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

A tangent $AB$ at a point $A$ of a circle of radius $5$ cm. meets a line through the centre $O$ at a point $B$, so that $OB$ $=$ $12$ cm. Then find the length of $AB$.
  1. $\sqrt{119}$ cm
  2. $119$ cm
  3. $\sqrt{126}$ cm
  4. $112$ cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The radius OA is perpendicular to the tangent AB at point A, forming a right-angled triangle OAB. By the Pythagorean theorem, AB^2 + OA^2 = OB^2, so AB^2 + 5^2 = 12^2. AB^2 = 144 - 25 = 119, so AB = sqrt(119).

Multiple choice

If pair of tangents $PA$ and $PB$ are draw from point $P(2,3)$ to the circle ${x}^{2}+{y}^{2}+4x+2y+1=0$ area $PAOB$, $O$ is centre.
  1. $2\sqrt{7}$
  2. $\sqrt{7}$
  3. $4\sqrt{7}$
  4. $8\sqrt{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Circle: x^2 + y^2 + 4x + 2y + 1 = 0. Center O = (-2, -1). Radius r = sqrt(2^2 + 1^2 - 1) = 2. Point P = (2, 3). Distance OP = sqrt((2 - (-2))^2 + (3 - (-1))^2) = sqrt(4^2 + 4^2) = sqrt(32) = 4 * sqrt(2). Length of tangent PA = sqrt(OP^2 - r^2) = sqrt(32 - 4) = sqrt(28) = 2 * sqrt(7). Area of quadrilateral PAOB = 2 * Area(triangle OAP) = 2 * (1/2 * base * height) = PA * r = 2 * sqrt(7) * 2 = 4 * sqrt(7).

Multiple choice

Tangents drawn from the point $P(1,8)$ to the circle $\displaystyle x^{2}+y^{2}-6x-4y-11=0$ touch the circle at the point A and B. The equation of the circumcentre of the $\displaystyle \triangle PAB$ is
  1. $\displaystyle x^{2}+y^{2}+4x-6y+19=0$
  2. $\displaystyle x^{2}+y^{2}-4x-10y+19=0$
  3. $\displaystyle x^{2}+y^{2}-2x+6y-29=0$
  4. $\displaystyle x^{2}+y^{2}+6x-4y+19=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The circumcircle of triangle PAB, where P is an external point and A, B are points of tangency, has the segment OP as diameter, where O is the center of the circle. Center of circle x^2+y^2-6x-4y-11=0 is (3,2). P is (1,8). Diameter endpoints are (3,2) and (1,8). Midpoint (center) = (2,5). Radius squared = (3-2)^2 + (2-5)^2 = 1+9 = 10. Equation: (x-2)^2 + (y-5)^2 = 10 => x^2-4x+4 + y^2-10y+25 = 10 => x^2+y^2-4x-10y+19=0.

Multiple choice

If $ OA$ and $OB$ are the tangents from the origin to the circle $x^{2} + y^{2} + 2gx + 2fy + c = 0,$ and $C$ is the centre of the circle, the area of the quadrilateral $OACB$ is
  1. $\displaystyle \frac{1}{2} \sqrt{c(g^{2}+f^{2}-c}$
  2. $\sqrt {c (g^{2} + f^{2} -c)}$
  3. $ c \sqrt { (g^{2} + f^{2} -c)}$
  4. $\displaystyle \frac{\sqrt{g^{2}+f^{2}-c}}{c}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The centre is C(-g, -f), and the radius is sqrt(g^2 + f^2 - c). The tangent length from O is sqrt(c). The quadrilateral consists of two right triangles, so its area is radius times tangent length, giving sqrt(c(g^2 + f^2 - c)).

Multiple choice

AB and CD are two common tangents to circles which touch each other at C. If D lies on AB such that CD = 4 cm, then AB is equal to

  1. 4 cm

  2. 6 cm

  3. 8 cm

  4. 12 cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For two circles touching externally, the length of the common tangent segment between the points of contact is 2 * sqrt(R * r). The property of the common tangent segment AB where D is the point of contact on the tangent implies AB = 2 * CD = 2 * 4 = 8 cm.