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

AB and CD are two parallel chords of a circle such that AB = 10 cm and CD = 24 cm. If the chords are on the opposite sides of the centre and distance between them is 17 cm, then the area of the circle : AB तथा CD, एक वृत्त की दो समान्तर जीवाएँ हैं, जिनमें AB = 10 सेमी, तथा CD = 24 सेमी है। तदनुसार, यदि वे जीवाएँ केन्द्र के विपरीत दिशाओं में हों और उनके बीच की दूरी 17 सेमी हो, तो उस वृत्त का क्षेत्रफल है :

  1. 144 π cm.2/सेमी.2

  2. 169 π cm.2/सेमी.2

  3. 100 π cm.2/सेमी.2

  4. 121 π cm.2/सेमी.2
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For chord length L at distance d from center in a circle of radius r: L = 2√(r²-d²). AB=10 gives d1=√(r²-25), CD=24 gives d2=√(r²-144). Since chords are on opposite sides: d1+d2=17. Solving gives r=13, so area = πr² = 169π.

Multiple choice

The length of two parallel chords of a circle are 6 cm. and 8 cm. If the smaller chord is at a distance of 4 cm. from the centre, the distance of the other chord from the centre is - एक वृत्त की दो समांतर जीवाएं 6 सेमी. तथा 8 सेमी. लम्बी है। यदि उनमें छोटी जीवा केंद्र से 4 सेमी. की दूरी पर हो तो अन्य जीवा की केंद्र से दूरी कितनी होगी?

  1. 5 cm./सेमी.

  2. 6 cm./सेमी.

  3. 4 cm./सेमी.

  4. 3 cm./सेमी.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For a chord of length L at distance d from the center: (L/2)² + d² = R². For the 6 cm chord at 4 cm: 3² + 4² = R², so R² = 25. For the 8 cm chord: 4² + d² = 25, giving d² = 9 and d = 3 cm. Option A (5 cm) incorrectly uses the radius itself. The larger chord is closer to the center.

Multiple choice

Two equal chords of two circles make angles (60^{\circ}) and (75^{\circ}). Then the ratio of the radii of the circles is?

  1. 5 : 2

  2. 5 : 4

  3. 3 : 2

  4. 2 : 1
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For equal chords of lengths c in circles with radii r1 and r2, making angles θ1 and θ2: c = 2r1sin(θ1/2) = 2r2sin(θ2/2). Therefore r1/r2 = sin(θ2/2)/sin(θ1/2) = sin(75°/2)/sin(60°/2) = sin(37.5°)/sin(30°) = sin(37.5°)/(1/2) = 2sin(37.5°). Using sin(37.5°) = sin(75°/2) and the half-angle formula, sin(37.5°) ≈ 0.608, giving ratio ≈ 1.216, or 5:4.

Multiple choice

The length of the common chord of two circles of radii 30 cm and 40 cm whose centers are 50 cm apart, is (in cm)-

  1. 48

  2. 26

  3. 24

  4. 16
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For two circles with radii 30 and 40 cm and center distance 50 cm, note that 30² + 40² = 50², so the circles intersect at right angles. The common chord length = 2 × (product of radii / center distance) = 2 × (30 × 40)/50 = 48 cm. This can also be derived using perpendicular distance from centers to the chord.

Multiple choice

Two parallel chords of a circle of diameter 20 cm lying on the opposite sides of the center are of lengths 12 cm and 16 cm. The distance between the chords is -

  1. 12

  2. 14

  3. 16

  4. 20
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a chord of length L in a circle of radius r, perpendicular distance from center = sqrt(r² - (L/2)²). With radius 10 cm: for 12 cm chord, distance = sqrt(100-36)=8 cm; for 16 cm chord, distance = sqrt(100-64)=6 cm. Since chords are on opposite sides of center, total distance = 8+6=14 cm.

Multiple choice

If a chord of length 30 cm is at a distance of 8 cm from the center of the circle, then the length of the chord of the same circle which is at a distance of 15 cm from the center is equal to

  1. 12 cm.

  2. 18 cm.

  3. 16 cm.

  4. 20 cm.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a chord at distance d from center: chord length = 2√(r² - d²). From chord 1: 30 = 2√(r² - 64), so r² - 64 = 225, r² = 289. For chord 2 at d = 15: length = 2√(289 - 225) = 2√64 = 2 × 8 = 16 cm. Option C is correct.

Multiple choice

Two tangents are drawn from a point P to a circle at A and B .O is the centre of the circle .If ∠AOP = 600 ,then ∠APB is -

  1. 600

  2. 300

  3. 1200

  4. 900
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Tangents from external point P are equal, so PA = PB. Triangle OAP is right-angled at A (radius perpendicular to tangent). In triangle OAP: angle OAP = 90°, angle AOP = 60°, so angle APO = 30°. Triangle APB is isosceles with PA = PB, so angle APB = 2 × angle APO = 60°. Assuming '600' means 60°.

Multiple choice

Find the area of a right triangle whose base is 3cm and hypotenuse 5 cm.
  1. $6 cm2$
  2. $8 cm2$
  3. $12 cm2$
  4. $8.4 cm2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using Pythagoras theorem, height = sqrt(5^2 - 3^2) = sqrt(25-9) = sqrt(16) = 4 cm. Area = (1/2) * base * height = (1/2) * 3 * 4 = 6 cm^2. For a right triangle with hypotenuse 5 cm and one leg 3 cm, the other leg must be 4 cm (3-4-5 right triangle).

Multiple choice

The length of the common chord of two intersecting circles is 24 cm. If the diameters of the circles are 30 cm. and 26 cm., then the distance between the centres in cm. is -

  1. 13

  2. 14

  3. 25

  4. 16
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For intersecting circles with common chord length L and radii r1 and r2, the distance between centers d = sqrt(r1^2 - (L/2)^2) + sqrt(r2^2 - (L/2)^2). Here r1=15, r2=13, L=24. So d = sqrt(225-144) + sqrt(169-144) = sqrt(81) + sqrt(25) = 9 + 5 = 14 cm.

Multiple choice

In a circle with centre O, PT and PS are tangents drawn to it from point P. If PT = 24 cm and OT = 10 cm Quantity I: The length of PO. Quantity II: Double the length of the hypotenuse of a right angled triangle the other two sides of which are 8 cm and 15 cm respectively.

  1. Quantity II > Quantity I

  2. Quantity I ≥ Quantity II

  3. Quantity I > Quantity II

  4. Quantity I ≤ Quantity II

  5. Quantity I = Quantity II or relationship cannot be established
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Tangents from an external point to a circle are equal. So PT = PS = 24 cm. In right triangle OTP, PT is tangent and OT is radius. By Pythagoras: PO² = PT² + OT² = 24² + 10² = 576 + 100 = 676, so PO = 26 cm. This is Quantity I. For Quantity II: In a right triangle with sides 8 cm and 15 cm, hypotenuse = √(8² + 15²) = √(64 + 225) = √289 = 17 cm. Doubling this gives 34 cm. Since 34 > 26, Quantity II > Quantity I.

Multiple choice

The difference between the areas of the circumcircle and incircle of a regular polygon of n sides with each side of length 2a is.

  1. 2pna2

  2. pna2

  3. (2n+1)pa2

  4. pa2
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For a regular polygon of n sides with side length 2a, the difference between circumcircle area and incircle area equals pi*a^2. The circumcircle radius R = a/sin(pi/n) and incircle radius r = a*cos(pi/n)/sin(pi/n). The difference pi(R^2 - r^2) simplifies to pi*a^2 through trigonometric identity.