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

Three circles, whose radii are $a,b,c$ touch each other externally and the tangents at their points of contact meet in a point. Then, the distance of this point from either of points of contact is
  1. $\displaystyle \sqrt { \frac { abc }{ a+b+c }  } $
  2. $\displaystyle \sqrt { \frac { abc }{ a-b-c }  } $
  3. $\displaystyle 2\sqrt { \frac { abc }{ a+b+c }  } $

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

This is a standard geometry result for the radical center of three mutually tangent circles. The distance from the radical center to the points of contact is sqrt(abc / (a+b+c)).

Multiple choice

$\bigtriangleup$ABC is a right angled triangle with $\angle A = 90^{\circ}$.A circle is inscribed in it. The lengths of the sides containing the right angle are 6 cm and 8 cm.Find the radius of the circle.
  1. $3 cm$
  2. $3.5 cm$
  3. $2.5 cm$
  4. $2 cm$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For a right triangle with legs a and b and hypotenuse c, the inradius r = (a + b - c) / 2. Here, legs are 6 and 8, so hypotenuse c = 10. r = (6 + 8 - 10) / 2 = 4 / 2 = 2.

Multiple choice

The sides of a triangle inscribed in a given circle subtend angles $\alpha, \beta, \gamma$ at the centre. The minimum value of the A.M. of $\cos\left(\alpha +\dfrac{\pi}{2}\right), \cos \left(\beta +\dfrac{\pi}{2}\right)$ and $\cos\left(\gamma +\dfrac{\pi}{2}\right)$ is equal to?
  1. $-\dfrac{\sqrt{3}}{2}$
  2. $\dfrac{\sqrt{3}}{2}$
  3. $-\dfrac{2}{\sqrt{3}}$
  4. $\sqrt{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The angles subtended at the center by sides of an inscribed triangle satisfy alpha + beta + gamma = 2*pi. The expression is (1/3) * (cos(alpha + pi/2) + cos(beta + pi/2) + cos(gamma + pi/2)) = -(1/3) * (sin(alpha) + sin(beta) + sin(gamma)). For a triangle inscribed in a circle, the sum of sines of central angles is maximized for an equilateral triangle (alpha=beta=gamma=2*pi/3). sin(2*pi/3) = sqrt(3)/2. Sum = 3 * sqrt(3)/2. Average = -(1/3) * 3 * sqrt(3)/2 = -sqrt(3)/2.

Multiple choice

2 perpendicular chords AB and CD of a circle intersect at O (inside the circle). AB is bisected at O. If AO = 4 units and OD = 6 units. Find circumradius of $\displaystyle \bigtriangleup ACO $.
  1. $\displaystyle \sqrt{13} $
  2. $\displaystyle \frac{2\sqrt{13}}{3}$
  3. $\displaystyle \frac{\sqrt{15}}{2}$
  4. $\displaystyle \frac{2\sqrt{15}}{4}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In a circle, if two chords intersect at O, AO * OB = CO * OD. Since AB is bisected at O, AO = OB = 4. Thus, 4 * 4 = CO * 6, so CO = 16/6 = 8/3. Triangle ACO is right-angled at O. The hypotenuse AC = sqrt(AO^2 + CO^2) = sqrt(16 + 64/9) = sqrt(208/9) = 4*sqrt(13)/3. The circumradius of a right triangle is half the hypotenuse, which is 2*sqrt(13)/3.

Multiple choice

Given a triangle ABC with circumcentre at origin. If a $\leq$ 4 sin A then for any point (a, y) inside the circumcircle of $\Delta$ABC
  1. $|xy| < 2$
  2. $2 < |xy| < 4$
  3. $\dfrac{1}{2}$ < |xy| < $\dfrac{3}{4}$
  4. $|xy| > 4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the sine rule, the side length a is given by 2R sin A, where R is the circumradius. The condition a <= 4 sin A simplifies to 2R sin A <= 4 sin A, which implies R <= 2. Since the circumcentre is at the origin, the equation of the circumcircle is x^2 + y^2 = R^2 <= 4. For any point (x, y) inside this circle, we have x^2 + y^2 < 4, and by the AM-GM inequality, |xy| <= (x^2 + y^2)/2 < 2.

Multiple choice

$\displaystyle \Delta ABC$ is an equilateral $\displaystyle \Delta $ with area $\displaystyle 36\sqrt{3}cm^{2}$ The area of the inscribed circle is
  1. $\displaystyle 12\pi \: cm^{2}$
  2. $\displaystyle 48\pi \: cm^{2}$
  3. $\displaystyle 24\pi \: cm^{2}$
  4. $\displaystyle 36\pi \: cm^{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Area of equilateral triangle = (sqrt(3)/4) * side^2 = 36*sqrt(3). side^2 = 144, side = 12. Inradius r = side / (2*sqrt(3)) = 12 / (2*sqrt(3)) = 6/sqrt(3) = 2*sqrt(3). Area of inscribed circle = pi * r^2 = pi * (2*sqrt(3))^2 = pi * 12 = 12*pi.

Multiple choice

Three circles touch one another externally. The tangents at their points of contact meet at a point whose distance from a point of contact is $4$. Then the ratio of the product of the radii to the sum of the radii of the circles is
  1. $16$
  2. $4$
  3. $8$
  4. $\displaystyle\frac{1}{16}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For three circles with radii r1, r2, r3 touching externally, the distance from the point of intersection of tangents to the point of contact is given by sqrt(r1*r2*r3 / (r1+r2+r3)). Given this distance is 4, then 16 = r1*r2*r3 / (r1+r2+r3).

Multiple choice

If S is the circumcentre of a triangle ABC and $\angle SAB=\angle SBC=\angle SCA=\theta$, then $\sin\theta=$

  1. 1
  2. $\displaystyle \frac{1}{2}$
  3. $\displaystyle \frac{1}{\sqrt{2}}$
  4. $\displaystyle \frac{\sqrt{3}}{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In a triangle where the circumcentre S satisfies the given angle conditions, the triangle must be equilateral. For an equilateral triangle, the circumcentre coincides with the centroid, and the angles SAB, SBC, and SCA are each 30 degrees. Since sin(30 degrees) = 1/2, the result follows.

Multiple choice

In a triangle ABC, $I$ be the incentre and $I_1, I_2$ and $I_3$ be the excetres. If 5 and 2 be the radius of circum circle and incircle respectively. Then $II_1.II_2.II_3=$

  1. 900

  2. 600

  3. 400

  4. 800
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The product of the distances from the incenter to the excenters is given by the formula II_1 * II_2 * II_3 = 16 * R^2 * r, where R is the circumradius and r is the inradius. Substituting the given values R = 5 and r = 2 into this formula, we get 16 * 5^2 * 2 = 800.