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

In a circle of radius $ 5\ cm$ and $AB$ and $AC$ are two chords such that $AB=AC=6 cm$. The distance of the chord $BC$ from the center will be equal to:
  1. $3.6 cm$
  2. $1.4 cm$
  3. $1 cm$
  4. $4 cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In triangle ABC, AB=AC=6, radius=5. Let O be the center. Triangle ABC is isosceles. The altitude from A to BC passes through O. Let M be the midpoint of BC. In triangle ABM, AM = sqrt(AB^2 - BM^2). This requires finding BC. Using the circle properties, the distance from center to chord AB is sqrt(5^2 - 3^2) = 4. The distance from center to chord BC is 1.4 cm.

Multiple choice

Through a point $\left( h,k \right)$ secants are drawn to the circle $\left( { x }^{ 2 }+{ y }^{ 2 } \right) ={ r }^{ 2 }$. The mid-points of the corresponding chords describe the curve
  1. ${ x }^{ 2 }+{ y }^{ 2 }+hx+ky=0$
  2. ${ x }^{ 2 }+{ y }^{ 2 }-hx-ky=0$
  3. ${ x }^{ 2 }+{ y }^{ 2 }-2hx-2ky=0$
  4. ${ x }^{ 2 }+{ y }^{ 2 }+2hx+2ky=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The midpoint (x, y) of a chord of the circle x^2 + y^2 = r^2 that passes through (h, k) satisfies the condition that the vector from the origin to the midpoint is perpendicular to the chord. This leads to the equation x(x-h) + y(y-k) = 0, which simplifies to x^2 + y^2 - hx - ky = 0.

Multiple choice

If two equal circles are such that the centre of one lies on the circumference of the other, then the ratio of the length of the common chord of two circles to the radius of any of the circle is:
  1. $\sqrt3 : 2$
  2. $\sqrt3 : 1$
  3. $\sqrt5 : 1$

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

For two circles of radius r with centers at (0,0) and (r,0), the intersection points are at x = r/2. y^2 = r^2 - (r/2)^2 = 3r^2/4, so y = sqrt(3)r/2. The chord length is 2y = sqrt(3)r. Ratio of chord to radius = sqrt(3)r / r = sqrt(3) : 1.

Multiple choice

Choose the correct answer from the alternative given. PQ and RS are two parallel chords of a circle. If PQ = 30 cm, RS = 16 cm and distance between PQ and RS is 23 cm, then the radius of the circle.
  1. $12$
  2. $17$
  3. $34$
  4. $19$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let center be O. Distance from O to PQ is d1, to RS is d2. d1^2 + 15^2 = r^2. d2^2 + 8^2 = r^2. d1+d2 = 23. Solving d1^2 + 225 = d2^2 + 64 => d2^2 - d1^2 = 161 => (d2-d1)(23) = 161 => d2-d1 = 7. d2+d1 = 23, so 2*d2 = 30, d2=15, d1=8. r^2 = 15^2 + 8^2 = 225 + 64 = 289. r = 17.

Multiple choice

Two chords AB and CD of a circle with centre 0 intersect each other at the point P. If $\angle $ AOD = $20^{\circ}$ and $\angle BOC = 30^{\circ}$, then $\angle $BPC is equal to:
  1. $50^{\circ}$
  2. $20^{\circ}$
  3. $25^{\circ}$
  4. $30^{\circ}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Angle at center AOD = 20, BOC = 30. Angle at circumference subtended by arc AD is 10, arc BC is 15. The angle between chords intersecting at P is half the sum of angles subtended by the arcs at the center. Angle BPC = 1/2 * (Angle AOD + Angle BOC) = 1/2 * (20 + 30) = 25 degrees.

Multiple choice

Two equal chords $AB$ and $AC$ of the circle $x^{2}+y^{2}-6x-8y-24=0$ are drawn from the point $A(\sqrt{33}+3, 0)$. Another chord $PQ$ is drawn intersecting $AB$ and $AC$ at points $R$ and $S$, respectively given that $AR=SC=7$ and $RB=AS=3$. The value of $PR/QS$ is
  1. $1$
  2. $1.5$
  3. $2$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

By the power of a point theorem for intersecting chords, AR * RB = PR * RS and AS * SC = QS * SR. Given AR=7, RB=3, AS=3, SC=7, we have 7*3 = PR * RS and 3*7 = QS * SR. Thus, PR * RS = QS * SR, which implies PR = QS, so PR/QS = 1.

Multiple choice

Choose the correct answer from the alternatives given. AB is a diameter of a circle with centre at O. DC is a chord of it such that DC || AB. If $\angle BAC \, = \, 20^\circ$, then $\angle ADC$ is equal to
  1. 120$^\circ$
  2. 110$^\circ$
  3. 115$^\circ$
  4. 100$^\circ$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Since DC || AB, angle BAC = angle ACD = 20 degrees (alternate interior angles). In circle, OA = OC (radii), so triangle OAC is isosceles. Angle OAC = angle OCA = 20 degrees. Angle AOC = 180 - 20 - 20 = 140 degrees. Angle ADC = 1/2 * (360 - 140) = 110 degrees (angle at circumference is half the reflex angle at center).