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

AD is a diameter of a circle and AB is a chord. If $AD = 34 cm\ and\ AB = 30$ cm, the distance of AB from the centre of the circle is

  1. 17cm

  2. 15cm

  3. 4cm

  4. 8cm
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The radius r = 34 / 2 = 17 cm. A chord AB of 30 cm is at a distance d from the center. Using the Pythagorean theorem in the triangle formed by the radius, half the chord, and the distance d: d^2 + (30/2)^2 = 17^2. d^2 + 15^2 = 17^2, d^2 + 225 = 289, d^2 = 64, d = 8 cm.

Multiple choice

The hypotenuse of a right triangle is $6$ m more than twice the shortest side. If the third side is $2$ m less than the hypotenuse, find the hypotenuse of the triangle.
  1. $24$ m
  2. $34$ m
  3. $26$ m
  4. $10$ m
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the shortest side be x. Hypotenuse = 2x + 6. Third side = (2x + 6) - 2 = 2x + 4. By Pythagoras theorem, x^2 + (2x + 4)^2 = (2x + 6)^2. Solving x^2 + 4x^2 + 16x + 16 = 4x^2 + 24x + 36 gives x^2 - 8x - 20 = 0, so (x-10)(x+2) = 0. Thus x = 10, and the hypotenuse is 2(10) + 6 = 26.

Multiple choice

$ABCD$ is a rectangle inscribed in a quadrant of a circle having radius $10cm$. If $AD=2\sqrt { 5 } cm$, find the area of $ABCD$.
  1. $30{ cm }^{ 2 }$
  2. $50{ cm }^{ 2 }$
  3. $40{ cm }^{ 2 }$
  4. $35{ cm }^{ 2 }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the rectangle have sides x and y. Since it is inscribed in a quadrant of radius 10, the diagonal from the origin to the opposite corner is 10. Thus, x^2 + y^2 = 10^2 = 100. Given one side is 2*sqrt(5), (2*sqrt(5))^2 + y^2 = 100 => 20 + y^2 = 100 => y^2 = 80 => y = sqrt(80) = 4*sqrt(5). Area = x * y = 2*sqrt(5) * 4*sqrt(5) = 8 * 5 = 40.

Multiple choice

Two chords $AB$ and $CD$ of lengths $5\ cm$ and $11\ cm$ respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between $AB$ and $CD$ is $6\ cm$, find the radius of the circle.
  1. $\dfrac{5\sqrt{5}}{2}$
  2. $\dfrac{11\sqrt{5}}{2}$
  3. $\dfrac{5\sqrt{5}}{3}$
  4. $\dfrac{5\sqrt{3}}{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the radius be r. Distance from center to chord AB (length 5) is d1 = sqrt(r^2 - 2.5^2). Distance to chord CD (length 11) is d2 = sqrt(r^2 - 5.5^2). Since they are on opposite sides, d1 + d2 = 6. Solving sqrt(r^2 - 6.25) + sqrt(r^2 - 30.25) = 6 leads to r = sqrt(125/4) = 5*sqrt(5)/2.