: 5 (8,8,9,8,7), 10-14: 0, 15-19: 0, 20-24: 0, 25-29: 2 (27,27), 30-34: 2 (33,30), 35-39: 1 (35), 40-44: 3 (44,43,43), 45-49: 0, ..., 75-79: 1 (77), 80-84: 0, 85-89: 0, 90-94: 1 (94), 95-99: 1 (99).

24. Find the cost of making two rectangular boards at the rate of Rs. 10 per square centimeter. Size of rectangular informational sign board is 25 cm × 9 cm.

    Solution: Area of one board = 25 × 9 = 225 cm². Area of two boards = 2 × 225 = 450 cm². Cost = 450 × 10 = Rs. 4500.

Solved by Graph Method