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

A circle is inscribed in a square and then a smaller square is inscribed in the circle. The ratio of the area of the smaller square to that of the larger square is
  1. $1:4$
  2. $\sqrt{2} :2$
  3. $1:2$
  4. $1:\sqrt{2}$
  5. $2:3$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the larger square have side 2s. The inscribed circle has diameter 2s, so radius s. The smaller square inscribed in this circle has a diagonal equal to the circle's diameter (2s). Side of smaller square = 2s / sqrt(2) = s * sqrt(2). Area of larger square = (2s)^2 = 4s^2. Area of smaller square = (s * sqrt(2))^2 = 2s^2. Ratio = 2s^2 / 4s^2 = 1:2.

Multiple choice

A right-angled triangle whose sides are 15 cm and 20 cm, is made to revolve about its hypotenuse. Find the volume and the surface area of the double cone so formed. [Take $\pi \simeq 3.14]$
  1. 3878 $cm^3$,1315.8$cm^2$
  2. 3777 $cm^3$,1312.8 $cm^2$
  3. 3788 $cm^3$ ,1310.8 $cm^2$
  4. 3768 $cm^3 $,1318.8 $cm^2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Sides 15, 20 form a hypotenuse of 25. The double cone has a radius equal to the altitude to the hypotenuse: r = (15 * 20) / 25 = 12. The two cones have heights h1 and h2 such that h1 + h2 = 25. Volume = (1/3) * pi * r^2 * (h1 + h2) = (1/3) * 3.14 * 144 * 25 = 3768. Surface area = pi * r * (l1 + l2) = 3.14 * 12 * (15 + 20) = 3.14 * 12 * 35 = 1318.8.

Multiple choice

Two circles touch each other externally at P. AB is a common tangent to the circles touching them at A and B. The value of $\displaystyle \angle APB$ is
  1. $\displaystyle { 30 }^{ \circ }$
  2. $\displaystyle { 45 }^{ \circ }$
  3. $\displaystyle { 60 }^{ \circ }$
  4. $\displaystyle { 90 }^{ \circ }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In two circles touching externally at P with a common tangent AB, the angle APB is always 90 degrees because the common tangent at P bisects the common tangent AB at a point that is the center of a circle passing through A, B, and P.

Multiple choice

An isosceles triangle with base $24$ and legs of $15$ is inscribed in a circle. Find the radius.
  1. $7$
  2. $12\dfrac{1}{2}$
  3. $25$
  4. $25\dfrac{1}{2}$

  5. cannot be determined
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In an isosceles triangle with sides 15, 15, 24, the height to the base is sqrt(15^2 - 12^2) = sqrt(225 - 144) = sqrt(81) = 9. Using the circumradius formula R = (abc) / (4 * Area), Area = 0.5 * 24 * 9 = 108. R = (15 * 15 * 24) / (4 * 108) = 5400 / 432 = 12.5.

Multiple choice

The two shorter sides of a triangle measure $9$ and $18$. If the internal angle bisector drawn to the longest side measures $8$, find the measure of the longest side of the triangle.
  1. $18$
  2. $23$
  3. $22$
  4. $21$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Using the formula for the length of an internal angle bisector d = 2ab cos(C/2) / (a+b), where a=9, b=18, d=8. This leads to cos(C/2) = 8 * 27 / (2 * 9 * 18) = 216 / 324 = 2/3. Using the law of cosines, c^2 = a^2 + b^2 - 2ab cos(C). Since cos(C) = 2cos^2(C/2) - 1 = 2(4/9) - 1 = -1/9, c^2 = 81 + 324 - 2(9)(18)(-1/9) = 405 + 36 = 441. c = 21.