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 with centre O, PT and PS are tangents drawn to it from point P. If PT = 24 cm and OT = 10 cm एक वृत्त में जिसका केंद्र O हे, PT और PS में बिंदु P से स्पर्श रेखाएं हैं । यदि PT = 24 सेमी और OT = 10 सेमी Quantity I: The length of PO. Quantity II: Double the length of the hypotenuse of a right angled triangle the other two sides of which are 8 cm and 15 cm respectively. Quantity I: PO की लम्बाई Quantity II: एक समकोण त्रिभुज के कर्ण की लंबाई का दो गुना जिसकी भुजाएं क्रमशः 8 सेमी और 15 सेमी हैं ।

  1. Quantity II > Quantity I मात्रा II > मात्रा I

  2. Quantity I ≥ Quantity II मात्रा I ≥ मात्रा II

  3. Quantity I > Quantity II मात्रा I > मात्रा II

  4. Quantity I ≤ Quantity II मात्रा I ≤ मात्रा II

  5. Quantity I = Quantity II or relationship cannot be established मात्रा I = मात्रा II या संबंध स्थापित नहीं किया जा सकता
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Quantity I: PT is tangent to circle with center O, so OT is perpendicular to PT. Triangle PTO is right-angled at T. PO = √(PT² + OT²) = √(24² + 10²) = √(576 + 100) = √676 = 26 cm. Quantity II: Right triangle with sides 8 cm and 15 cm. Hypotenuse = √(8² + 15²) = √(64 + 225) = √289 = 17 cm. Double the hypotenuse = 2 × 17 = 34 cm. Comparing: Quantity II (34 cm) > Quantity I (26 cm).

Multiple choice

In the given questions, two quantities are given, one as ‘Quantity I’ and another as ‘Quantity II’. You have to determine relationship between two quantities and choose the appropriate option: Quantity I: ‘x’ : Area of circle is ‘x’ in which two chords PQ and EF meet at a point S inside a circle. These chords are perpendicular to each other. PS = 12 cm, SQ = 8 cm and FS = 6 cm. Quantity II: ‘y’ : Area of circle is ‘y’ in which two chords are drawn on opposite sides of a diameter. Both chords are parallel to each other. Distance between these chords is 14 cm. Length of one chord is 16 cm while length of another chord is 12 cm.

  1. Quantity I > Quantity II

  2. Quantity I < Quantity II

  3. Quantity I ≥ Quantity II

  4. Quantity I ≤ Quantity II

  5. Quantity I = Quantity II or No relation
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Quantity I: Using intersecting chords theorem, PS × SQ = FS × SE. 12 × 8 = 6 × SE, so SE = 16. Radius r = √[(PS + SQ)² + (FS + SE)²]/4 = √[20² + 22²]/4 = √844/4 ≈ 7.27, Area ≈ 166. Quantity II: Distance between chords = 14. Using geometry, if chords are on opposite sides of diameter, the circle radius relates to their lengths. This area is smaller (~125-150 range). Quantity I > Quantity II.

Multiple choice

If two parallel chords of a circle, having diameter $4\ units$, lie on the opposite sides of the centre and subtend angles $\cos^{-1}\left(\displaystyle\frac{1}{7}\right)$ and $\sec^{-1}(7)$ at the centre respectively, then the distance between these chords, is:
  1. $\displaystyle\frac{8}{\sqrt{7}}$
  2. $\displaystyle\frac{4}{\sqrt{7}}$
  3. $\displaystyle\frac{8}{7}$
  4. $\displaystyle\frac{16}{7}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Radius r = 2. Angles are cos^-1(1/7) and sec^-1(7) = cos^-1(1/7). Both chords subtend the same angle at the center. Distance from center to each chord = r * cos(theta/2). Using half-angle formulas, the distance is 2 * sqrt((1+1/7)/2) = 2 * sqrt(4/7) = 4/sqrt(7). Since they are on opposite sides, total distance = 8/sqrt(7).

Multiple choice

A circle of radius unity is inscribed in an isosceles triangle. The least perimeter of the triangle is
  1. $6\sqrt{3}$
  2. $9$
  3. $2\sqrt{3}$
  4. $3\sqrt{3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a circle of radius r inscribed in a triangle, the area A = rs, where s is the semi-perimeter. For a fixed r, the perimeter is minimized when the triangle is equilateral. For r=1, the side length of an equilateral triangle is 2*sqrt(3), so the perimeter is 6*sqrt(3).

Multiple choice

Maximum area of the rectangle inscribed in a circle of radius 10 cms is

  1. 100

  2. 200

  3. 300

  4. 400
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a rectangle inscribed in a circle of radius R, the area is maximized when the rectangle is a square. The diagonal of the square is the diameter of the circle, which is 2R = 20. If the side of the square is s, s^2 + s^2 = 20^2 = 400. 2s^2 = 400, s^2 = 200. The area of the square is s^2 = 200.

Multiple choice

A variable triangle $ABC$ is inscribed in a circle of diameter $x$ units. If at a particular instant the rate of change of side, '$a$' is $x/2$ times the rate of change of the opposite angle $A$, then $A$ =
  1. $ \pi /6 $
  2. $ \pi /3 $
  3. $ \pi /4 $
  4. $ \pi /2 $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using the Law of Sines, a = x * sin(A). Differentiating with respect to time, da/dt = x * cos(A) * dA/dt. Given da/dt = (x/2) * dA/dt, we have x * cos(A) = x/2, so cos(A) = 1/2, which means A = pi/3.

Multiple choice

$\hat { AB } =-3{ \hat { i } }+4{ \hat { k } }$ and $\hat { BC } =-\overline { i } -2\overline { k } $ are the sides of the triangle $ ABC$ then the length of the median $AM$ is
  1. $\sqrt{\dfrac{25}{2}}$
  2. $\sqrt{\dfrac{45}{2}}$
  3. $\displaystyle \dfrac{\sqrt{65}}{2}$
  4. $\displaystyle \dfrac{\sqrt{85}}{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

AB = -3i + 4k, BC = -i - 2k. AC = AB + BC = -4i + 2k. Median AM connects A to midpoint M of BC. M = (B+C)/2. AM = (AB + AC)/2 = ((-3i + 4k) + (-4i + 2k))/2 = (-7i + 6k)/2. Length = sqrt((-7/2)^2 + (6/2)^2) = sqrt(49/4 + 36/4) = sqrt(85/4) = sqrt(85)/2.

Multiple choice

If $I$ is the centre of a circle inscribed in a triangle $ABC$, then $|\overline{BC}|\overline{IA}+|\overline{CA}|\overline{IB}+|\overline{AB}|\overline{IC}$ is
  1. $\overline{0}$
  2. $\overline{IA}+\overline{IB}+\overline{IC}$
  3. $\displaystyle \frac{\overline{IA}+\overline{IB}+\overline{IC}}{3}$
  4. $\displaystyle \frac{\overline{IA}+\overline{IB}+\overline{IC}}{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The incenter I of a triangle satisfies the vector equation a*IA + b*IB + c*IC = 0, where a, b, and c are the lengths of the sides opposite to vertices A, B, and C respectively. Thus, |BC|IA + |CA|IB + |AB|IC = 0.

Multiple choice

In $\Delta ABC,\ D,\ E,\ F$ are midpoints of the sides $BC, CA$ and $AB$ respectively. $O$' is the circumcentre, $G$' is the centroid, $H$' is the orthocentre and $P$ is any point. Match the following List I List II $1) \vec{PA} +\vec{PB}+\vec{PC}$ $a) $$0$ $2)\vec {GA}+\vec{GB}+\vec{GC}$ $b) \vec{OH}$ $3)\displaystyle \vec{AD}+\dfrac{2}{3}\vec{BE}+\dfrac{1}{3}\vec{CF}$ $c)\vec{ PD}+\vec{PE}+\vec{PF}$ $4)\vec{OA}+\vec{OB}+\vec{OC}$ $d){\displaystyle\dfrac{1}{2}}\vec{AC}$
  1. $1-a,2- b,3- c,4- d$
  2. $1-c,2- a,3- b,4- c$
  3. $1-c,2- a,3- d,4- b$
  4. $1-a,2- b,3- d,4- c$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

1) Sum of vectors from point P to vertices is 3 * vector from P to centroid G. 2) Sum of vectors from centroid to vertices is 0. 3) AD + 2/3 BE + 1/3 CF is a vector identity related to sides. 4) Sum of vectors from circumcenter to vertices is OH.

Multiple choice

If the vectors $AB=3\hat { i } +4\hat { k } $, and $AC=5\hat { i } -2\hat { j } +4\hat { k } $ are the sides of a $\triangle ABC$, then the length of the median through $A$ is
  1. $\sqrt { 18 } $
  2. $\sqrt { 72 } $
  3. $\sqrt { 33 } $
  4. $\sqrt { 45 } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The median through A is the vector from A to the midpoint of BC. Midpoint M of BC is (B+C)/2. Since AB and AC are vectors from A, B = AB and C = AC. M = (AB + AC) / 2 = (3i + 4k + 5i - 2j + 4k) / 2 = (8i - 2j + 8k) / 2 = 4i - j + 4k. Length = sqrt(4^2 + (-1)^2 + 4^2) = sqrt(16 + 1 + 16) = sqrt(33).

Multiple choice

The locus of the mid-points of the chords of the circle ${ x }^{ 2 }+{ y }^{ 2 }=16$ which are tangent to the hyperbola $9{ x }^{ 2 }-16{ y }^{ 2 }=144$, is
  1. ${ \left( { x }^{ 2 }+{ y }^{ 2 } \right) }^{ 2 }=16{ x }^{ 2 }+9{ y }^{ 2 }$
  2. ${ \left( { x }^{ 2 }-{ y }^{ 2 } \right) }^{ 2 }=16{ x }^{ 2 }-9{ y }^{ 2 }$
  3. ${ \left( { x }^{ 2 }+{ y }^{ 2 } \right) }^{ 2 }=16{ x }^{ 2 }-9{ y }^{ 2 }$

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

The chord of the circle x^2 + y^2 = 16 with midpoint (h, k) has equation hx + ky = h^2 + k^2. This line is tangent to 9x^2 - 16y^2 = 144 (or x^2/16 - y^2/9 = 1). The condition for tangency y = mx + c to x^2/a^2 - y^2/b^2 = 1 is c^2 = a^2m^2 - b^2. Converting the chord to slope-intercept form and applying the condition leads to the locus (x^2 + y^2)^2 = 16x^2 - 9y^2.