Quantitative Aptitude

Geometry

1,950 Questions
All Circle Tangents and Chords 772 Coordinate Geometry 360 Circle and Arc Measures 216 Triangle Centers and Medians 175 Composite and Inscribed Shapes 174 Similarity, Congruence, and Proportionality 118 Right Triangles and Pythagorean Theorem 100 Standard 2D Mensuration 22 Polygon and Angle Sums 10 3D Solid Mensuration 3

Geometry Questions

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

An arc AB of a circle subtends an angle x radians at the centre O of the circle. Given that the area of the sector AOB is equal to the square of the length of the arc AB, then the value of x?

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

Area of sector = (1/2) * r^2 * x. Arc length = r * x. Given: (1/2) * r^2 * x = (r * x)^2. This simplifies to (1/2) * r^2 * x = r^2 * x^2. Dividing by r^2 * x (assuming x is not 0), we get 1/2 = x.

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

The radius of a circle is $3.5$ cm and area of the sector is $3.85$ $cm^2$. Find the length of the corresponding arc.

  1. $2.2cm$
  2. $4.2cm$
  3. $5.1cm$
  4. $6.2cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the angle of centre made by the sector be $\theta$
Therefore,
Area of the sector=$\pi r^2\dfrac{\theta }{360}$
                        $=>3.85=\dfrac{\pi(3.5)^2\theta}{360}$


                        $=>\theta=\dfrac{3.8\times 360\times 7}{(3.5)^2\times 22}$
                        $=35.5$
                        $=36$
Thus length of the arc =$2\pi r\dfrac{\theta}{360}$
                                   =$2\times \dfrac{22}{7}\times 3.5\times \dfrac{36}{360}$
                                   =$2.2cm$

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

A chord of a circle of radius 6 cm subtends an angle of $\displaystyle 60^{\circ}$ at the centre of the circle. The area of the minor segment is
(use $\displaystyle \pi =3.14$)

  1. 6.54 $\displaystyle cm^{2}$
  2. 0.327 $\displaystyle cm^{2}$
  3. 7.25 $\displaystyle cm^{2}$
  4. 3.27 $\displaystyle cm^{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\displaystyle \theta =60^{\circ}$, r = 6 cm
Area of minor segment = $\displaystyle \dfrac{36}{2}\left [ \dfrac{60\times 3.14}{180}-\dfrac{\sqrt{3}}{2} \right ]$
                                     = 3.27 $\displaystyle cm^{2}$

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

The areas of two similar triangles are $16cm^2$ and $36cm^2$ respectively. If the altitude of the first triangle is $3cm$, then the corresponding altitude of the other triangle is:

  1. $4cm$
  2. $6.5cm$
  3. $4.5cm$
  4. $6cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Let ${A} _{1}$ and ${A} _{2}$ be the areas of the similar triangles.

$\Rightarrow \dfrac{{A} _{1}}{{A} _{2}}=\dfrac{{s} _{1}^{2}}{{s} _{2}^{2}}$

$\Rightarrow \dfrac{16}{36}=\dfrac{{\left(3\right)}^{2}}{{s} _{2}^{2}}$ given $({s} _{1}=3 \ cm )$

$\Rightarrow {s} _{2}^{2}=\dfrac{36\times 9}{16}$

$\Rightarrow {s} _{2}=\dfrac{6\times 3}{4}=4.5 \ cm$  

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

The areas of two similar triangles are $12$ ${cm}^{2}$ and $48$ ${cm}^{2}$. If the height of the smaller one is $2.1$ $cm$, then the corresponding height of the bigger one is:

  1. $4.41$ $cm$
  2. $8.4$ $cm$
  3. $4.2$ $cm$
  4. $0.525$ $cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Areas of two similar triangles are $12 cm^2$ and $48 cm^2$
For similar triangles the ratio of areas is equal to the ratio of square of corresponding heights
Hence, $\dfrac{A _1}{A _2} = \dfrac{(h _1)^2}{(h _2)^2}$

$\dfrac{12}{48} = \dfrac{(2.1)^2}{(h _2)^2}$

$(h _2)^2= 4 \times (2.1)^2$

$h _2 = 2 \times 2.1$

$h _2 = 4.2 cm$

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

The areas of two similar triangles are $\displaystyle 9\ { cm }^{ 2 }$ and $\displaystyle 16\ { cm }^{ 2 }$, respectively. The ratio of their corresponding heights is

  1. $3 : 4$
  2. $4 : 3$
  3. $2 : 3$
  4. $4 : 5$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
In similar traingles: -

${(\dfrac{{h1}}{{h2}})^2} = \dfrac{{S1}}{{S2}}$

Where h1 and h2 are the heights 

and S1, S2 are the areas of similar traingles

${{\rm{(}}\dfrac{{h1}}{{h2}})^2} = \dfrac{{9c{m^2}}}{{16c{m^2}}}$

$\dfrac{{h1}}{{h2}} = \sqrt {\dfrac{9}{{16}}} $

$\dfrac{{h1}}{{h2}} = \dfrac{3}{4}\ or\  {3:4}$
Multiple choice maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

The areas of two similar triangles are $121 cm^2$ and $81 cm^2$ respectively. Find the ratio of their corresponding heights.

  1. $\dfrac{11}{9}$
  2. $\dfrac{10}{9}$
  3. $\dfrac{9}{11}$
  4. $\dfrac{9}{10}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given the areas of two similar triangles are $121$ sq cm and $81$ sq cm

We know that, the ratio of areas of two similar triangles is equal to the ratio of the squares of the corresponding heights.

The ratio of area of triangles $= \dfrac{121}{81}=\dfrac{(11)^{2}}{(9)^{2}}$
Then ratio of height of triangle $=\sqrt{\left [ \dfrac{11}{9} \right ]^{2}}=\dfrac{11}{9}$

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

The area of two similar triangles ABC and PQR are $25\ cm^{2}\ & \  49\ cm^{2}$, respectively. If QR $=9.8$ cm, then BC is:

  1. 9.8 cm

  2. 7 cm

  3. 49 cm

  4. 25 cm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\dfrac { ar(ABC) }{ ar(PQR) } =\dfrac { 25 }{ 49 } $

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

$\Rightarrow { \left( \dfrac { BC }{ QR }  \right)  }^{ 2 }=\dfrac { 25 }{ 49 } \\ \Rightarrow \dfrac { BC }{ QR } =\dfrac { 5 }{ 7 } \\ \Rightarrow \dfrac { BC }{ 9.8 } =\dfrac { 5 }{ 7 } \\ \Rightarrow BC=\dfrac { 5 }{ 7 } \times 9.8=7$

 

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

Area of similar triangles are in the ratio $25:36$ then ratio of their similar sides is _________?

  1. $5:7$
  2. $5:6$
  3. $6:5$
  4. $6:7$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The areas and sides of similar triangles are related as 

$\dfrac{Ar(\Delta ABC)}{Ar(\Delta PQR)}=\left(\dfrac {AB}{PQ}\right)^2\\dfrac {25}{36}=\left(\dfrac {AB}{PQ}\right)^2\\dfrac{AB}{PQ}=\sqrt {\dfrac {25}{36}}=\dfrac 56$

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

The perimeter of two similar triangles is 30 cm and 20 cm. If one altitude of the former triangle is 12 cm, then length of the corresponding altitude of the latter triangle is 

  1. 8 cm

  2. 10 cm

  3. 12 cm

  4. 15 cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$\Delta$ABC and $\Delta$DEF be two similar triangle. Perimeter of first and second triangles are $30$cm and $20$cm respectively.
Then $\dfrac{AB}{DE}=\dfrac{BC}{EF}=\dfrac{AC}{DF}=k$ (say)
$\therefore AB=kDE, BC=kEF, AC=kDF$
$AB+BC+AC=k(DE+EF+DF)$
$\Rightarrow 30=k\times 20$
$\Rightarrow k=\dfrac{3}{2}$
$\Rightarrow \dfrac{AB}{DE}=\dfrac{3}{2}$
$\Rightarrow \dfrac{12}{DE}=\dfrac{3}{2}$
$\Rightarrow DE=8$.
Multiple choice maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

The perimeter of two similar triangles is 40 cm and 50 cm. Then the ratio of the areas of the first and second triangles is 

  1. 4 : 5

  2. 5 : 4

  3. 25 : 16

  4. 16 : 25
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
We know the ratio of perimeters of $2$ similar triangles are equal to ratio of corresponding sides

i.e., $\dfrac{perimeter \,of \,1^{st}}{perimeter\, of \,2^{nd}}=\dfrac{side\, of\, 1^{st}}{side\, of\, 2^{nd}}$

$\Rightarrow \dfrac{40}{50}=\dfrac{side\, of\, 1^{st}}{side\, of \,2^{nd}}=\dfrac{4}{5}$

As both the triangles are similar 

$\Rightarrow \dfrac{Area\, of\, 1^{st}}{Area\, of\, 2^{nd}}=\left(\dfrac{(side \,of\, 1^{st})^2}{(side\, of\, 2^{nd})^2}\right)=\dfrac{16}{25}$.
Multiple choice maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

The areas of two similar triangles are $49 \ {cm}^{2}$ and $64 \ {cm}^{2}$ respectively. The ratio of their corresponding sides is:

  1. $49:64$
  2. $7:8$
  3. $64:49$

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

Areas of two similar triangles are $49 $ cm $^2$ and $64$ cm $^2.$
For similar triangles the ratio of areas is equal to the ratio of square of corresponding sides.
Hence, $\dfrac{A _1}{A _2} = \dfrac{(s _1)^2}{(s _2)^2}$
$\Longrightarrow \dfrac{49}{64} = \dfrac{(s _1)^2}{(s _2)^2}$
$\Longrightarrow\dfrac{s _1}{s _2} = \dfrac{7}{8}$

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

If $\triangle ABC$ is similar to $\triangle DEF$ such that $BC=3$ cm, $EF=4$ cm and area of $\triangle ABC=54: \text{cm}^{2}.$ Find the area of $\triangle DEF.$ (in cm$^2$)

  1. $54$
  2. $36$
  3. $72$
  4. $96$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Since the ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides,
Therefore, $\displaystyle \frac{ar\left ( \triangle ABC \right )}{ar\left ( \triangle DEF \right )}=\frac{BC^{2}}{EF^{2}}$ 

$\Rightarrow $ $\displaystyle \frac{54}{ar\left ( \triangle DEF \right )}=\frac{3^{2}}{4^{2}}$ 
Thus $\displaystyle ar\left ( \triangle DEF \right )=\frac{54\times 16}{9}=96: \text{cm}^{2}$