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Geometry Questions for RRB NTPC Mathematics

Geometry questions test properties of triangles, circles, quadrilaterals and polygons — congruence, similarity, tangents, chords, angles — plus basic coordinate geometry. Parikshala's question bank has 339 exam-pattern Geometry questions for RRB NTPC, each with a step-by-step solution (103 easy, 165 medium, 71 hard). Mathematics carries 30 of 100 marks in RRB NTPC CBT 1.

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What Geometry Questions Test

  • Triangles: similarity, congruence, centroids and incentres
  • Circles: chords, tangents, and angles in segments
  • Quadrilaterals and polygon angle properties
  • Lines, angles and parallel-line configurations
  • Coordinate geometry: distance, section formula, slopes

Question bank coverage by sub-pattern

Special Points and Lines in Triangles · 25Important Triangle Theorems · 20Rectangle, Square, and Rhombus · 15Slope and Equation of Line · 15Tangent to a Circle · 15Triangle Basics · 15

How Geometry Appears in RRB NTPC

30
Mathematics questions in CBT 1
30
marks for the section (of 100 total)
90 min
CBT 1 duration
—
negative marks per wrong answer

Geometry is part of the Mathematics section (30% of the RRB NTPC syllabus).

Geometry Solved Examples for RRB NTPC

12 exam-pattern geometry questions from the practice bank, each with a full solution.

1Medium

A chord AB of a circle with center O subtends an angle of 110° at the center. If P is a point on the major arc AB, then angle APB is equal to:

  1. A.55°Correct
  2. B.110°
  3. C.125°
  4. D.70°

Solution

The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle. Therefore, angle APB = 1/2 * (angle AOB) = 1/2 * 110° = 55°.

2Medium

In a trapezium ABCD, AB || CD, AB = 20 cm, CD = 10 cm, and the area is 150 cm². What is the height of the trapezium?

  1. A.5 cm
  2. B.10 cmCorrect
  3. C.15 cm
  4. D.20 cm

Solution

Area of trapezium = (1/2) × (sum of parallel sides) × height. 150 = (1/2) × (20 + 10) × h. 150 = (1/2) × 30 × h. 150 = 15h. Therefore, h = 10 cm.

3Medium

In triangle ABC, AD is the internal bisector of angle A, meeting BC at D. If AB = 12 cm, AC = 15 cm, and BC = 18 cm, then find the length of BD.

  1. A.8 cmCorrect
  2. B.10 cm
  3. C.7 cm
  4. D.9 cm

Solution

By the Angle Bisector Theorem, BD/DC = AB/AC = 12/15 = 4/5. Let BD = 4x and DC = 5x. Then BD + DC = BC = 18 cm. So, 4x + 5x = 18, which means 9x = 18, and x = 2. Therefore, BD = 4x = 4 * 2 = 8 cm.

4medium

What is the slope of the line passing through the points (2, 3) and (5, 9)?

  1. A.2Correct
  2. B.3
  3. C.1/2
  4. D.6

Solution

Slope = (y₂ - y₁)/(x₂ - x₁) = (9 - 3)/(5 - 2) = 6/3 = 2.

5medium

Two tangents are drawn from an external point to a circle of radius 5 cm. If the angle between the two tangents is 60°, what is the distance from the external point to the centre?

  1. A.10 cmCorrect
  2. B.5√3 cm
  3. C.10√3 cm
  4. D.5√2 cm

Solution

The line from the external point to the centre bisects the angle between the tangents. So each half-angle = 30°. Using sin 30° = radius/distance: 1/2 = 5/d. Therefore d = 10 cm.

6medium

An angle inscribed in a semicircle is always equal to:

  1. A.90°Correct
  2. B.60°
  3. C.45°
  4. D.180°

Solution

By Thales' theorem, an angle inscribed in a semicircle (subtended by a diameter at the circumference) is always 90°.

7medium

The radius of a circle is 10 cm. What is the length of a chord that is 6 cm away from the centre?

  1. A.16 cmCorrect
  2. B.12 cm
  3. C.14 cm
  4. D.8 cm

Solution

Perpendicular from centre to chord bisects the chord. Let half chord = d. By Pythagoras: d² + 6² = 10². d² = 100 - 36 = 64. d = 8 cm. Full chord = 2 × 8 = 16 cm.

8medium

What is the midpoint of the line segment joining (2, 6) and (8, 4)?

  1. A.(5, 5)Correct
  2. B.(6, 5)
  3. C.(5, 10)
  4. D.(4, 5)

Solution

Midpoint = ((2+8)/2, (6+4)/2) = (10/2, 10/2) = (5, 5).

9medium

In a parallelogram ABCD, angle A = 70°. What is the measure of angle B?

  1. A.110°Correct
  2. B.70°
  3. C.90°
  4. D.140°

Solution

In a parallelogram, consecutive angles are supplementary. So angle B = 180° - 70° = 110°.

10medium

In triangle ABC, D and E are points on AB and AC respectively such that DE is parallel to BC. If AD = 4 cm, DB = 6 cm, and AE = 3 cm, then find EC.

  1. A.4.5 cmCorrect
  2. B.5 cm
  3. C.3.5 cm
  4. D.6 cm

Solution

By Basic Proportionality Theorem (BPT): AD/DB = AE/EC. So 4/6 = 3/EC. EC = 3 × 6/4 = 18/4 = 4.5 cm.

11medium

The sides of a triangle are 13 cm, 14 cm, and 15 cm. What is the area of the triangle?

  1. A.84 cm²Correct
  2. B.90 cm²
  3. C.78 cm²
  4. D.96 cm²

Solution

Using Heron's formula: s = (13+14+15)/2 = 21. Area = √(21 × 8 × 7 × 6) = √(7056) = 84 cm².

12medium

In a triangle, if two angles are 65° and 75°, then what is the third angle?

  1. A.40°Correct
  2. B.50°
  3. C.45°
  4. D.35°

Solution

Sum of angles in a triangle = 180°. Third angle = 180° - 65° - 75° = 40°.

Geometry — Frequently Asked Questions

How much geometry theory do I need before practising?

The core theorem set is small — around 30 properties covering triangles, circles and polygons. Learn those, then let practice questions teach you which property each question pattern wants.

Which geometry areas carry the most weight?

Triangles and circles dominate. Similarity ratios, tangent lengths and angle-in-segment results are the three most repeated question patterns.

How many Geometry practice questions does Parikshala have for RRB NTPC?

339 exam-pattern Geometry questions for RRB NTPC Mathematics, each with a step-by-step solution. Practice is free — no sign-up needed to start.

More RRB NTPC Mathematics Topics

Every topic links to its own practice set with solutions.

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Practice Geometry for RRB NTPC

339 exam-pattern questions with step-by-step solutions. Start free — no sign-up needed.