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

From a point P(16, 7), tangent PQ and PR are drawn to the circle ${ x }^{ 2 }+{ y }^{ 2 }-2x-4y-20=0$. If C be the center then area of the quadrilateral PQCR will be-

  1. 75

  2. 150

  3. 15

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

Circle: x^2 + y^2 - 2x - 4y - 20 = 0. Center C = (1, 2). Radius r = sqrt(1^2 + 2^2 - (-20)) = sqrt(25) = 5. Point P = (16, 7). Distance CP = sqrt((16-1)^2 + (7-2)^2) = sqrt(15^2 + 5^2) = sqrt(225 + 25) = sqrt(250). Area of triangle PQC = 1/2 * base * height = 1/2 * r * sqrt(CP^2 - r^2) = 1/2 * 5 * sqrt(250 - 25) = 1/2 * 5 * sqrt(225) = 1/2 * 5 * 15 = 37.5. Quadrilateral PQCR consists of two such triangles, so area = 2 * 37.5 = 75.

Multiple choice

If the perimeter of a right-angled isosceles triangle is $ \displaystyle \sqrt{2}+1$, then what is the length of the hypotenuse?
  1. $\dfrac{1}{\sqrt2}$ cm
  2. $\sqrt2$ cm
  3. $1$ cm
  4. $2$ cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the sides be x, x, and hypotenuse h = x*sqrt(2). Perimeter = 2x + x*sqrt(2) = x(2+sqrt(2)) = sqrt(2)+1. x = (sqrt(2)+1) / (2+sqrt(2)) = (sqrt(2)+1) / (sqrt(2)*(sqrt(2)+1)) = 1/sqrt(2). Hypotenuse = x*sqrt(2) = (1/sqrt(2)) * sqrt(2) = 1.

Multiple choice

$ABCD$ is a square inscribed in a circle of radius $14$ cm. $E, F, G$ and $H$ are the midpoints of the sides $DA, AB, BC$ and $CD$ respectively. The area of the square $EFGH$ will be equal to
  1. $89$ $\displaystyle cm^{2}$
  2. $196$ $\displaystyle cm^{2}$
  3. $98$ $\displaystyle cm^{2}$
  4. $392$ $\displaystyle cm^{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Square ABCD inscribed in circle radius 14. Diagonal of square = diameter = 28. Side of ABCD = 28 / sqrt(2) = 14 * sqrt(2). Area of ABCD = 196 * 2 = 392. EFGH is formed by midpoints, so its area is half of ABCD. Area = 392 / 2 = 196.

Multiple choice

If an arc of a circle subtends an angle of $ \displaystyle x^{\circ} $ at the centre then the length of the arc will be equal to - (Given radius of the circle=r)
  1. $ \displaystyle \frac{x}{360}(2\pi r) $
  2. $ \displaystyle \frac{360}{x}(2\pi r) $
  3. $ \displaystyle \frac{x}{360}\left ( \frac{1}{2\pi r} \right ) $

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

The length of an arc is a fraction of the circumference. The fraction is the angle x divided by the total degrees in a circle (360). Thus, Arc length = (x/360) * (2 * pi * r).

Multiple choice

In an equilateral triangle of side $24cm$, a circle is inscribed touching its sides. Find the area of the remaining portion of the triangle [Take $\sqrt 3=1.732$]
  1. $68.21cm^2$
  2. $98.55cm^2$
  3. $112.67cm^2$
  4. $154.12cm^2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The area of an equilateral triangle is (sqrt(3)/4) * side^2 = (1.732/4) * 24^2 = 249.408 cm^2. The radius of the inscribed circle is side / (2 * sqrt(3)) = 24 / (2 * 1.732) = 6.928 cm, so the area of the circle is pi * r^2 = 3.14159 * 6.928^2 = 150.77 cm^2. The remaining area is 249.408 - 150.77 = 98.638 cm^2, which is closest to 98.55 cm^2.