Mathematics · Quantitative Aptitude

Mensuration and Area

92 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 perimeter and area converting between units of areas the metric system metric system

The area of a square field is $325\ m^2$. Find the approximate length of one side of the field. (upto 2 places of decimals) (in $m^2$)

  1. $19.03$
  2. $18.02$
  3. $18.03$
  4. $17.03$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We know area of any square is its side x side 

Let us assume that side of square field is x 
Then area of square field $= x^2$
According to question

$x^2 = 325 m^2$

$\Rightarrow x^2 = 325$

$x = \sqrt{325}$

So $x = 18.03$

Hence side of square field is $18.03 m$

option (C)

Multiple choice maths mid-point and its converse application of the mid-point theorem mid point theorem mid-point theorem and its converse

In $\Delta ABC$, point P,Q and R are the mid points of the sides AB, BC and CA respectively. If area of $\Delta ABC$ is 32 sq units, then area of $\Delta PQR$ is

  1. $8$ sq cm
  2. $16$ sq cm
  3. $64$ sq cm
  4. $24$ sq cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
The line joining the midpoints of the sides of the triangle form four triangles, each of which is similar to the original triangle.

$ΔABC\sim ΔPQR$

In $ΔABC$, $P$ and $R$ are mid points of $AB$ and $AC$ respectively.

$∴ PR || BC$ (midpoint theorem)

In $ΔABC$ and $ΔAPR$:

$∠A$ is common and $∠APR = ∠ABC$ (corresponding angles)

Therefore, $ΔABC\sim ΔAPR$ (AA similarity)

In $ΔABC$ and $ΔPQR$, since $P, Q, R$ are the midpoints of $AB, BC$ and $AC$ respectively,

$PR =\dfrac {1}{2}BC$; (midpoint theorem)

$∴ ΔABC\sim ΔPQR$ (SSS similarity)

$\dfrac { Ar(ΔPQR) }{ Ar(ΔABC) } =\dfrac { PR^{ 2 } }{ BC^{ 2 } } =\left( \dfrac { 1 }{ 2 }  \right) ^{ 2 }=\dfrac { 1 }{ 4 }$ 

Now, it is given that area of $ΔABC$ is $32$ sq. units. Therefore, we have:

$\dfrac { Ar(ΔPQR) }{ 32 } =\dfrac { 1 }{ 4 } \\ Ar(ΔPQR)=\dfrac { 32 }{ 4 } \\ Ar(ΔPQR)=8$

Hence, the area of $ΔPQR$ is $8$ sq units.
Multiple choice maths mid-point and its converse proving the mid-point theorem the mid-point theorem mid point theorem

If $D, E, F$ are respectively the midpoints of the sides $AB, BC, CA$ of $\Delta ABC$ and the area of $\Delta ABC$ is $24\ sq.\ cm$, then the area of $\Delta DEF$ is:

  1. $24\ {cm}^{2}$
  2. $12\ {cm}^{2}$
  3. $8\ {cm}^{2}$
  4. $6\ {cm}^{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given: 

$\Delta ABC,\ \ D, E$ and $F$ are mid points of $AB, BC, CA$ respectively.

In $\Delta ABC$
$F$ is mid point of $AC$ and $D$ is mid point of $AB$. 
Thus, by Mid point theorem, we get
$FD = \dfrac{1}{2} CB$, 
$FD = CE$ and $FD \parallel CE$     ...(1)
Similarly,
$DE =  FC$ and $DE \parallel FC$   ...(2)
$FE = DB$ and $FE \parallel DB$    ...(3) 

From (1), (2) and (3)
$\Box ADEF$, $\Box DBEF$, $\Box DECF$ are parallelograms.

The diagonal of a parallelogram divides the parallelogram into two congruent triangles.
Hence, $\Delta DEF \cong \Delta ADF$
$\Delta DEF \cong \Delta DBE$
$\Delta DEF \cong \Delta FEC$
Or, $\Delta DEF \cong \Delta ADF \cong \Delta ECF \cong \Delta ADF$
Thus, mid points divide the triangle into $4$ equal parts.

Now, 
$A (\Delta DEF) = \dfrac{1}{4} A (\Delta ABC)$

$A (\Delta DEF) = \dfrac{1}{4} (24)$

$A (\Delta DEF) = 6\ {cm}^2$

Hence, option D.

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 length, mass and capacity problems on measurement choosing and converting between units conversion between different units

Convert the following into quintal:
$400\ $ ton

  1. $400$ quintal
  2. $4,000$ quintal
  3. $40$ quintal
  4. $4$ quintal
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We know that


$1$  $ton =10$  $quintal$

$1$  $quintal =\dfrac1{10}$  $ton$

Given That, we have to convert $400$  $ton$  to   $quintals$

$400$  $tons =400 \times 10$  $quintals$

                   
                   $=4,000$  $quintals$

So, option $B$ is correct

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 $cm^2$