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 chord 6 cm long is at distance of 4 cm from the centre of a circle. find the length of a chord which is at a distance of 3 cm from the centre of the circle

  1. 10 cm

  2. 6 cm

  3. 8 cm

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

The radius r of the circle is found using the Pythagorean theorem: r^2 = (chord/2)^2 + distance^2 = 3^2 + 4^2 = 25, so r = 5. For the second chord at distance d = 3, (chord/2)^2 = r^2 - d^2 = 5^2 - 3^2 = 16, so chord/2 = 4 and the chord length is 8 cm.

Multiple choice

In a circle of radius 25 cm, a chord is drawn at a distance of 7 cm from the centre. Find the length of the chord

  1. 24 cm

  2. 48 cm

  3. 50 cm

  4. 36 cm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In a circle, a perpendicular from the center to a chord bisects the chord. This forms a right triangle with the radius (25) as hypotenuse and the distance (7) as one leg. The half-chord length is sqrt(25^2 - 7^2) = sqrt(625 - 49) = sqrt(576) = 24. The full chord is 24 * 2 = 48 cm.

Multiple choice

AB and CD are two parallel chords of length 8 cm and 6 cm respectively. If they are 1 cm apart and lie on the same side of the centre of the circle. The radius of the circle is

  1. 3 cm

  2. 6cm

  3. 5cm

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

Let distance from center to chord of length 6 be d1 = sqrt(r^2 - 3^2) = sqrt(r^2 - 9). Distance to chord of length 8 is d2 = sqrt(r^2 - 4^2) = sqrt(r^2 - 16). Given d1 - d2 = 1. sqrt(r^2 - 9) - sqrt(r^2 - 16) = 1. Testing r=5: sqrt(25-9) - sqrt(25-16) = 4 - 3 = 1. Correct.

Multiple choice

The common chord of two intersecting circles $\displaystyle c_{1}: &: c_{2}$ can be seen from their centres at the angles of $\displaystyle 90^{\circ}: and: 60^{\circ}$ respectively. If the distance between their centres is equal to $\displaystyle \sqrt{3}+1$, then the radii of $\displaystyle c_{1}: &: c_{2}$ are
  1. $\displaystyle \sqrt{3}\: \&\: 3$
  2. $\displaystyle \sqrt{2}\: \&\: 2\sqrt{2}$
  3. $\displaystyle \sqrt{2}\: \&\: 2$
  4. $\displaystyle 2\sqrt{2}\: \&\: 4$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let d be the distance between centers. The common chord length L = 2*r1*sin(45) = r1*sqrt(2) and L = 2*r2*sin(30) = r2. So r2 = r1*sqrt(2). The distance between centers is r1*cos(45) + r2*cos(30) = r1/sqrt(2) + r2*sqrt(3)/2 = sqrt(3)+1. Substituting r2: r1/sqrt(2) + r1*sqrt(2)*sqrt(3)/2 = r1/sqrt(2) + r1*sqrt(6)/2 = r1(1/sqrt(2) + sqrt(3)/sqrt(2)) = r1(1+sqrt(3))/sqrt(2) = sqrt(3)+1. Thus r1 = sqrt(2). Then r2 = sqrt(2)*sqrt(2) = 2.

Multiple choice

If the radius of the circumcircle of the triangle TPQ where PQ is chord of contact corresponding to point T with respect to circle $\displaystyle x^{2}+y^{2}-2x+4y-11=0 $ is $6$ units then distance of T from the director circle of the given circle is
  1. $6$
  2. $12$
  3. $\displaystyle 6\sqrt{2} $
  4. $12$ - $\displaystyle 4\sqrt{2} $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

This involves circle geometry properties, specifically the chord of contact and the director circle. The distance of the point T from the director circle is derived from the power of the point and the radius of the circumcircle.

Multiple choice

The equation of the locus of the mid points of the chord of the circle $4\mathrm{x}^{2}+4\mathrm{y}^{2}-12\mathrm{x}+4\mathrm{y}+1=0$ that subtends an angle of $\displaystyle \frac{2\pi}{3}$ at its centre is
  1. $\displaystyle \mathrm{x}^{2}+\mathrm{y}^{2}-3\mathrm{x}+\mathrm{y}+\frac{16}{31}=0$
  2. $x^{2}+y^{2}-3x+y-\displaystyle \frac{31}{16}=0$
  3. $x^{2}+y^{2}+3x+y+\displaystyle \frac{31}{16}=0$
  4. $x^{2}+y^{2}-3x+y+\displaystyle \frac{31}{16}=0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The circle equation is x^2 + y^2 - 3x + y + 1/4 = 0. Center is (1.5, -0.5), radius squared is 2.25 + 0.25 - 0.25 = 2.25, so r=1.5. For a chord subtending 120 degrees, the distance from the center d = r * cos(60) = 1.5 * 0.5 = 0.75. The locus of the midpoint is a circle with radius d = 0.75. The equation is (x-1.5)^2 + (y+0.5)^2 = 0.75^2 = 0.5625 = 9/16. Expanding gives x^2 + y^2 - 3x + y + 2.25 + 0.25 - 0.5625 = 0, which is x^2 + y^2 - 3x + y + 1.9375 = 0, or 31/16.

Multiple choice

The locus of the midpoints of chords of the circle ${ x }^{ 2 }+{ y }^{ 2 }-2x-2y-2=0$ which makes an angle $120^{0}$ at the center is
  1. ${ x }^{ 2 }+{ y }^{ 2 }-2x-2y+1=0$
  2. ${ x }^{ 2 }+{ y }^{ 2 }-x-y+1=0$
  3. ${ x }^{ 2 }+{ y }^{ 2 }-2x-2y-1=0$

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

The circle is (x-1)^2 + (y-1)^2 = 4, center (1,1), radius r = 2. A chord subtending 120 degrees at the center has a distance d from the center given by d = r * cos(60) = 2 * 0.5 = 1. The locus of the midpoint (h,k) of a chord at distance d=1 from (1,1) is (h-1)^2 + (k-1)^2 = 1^2. Expanding: h^2 - 2h + 1 + k^2 - 2k + 1 = 1, so x^2 + y^2 - 2x - 2y + 1 = 0.

Multiple choice

lf the line $\displaystyle \frac{x}{\mathrm{a}}+\frac{y}{b}=1$ intersects the circle $\mathrm{x}^{2}+\mathrm{y}^{2}=\mathrm{c}^{2}$, then the length of the chord so formed is $(\sqrt{2}) \mathrm{c}$, then
  1. $\mathrm{a}^{2},\ \mathrm{b}^{2},\ \mathrm{c}^{2}$ are in $\mathrm{H}.\mathrm{P}$.
  2. $a, b, c$ are in $\mathrm{H}.\mathrm{P}$.
  3. $\mathrm{c}^{2},\ \mathrm{a}^{2},\ \mathrm{b}^{2}$ are in $\mathrm{H}.\mathrm{P}$.
  4. $\mathrm{a}^{2},\ \mathrm{c}^{2},\ \mathrm{b}^{2}$ are in $\mathrm{H}.\mathrm{P}$.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The distance from the origin to the line x/a + y/b = 1 is d = 1 / sqrt(1/a^2 + 1/b^2). For a chord of length sqrt(2)c in a circle of radius c, the distance from the center is sqrt(c^2 - (sqrt(2)c/2)^2) = c/sqrt(2). Thus, 1/(1/a^2 + 1/b^2) = c^2/2, which simplifies to 1/a^2 + 1/b^2 = 2/c^2. This implies 1/a^2, 1/c^2, 1/b^2 are in AP, so a^2, c^2, b^2 are in HP.

Multiple choice

From the origin, chords are drawn to the circle $\mathrm{x}^{2}+\mathrm{y}^{2}$ -2y $=0$. The locus of the middle point of these chords is
  1. $\mathrm{x}^{2}+\mathrm{y}^{2} -\mathrm{y} =0$
  2. $\mathrm{x}^{2}+\mathrm{y}^{2} -2\mathrm{x} =0$
  3. $\mathrm{x}^{2}+\mathrm{y}^{2} -\mathrm{x} =0$
  4. $\mathrm{x}^{2}+\mathrm{y}^{2} -\mathrm{x}-\mathrm{y} =

    0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the midpoint of the chord be (h, k). The equation of the chord with a given midpoint is T = S1, which gives x*h + y*k - (y + k) = h^2 + k^2 - 2k. Since the chord passes through the origin (0,0), substituting x = 0 and y = 0 yields h^2 + k^2 - k = 0, which represents the locus x^2 + y^2 - y = 0.

Multiple choice

The equation of a circle is $x^{2}+y^{2}=25$, The equation of its chord whose middle point is $(1,-2)$ is given by:
  1. $x+2y+5=0$
  2. $2x-y+5=0$
  3. $x-2y-5=0$
  4. $2x+y+5=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The equation of a chord of a circle x^2 + y^2 = r^2 with midpoint (x1, y1) is T = S1, where T = xx1 + yy1 - r^2 and S1 = x1^2 + y1^2 - r^2. Here x1=1, y1=-2, r^2=25. x(1) + y(-2) - 25 = 1^2 + (-2)^2 - 25. x - 2y - 25 = 1 + 4 - 25. x - 2y - 25 = -20. x - 2y - 5 = 0.

Multiple choice

AB and CD are two parallel chords of a circle such that AB = 5 cm and CD = 11 cm. If distance between them is 3 cm then radius of the circle is
  1. $\displaystyle 2\sqrt{146} cm$
  2. $\displaystyle \sqrt{146} cm$
  3. $\displaystyle \frac{\sqrt{146}}{2} cm$

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

Let the center be O and radius be r. Distance from center to chord AB is d1, to CD is d2. d1^2 + (5/2)^2 = r^2 and d2^2 + (11/2)^2 = r^2. Given d1 - d2 = 3 (or d1 + d2 = 3 if center is between chords). Assuming center is between, d1+d2=3. Solving the system yields r = sqrt(146)/2.