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
How Geometry Appears in SSC CPO
Geometry is part of the Quantitative Aptitude section (25% of the SSC CPO syllabus).
Geometry Solved Examples for SSC CPO
12 exam-pattern geometry questions from the practice bank, each with a full solution.
Two equal circles of radius 4 cm intersect such that each passes through the centre of the other. The length of the common chord is:
- A.2√3 cm
- B.4√3 cmCorrect
- C.4√2 cm
- D.8 cm
Solution
Let the centers of the circles be A and B, and the points of intersection be C and D. Since each circle passes through the center of the other, AB = AC = BC = radius = 4 cm. Triangle ABC is equilateral. Let M be the midpoint of the common chord CD. AM is perpendicular to CD and bisects CD. In triangle AMC, AM = AC * cos(30) = 4 * (√3/2) = 2√3. Also, MC = AC * sin(30) = 4 * (1/2) = 2. Since M is the midpoint of CD, CD = 2 * MC = 2 * 2 = 4√3 cm.
In a triangle ABC, if angle A = 70° and angle B = 50°, then angle C is:
- A.50°
- B.60°Correct
- C.70°
- D.80°
Solution
The sum of angles in a triangle is 180°. Therefore, angle C = 180° - (angle A + angle B) = 180° - (70° + 50°) = 180° - 120° = 60°.
ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and ∠ADC = 130°, then ∠BAC is equal to:
- A.40°Correct
- B.30°
- C.45°
- D.50°
Solution
Since ABCD is a cyclic quadrilateral, ∠ADC + ∠ABC = 180°. Given ∠ADC = 130°, so ∠ABC = 180° - 130° = 50°. Since AB is a diameter, ∠ACB = 90°. In triangle ABC, ∠BAC + ∠ABC + ∠ACB = 180°. Therefore, ∠BAC + 50° + 90° = 180°, which implies ∠BAC = 180° - 140° = 40°.
The circumcentre of a right-angled triangle lies at:
- A.The midpoint of the hypotenuseCorrect
- B.The vertex of the right angle
- C.Inside the triangle
- D.Outside the triangle
Solution
The circumcentre of a right-angled triangle is always at the midpoint of the hypotenuse, because the hypotenuse is the diameter of the circumscribed circle.
Two similar triangles have their corresponding sides in the ratio 3:5. What is the ratio of their areas?
- A.9:25Correct
- B.3:5
- C.6:10
- D.27:125
Solution
For similar triangles, the ratio of areas = square of the ratio of corresponding sides = 3² : 5² = 9 : 25.
In a triangle, if two angles are 65° and 75°, then what is the third angle?
- A.40°Correct
- B.50°
- C.45°
- D.35°
Solution
Sum of angles in a triangle = 180°. Third angle = 180° - 65° - 75° = 40°.
An angle inscribed in a semicircle is always equal to:
- A.90°Correct
- B.60°
- C.45°
- D.180°
Solution
By Thales' theorem, an angle inscribed in a semicircle (subtended by a diameter at the circumference) is always 90°.
What is the distance between the points (3, 4) and (7, 1)?
- A.5Correct
- B.6
- C.7
- D.4
Solution
Distance = √((7-3)² + (1-4)²) = √(16 + 9) = √25 = 5.
In a circle, if an arc subtends an angle of 60° at the centre, what angle does it subtend at any point on the remaining part of the circle?
- A.30°Correct
- B.60°
- C.120°
- D.90°
Solution
The angle subtended by an arc at the centre is twice the angle subtended at any point on the remaining circumference. So the inscribed angle = 60°/2 = 30°.
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?
- A.10 cmCorrect
- B.5√3 cm
- C.10√3 cm
- 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.
In a triangle, an exterior angle is 120°. If one of the interior opposite angles is 50°, then what is the other interior opposite angle?
- A.70°Correct
- B.60°
- C.50°
- D.80°
Solution
Exterior angle = sum of the two interior opposite angles. So 120° = 50° + other angle. Other angle = 120° - 50° = 70°.
What is the midpoint of the line segment joining (2, 6) and (8, 4)?
- A.(5, 5)Correct
- B.(6, 5)
- C.(5, 10)
- D.(4, 5)
Solution
Midpoint = ((2+8)/2, (6+4)/2) = (10/2, 10/2) = (5, 5).
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 SSC CPO?
344 exam-pattern Geometry questions for SSC CPO Quantitative Aptitude, each with a step-by-step solution. Practice is free — no sign-up needed to start.