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 how many squares area of rectangular paths comparing areas spaces and boundaries - 2

The side of a square is 2 cm Semicircles are constructed on two sides of the square then the area of the whole figure is

  1. $ \displaystyle (4+\pi )cm^{2} $
  2. $ \displaystyle (4+4\pi )cm^{2} $
  3. $ \displaystyle 4\pi cm^{2} $
  4. $ \displaystyle 8\pi cm^{2} $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The side of square is 2 cm 

Then area of square =$(2)^{2}=4 cm^{2}$
The Semicircle constructed on two side diameter 2 cm then radius =1 cm
Then area of one semicircle =$\frac{\pi r^{2}}{2}=\frac{\pi (1)^{2}}{2}=\frac{\pi }{2}$
Then  area of two semicircle=$2\times \frac{\pi }{2}=\pi $
So total area of whole figure=$(4+\pi )cm^{2}$

Multiple choice maths enlargement and scale drawing dilation enlargement similarity as a size transformation mapping mapping space around us bearing and drawings

If the area of square is $36\pi \ \text{cm}^2$. If its length is scaled three times, what would be its new area?

  1. $342\pi$
  2. $324\pi$
  3. $352\pi$
  4. $322\pi$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Area of square whose side is $a =a^2$ 

If its length is scaled three times, then area $=9a^2$
Therefore, new area $=9\times 36\pi \text{cm}^2=324\pi \text{cm}^2$

Multiple choice maths construction of polygons construction of parallelograms and rectangles construction of special quadrilaterals constructions related to a quadrilateral

You are given the length of a diagonal of a rhombus and one of the angles of the rhombus. Which property of the rhombus will be used in the construction of this rhombus?

  1. The lengths of the sides of a rhombus are equal.
  2. The angles of a rhombus are $90^\circ$

  3. Diagonal of a rhombus bisects the opposite angles.

  4. Diagonals of a rhombus are perpendicular bisectors of each other.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\Rightarrow$   We have given the length of diagonal of rhombus and one of angles of rhombus.

$\Rightarrow$  To construct an rhombus we will use the property that the diagonal of a rhombus bisect the opposite angle.
Because we know opposite angles of rhombus are equal, so it will be easier to construct rhombus.

Multiple choice maths construction of polygons construction of parallelograms and rectangles construction of special quadrilaterals constructions related to a quadrilateral

What would be the length of side $BC$ in Square $ABCD$ if the diagonal of the square given is $10$ cm?

  1. $5$ cm
  2. $5\sqrt2$ cm
  3. $10$ cm
  4. $10\sqrt2$ cm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The side of a square is $\dfrac{1}{\sqrt2}$ times of the diagonal.


If the length of diagonal $=10$ cm

Then length of side $=10\times \dfrac{1}{\sqrt2}=5\sqrt2$ cm.

Multiple choice maths construction of polygons construction of parallelograms and rectangles construction of special quadrilaterals constructions related to a quadrilateral

The side of a regular hexagon is 'p' cm then its area is

  1. $ \displaystyle \frac{\sqrt{3}}{2}p^{2}cm^{2} $
  2. $ \displaystyle \frac{3\sqrt{3}}{2}p^{2}cm^{2} $
  3. $ \displaystyle 2\sqrt{3}p^{2}cm^{2} $
  4. $ \displaystyle 6p^{2}cm^{2} $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given side of hexa gon is p cm 

Then area of hexagon =$\frac{(side)^{2}\times  n}{4tan\frac{180}{n}}=\frac{p^{2}\times 6}{4tan\frac{180}{6}}=\frac{6p^{2}}{4tan30^{0}}=\frac{3p^{2}}{2\times \frac{1}{\sqrt{3}}}=\frac{3\sqrt{3}p^{2}}{2} cm^{2}$

Multiple choice maths perimeter and area of rectilinear figures region enclosed by a plane figure area of square and rectangle mensuration-i (area)

The perimeter of a rhombus is 100 cm and one of the diagonals is 40 cm, Then the area of the rhombus is

  1. 1000 cm$^2$
  2. 500 cm$^2$
  3. 1200 cm$^2$
  4. 600 cm$^2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given that, perimeter of rhombus is $100cm$ and one diagonal $40cm$

Perimeter of rhombus=4$\times $side

So side$=25cm$

And we know that in a rhombus diagonal divides rhombus in two eauilateral triangle.

Area of one triangle:- s$=(25+25+40)/2 =45$

So area=(s$\times $(s-a)$\times $ (s-b)$\times $(s-c))^1/2

= (45$\times $(45-40)$\times $(45-25)$\times $(45-25))^1/2

$= 300cm^2$

So total area is 300$\times $2=600cm^2

 

Hence, this is the answer.

Multiple choice maths perimeter and area converting between units of areas the metric system metric system

The area of a square field is $30\dfrac {1}{4}m^2$. Calculate the length of the side of the square.

  1. $5\dfrac {1}{3}m$
  2. $5\dfrac {1}{2}m$
  3. $5\dfrac {2}{5}m$
  4. $5\dfrac {1}{4}m$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let side of square field be $x$.


Then area of square field $= x^2$


According to question

$x^2 = 30 \dfrac{1}{4} m^2$

$x^2 = \dfrac{121}{4} m^2$

$x = \dfrac{\sqrt{121}}{\sqrt{4}} m$

$x = \dfrac{11}{2}$

Hence side of square field is $5 \dfrac{1}{2} m$

Option (B)

Multiple choice maths perimeter and area converting between units of areas the metric system metric system

The area of a square field is $80\dfrac {244}{729}$ square metres. Find the length of each sides field.

  1. $8\dfrac {25}{27}m$
  2. $8\dfrac {24}{27}m$
  3. $8\dfrac {26}{27}m$
  4. $8\dfrac {22}{27}m$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given area of square field $= 80 \dfrac{244}{729} m^2$


Let us assume that side of square field is x 


Then area of field $= x^2$

According to question

$x^2 = 80 \dfrac{244}{729} m^2$

$x^2 = \dfrac{58,564}{729} m^2$

$x = \dfrac{\sqrt{58,564}}{\sqrt{729}} m$

$x = \dfrac{242}{27} m$

Hence side of field is $8 \dfrac{26}{27} m$.

Option (C)

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.