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

If the common chord of the circle ${x^2} + {(y - \lambda )^2} = 16$ and $x^2 + y^2 = 16$ subtend a right angle at the origin then $\lambda$ is equal to

  1. 4
  2. $4\sqrt{2}$
  3. $\pm 4\sqrt{2}$

  4. 8
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The common chord of x^2 + (y-lambda)^2 = 16 and x^2 + y^2 = 16 is found by subtracting the equations: x^2 + y^2 - 2y*lambda + lambda^2 = 16 and x^2 + y^2 = 16. This yields -2y*lambda + lambda^2 = 0, so y = lambda/2. The chord is a horizontal line. For it to subtend 90 degrees at the origin, the distance from the origin to the chord must be related to the radius.

Multiple choice

The area of the circle inscribed in a square of side $a$, is $a^2cm^2$. With justification, state whether the statement is:
  1. true, since the radius of the circle is $\cfrac{a}{2}$
  2. false, since the radius of the circle is $\cfrac{a}{2}$
  3. true, since the radius of the circle is $\cfrac{a}{4}$
  4. false, since the radius of the circle is $\cfrac{a}{4}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

A circle inscribed in a square of side a has a diameter equal to a, so its radius is a/2. The area of this circle is pi * (a/2)^2 = pi * a^2 / 4, which is not a^2.

Multiple choice

Two circles with centres $A$ and $B$ of radii $3\ cm$ and $4\ cm$, respectively intersect at two points $C$ and $D$ such that $AC$ and $BC$ are tangents to the two circles. Find the 10 times length of the common chord $CD$

  1. 48

  2. 58

  3. 56

  4. 54
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In triangle ABC, AC=3, BC=4. Since AC and BC are tangents, they are perpendicular to the radii at the points of contact. This implies angle ACB = 90 degrees. AB = sqrt(3^2 + 4^2) = 5. The common chord CD is perpendicular to AB. Area of triangle ABC = 1/2 * 3 * 4 = 6. Also area = 1/2 * AB * h = 1/2 * 5 * h. h = 12/5. CD = 2 * h = 24/5 = 4.8. 10 * CD = 48.

Multiple choice

The radius of a circle with centre O is $13$cm. The distance of a chord from the centre is $5$ cm.Find the length of the chord.
  1. $24$cm
  2. $12$cm
  3. $13$cm
  4. $26$cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A perpendicular from the center to a chord bisects the chord. This forms a right triangle with the radius (13) as the hypotenuse and the distance (5) as one leg. The other leg (half the chord) is sqrt(13^2 - 5^2) = sqrt(169 - 25) = sqrt(144) = 12. The total chord length is 12 * 2 = 24.

Multiple choice

If two chords of the circle ${ x }^{ 2 }+{ y }^{ 2 }-ax-by=0$, drawn from the point $\left( a,b \right)$ is divided by the x-axis in the ratio $2:1$ then:
  1. ${ a }^{ 2 }>3{ b }^{ 2 }$
  2. ${ a }^{ 2 }v<3{ b }^{ 2 }$
  3. ${ a }^{ 2 }>{ 4b }^{ 2 }$
  4. ${ a }^{ 2 }<4{ b }^{ 2 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Parametrize the chord from (a,b) to its other intersection with the circle; for the chord to meet the x-axis in ratio 2:1, the point dividing it in ratio 2:1 must have y = 0. Enforcing the real intersection condition yields a^2 > 3b^2.

Multiple choice

$A,B,C$ are three points on a circle such that $AB$ is the chord and $CP$ is the perpendicular to $OP$, where $O$ is the centre and $P$ is any point on $AB$. The radius $r$ of the circle is given by
  1. ${ r }^{ 2 }={ OP }^{ 2 }+AP\times CP$
  2. ${ r }^{ 2 }={ OP }^{ 2 }+AP\times PB$
  3. ${ r }^{ 2 }={ OP }^{ 2 }+PB\times PC$
  4. ${ r }^{ 2 }={ OP }^{ 2 }+{ PB }^{ 2 }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let O be the origin. P is on AB. CP is perpendicular to OP. This is a property of chords in a circle. By the power of a point theorem or geometric construction, the relation r^2 = OP^2 + AP*PB holds.

Multiple choice

In a circle with centre $O$ chord $AB=$ chord $AC$. Chord $AD$ cuts $BC$ in $E$. If $\overline { AC } =12$ and $\overline { AE } =8$, then $\overline { AD } $ equals:
  1. $27$
  2. $24$
  3. $21$
  4. $18$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In a circle, if chords AB and AC are equal, the triangle ABC is isosceles. Using the property of intersecting chords or similar triangles formed by the secant AD, we find the length of AD based on the given segments AE and AC.

Multiple choice

Two chords of lengths16 cm and 17 cm are drawn perpendicular to each other in a circle of radius 10 cm. The distance of their point of intersection from the centre is approximately

  1. 6.5 cm

  2. 7.2 cm

  3. 7.6 cm

  4. 8 cm
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let the chords be AB and CD intersecting at P. Let the center be O. The distance of a chord from the center is d = sqrt(r^2 - (L/2)^2). For L=16, d1 = sqrt(100 - 64) = 6. For L=17, d2 = sqrt(100 - 72.25) = sqrt(27.75) = 5.27. If the chords are perpendicular, the distance from the center to the intersection point P is sqrt(d1^2 + d2^2) = sqrt(36 + 27.75) = sqrt(63.75) = 7.98, which is approximately 8.