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

The point of intersection of the tangents drawn at the ends of the chord joining the points $\alpha$ and $\beta$ on the circle $\mathrm{x}^{2}+\mathrm{y}^{2}=\mathrm{a}^{2}$ is
  1. <p class="MsoNormal">$(\displaystyle \dfrac {\mathrm{a}\sin\dfrac {\alpha+\beta}{2}}{\sin\dfrac {\alpha-\beta}{2}},\dfrac {\mathrm{a}\mathrm{c}\mathrm{o}\mathrm{s}\dfrac {\alpha+\beta}{2}}{\mathrm{c}\mathrm{o}\mathrm{s}\dfrac {\alpha-\beta}{2}})$</p>
  2. $(\displaystyle \dfrac{\mathrm{a}\cos\dfrac{\alpha+\beta}{2}}{\cos\dfrac{\alpha-\beta}{2}},\dfrac{\mathrm{a}\sin\dfrac{\alpha+\beta}{2}}{\cos\dfrac{\alpha-\beta}{2}})$
  3. <p class="MsoNormal">$(\displaystyle \dfrac {\mathrm{a}\cos\dfrac {\alpha-\beta}{2}}{\mathrm{c}\mathrm{o}\mathrm{s}\dfrac {\alpha+\beta}{2}}\dfrac {\mathrm{a}\cos\dfrac {\alpha-\beta}{2}}{\sin\dfrac {\alpha+\beta}{2}})$</p>
  4. <p class="MsoNormal">$(\displaystyle \dfrac {\mathrm{a}\cos\dfrac {\alpha-\beta}{2}}{\cos\dfrac {\alpha+\beta}{2}},\dfrac {\mathrm{a}\sin\dfrac {\alpha-\beta}{2}}{\sin\dfrac {\alpha+\beta}{2}})$</p>
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The tangent at a point with parameter angle t on x^2 + y^2 = a^2 is x cos t + y sin t = a. Solving the two tangent equations for t = alpha and t = beta gives the coordinates in option B.

Multiple choice

If chord of contact of the tangent drawn from the point $(\alpha,\, \beta)$ to the ellipse $\displaystyle \frac {x^2}{a^2}\, +\, \displaystyle \frac {y^2}{b^2}\, =\, 1$ touches the circle $x^2 + y^2 = k^2$, then find the locus of the point $(\alpha,\, \beta)$.
  1. $\displaystyle \frac {x^2}{a^2}\, +\, \displaystyle \frac {y^2}{b^2}\, =\, \displaystyle \frac {1}{k^2}$
  2. $\displaystyle \frac {x^2}{a^4}\, +\, \displaystyle \frac {y^2}{b^4}\, =\, \displaystyle \frac {1}{k^2}$
  3. $\displaystyle \frac {x^2}{a^2}\, +\, \displaystyle \frac {y^2}{b^2}\, =\, \displaystyle k^2$
  4. $\displaystyle \frac {x^2}{a^4}\, +\, \displaystyle \frac {y^2}{b^4}\, =\, \displaystyle k^2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Chord of contact from (alpha, beta) to x^2/a^2 + y^2/b^2 = 1 is (alpha*x)/a^2 + (beta*y)/b^2 = 1. This line touches x^2 + y^2 = k^2. The condition for a line lx + my = 1 to touch x^2 + y^2 = k^2 is k^2(l^2 + m^2) = 1. Here l = alpha/a^2 and m = beta/b^2. So k^2(alpha^2/a^4 + beta^2/b^4) = 1. Thus, alpha^2/a^4 + beta^2/b^4 = 1/k^2.

Multiple choice

In a circle, the height of an arc is 21 cm and the diameter is 84 cm. Find the chord of half of the arc.

  1. 45 cm

  2. 40 cm

  3. 42 cm

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

In a circle, let R be the radius. Diameter = 84, so R = 42. Height of arc h = 21. Distance from center to chord = R - h = 42 - 21 = 21. The chord of the arc is 2 * sqrt(R^2 - (R-h)^2) = 2 * sqrt(42^2 - 21^2) = 2 * sqrt(1764 - 441) = 2 * sqrt(1323) = 2 * 21 * sqrt(3) = 42*sqrt(3). The question asks for the chord of half the arc. Using the formula for chord length c = 2R sin(theta/2), the chord of half the arc is 2R sin(theta/4). Given the geometry, the chord of half the arc is 42 cm.