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

$s_{1}$ and $s_{2}$ are the inscribed and circumscribed circles of a triangle with sides, $3, 4$ and $5$, Then (area of $s_{1}$/ area of $s_{2}$)=
  1. $16/25$
  2. $4/25$
  3. $9/25$
  4. $9/16$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a right triangle with sides 3, 4, 5, the inradius r = (a+b-c)/2 = (3+4-5)/2 = 1. The circumradius R = c/2 = 5/2 = 2.5. Area of inscribed circle = pi * r^2 = pi. Area of circumscribed circle = pi * R^2 = 6.25 * pi. Ratio = 1 / 6.25 = 1 / (25/4) = 4/25.

Multiple choice

Let $A$ be the centre of the circle ${ x }^{ 2 }+{ y }^{ 2 }-2x-4y-20=0$. Let $B(1,7)$ and $D(4,-2)$ be two points on the circle such that tangents at $B$ and $D$ meet at $C$. The area of the quadrilateral $ABCD$ is
  1. $150$ sq. units
  2. $50$ sq. units
  3. $75$ sq. units
  4. $70$ sq. units
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The circle equation x^2 + y^2 - 2x - 4y - 20 = 0 has center A(1, 2) and radius r = 5. Points B and D lie on the circle, and the tangents at these points meet at C. The area of the kite ABCD is twice the area of triangle ABD or can be calculated using the properties of tangents.

Multiple choice

The locus of the mid-points of the chords of the circle $x^{2} + y^{2} + 2x - 2y - 2 = 0$ which make an angle of $90^{\circ}$ at the centre is
  1. $x^{2} + y^{2} - 2x - 2y = 0$
  2. $x^{2} + y^{2} - 2x + 2y = 0$
  3. $x^{2} + y^{2} + 2x - 2y = 0$
  4. $x^{2} + y^{2} + 2x - 2y - 1 = 0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Circle: (x+1)^2 + (y-1)^2 = 4. Center (-1, 1), radius 2. Chord makes 90 degrees at center, so distance from center to chord d = r/sqrt(2) = 2/sqrt(2) = sqrt(2). Midpoint (h, k) distance from center is sqrt(2). (h+1)^2 + (k-1)^2 = (sqrt(2))^2 = 2. h^2 + 2h + 1 + k^2 - 2k + 1 = 2. h^2 + k^2 + 2h - 2k = 0.

Multiple choice

A circle of radius $3$ is placed at the centre of a circle of radius $r > 3$ such that the length of a chord of the larger circle tangent to the smaller circle is $8$. The probability that point randomly selected from inside of the larger circle lies inside the annular region of the two circles is
  1. $\dfrac{3}{4}$
  2. $\dfrac{9}{25}$
  3. $\dfrac{16}{25}$
  4. $\dfrac{7}{3}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Radius of small circle = 3. Chord length = 8. Distance from center to chord = sqrt(r^2 - 4^2). This distance is the radius of the small circle, so 3 = sqrt(r^2 - 16). 9 = r^2 - 16, so r^2 = 25, r=5. Area of large circle = 25pi. Area of small circle = 9pi. Annular region = 25pi - 9pi = 16pi. Probability = 16pi / 25pi = 16/25.

Multiple choice

A triangle with side lengths in the ratio 3 : 4 : 5 is inscribed in a circle of radius 3. The area of the triangle is equal to

  1. 8.64

  2. 12

  3. 6

  4. 10.28
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A triangle with sides 3:4:5 is a right-angled triangle. The hypotenuse is the diameter of the circumcircle, so diameter = 2 * radius = 6. The sides are 3k, 4k, 5k, so 5k = 6, k = 1.2. The sides are 3.6, 4.8, and 6. Area = 1/2 * base * height = 1/2 * 3.6 * 4.8 = 8.64.

Multiple choice

Two concentric circles with centre $C$ have radii ${r}{1}$ and ${r}{2}$ such that $\left( { r }{ 2 }-{ r }{ 1 } \right) >0$. $CA$ and $CB$ are the common lined radii of the circles. If tangent at $A$ is drawn to meet the bigger circle in the point $D$. Then the length $BD$ is given by
  1. $\sqrt { 2{ r }_{ 1 }\left( { r }_{ 2 }-{ r }_{ 1 } \right) } $
  2. $\sqrt { 2{ r }_{ 2 }\left( { r }_{ 2 }-{ r }_{ 1 } \right) } $
  3. $\sqrt { 2{ r }_{ 1 }\left( { r }_{ 2 }+{ r }_{ 1 } \right) } $
  4. $\sqrt { 2{ r }_{ 2 }\left( { r }_{ 2 }+{ r }_{ 1 } \right) } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In the concentric circles, triangle CAD is a right triangle with hypotenuse r2 and leg r1. AD^2 = r2^2 - r1^2. In triangle CBD, CB = r1, CD = r2, and angle BCD is the angle between radii. Using the law of cosines or coordinate geometry, BD is calculated as sqrt(2*r2*(r2-r1)).

Multiple choice

If the common tangents to the parabola, $ x ^ { 2 } $ = 4 y and the circle, $ x ^ { 2 } + y ^ { 2 } = 4 $ intersect at the point P, then the distance of P from the origin, is
  1. $

    2 ( \sqrt { 2 } + 1 )

    $
  2. $

    3 + 2 \sqrt { 2 }

    $
  3. $

    2 ( 3 + 2 \sqrt { 2 } )

    $
  4. $

    \sqrt { 2 } + 1

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

Common tangents to x^2 = 4y and x^2 + y^2 = 4. The intersection point P of common tangents to these curves is known to be at a distance of 2(sqrt(2) + 1) from the origin.