Quantitative Aptitude

Mensuration

2,507 Questions
All Direct Formula Application 1067 Shape Transformations 343 Mensuration Ratio Equations 277 Combined & Hollow Solids 247 Successive Percentage Changes 178 Inscribed & Circumscribed Figures 165 Cost Estimation Problems 100 Liquid Pour & Submergence 67 Unit Conversions 35 Path & Border Calculations 17 Time & Work Integration 7 Scale & Model Proportions 2

Mensuration Questions

Multiple choice

Find the area of the portion of the circle $\displaystyle { x }^{ 2 }+{ \left( y-2 \right) }^{ 2 }=4$ in Quadrant 1.

  1. 0
  2. $\displaystyle \pi $
  3. $\displaystyle 2\pi $
  4. $\displaystyle 4\pi $
  5. $\displaystyle 16\pi $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The circle is x^2 + (y-2)^2 = 4, centered at (0,2) with radius 2. It is tangent to the x-axis at (0,0). The portion in Quadrant 1 is the right half of the circle, which is a semicircle. Area = (1/2) * pi * r^2 = (1/2) * pi * 4 = 2pi.

Multiple choice

The volume of a sphere of radius $r$ is
  1. $\cfrac { 4 }{ 2 } \pi { r }^{ 3 }$
  2. $\cfrac { 4 }{ 3 } \pi { r }^{ 3 }$
  3. $\cfrac { 3 }{ 4 } \pi { r }^{ 3 }$

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

The standard formula for the volume of a sphere with radius r is (4/3) * pi * r^3.

Multiple choice

A metal wire, when bent in the form of an equilateral triangle of largest area, encloses an area of $484\sqrt{3}: cm^{2}$ . If the same wire is bent into the form of a circle of largest area, find the area of this circle.
  1. $1386 \: cm^{2}$
  2. $1388 \: cm^{2}$
  3. $1286 \: cm^{2}$
  4. $1186 \: cm^{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Area of equilateral triangle = (sqrt(3)/4) * s^2 = 484*sqrt(3). s^2 = 1936, s = 44. Perimeter = 3 * 44 = 132. Wire length = 132. For circle, 2 * pi * r = 132. r = 132 / (2 * 22/7) = 21. Area = pi * r^2 = 22/7 * 21 * 21 = 1386.

Multiple choice

A figure is in the form of a quadrilateral $ABCD$ its area is $165\;cm^2$. Find the length of the perpendicular drawn from $D$ on $AC$ if $AC=15\;cm$ and length of perpendicular from $B$ on $AC$ is $12\;cm$.
  1. $10\;cm$
  2. $20\;cm$
  3. $30\;cm$
  4. $40\;cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Area of quadrilateral ABCD = Area(ADC) + Area(ABC). Area = (1/2) * AC * h1 + (1/2) * AC * h2 = (1/2) * AC * (h1 + h2). 165 = (1/2) * 15 * (h1 + 12). 165 = 7.5 * (h1 + 12). 22 = h1 + 12, so h1 = 10.

Multiple choice

Two circles touch each other externally. The sum of their areas is $130 \pi \, \ \text{cm}^2$ and the distance between their centres is $14$ cm. Find the diameters of the circle.
  1. $22, 8$
  2. $22, 6$
  3. $26, 8$
  4. $22, 4$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let radii be r1, r2. r1+r2 = 14 (distance between centers). Area sum: pi*r1^2 + pi*r2^2 = 130*pi. r1^2 + r2^2 = 130. (r1+r2)^2 - 2*r1*r2 = 130. 196 - 2*r1*r2 = 130. 2*r1*r2 = 66. r1*r2 = 33. Solving r^2 - 14r + 33 = 0 gives r=11 and r=3. Diameters are 22 and 6.

Multiple choice

Two circles which are not congruent touch externally. The sum of their areas is ${ 130\pi cm }^{ 2 }$ and the distance between their centres is $14\,cm$. The sum of radii of the circles is
  1. $14cm$
  2. $21cm$
  3. $7cm$
  4. $28cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let radii be r1, r2. r1+r2 = 14 (distance between centers). Area sum = pi(r1^2 + r2^2) = 130pi. r1^2 + r2^2 = 130. (r1+r2)^2 = r1^2 + r2^2 + 2r1r2. 14^2 = 130 + 2r1r2. 196 - 130 = 66 = 2r1r2. r1r2 = 33. We know r1+r2 = 14. The sum of radii is given directly as the distance between centers for externally touching circles.

Multiple choice

Two circles touch externally. The sum of their area is $130\;\pi$ sq. cm and the distance between their centres is $14$ cm. The radii of the circles are
  1. $11$ cm and $3$ cm
  2. $10$ cm and $2$ cm
  3. $9$ cm and $1$ cm
  4. $8$ cm and $1$ cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let radii be r1 and r2. r1 + r2 = 14. Area sum = pi(r1^2 + r2^2) = 130pi, so r1^2 + r2^2 = 130. (r1+r2)^2 = r1^2 + r2^2 + 2r1r2. 196 = 130 + 2r1r2, so 2r1r2 = 66, r1r2 = 33. Solving r^2 - 14r + 33 = 0 gives (r-11)(r-3)=0. Radii are 11 and 3.

Multiple choice

If $K$ is a diameter of a circle and $A$ is the area of sector of the circle whose vertical angle is $\theta$ then $\frac {dA}{dt}=$
  1. $\dfrac{{{k^2}}}{8}\left( {\dfrac{{d\theta }}{{dt}}} \right)$
  2. $\left( {\dfrac{{{k^2}}}{4}} \right)\left( {\dfrac{{d\theta }}{{dt}}} \right)$
  3. $\dfrac{d{\theta}}{dt}$
  4. $k\left( {\dfrac{{d\theta }}{{dt}}} \right)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Area of sector A = (1/2) * r^2 * theta. Since diameter K = 2r, r = K/2. A = (1/2) * (K/2)^2 * theta = (K^2 / 8) * theta. Differentiating with respect to t: dA/dt = (K^2 / 8) * (d(theta)/dt).

Multiple choice

$AB$ is the diameter of a circle, centre $O$. $C$ is a point on the circumference such that $\angle COB=\theta $. The area of the minor segment cut off by $AC$ is equal to twice, the area of the sector $BOC$. Find whether the statement $\sin { \cfrac { \theta }{ 2 } } \cos { \cfrac { \theta }{ 2 } } =\pi \left( \cfrac { 1 }{ 2 } -\cfrac { \theta }{ 120 } \right) $ is True/False

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is a geometric identity verification. The area of the minor segment is (1/2)R^2(theta - sin(theta)). The area of the sector is (1/2)R^2(theta). The condition leads to the stated trigonometric identity.