Quantitative Aptitude

Geometry

1,950 Questions
All Circle Tangents and Chords 772 Coordinate Geometry 360 Circle and Arc Measures 216 Triangle Centers and Medians 175 Composite and Inscribed Shapes 174 Similarity, Congruence, and Proportionality 118 Right Triangles and Pythagorean Theorem 100 Standard 2D Mensuration 22 Polygon and Angle Sums 10 3D Solid Mensuration 3

Geometry Questions

Multiple choice

An equilateral ΔABC has AD as its median. Find the length of OD where ‘O’ is the centroid if the area of the triangle is √3 cm2. एक समबाहु त्रिभुज ΔABC की माध्यिका AD है | OD की लम्बाई ज्ञात कीजिये, जहाँ ‘O’ केन्द्रक है, यदि त्रिभुज का क्षेत्रफल √3 सेमी2 है|

  1. 1/2

  2. √3

  3. 1/√3

  4. 2/√3
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Area of equilateral triangle = (√3/4) × side². Given area = √3, so (√3/4) × a² = √3. Dividing by √3: a²/4 = 1, so a² = 4 and a = 2 cm. In equilateral triangle, median = altitude = (√3/2) × side = (√3/2) × 2 = √3 cm. Centroid divides median in ratio 2:1 (vertex to centroid : centroid to midpoint). OD = (1/3) × AD = (1/3) × √3 = 1/√3 cm.

Multiple choice

A line through the circle O of a circle of radius 7 cm cuts the tangent, at a point P on the circles, at Q such that PQ = 24 cm.Find OQ.

  1. 12 cm

  2. 23 cm

  3. 12.5 cm

  4. 25 cm
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The tangent at P is perpendicular to radius OP, so triangle OPQ is right-angled at P. OP = 7 cm (radius), PQ = 24 cm (given). Using Pythagoras theorem: OQ² = OP² + PQ² = 7² + 24² = 49 + 576 = 625. Therefore OQ = sqrt(625) = 25 cm. This is a standard application of tangent-radius property and Pythagoras theorem.

Multiple choice

In a triangle ABC, AB = 6 cm, BC = 8 cm, and AC = 10 cm. The length of the median bisecting the shortest side is (approx.) एक त्रिभुज ABC में, AB = 6 सेमी, BC = 8 सेमी, और AC = 10 सेमी । सबसे छोटी भुजा को समद्विभाजित करने वाली माध्यिका की लंबाई है (लगभग)

  1. 10 cm./सेमी.

  2. 8.5 cm./सेमी.

  3. 9 cm./सेमी.

  4. 9.5 cm./सेमी.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Shortest side is 6 cm. Using Apollonius theorem for median: m² = (2b² + 2c² - a²)/4 where a is the side with median. Here m² = (2×64 + 2×100 - 36)/4 = 292/4 = 73. So m = √73 ≈ 8.54 cm ≈ 8.5 cm.

Multiple choice

The length of the diameter of the circle is 10 cm and length of the chord drawn in a circle is 6 cm. Find the distance (in cm) of chord from the center. वृत्त के व्यास की लंबाई 10 सेमी है तथा एक वृत्त में खींची गई जीवा की लम्बाई 6 सेमी है । केंद्र से जीवा की दूरी (सेमी में) ज्ञात कीजिये।

  1. 10

  2. 8

  3. 6

  4. 4
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For a circle of radius 5 cm (diameter 10 cm), a chord of length 6 cm is 3 cm from each end (half of chord length). Using the perpendicular from center to the chord, we get a right triangle with hypotenuse = 5 cm and one leg = 3 cm. The distance from center = √(5² - 3²) = √(25 - 9) = √16 = 4 cm. This uses the property that the perpendicular from the center bisects the chord.

Multiple choice

Two circles touch each other externally at a point. Then the number of direct common tangents, the circles may have is -

  1. 2

  2. 4

  3. 3

  4. 1
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

When two circles touch externally at one point, there are exactly 3 common tangents: 2 direct common tangents (which do not intersect the line joining centers) and 1 transverse common tangent (which passes through the point of contact). The question asks for direct common tangents, but the expected answer appears to include all common tangents.

Multiple choice

Two chords PQ and RS of a circle intersect at D inside a circle. If PD = 8.4 cm, DQ = 2.4 cm, SD = 1.2 cm, the length of RS is equal to / एक वृत्त के अंदर D बिंदु पर दो जिह्वा PQ और RS एक दुसरे को काटते हैं। यदि PD = 8.4 सेमी, DQ = 2.4 सेमी, SD = 1.2 सेमी, RS की लंबाई बराबर है

  1. 18 cm

  2. 11 cm

  3. 16.8 cm

  4. 17 cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

By the intersecting chords theorem: PD × DQ = SD × DR. Given PD = 8.4 cm, DQ = 2.4 cm, SD = 1.2 cm. So 8.4 × 2.4 = 1.2 × DR, giving DR = 20.16/1.2 = 16.8 cm. Total RS = SD + DR = 1.2 + 16.8 = 18 cm. Option A is correct.

Multiple choice

(\text{In }\triangle ABC, \angle B = 90^{\circ}\text{ and AC is the hypotenuse. D is its circumcenter, and } AB = 3 \text{ cms, } BC = 4 \text{ cms. Find the value of BD.})

  1. 3 cm

  2. 4 cm

  3. 5 cm

  4. 2.5 cm
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In a right triangle, the circumcenter is the midpoint of the hypotenuse. For triangle ABC with angle B = 90°, AC is the hypotenuse. AC = sqrt(3² + 4²) = 5 cm by Pythagoras theorem. The circumcenter D is the midpoint of AC, so BD = half of AC = 5/2 = 2.5 cm. This is the circumradius of a right triangle.

Multiple choice

AB is the chord of a circle with centre O and DOC is a line segment originating from a point D on the circle and intesecting AB produced at C such that BC = OD. If ∠BCD = 200, then ∠AOD = ?

  1. 200

  2. 300

  3. 400

  4. 600
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

This geometry problem requires analyzing angles in a circle with chord AB and point D on the circumference. Given BC = OD and ∠BCD = 20°, we can work through isosceles triangles and circle theorems to find that ∠AOD = 60°. The large values (200°, 300°, 400°, 600°) suggest this may use a different notation or the diagram is needed for clarity.

Multiple choice

A inscribe circle in a ΔABC in such way it can touch the sides AB, BC,CA at P,Q, and R respectively and AP = 7cm , CR = 6cm and BC = 12 cm then find the perimeter of the ΔABC. एक Δ ABC में एक अंत:वृत्त इसप्रकार है की वह भुजा AB,BC और CA को क्रमशः P, Q, और R पर स्पर्श करता हैं और AP = 7cm, CR = 6 सेमी और BC = 12 सेमी है तो Δ ABC की परिमाप ज्ञात कीजिए।

  1. 28

  2. 38

  3. 12

  4. 13
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For an incircle touching sides AB, BC, CA at P, Q, R respectively: AP = AR = 7 cm (tangents from A), CR = CQ = 6 cm (tangents from C), BQ = BP = 12 cm (tangents from B, since BC = BQ+CQ = BQ+6 = 12, so BQ = 6). Wait, BC = 12 is given. BQ+CQ = 12. If CQ=6, then BQ=6. So BP=6, BQ=6. Perimeter = AB+BC+CA = (AP+PB)+(BQ+CQ)+(CR+RA) = (7+6)+(6+6)+(6+7) = 13+12+13 = 38 cm. ✓

Multiple choice

An arc of (60^o) in one circle is double the arc in a second circle whose radius is three times that of the first circle. What is the degree measure of the arc of the second circle?
  1. $\(30^o\)$
  2. $\(20^o\)$
  3. $\(10^o\)$
  4. $\(15^o\)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Arc length = radius × angle (in radians). Let first circle have radius r and angle 60°. Its arc length = r × 60°. Second circle has radius 3r and unknown angle θ. Its arc length = 3r × θ. Given that first arc is double the second: r × 60° = 2(3r × θ). Solving: 60° = 6θ, so θ = 10°.