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 is the chord of contact of tangents drawn from a point $(6,8)$ to the circle $\displaystyle x^{2}+y^{2}= r^{2}.$ If the area of the triangle $PAB$ be maximum, then radius $r$ of the circle is ?
  1. $5$
  2. $ 5\sqrt 3$
  3. $ 10 $
  4. $ 8 $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a point (x1, y1) outside a circle x^2 + y^2 = r^2, the chord of contact is xx1 + yy1 = r^2. Here, 6x + 8y = r^2. The distance from the origin to this line is d = r^2 / sqrt(6^2 + 8^2) = r^2 / 10. The length of the chord AB is 2 * sqrt(r^2 - d^2). Area = 1/2 * base * height = 1/2 * (2 * sqrt(r^2 - d^2)) * (sqrt(x1^2 + y1^2) - d) = sqrt(r^2 - d^2) * (10 - d). Substituting d = r^2/10, Area = sqrt(r^2 - r^4/100) * (10 - r^2/10) = (r/10) * sqrt(100 - r^2) * (10 - r^2/10) = (r/100) * (100 - r^2)^(3/2). Maximize by setting derivative to 0, which occurs at r = 5.

Multiple choice

$I:$ The sides of a rectangle are $(6-x)\ cm$ and $(x-3)\ cm$. If its area is maximum, then $x =4.5$ $II:$ The maximum area of the rectangle inscribed in a circle of radius $5\ cm$ is $50\ sq.cm$. Which of the above statements are true?
  1. only $I$
  2. only $II$
  3. both $I$ and $II$
  4. neither $I$ nor $II$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

I: Area = (6-x)(x-3) = -x^2 + 9x - 18. Max area at x = -b/2a = -9/-2 = 4.5. II: For a rectangle inscribed in a circle of radius R, max area is a square with side R*sqrt(2). Area = (5*sqrt(2))^2 = 50. Both are true.

Multiple choice

Let position vector of points A,B and C of triangle ABC respectively will be $\hat { i } +\hat { j } 2+\hat { k } $ , $\hat { i } +2\hat { j } +\hat { k } $ and $2\hat { i } +\hat { j } +\hat { k } $ . let ${ l }{ 1 }.{ l }{ 2 }\quad and\quad { l }{ 3 }$ be the lengths of perpendiculars drawn from the orthocentre 'O' on the side AB . BC and CA . then $({ l }{ 1 }+{ l }{ 2 }+{ l }{ 3 })$ equals-
  1. $\frac { 2 }{ \sqrt { 6 } } $
  2. $\frac { 3 }{ \sqrt { 6 } } $
  3. $\frac { \sqrt { 6 } }{ 2 } $
  4. $\frac { \sqrt { 6 } }{ 3 } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice

A point $O$ is the centre of a circle circumscribed about a triangle $ABC$, then $\vec{OA}\sin 2A + \vec{OB}\sin 2B + \vec{OC} \sin 2C $ is equal to
  1. $(\vec{OA} + \vec{OB} + \vec{OC})\sin 2A$
  2. $3\vec{OG}$, where $G$ is the centroid of triangle $ABC$
  3. $\vec 0$

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

Using the properties of a circumcenter O, the vector OA sin 2A + OB sin 2B + OC sin 2C is a known identity in triangle geometry. Since sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C, the vector sum simplifies to the zero vector.

Multiple choice

If points represented by the complex numbers a,b,c lie on a circle with centre O and radius r. The tangent at c cuts the chord joining the points a,b at z. Then the value of $\frac { { a }^{ -1 }+{ b }^{ -1 }-2{ c }^{ -1 } }{ { a }^{ -1 }{ b }^{ -1 }-{ c }^{ -2 } } $.

  1. z

  2. 2z

  3. 3z

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

This is a known property in complex geometry where the tangent at a point on a circle relates to the inversion of points. The expression simplifies to the complex number z representing the intersection point.

Multiple choice

In a triangle ABC, bisector of angle C meets the side AB at D and circumcentre at E. The maximum value of CD. DE is equal to
  1. $a^2 /4$
  2. $b^2 /4$
  3. $c^2 /4$
  4. $(a + b)^2/4$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In a triangle, the product of the segment of the angle bisector (CD) and the segment from the bisector to the circumcircle (DE) is related to the sides of the triangle. By the power of a point theorem, CD * DE = AD * DB. For a triangle, this product is maximized as c^2 / 4.

Multiple choice

In $\triangle ABC,\quad AD\, \, \, $ is median and E is the mid-point of AD. If BE is extended. it meets AC in F. AB=8 cm. BC = 21 cm and AC = 15 cm. then AF=

  1. 7 cm

  2. 3 cm

  3. 12 cm

  4. 5 cm
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In triangle ABC, AD is a median. E is midpoint of AD. BE extended meets AC at F. By Van Schooten's or Menelaus theorem, or using mass point geometry: assign mass 1 to A and 1 to C (since AD is median, D is midpoint of BC, mass at D is 2). E is midpoint of AD, so mass at E is 2. Since BEF is a line, mass at F must balance. AF/FC = 1/2. Since AC = 15, AF = 15 * (1/3) = 5 cm.