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 triangle has vertices $A,\ B$ and $C$ and the respective opposite sides have lengths $a,b$ and $c$. This triangle is inscribed in a circle of radius $R$. If $b=c=1$ and the altitude from $A$ to side $BC$ has length $\sqrt{\dfrac{2}{3}}$, then $R$ equals.
  1. $\dfrac{1}{\sqrt{3}}$
  2. $\dfrac{2}{\sqrt{3}}$
  3. $\dfrac{\sqrt{3}}{2}$
  4. $\dfrac{\sqrt{3}}{2\sqrt{2}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In triangle ABC, b = c = 1. Let the base BC = a. The altitude h = sqrt(2/3). Area = 0.5 * a * h = 0.5 * a * sqrt(2/3). Also, Area = (abc) / (4R) = (1 * 1 * a) / (4R) = a / (4R). Equating: 0.5 * a * sqrt(2/3) = a / (4R). 0.5 * sqrt(2/3) = 1 / (4R). R = 1 / (2 * sqrt(2/3)) = sqrt(3) / (2 * sqrt(2)).

Multiple choice

While constructing a segment of a circle containing and angle $\theta$, which of the following property of tangents is used?

  1. Tangent is perpendicular to the radius.

  2. The angle between a tangent and a chord is equal to the inscribed angle on the opposite side of the chord.

  3. A tangent intersects the circle at only one point.

  4. Two tangents can be drawn from an external point to a given circle and the lengths of both the tangents are equal.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The property used to construct a segment of a circle containing an angle theta is the alternate segment theorem, which states that the angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment.

Multiple choice

Two circles of radii $r_{1}, r_{2}$ intersect orthogonally. The length of their common chord is
  1. $\dfrac {2r_{1}r_{2}}{\sqrt {r_{1}^{2} + r_{2}^{2}}}$
  2. $\dfrac {r_{1}r_{2}}{\sqrt {r_{1}^{2} + r_{2}^{2}}}$
  3. $\dfrac {2r_{1}^{2}r_{2}}{\sqrt {r_{1}^{2} + r_{2}^{2}}}$
  4. $\dfrac {2r_{2}^{2}r_{1}}{\sqrt {r_{1}^{2} + r_{2}^{2}}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For orthogonal circles, the distance between centers d^2 = r1^2 + r2^2. The length of the common chord is 2 * (r1 * r2) / d. Substituting d = sqrt(r1^2 + r2^2) gives 2 * r1 * r2 / sqrt(r1^2 + r2^2).

Multiple choice

The radii of two circles area in the ratio $1:2,C$ is the centre of the smaller circle and $D$ that of the bigger circle. They touch externally at $A.PAQ$ is a straight line with $P$ on the smaller circle and $Q$ on the bigger circle. If the line $PAQ$ does not pass through $C$, the angle between the tangent at $Q$ to the bigger circle and the diameter through $P$ of the smaller circle (produced if neassary) is
  1. $60^{o}$
  2. $75^{o}$
  3. $80 ^{o}$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice

Two circles whose radii are equal to 4 and 8 intersect at right angles. The length of their common chord is
  1. $\displaystyle \frac{16}{\sqrt{5}}$
  2. $8$
  3. $\displaystyle 4\sqrt{6}$
  4. $\displaystyle \frac{8\sqrt{5}}{5}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For two intersecting circles with radii r1, r2 and distance d between centers, if they intersect at right angles, d^2 = r1^2 + r2^2. Here d^2 = 4^2 + 8^2 = 16 + 64 = 80, so d = sqrt(80) = 4*sqrt(5). The length of the common chord is (2 * r1 * r2) / d = (2 * 4 * 8) / (4 * sqrt(5)) = 16 / sqrt(5).

Multiple choice

The square of the tangent that can be drawn from any point on one circle to another circle is equal to twice the product of perpendicular distance of the point from the radical axis of two circles and distance between their centres.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The power of a point P with respect to a circle is the square of the tangent length. For two circles, the radical axis is the locus of points with equal power. The property stated is a standard theorem in geometry regarding radical axes.

Multiple choice

The equation of two circles are $x^{2}+y^{2}+2\lambda x+5=0$ and $x^{2}+y^{2}+2\lambda y+5=0$. $P$ is any point on the line $x-y=0$. If $PA$ and $PB$ are the lengths of the tangents from $P$ to the two circles and $PA=3$ then $PB$ is equal to
  1. $1.5$
  2. $6$
  3. $3$

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

The length of the tangent from a point to a circle is given by sqrt(S1). For both circles, the power of point P(h, h) is h^2 + h^2 + 2*lambda*h + 5. Since the power is the same for both circles, the tangent lengths PA and PB must be equal. Given PA = 3, PB must also be 3.

Multiple choice

lf $S_{1}=0,\ S_{2}=0$ and $S_{3}=0$ are the three circles whose radical centre is the point $\mathrm{P}$, then the lengths $l_{1},\ l_{2},\ l_{3}$ of the tangent from $\mathrm{P}$ to the three circles are such that
  1. $l_{1}=21_{2}=3l_{3}$
  2. $l_{1}=l_{2}=l_{3}$
  3. $l_{1}\neq l_{2}\neq l_{3}$
  4. $l_{1}=l_{2}\neq l_{3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

At the radical centre, the powers with respect to all three circles are equal. Tangent length squared equals the corresponding power, so the three tangent lengths are equal.

Multiple choice

The point from which the lengths of tangents to the three circles $\mathrm{x}^{2}+\mathrm{y}^{2}-4=0,\ \mathrm{x}^{2}+\mathrm{y}^{2}$ -2x $+3\mathrm{y}=0$ and $\mathrm{x}^{2}+\mathrm{y}^{2}+7\mathrm{y}-18=0$ are equal is
  1. $(2, 5)$
  2. $(3, 4)$
  3. $(4, 3)$
  4. $(5, 2)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The point (x, y) from which tangent lengths are equal is the radical center of the three circles. The radical axis of the first two circles is x^2+y^2-4 = x^2+y^2-2x+3y => 2x-3y=4. Testing (5, 2): 2(5)-3(2) = 10-6 = 4. This satisfies the first radical axis. Checking the third circle: x^2+y^2+7y-18 = x^2+y^2-4 => 7y-18 = -4 => 7y=14 => y=2. If y=2, 2x-3(2)=4 => 2x=10 => x=5.

Multiple choice

ABC is triangle with side lenghts 13, 14, and 15 units. if is the incentre and R is its circum radius, then the value of $frac{{AI \times BI \times CI}}{R}$ is equal to (in equal to Sq. units)

  1. 130

  2. 64

  3. 126

  4. 100
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The triangle has semiperimeter 21 and area 84, so its inradius is 4 and circumradius is 65/8. Using AI = r/sin(A/2), BI = r/sin(B/2), and CI = r/sin(C/2), the product AI × BI × CI divided by R simplifies to 64.