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

Choose the correct answers from the alternatives given. If length of three medians of a triangle are respectively $9$ cm, $12$ cm and $15$ cm then what is the area of the triangle in $cm^2$?
  1. $48$
  2. $72$
  3. $96$
  4. $36$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The area of a triangle with medians m1, m2, m3 is (4/3) * sqrt(s(s-m1)(s-m2)(s-m3)), where s = (m1+m2+m3)/2. Here s = (9+12+15)/2 = 18. Area = (4/3) * sqrt(18 * 9 * 6 * 3) = (4/3) * sqrt(2916) = (4/3) * 54 = 72.

Multiple choice

In a $\triangle ABC, AB = 10\ cm, BC = 12\ cm$ and $AC = 14\ cm$. Find the length of median $AD$. If $G$ is the centroid, find length of $GA$.
  1. $\dfrac {5}{3}\sqrt {7}, \dfrac {5}{9}\sqrt {7}$
  2. $5\sqrt {7}, 4\sqrt {7}$
  3. $\dfrac {10}{\sqrt {3}}, \dfrac {8}{3}\sqrt {7}$
  4. $4\sqrt {7}, \dfrac {8}{3}\sqrt {7}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Using Apollonius theorem for median AD: AB^2 + AC^2 = 2(AD^2 + BD^2). 100 + 196 = 2(AD^2 + 36). 296 = 2(AD^2 + 36), 148 = AD^2 + 36, AD^2 = 112, AD = sqrt(112) = 4*sqrt(7). The centroid G divides the median in 2:1 ratio, so GA = 2/3 * AD = 2/3 * 4*sqrt(7) = 8/3*sqrt(7).

Multiple choice

In $\triangle{ABC}$, $AB=9$cm , $BC=10$cm and $AC=13$cm.If $G$ is centroid of $\triangle{ABC}$ and $D$ is midpoint of $BC$ then the length of $GD$ is
  1. $25$
  2. $20$
  3. $\dfrac{10}{3}$
  4. $\dfrac{20}{3}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In triangle ABC, D is the midpoint of BC, so AD is the median to side BC. Using Apollonius theorem: AB^2 + AC^2 = 2(AD^2 + BD^2). With AB=9, AC=13, BC=10 (BD=5): 81 + 169 = 2(AD^2 + 25) => 250 = 2(AD^2 + 25) => 125 = AD^2 + 25 => AD^2 = 100 => AD = 10. The centroid G divides the median AD in ratio 2:1, so GD = 1/3 * AD = 10/3.

Multiple choice

In a triangle ABC, which of the option does not represent the length of the median in terms of its sides.
  1. $\dfrac{\sqrt{2b^2+2c^2-a^2}}{2}$
  2. $\dfrac{\sqrt{2b^2+2c^2-2a^2}}{2}$
  3. $\dfrac{\sqrt{2b^2+2a^2-c^2}}{2}$
  4. $\dfrac{\sqrt{2c^2+2a^2-b^2}}{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The length of the median m_a to side a is given by the formula m_a = 0.5 * sqrt(2b^2 + 2c^2 - a^2). Option B incorrectly subtracts 2a^2 instead of a^2, making it the incorrect representation.

Multiple choice

A square is inscribed in a circle. If ${ p }{ 1 }$ is the probability that a randomly chosen point of the circle lies within the square and ${ p }{ 2 }$ is the probability taht the point lies outside the square, then
  1. ${ p }_{ 1 }={ p }_{ 2 }$
  2. ${ p }_{ 1 }>{ p }_{ 2 }$ and $\displaystyle { { p }_{ 1 } }^{ 2 }-{ { p }_{ 2 } }^{ 2 }<\frac { 1 }{ 3 } $
  3. ${ p }_{ 1 }<{ p }_{ 2 }$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Area of circle = pi*r^2. Side of inscribed square = r*sqrt(2), Area = 2*r^2. p1 = 2/pi approx 0.636. p2 = 1 - 2/pi approx 0.364. Clearly p1 > p2. p1^2 - p2^2 = (p1-p2)(p1+p2) = (4/pi - 1)(1) = 4/pi - 1 approx 1.27 - 1 = 0.27. Since 0.27 < 0.33, the condition holds.

Multiple choice

The equation of a circle whose centre is $(3, -1)$ and which intercept chord of $6$ units length on straight line $2x - 5y + 18 = 0$ is
  1. $x^2 + y^2 - 6x + 2y - 28 = 0$
  2. $x^2 + y^2 + 6x - 2y - 28 = 0$

  3. Both of them

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The distance from the center (3, -1) to the line 2x - 5y + 18 = 0 is d = |2(3) - 5(-1) + 18| / sqrt(2^2 + (-5)^2) = |6 + 5 + 18| / sqrt(29) = 29 / sqrt(29) = sqrt(29). The chord length is 6, so half-length is 3. Radius r satisfies r^2 = d^2 + 3^2 = 29 + 9 = 38. Equation: (x-3)^2 + (y+1)^2 = 38, which is x^2 - 6x + 9 + y^2 + 2y + 1 = 38, or x^2 + y^2 - 6x + 2y - 28 = 0.