IX MAT SP1

KENDRIYA VIDYALAYA STEEL PLANT - CLASS IX MATHS SAMPLE PAPER

TIME: 1.5 Hours | TOTAL MARKS: 40

Section – A: Multiple Choice Questions (8 x 1 = 8 Marks)

1. According to Theorem 8.1, a diagonal of a parallelogram divides it into two:

(a) Similar triangles   (b) Equilateral triangles   (c) Congruent triangles   (d) Isosceles triangles

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Answer: (c) Congruent triangles

Solution:
Step 1: Refer to Theorem 8.1 which states a diagonal divides the parallelogram into two congruent triangles.

2. In Heron's formula, the semi-perimeter 's' is calculated as:

(a) (a+b+c)/3   (b) (a+b+c)/2   (c) a+b+c   (d) abc/2

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Answer: (b) (a+b+c)/2

Formula: s = (a + b + c) / 2
Step 1: 's' stands for semi-perimeter, which is half of the total perimeter.

3. The sum of either pair of opposite angles of a cyclic quadrilateral is:

(a) 90°   (b) 180°   (c) 270°   (d) 360°

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Answer: (b) 180°

Formula: ∠A + ∠C = 180°
Step 1: Theorem 9.10 states opposite angles of a cyclic quadrilateral sum to 180°.

4. Equal chords of a circle subtend equal angles at the:

(a) Circumference   (b) Centre   (c) Tangent   (d) Segment

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Answer: (b) Centre

Solution:
Step 1: Theorem 9.1 states equal chords of a circle subtend equal angles at the centre.

5. The perpendicular from the centre of a circle to a chord ________ the chord.

(a) Trisects   (b) Doubles   (c) Bisects   (d) Parallel to

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Answer: (c) Bisects

Solution:
Step 1: Theorem 9.3 states the perpendicular from the centre bisects the chord.

6. Diagonals of a rectangle are:

(a) Perpendicular   (b) Unequal   (c) Equal   (d) Parallel

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Answer: (c) Equal

Solution:
Step 1: According to the properties of quadrilaterals, diagonals of a rectangle are equal and bisect each other.

7. Calculate the semi-perimeter of a triangle with sides 40 m, 32 m, and 24 m.

(a) 96 m   (b) 48 m   (c) 80 m   (d) 64 m

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Answer: (b) 48 m

Formula: s = (a + b + c) / 2
Step 1: s = (40 + 32 + 24) / 2 = 96 / 2 = 48 m.

8. The angle subtended by a diameter at any point on the circle is:

(a) 45°   (b) 60°   (c) 90°   (d) 180°

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Answer: (c) 90°

Solution:
Step 1: Angle in a semicircle is a right angle (90°).

Section – B: Short Answer Questions (4 x 2 = 8 Marks)

9. Find the area of a triangle, two sides of which are 8 cm and 11 cm and the perimeter is 32 cm.

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Formula: Area = √[s(s-a)(s-b)(s-c)]
Step 1: Find third side c = 32 - (8 + 11) = 13 cm.
Step 2: s = 32/2 = 16 cm. Area = √[16(16-8)(16-11)(16-13)] = 8√30 cm².

10. Prove that in a parallelogram, opposite sides are equal.

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Property: Theorem 8.2
Step 1: Diagonal AC divides parallelogram ABCD into ΔABC and ΔCDA.
Step 2: Since ΔABC ≅ ΔCDA (ASA rule), corresponding sides AB = DC and AD = BC.

11. Show that the angle subtended by an arc at the centre is double the angle subtended at the remaining part.

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Formula: ∠POQ = 2∠PAQ
Step 1: exterior angle of a triangle equals the sum of two interior opposite angles.
Step 2: In ΔOAQ, OA = OQ, so ∠OAQ = ∠OQA. Thus ∠BOQ = 2∠OAQ.

12. State Theorem 8.9 (Converse of Mid-point Theorem).

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Statement: Theorem 8.9
Step 1: The line drawn through the mid-point of one side of a triangle, parallel to another side, bisects the third side.

Section – C: Short Answer Questions (4 x 3 = 12 Marks)

13. Show that the diagonals of a rhombus are perpendicular to each other.

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Property: Linear Pair Axiom
Step 1: In ΔAOD and ΔCOD, OA = OC, OD = OD, and AD = CD.
Step 2: By SSS rule, ΔAOD ≅ ΔCOD. So, ∠AOD = ∠COD.
Step 3: ∠AOD + ∠COD = 180° (Linear pair), thus ∠AOD = 90°.

14. Find the area of an equilateral triangle with side 10 cm.

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Formula: Area = √[s(s-a)(s-b)(s-c)]
Step 1: s = (10+10+10)/2 = 15 cm.
Step 2: Area = √[15(15-10)(15-10)(15-10)] = √[15 × 5 × 5 × 5].
Step 3: Area = 25√3 cm².

15. Prove that the line segment joining the mid-points of two sides of a triangle is parallel to the third side.

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Theorem: Theorem 8.8
Step 1: Draw CD || BA. Show ΔAEF ≅ ΔCDF (ASA Rule).
Step 2: BE = AE = DC, so BCDE is a parallelogram.
Step 3: This implies EF || BC and EF = 1/2 BC.

16. Prove that angles in the same segment of a circle are equal.

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Formula: ∠POQ = 2∠PAQ = 2∠PCQ
Step 1: Let A and C be points in the same segment.
Step 2: From Theorem 9.7, the angle at the centre is double the angle at the circumference.
Step 3: Thus, 2∠PAQ = 2∠PCQ, so ∠PAQ = ∠PCQ.

Section – D: Long Answer Questions (3 x 4 = 12 Marks)

17. A triangular park ABC has sides 120m, 80m and 50m. Dhania has to fence it with wire leaving 3m for a gate. Find the area and cost of fencing at `20 per metre.

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Formula: s = (a+b+c)/2 ; Area = √[s(s-a)(s-b)(s-c)]
Step 1: 2s = 50 + 80 + 120 = 250m, so s = 125m.
Step 2: Area = √[125(125-120)(125-80)(125-50)] = 375√15 m².
Step 3: Perimeter = 250m. Wire needed = 250 - 3 = 247m.
Step 4: Cost = 247 × 20 = `4940.

18. Show that the bisectors of angles of a parallelogram form a rectangle.

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Formula: ∠A + ∠D = 180°
Step 1: In ΔASD, ∠DAS + ∠ADS = 1/2(∠A + ∠D) = 1/2(180°) = 90°.
Step 2: By angle sum property, ∠DSA = 90°. So ∠PSR = 90° (vertically opposite).
Step 3: Similarly, all other angles are shown to be 90°.
Step 4: A quadrilateral with all angles as right angles is a rectangle.

19. In parallelogram ABCD, E and F are mid-points of AB and CD. Show that AF and EC trisect diagonal BD.

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Property: Mid-point Theorem
Step 1: Show AECF is a parallelogram because AE = FC and AE || FC.
Step 2: In ΔDQC, F is mid-point of DC and FP || CQ, so P is mid-point of DQ (DP = PQ).
Step 3: In ΔABP, E is mid-point of AB and EQ || AP, so Q is mid-point of PB (PQ = QB).
Step 4: Thus DP = PQ = QB, hence the diagonal is trisected.