IX MAT SP2

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. In a parallelogram, opposite angles are always:

(a) Supplementary   (b) Complementary   (c) Equal   (d) Right angles

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Answer: (c) Equal

Solution:
Step 1: Theorem 8.4 states that in a parallelogram, opposite angles are equal.

2. The area of a triangle with sides a, b, c is given by √[s(s-a)(s-b)(s-c)]. This is:

(a) Euclid's Formula   (b) Heron's Formula   (c) Newton's Formula   (d) Pythagoras Formula

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Answer: (b) Heron's Formula

Formula: √[s(s-a)(s-b)(s-c)]

3. If a chord is equal to the radius of the circle, the angle subtended at the centre is:

(a) 30°   (b) 60°   (c) 90°   (d) 120°

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Answer: (b) 60°

Solution:
Step 1: If chord = radius, the triangle formed with radii is equilateral. Angles in equilateral triangle are 60°.

4. The segment joining mid-points of two sides of a triangle is ____ to the third side.

(a) Perpendicular   (b) Parallel   (c) Oblique   (d) Intersecting

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Answer: (b) Parallel

Solution:
Step 1: This is a direct consequence of the Mid-point Theorem.

5. Chords equidistant from the centre of a circle are:

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

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Answer: (c) Equal in length

Solution:
Step 1: Theorem 9.6 states chords equidistant from the centre are equal.

6. A quadrilateral is cyclic if the sum of opposite angles is 180°. This is:

(a) Th. 9.1   (b) Th. 9.11   (c) Th. 8.8   (d) Th. 8.1

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Answer: (b) Th. 9.11

Solution:
Step 1: Theorem 9.11 is the converse property for cyclic quadrilaterals.

7. Congruent arcs of a circle subtend ______ angles at the centre.

(a) Equal   (b) Different   (c) Complementary   (d) Reflex

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Answer: (a) Equal

Solution:
Step 1: According to point 9 in the Circle summary, congruent arcs subtend equal angles at the centre.

8. Find the area of an isosceles triangle with sides 5 cm, 5 cm and 8 cm.

(a) 24 cm²   (b) 12 cm²   (c) 15 cm²   (d) 20 cm²

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Answer: (b) 12 cm²

Formula: s = (a + b + c) / 2
Step 1: s = (8+5+5)/2 = 9 cm. Area = √[9(9-8)(9-5)(9-5)] = 12 cm².

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

9. Show that each angle of a rectangle is a right angle.

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Formula: ∠A + ∠B = 180°
Step 1: Let ∠A = 90°. Since AD || BC, ∠A + ∠B = 180° (Interior angles).
Step 2: ∠B = 180° - 90° = 90°. Since opposite angles are equal, all angles are 90°.

10. Two circles of radii 5 cm and 3 cm intersect at two points. If distance between centres is 4 cm, find length of common chord.

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Property: Pythagoras Theorem check
Step 1: Since 3² + 4² = 5², the centre of the smaller circle lies on the chord.
Step 2: The radius of the smaller circle (3 cm) is half the chord. Chord length = 6 cm.

11. Find the area of a triangle, two sides of which are 18cm and 10cm and the perimeter is 42cm.

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Formula: s = Perimeter/2
Step 1: Third side = 42 - (18 + 10) = 14 cm. s = 42/2 = 21 cm.
Step 2: Area = √[21(21-18)(21-10)(21-14)] = 21√11 cm².

12. State Theorem 8.6 regarding diagonals of a parallelogram.

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Statement: Theorem 8.6
Step 1: The diagonals of a parallelogram bisect each other.

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

13. Show that ΔABC is divided into four congruent triangles by joining the mid-points D, E, and F.

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Property: Theorem 8.8
Step 1: DE || AC, DF || BC and EF || AB by mid-point theorem.
Step 2: ADEF, BDFE and DFCE are parallelograms.
Step 3: Diagonal of parallelogram divides it into congruent triangles. Thus all 4 are congruent.

14. Prove that if two intersecting chords make equal angles with the diameter through intersection, then chords are equal.

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Property: Congruence of triangles
Step 1: Draw OL ⊥ AB and OM ⊥ CD. In ΔOLE and ΔOME, ∠LEO = ∠MEO (given).
Step 2: EO = EO (common) and ∠OLE = ∠OME = 90°. So ΔOLE ≅ ΔOME.
Step 3: OL = OM (CPCT). Chords equidistant from centre are equal, so AB = CD.

15. A traffic signal board is an equilateral triangle with side 'a'. Find area using Heron's formula.

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Formula: s = 3a/2
Step 1: Area = √[s(s-a)(s-a)(s-a)] = √[(3a/2)(a/2)(a/2)(a/2)].
Step 2: Area = (a²/4)√3.

16. In cyclic quad ABCD, if ∠DBC = 55° and ∠BAC = 45°, find ∠BCD.

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Property: ∠DAB + ∠BCD = 180°
Step 1: ∠CAD = ∠DBC = 55° (Angles in same segment).
Step 2: ∠DAB = 55° + 45° = 100°.
Step 3: ∠BCD = 180° - 100° = 80°.

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

17. Triangular walls of a flyover are 122m, 22m and 120m. Ad earnings are `5000/m²/year. Find rent paid for 3 months.

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Formula: Area = √[s(s-a)(s-b)(s-c)]
Step 1: s = (122+22+120)/2 = 132m.
Step 2: Area = √[132(132-122)(132-22)(132-120)] = 1320 m².
Step 3: Yearly rent = 1320 × 5000 = `66,00,000.
Step 4: Rent for 3 months = 66,00,000 × (3/12) = `16,50,000.

18. ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD.

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Property: Concyclic points
Step 1: Since ∠ABC = ∠ADC = 90° (given), both triangles are in a circle with diameter AC.
Step 2: Points A, B, C, D are concyclic.
Step 3: ∠CAD and ∠CBD are angles subtended by the same arc CD.
Step 4: Angles in the same segment are equal, thus ∠CAD = ∠CBD.

19. Show that three parallel lines l, m, n cutting equal intercepts AB and BC on p, will cut equal intercepts DE and EF on q.

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Property: Mid-point Theorem
Step 1: Join A to F intersecting m at G.
Step 2: In ΔACF, B is mid-point of AC and BG || CF, so G is mid-point of AF.
Step 3: In ΔAFD, G is mid-point of AF and GE || AD.
Step 4: Thus E is mid-point of DF, meaning DE = EF.