Mathematics · Quantitative Aptitude

Mensuration and Area

83 Questions

Mensuration and area problems cover calculating dimensions of 2D geometric figures. Questions involve finding areas for trapeziums, squares, and hexagons using standard formulas. This arithmetic topic consistently appears in quantitative aptitude tests for competitive exams.

Trapezium areaSquare dimensionsHexagon propertiesParallelogram areaGeometric conversions

Mensuration and Area Questions

Multiple choice maths understanding 3d and 2d shapes defining regular polygons sum of exterior angles of a polygon regular polygons

For a regular hexagon with apothem $5m$, the side length is about $5.77m$. The area of the regular hexagon is (in $m^2$).

  1. $75.5$
  2. $85.5$
  3. $76.5$
  4. $86.5$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For a regular hexagon with apothem $5 m$ the side length is about $5.77m$.
Use the formula
$A = \frac{1}{2}pa$
to find the area of the hexagon.
The perimeter of the hexagon is about $6(5.77)$ or $34.62m$.
Now substitute the values.
$A = \frac{1}{2} (34.62)(5)$
Simplify.
$A = \frac{1}{2}(173.1)$
= $86.5$
$A = \frac{1}{2}(173.1)$
= $86.5$
Therefore, the area of the regular hexagon is about $86.5 m^2$.

Multiple choice physics measurements and units measurement of area and volume measurement of volume measurement of area, volume and density

A public park, in the form of a square, has an area of $(100 \pm 0.2 )m^2 $ .The side of park is :

  1. $ (10 \pm 0.01 ) m $
  2. $ (10 \pm 0.1 ) m $
  3. $ (10 \pm 0.02 ) m $
  4. $ (10 \pm 0.2 ) m $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the side length of the square park is $l$.

The area of a square is given as,

$A = {l^2}$

$100 = {l^2}$

$l = 10\;{\rm{m}}$

The error in length is given as,

$\dfrac{{\Delta A}}{A} = 2\dfrac{{\Delta l}}{l}$

$\dfrac{{0.2}}{{100}} = 2\dfrac{{\Delta l}}{{10}}$

$\Delta l = 0.01\;{\rm{m}}$

Thus, the side of the park is $\left( {10 \pm 0.01} \right)\;{\rm{m}}$.

Multiple choice maths perimeter and area of rectilinear figures parallelogram and rectangle area of parallelogram area of a parallelogram

Find the area of the parallelogram whose base is $17\ cm$ and height $0.8\ m$?

  1. $\displaystyle 13.6\:cm^{2}$
  2. $\displaystyle 1360\:cm^{2}$
  3. $\displaystyle 13.6\:m^{2}$
  4. $\displaystyle 1360\:m^{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Base $= 17\ cm$
Height $= 0.8\ m =0.8 \times 100 = 80\ cm$
Area of parallelogram 
$= b \times h$
$= 17\times 80$
$= 1360\ cm^2$

Multiple choice maths perimeter and area of rectilinear figures parallelogram and rectangle area of parallelogram area of a parallelogram

In a trapezium whose parallel sides measure 12 cm and 10 cm and the distance between them is 8 cm. Find the area of trapezium---

  1. 84 $cm^2$
  2. 48 $cm^2$
  3. 88 $cm^2$
  4. 188 $cm^2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Area of trapezium = $\displaystyle \frac{h}{2} (a + b)$
                                = $\displaystyle \frac{8}{2} (12 + 10)$
                                = 4 $\times$ 22 = 88 $cm^2$

Multiple choice maths area and volume of cylinder hollow cylinder volume of cylinder volume of cylinder and cone

An iron pipe $20\space cm$ long has exterior diameter equal to $25\space cm$. If the thickness of the pipe is $1\space cm$, find the whole surface area of the pipe.

  1. $3167\space cm^2$
  2. $3160\space cm^2$
  3. $3068\space cm^2$
  4. $3268\space cm^2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

TSA of pipe $=$ $2\pi h(R+r)+2\pi ({ R }^{ 2 }-{ r }^{ 2 })$


                     $=$ $2\times \dfrac { 22 }{ 7 } \times 20(12.5+11.5)+2\times \dfrac { 22 }{ 7 } \left( { \left( 12.5 \right)  }^{ 2 }-{ \left( 11.5 \right)  }^{ 2 } \right) $


                     $=$ $\dfrac { 44\times 480 }{ 7 } +\dfrac { 44\times 24 }{ 7 } $

                    $ =$ $\dfrac { 21120 }{ 7 } +\dfrac { 1056 }{ 7 } =\dfrac { 22176 }{ 7 } $

                     $=$ $3167$ ${ cm }^{ 2 }$

Multiple choice maths circle measures area of a sector of a circle sector and arc of a circle area of sectors and segments

Find the area of sector whose length is $30\ \pi$ cm and angles of the sector is $40^o$.

  1. $2125\ \pi $ sq. cm
  2. $2225\ \pi $ sq. cm
  3. $2025\ \pi $ sq. cm
  4. $2200\ \pi $ sq. cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
As we know that,
$1° = \cfrac{\pi}{180}$

$\therefore 40° = \cfrac{\pi}{180} \times 40 = \cfrac{2 \pi}{9}$

Let $S$ be the length of the arc and $A$ be the area of the corresponding sector.

Given that length of arc $\left( S \right) = 30 \pi \; cm$

As we know,
$S = r \theta$

$\Rightarrow 30 \pi = r \left( \cfrac{ \pi}{9} \right)$

$\Rightarrow r = 135 \; cm$

$\therefore$ Area of corresponding seector $\left( A \right) = \cfrac{1}{2} {r}^{2} \theta$

$\Rightarrow A = \cfrac{1}{2} {\left( 135 \right)}^{2} \left( \cfrac{2 \pi}{9} \right) = 2025 \pi \; {cm}^{2}$
Multiple choice maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

If $\Delta ABC\sim \Delta DEF$ such that area of $\Delta ABC$ is $9 cm^2$ and area of $\Delta DEF$ is $16 cm^2$ and $BC=1.8 cm$, then EF is

  1. 2.4 cm

  2. 1.35 cm

  3. 2.1 cm

  4. 3.2 cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$ar(\triangle ABC)=9cm^2,\,ar(\triangle DEF)=16cm^2$ and $BC=1.8cm$

$\triangle ABC\sim\triangle DEF$                [ Given ]

$\Rightarrow$  $\dfrac{ar(\triangle ABC)}{ar(\triangle DEF)}=\dfrac{(BC)^2}{(EF)^2}$                    [ Area of similar triangle theorem ]

$\Rightarrow$  $\dfrac{9}{16}=\dfrac{(1.8)^2}{(EF)^2}$
Taking square root on both sides,

$\Rightarrow$  $\dfrac{3}{4}=\dfrac{1.8}{EF}$

$\Rightarrow$  $EF=\dfrac{1.8\times 4}{3}$

$\Rightarrow$  $EF=\dfrac{7.2}{3}$

$\Rightarrow$  $EF=2.4\,cm$


Multiple choice maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

In $\displaystyle \Delta ABC\sim \Delta DEF$ and their areas are $\displaystyle { 36cm }^{ 2 }$ and $\displaystyle { 64cm }^{ 2 }$ respectively.If side AB=3 cm. Find DE.

  1. 3 cm

  2. 2 cm

  3. 5 cm

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

In $\displaystyle \Delta ABC\sim \Delta DEF$
$\displaystyle \frac { ar.\left( \Delta ABC \right)  }{ ar.\left( \Delta DEF \right)  } =\frac { { AB }^{ 2 } }{ { DE }^{ 2 } } =\frac { { AC }^{ 2 } }{ { DF }^{ 2 } } =\frac { { BC }^{ 2 } }{ { EF }^{ 2 } } $
$\displaystyle \frac { 36 }{ 64 } =\frac { { AB }^{ 2 } }{ { DE }^{ 2 } } $
$\displaystyle \frac { 6 }{ 8 } =\frac { 3 }{ DE } $
$\displaystyle DE=\frac { 8\times 3 }{ 6 } =\frac { 24 }{ 6 } =4cm$
Therefore, D is the correct answer.

Multiple choice maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

$\Delta ABC\sim\Delta PQR.$ If area$\left (ABC \right)= 2.25 m^{2}$, area$ \left (PQR \right)= 6.25 m^{2}$, $ PQ = 0.5 m $, then length of AB is:

  1. 30 cm

  2. 0.5 m

  3. 50 cm

  4. 3 m
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\triangle ABC\sim \triangle DEF$

In two similar triangles, the ratio of their areas is the square of the ratio of their sides

$\Rightarrow \dfrac { ar(ABC) }{ ar(PQR) } ={ \left( \dfrac { AB }{ PQ }  \right)  }^{ 2 }\ \Rightarrow \dfrac { 2.25 }{ 6.25 } ={ \left( \dfrac { AB }{ .5 }  \right)  }^{ 2 }\ \Rightarrow \dfrac { AB }{ .5 } =\dfrac { 15 }{ 25 } \ \Rightarrow AB=.3m\ \Rightarrow AB=.3\times 100=30cm$