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

Calculate the area enclosed by the inscribed and circumscribed circles to a square with a diagonal of $8$ m in length.
  1. $16.89\ m^2$
  2. $25.13\ m^2$
  3. $15.64\ m^2$

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

For a square with diagonal d = 8 m, the circumscribed circle has a diameter equal to the diagonal, so its radius R is 4 m. The inscribed circle has a diameter equal to the side of the square, which is 8 / sqrt(2) m, so its radius r is 2 * sqrt(2) m. The enclosed area is the difference between the two circle areas, which is pi * (R^2 - r^2) = pi * (16 - 8) = 8 * pi, approximately 25.13 square meters.

Multiple choice

Find the area of circle in sq. cms, in which the chord of length 18 cm is at the distance of half of the radius of circle from the centre of circle:
  1. $108 \pi$
  2. $54 \pi$
  3. $27 \pi$
  4. $81 \pi$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let radius be R. Distance from center to chord is R/2. Using Pythagorean theorem on the triangle formed by radius, half-chord, and distance: R^2 = (R/2)^2 + 9^2. R^2 = R^2/4 + 81. 3/4 * R^2 = 81. R^2 = 108. Area = pi * R^2 = 108 * pi.

Multiple choice

The areas of two sectors of two different circles are equal. Is it necessary that their corresponding arc lengths are equal? Why?
  1. False, since it requires the condition $2{ { r }_{ 1 } }^{ 2 }<{ { r }_{ 2 } }^{ 2 }\\ $
  2. True, since given that $2{ { r }_{ 1 } }^{ 2 }={ { r }_{ 2 } }^{ 2 }.\\ $
  3. False, since it requires the condition $2{ { r }_{ 1 } }^{ 2 }={ { r }_{ 2 } }^{ 2 }.\\ $
  4. True, since given that $2{ { r }_{ 1 } }^{ 2 }>{ { r }_{ 2 } }^{ 2 }.\\ $
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice

Consider a rhombus with diagonals of length $6$ and $8$ and the inscribed circle $C_{1}$. The vertices of the rhombus are the midpoints of the sides of a rectangle, inscribed in a circle $C_{2}$. The ratio of the radius of $C_{1}$ and the radius of $C_{2}$ is
  1. Less than $0.48$
  2. $0.48$
  3. $0.56$
  4. $0.64$
  5. Greater than $0.64$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Rhombus diagonals 6 and 8 imply side = sqrt(3^2 + 4^2) = 5. Area = 0.5*6*8 = 24. Inradius r1 = Area / semi-perimeter = 24 / 10 = 2.4. The rhombus vertices are midpoints of a rectangle, so the rectangle sides are 6 and 8. The circumradius of this rectangle is R2 = sqrt(3^2 + 4^2) = 5. Ratio r1/R2 = 2.4/5 = 0.48.

Multiple choice

If $p$ is the perimeter of an equilateral triangle inscribed in a circle, the area of the circle is:
  1. $\cfrac { \pi { p }^{ 2 } }{ 3 } $
  2. $\cfrac { \pi { p }^{ 2 } }{ 9 } $
  3. $\cfrac { \pi { p }^{ 2 } }{ 27 } $
  4. $\cfrac { \pi { p }^{ 2 } }{ 81 } $
  5. $\cfrac { \pi { p }^{ 2 }\sqrt { 3 } }{ 27 } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For an equilateral triangle with side s, perimeter p = 3s, so s = p/3. The radius of the circumcircle R = s / sqrt(3) = p / (3 * sqrt(3)). Area of circle = pi * R^2 = pi * (p^2 / (9 * 3)) = pi * p^2 / 27.

Multiple choice

Find the area of the circle if the area of an isosceles right triangle inscribed in it is 18 $\displaystyle cm^{2}$
  1. 16$\displaystyle \pi $
  2. 21$\displaystyle \pi $
  3. 18$\displaystyle \pi $
  4. 24$\displaystyle \pi $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For an isosceles right triangle inscribed in a circle, the hypotenuse is the diameter. Area = 1/2 * base * height. If legs are 'a', area = a^2/2 = 18, so a^2 = 36, a = 6. Hypotenuse = sqrt(6^2 + 6^2) = sqrt(72) = 6*sqrt(2). Radius = 3*sqrt(2). Area of circle = pi * r^2 = pi * (3*sqrt(2))^2 = 18pi.