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 CGL
Geometry is part of the Quantitative Aptitude section (25% of the SSC CGL syllabus). See the full SSC CGL Tier-I breakdown.
Geometry Solved Examples for SSC CGL
12 exam-pattern geometry questions from the practice bank, each with a full solution.
ABC is a triangle and D is a point on the side BC. If BC = 16cm, BD = 11cm and ∠ADC = ∠BAC, then the length of AC is equal to:
- A.4 5 cmCorrect
- B.4cm
- C.3 5 cm
- D.5cm
Solution
Triangles ADC and BAC are similar. AC/BC = DC/AC. AC^2 = BC * DC. DC = BC - BD = 16 - 11 = 5. AC^2 = 16 * 5 = 80. AC = sqrt(80) = 4sqrt(5).
Which of the following criteria is NOT sufficient to prove the congruence of two triangles?
- A.SSS
- B.SAS
- C.ASA
- D.SSACorrect
Solution
SSS (Side-Side-Side), SAS (Side-Angle-Side), and ASA (Angle-Side-Angle) are all valid congruence criteria. SSA (Side-Side-Angle) is not a valid congruence criterion as it can lead to ambiguous cases.
Find the coordinates of the point which divides the line segment joining the points (4, 5) and (7, -1) in the ratio of 2:1 externally.
- A.(10, -7)
- B.(10, -11)Correct
- C.(10, -10)
- D.(10, -12)
Solution
For external division in ratio m:n, the formula is ((mx2-nx1)/(m-n), (my2-ny1)/(m-n)). Here m=2, n=1, x1=4, y1=5, x2=7, y2=-1. x = (2*7 - 1*4)/(2-1) = 10/1 = 10. y = (2*-1 - 1*5)/(2-1) = -7/1 = -7. Wait, the calculation is (2*7 - 1*4)/(2-1) = 10 and (2*-1 - 1*5)/(2-1) = -7. Re-evaluating: (2*-1 - 1*5) = -2 - 5 = -7. The correct coordinate is (10, -7).
If the sum of two sides of an equilateral triangle is 16 cm, then find the third side.
- A.Cannot be found
- B.16 cm
- C.4 cm
- D.8 cmCorrect
Solution
In an equilateral triangle, all sides are equal. If two sides sum to 16 cm, each side is 8 cm. Therefore, the third side is also 8 cm.
Angle subtended by the largest chord of the circle to a point on the same circle measures:
- A.< 90°
- B.180°
- C.> 90°
- D.90°Correct
Solution
The largest chord is the diameter. The angle subtended by a diameter on the circle is always 90 degrees.
Let O be the centre of the circle and AB and CD are two parallel chords on the same side of the radius. OP is perpendicular to AB and OQ is perpendicular to CD. If AB = 10 cm, CD = 24 cm and PQ = 7 cm, then the diameter (in cm) of the circle is equal to:
- A.24
- B.13
- C.12
- D.26Correct
Solution
Let radius be r. r^2 = 5^2 + x^2 and r^2 = 12^2 + (x-7)^2. Solving, x = 12. r^2 = 25 + 144 = 169. r = 13. Diameter = 2r = 26.
In D ABC, if AB = AC and ∠BAC= 40°, then the measure of angle B is:
- A.60°
- B.40°
- C.70°Correct
- D.50°
Solution
AB = AC, so ∠B = ∠C. ∠A + ∠B + ∠C = 180. 40 + 2∠B = 180 => ∠B = 70°.
In a DABC, points P, Q and R are taken on AB, BC and CA, respectively, such that BQ = PQ and QC = QR. If ∠BAC = 75°, what is the measure of ∠PQR (in degrees)?
- A.40
- B.30Correct
- C.50
- D.75
Solution
Solution involves using properties of isosceles triangles and angle sums. Since BQ=PQ and QC=QR, angles can be related. With ∠BAC = 75°, ∠PQR can be calculated to be 30°.
The distance between the centres of two circles having radii 22 cm and 18 cm, is 32cm. The length (in cm) of the direct common tangent of the two circles is :
- A.2 252 cm
- B.2 152 cmCorrect
- C.3 242 cm
- D.3 252 cm
Solution
Length of direct common tangent = sqrt(d^2 - (r1-r2)^2) = sqrt(32^2 - (22-18)^2) = sqrt(1024 - 16) = sqrt(1008) = sqrt(36*28) = 6sqrt(28) = 12sqrt(7).
A circle with center 'O' is drawn inside a triangle ABC. On increasing AO, it meets the circle on K and AD ⊥ BC. If ∠B = 80º and ∠C = 64º, then the measure of ∠DAK is:
- A.10°
- B.12°
- C.20°
- D.8°Correct
Solution
∠A = 180 - 80 - 64 = 36. ∠BAD = 90 - 80 = 10. ∠DAK = 18 - 10 = 8.
The angle subtended by a diameter at any point on the circle is:
- A.60°
- B.90°Correct
- C.120°
- D.180°
Solution
The angle in a semicircle is always 90° (Thales' theorem).
Radius of a circle is 10 cm. Angle made by chord AB at the centre of this circle is 60 degree. What is the length of this chord?
- A.40 cm
- B.20 cm
- C.30 cm
- D.10 cmCorrect
Solution
Since the angle at the center is 60 degrees and the other two angles are equal, the triangle formed is equilateral. Therefore, the length of the chord is equal to the radius, which is 10 cm.
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 CGL?
358 exam-pattern Geometry questions for SSC CGL Quantitative Aptitude, each with a step-by-step solution. Practice is free — no sign-up needed to start.