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 circle measures area between two concentric circles the area of ring semicircle and ring

If the perimeter of a semi-circle is $36\ cm$. What will be its diameter?

  1. $88\ cm$
  2. $22\ cm$
  3. $28\ cm$
  4. $14\ cm$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Perimeter of semicircle$=36cm$
$\pi r+d=36$
$\Rightarrow \pi r+2r=36$
$\Rightarrow r\left( \cfrac { 22 }{ 7 } +2 \right) =36$
$\Rightarrow \cfrac { 36 }{ 7 } \times r=36$
$\Rightarrow r=7$
$\therefore $ Diameter $2r=2\times 7=14cm$
Multiple choice maths circle measures area between two concentric circles the area of ring semicircle and ring

If a wire is bent into the shape of a square the area of the square is 81 sq cm .When the wire is bent into a semi circular shape; what is the area of the semicircle? $\displaystyle \left ( \pi =\frac{22}{7} \right )$

  1. $\displaystyle 77\:\text{cm}^{2}$
  2. $\displaystyle 73\:\text{cm}^{2}$
  3. $\displaystyle 37\:\text{cm}^{2}$
  4. $\displaystyle 33\ \text{cm}^{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let '$a$' be the length of each side of the square
Then $\displaystyle a^{2}=81\Rightarrow a=9$ cm
Length of wire $=$ Perimeter of square
=$ 4a= 36$ cm
$\displaystyle \Rightarrow $ Circuference of semicircle $= 36$ cm
$\displaystyle \Rightarrow \pi r+2r=36$

$ \displaystyle \Rightarrow r\left ( \pi +2 \right )=36$
$\displaystyle \Rightarrow  r =\dfrac{36}{\pi +2}=\dfrac{36}{\dfrac{22}{7}+2}=\dfrac{36\times 7}{\left ( 22+14 \right )}=\dfrac{36\times 7}{36}$ cm $=7$ cm
$\displaystyle \therefore \ \text{Area of the semicircle}=\frac{1}{2}\pi r^{2}$
$\displaystyle =\dfrac{1}{2}\times \dfrac{22}{7}\times 7\times \text{cm}^{2}=77\text{cm}^{2}$

Multiple choice maths circle measures area between two concentric circles the area of ring semicircle and ring

If a wire is bent into the shape of a square, then the area of the square is $81\ cm^{2}$. When the same wire is bent into a semi-circular shape, then the area of the semi circle will be

  1. $22\ cm^{2}$
  2. $44\ cm^{2}$
  3. $77\ cm^{2}$
  4. $154\ cm^{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the side of the square be $a$ cm

Thus, perimeter is $4a$ and area of the square is $a^2$
$\therefore a^2=81$
$\Rightarrow a=\sqrt{81}$
$\Rightarrow a=9$
Thus, perimeter $=4 \times 9$ $=36$ cm
Now, wire is bent into a semicircle.
Therefore, $2r+πr=36$
$⇒r(π+2)=36$ $
$⇒r=\cfrac{36}{\dfrac{22}{7}+2}$

$⇒r=\cfrac { 36 }{\dfrac{36}{7}}$
$⇒r=7$ cm

Area of semicircle $=\cfrac{1}{2}πr^2$
$=\cfrac{1}{2}×\cfrac{22}{7}×7×7$
$=77\ cm^2$

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 area and volume of cylinder hollow cylinder volume of cylinder volume of cylinder and cone

A cistern $6$ m long and $4$ m wide contains water to a depth of $1.25$ m. What is the area of wetted surface?

  1. $40$ sq. m
  2. $45$ sq. m
  3. $49$ sq. m
  4. $73$ sq. m
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given, $l = 6, b = 4$

Also given depth i.e., $ h = 1.25$
Area of the wetted surface $= 2[lb + bh + hl]$
$=2[(6\times 4)+(4\times 1.25)+(1.25\times 6)]$
$=2[24+5+7.5]$
$= 73$ sq. m
Therefore, the area of wetted surface is $73$ sq. m.

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 introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

If the extremities of a diagonal of a square are $(1, -2, 3)$ and $(2, -3, 5)$, then area of the square is

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

Let the extremities of the diagonal of a square be $(1,-2,3)$ and $B(2,-3,5)$.
Then $AB$ is given by $ {({1}^{2} + {1}^{2} + {2}^{2})}^{0.5} $ = $ \sqrt{6} $
Hence, length of the side $ = \sqrt {3} $
So, area of square will be $ \sqrt{3} \times \sqrt{3}  = 3$

Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

If the extremities of a diagonal of a square are $(1, -2, 3)$ and $(4, 2, 3)$ then the area of the square is

  1. $25$
  2. $50$
  3. $\displaystyle \frac{25}{2}$
  4. $\sqrt{50}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If $a$ is the length of a side of square then the length of diagonal is given by $\sqrt{2}a$. Distance between two given  points is $\sqrt{(1-4)^2+(-2-2)^2+(3-3)^2}=5=\sqrt{2}a$. Hence the area is given by $a^2=\dfrac{25}{2}$.

Multiple choice maths equation reducing simple equations to simpler form solving linear equations solution of a linear equation in one variable

The area of square $ABCD$ is three-fourths the area of parallelogram $EFGH$. The area of parallelogram $EFGH$ is one-third the area of trapezoid $IJKL$. If square $ABCD$ has an area of $125$ square feet, calculate the area of trapezoid $IJKL$, in square feet.

  1. $75$
  2. $225$
  3. $350$
  4. $500$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given, area of square $ABCD$ is three fourth of area of parallelogram $EFGH$,

And the area of parallelogram $EFGH$ is one-third of the area of trapezoid $IJKL$ and area of square $ABCD$ is $125$.
Let the area of trapezoid $IJKL$ is $x$
Then  area of  parallelogram $EFGH =$ $\dfrac{1}{3}x$
And  area of square $ABCD=$ $\dfrac{3}{4}$ area of  parallelogram $EFGH=$ $\dfrac{3}{4}\times \dfrac{1}{3}x=\dfrac{1}{4}x$
But area of square $ABCD =125$
$\therefore \dfrac{1}{4}x=125$
$\Rightarrow x=500$
So, area of trapezoid $IJKL=500$.

Multiple choice maths calculations and mental strategies 1 equations from statements forming equations from statements writing mathematical statements

The area of a field in the shape of a trapezium measures $1440{m}^{2}$. The perpendicular distance between its parallel sides is $24m$. If the ratio of the parallel sides is $5:3$, the length of the longer parallel side is:

  1. $45m$
  2. $60m$
  3. $75m$
  4. $120m$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Parallel sides $= 5x,\, 3x$
area $=\dfrac{24}{2}(5x+3x)=1440 $
$12(8x)=1440$
$x=\dfrac{120}{8}=15$ 
$5x=15\times 5$ 
     $=75m$
Multiple choice maths congruence introduction to shapes similarity of triangles introduction to similar triangles

If area $(\Delta ABC)=36 cm^2, area (\Delta DEF)=64 cm^2$ and $DE=6.4 cm$. Find AB if $\Delta ABC\sim \Delta DEF$

  1. $3.6$ cm
  2. $7.2$ cm
  3. $4.8 $cm

  4. None
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In similar triangles, $\dfrac {area\Delta ABC}{area \Delta DEF}=\dfrac {AB^2}{DE^2}=\dfrac {36}{64}$


$\Rightarrow \dfrac {AB}{6.4}=\dfrac {3}{4}\Rightarrow AB=4.8$.

Multiple choice physics units and measurement: error analysis significant figures significant figures and rounding of digits units and measurements

Area of a square is $(100\pm 2)m^2$. Its side is:

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

$Area=(Length)^2$
$Length =(Area)^{1/2}$
             $=(100\pm 2)^{1/2}$
             $=(100)^{1/2}\pm \dfrac {1}{2}\times 2$
             $=(10\pm 1)m$

Multiple choice maths geometry similarity and right angled triangle angle theorems for a right angled triangle apollonius's theorem

Find the side of the square whose diagonal is $16 \sqrt 2$ cm.

  1. $4$ cm
  2. $16$ cm
  3. $8$ cm
  4. $16\sqrt 2$ cm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We know that,

1) All angles of a square are congruent. i.e $90^o$
2) Diagonal of a square bisects each of its angles.
Therefore, the square gets divided into $2$ triangles of degrees $45^o-45^o-90^o$
$\therefore \sin 45^o = \cfrac {\text {side}}{\text {hyp}}$ 
$\therefore \cfrac {1}{\sqrt 2} = \cfrac {\text {side}}{16 \sqrt 2}$
$\therefore$ side of the square $= 16$ cm.