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 length of an arc area of a sector of a circle sector and arc of a circle
The diameter of a circle is $10$ cm, then find the length of the arc, when the corresponding central angle is $180^{\circ}$.  $(\pi =3.14)$
  1. $15.7$
  2. $16$
  3. $3.14$
  4. $18$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Radius of the circle $ = \dfrac {\text{Diameter}}{2} = 5 $ cm 


Length of an arc subtending an angle $ \theta  = \dfrac { \theta  }{ 360 }

\times 2\pi R $, where $R$ is the radius of the circle. 

So, length of the arc $ = \dfrac {180}{360} \times 2 \times 3.14\times 5  = 15.7 $ cm

Multiple choice maths circle measures length of an arc area of a sector of a circle sector and arc of a circle
The diameter of a circle is $10$ cm, then find the length of the arc, when the corresponding central angle is $144^{\circ}$.$(\pi =3.14)$
  1. $44$ cm
  2. $12.56$ cm
  3. $12$ cm
  4. $88$ cm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Radius of the circle $ = \dfrac {Diameter}{2} = 5  cm $

Length of an arc subtending an angle $ \theta  = \dfrac { \theta  }{ 360 }

\times 2\pi R $ where R is the radius of the circle. 



So, length of the arc $ = \dfrac {144}{360} \times 2 \times 3.14 \times 5  = 12.56  cm $

Multiple choice maths circle measures length of an arc area of a sector of a circle sector and arc of a circle

A sector is cut from a circle of radius $21$ cm. The angle of the sector is $150^o$. Find the length of its arc and area.

  1. $27$ cm and $412.7cm^2$
  2. $36$ cm and $436.9cm^2$
  3. $45$ cm and $517.5cm^2$
  4. $55$ cm and $577.5cm^2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The length of arc $l$ and area $A$ of a sector of angle $\theta$ in a circle of radius $r$ are given by,


$l=\displaystyle\frac{\theta}{360^o}\times 2\pi r$

and $A=\displaystyle\frac{\theta}{360^o}\times \pi r^2$ respectively.

Here, $r=21$ cm and $\theta=150^0$


$\therefore l = \dfrac{150}{360}\times2\times\dfrac{22}7\times21 = 55$ cm

and 

$A = \dfrac{150}{36}\times\dfrac{22}7\times21^2 = \dfrac{1155}2 = 577.5\ {cm}^2$

Multiple choice maths circle measures length of an arc area of a sector of a circle sector and arc of a circle

A sector is cut off from a circle of radius $21$ cm The angle of the sector is $\displaystyle 120^{\circ} $ The length of its arc is [Take $\displaystyle \pi =\frac{22}{7} $]

  1. $40 cm$
  2. $44 cm$
  3. $35 cm$
  4. $28 cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given radius$(r)=21cm$ and the angle$(\theta)=120^\circ$

length of arc$=r\times \theta$

here $r=21cm,$ $\theta=120^\circ=\dfrac{120}{360}\times 2\pi$

length of arc = $\dfrac { \theta  }{ { 360 }^{ 0 } } \times 2\pi r=\dfrac { { 120 }^{ 0 } }{ { 360 }^{ 0 } } \times 2\times \dfrac { 22 }{ 7 } \times 21=44cm$

Multiple choice maths circle measures length of an arc area of a sector of a circle sector and arc of a circle

What is the length of an arc of a circle with a radius of $5$ if it subtends an angle of ${60}^{o}$ at the center?

  1. $3.14$
  2. $5.24$
  3. $10.48$
  4. $2.62$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given:
Radius (r)$=5$
Angle $=60^o$

Arc length for a particular angle we can write as -
$=\dfrac{\theta}{360}\times (2\pi r)$

$=\dfrac{60}{360}\times 2\pi \times 5$

$=\dfrac{10\pi}{6}=5.24$

Option 'B'.
Multiple choice maths circle measures length of an arc area of a sector of a circle sector and arc of a circle

If the sector of a circle of diameter $10$ cm subtends an angle of $144^{\circ}$ at the centre, then the length of the arc of the sector is

  1. $2\pi $ cm
  2. $4\pi $ cm
  3. $5\pi$ cm
  4. $6\pi $ cm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given, diameter $=10$ cm, $\theta=144^o$
Length of an arc of a circle $=\dfrac { \theta  }{ 360 } \times 2\pi { r }=\dfrac { 144 }{ 360 } \times 2\pi \times \dfrac { 10 }{ 2 } =4\pi $ cm 
Hence, option B is correct.
Multiple choice maths circle measures length of an arc area of a sector of a circle sector and arc of a circle

A circular wire of radius $7$ cm is cut and bend again into an arc of a circle of radius $12$ cm. The angle subtended by the arc at the centre is

  1. $50^\circ$
  2. $210^\circ$
  3. $100^\circ$
  4. $60^\circ$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given, radius of circular wire $= 7$ cm

Circumference of wire $= 2 \pi r = 2 \pi (7) = 14 \pi$
Radius of arc $= 12$ cm

Angle subtended by the arc $= \dfrac{\text{arc}}{\text{radius}} = \dfrac{14 \pi}{12} = \dfrac{7 \pi}{6}$

Angle subtended by arc $=\cfrac{7\pi}{6}\times \cfrac{180}{\pi}= 210^{\circ}$

Multiple choice maths geometric constructions circumscribing and inscribing a circle on a regular hexagon construction of tangent to a circle construction of tangents

A circle is inscribed in a quadrilateral ABCD in which $\angle B = 90^o$. If $AD = 23 cm$, $AB = 29 cm$ and $DS = 5 cm$. Find the radius of the circle.

  1. $11$ cm
  2. $13$ cm
  3. $9$ cm

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$AS$ and $AP$ are tangents drawn to the circle at $A$

$\implies AS = AP$

Similarly

$BP = BQ$

$QC = CR$

$RD = DS$

Given

$AD = 23$

$\implies AS + SD = 23$

$AS = 23 – 5 = 18 = AP$

$AB = 29 \implies AP + BP = 29$

$\implies 18 + BP = 29 \implies BP = 11cm$

Now consider rectangle $PBQO$

$PB – BQ , OP = OQ = radius$

$\angle PBQ = 90$    

WKT

$OP \perp BP $ and $OQ \perp BQ$

Since radius is perpendicular to tangent at point of contact

$\implies$ All the angles are 90 degree and adjacent sides are equal

So, It is a square

$\implies r = BP = 11cm$

Multiple choice maths geometric constructions circumscribing and inscribing a circle on a regular hexagon construction of tangent to a circle construction of tangents

Given are the steps are construction of a pair of tangents to a circle of radius $4$cm from a point on the concentric circle of radius $6$cm. Find which of the following step is wrong?
(P) Take a point O on the plane paper and draw a circle of radius OA$=4$cm. Also, draw a concentric circle of radius OB$=6$cm.
(Q) Find the mid-point A of OB and draw a circle of radius BA$=$AO. Suppose this circle intersects the circle of radius $4$cm at P and Q.
(R) Join BP and BQ to get the desired tangents from a point B on the circle of radius $6$ cm.

  1. Only (P)

  2. Only (Q)

  3. Both (P) & (Q)

  4. Both (Q) & (R)
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice maths mid-point and its converse proving the mid-point theorem the mid-point theorem mid point theorem

If the lengths of the medians $AD, BE$ and $CF$ of the triangle $ABC$, are $6,8,10$ respectively, then

  1. $AD$ and $BE$ are perpendicular
  2. $BE$ and $CF$ are perpendicular
  3. area of $\Delta ABC=32$
  4. area of $\Delta DEF=8$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\begin{array}{l}AD = \sqrt {2A{B^2} + 2A{C^2} - B{C^2}}  = 6\2A{B^2} + 2A{C^2} - B{C^2} = 36\BE = \sqrt {2A{B^2} + 2B{C^2} - A{C^2}}  = 8\2A{B^2} + 2B{C^2} - A{C^2} = 64\CF = \sqrt {2A{C^2} + 2B{C^2} - A{B^2}}  = 10\2A{C^2} + 2B{C^2} - A{B^2} = 100\A{B^2} = x\A{C^2} = y\B{C^2} = z\2x + 2y - z = 36\2x + 2z - y = 64\2y + 2z - x = 100\x = A{B^2} = \frac{{100}}{9}\y = A{C^2} = \frac{{208}}{9}\z = B{C^2} = \frac{{292}}{9}\AD = 6,BE = 8,CF = 10\in,\Delta ABE\AD \bot BE\area,\Delta BEC = 16\area,\Delta ABE = 16 + 16 = 32\\frac{{area,\Delta ABE}}{{area,\Delta DEF}} = 4\\frac{{32}}{{area,\Delta DEF}} = 4\area,\Delta DEF = 8\end{array}$

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

If $R$ is the radius of circumscribing circle of a regular polygon of $n$ sides, then $R =?$

  1. $\dfrac{a}{2} sin (\dfrac{\pi}{n})$
  2. $\dfrac{a}{2} cos (\dfrac{\pi}{n})$
  3. $\dfrac{a}{2} cosec (\dfrac{\pi}{n})$
  4. $\dfrac{a}{2} cosec (\dfrac{\pi}{2n})$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
since, it is a regular polygon so its interior angle will be equal  

Hence, $nA=\pi\Rightarrow A=\dfrac{\pi}{n}$

and we know that 
$\dfrac{a}{sinA}=2R\Rightarrow R=\dfrac{a}{2}cosec(\dfrac{\pi}{n})$

therefore,Answer is $C$
Multiple choice maths understanding 3d and 2d shapes defining regular polygons sum of exterior angles of a polygon regular polygons

The ratio of the areas of two regular octagons which are respectively inscribed and circumscribed to a circle of radius $r$ is

  1. $\cos{\dfrac{\pi}{8}}$
  2. ${\sin}^{2}{\dfrac{\pi}{8}}$
  3. ${\cos}^{2}{\dfrac{\pi}{8}}$
  4. ${\tan}^{2}{\dfrac{\pi}{8}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Inscribed circle of a regular polygon of $n$ sides
${A} _{1}=n{r}^{2}\tan{\dfrac{\pi}{n}}$
Here $n=8$
$\therefore {A} _{1}=8{r}^{2}\tan{\dfrac{\pi}{8}}$
Circumscribed circle of a regular polygon of $n$ sides is
${A} _{2}=\dfrac{n{R}^{2}}{2}\sin{\dfrac{2\pi}{n}}$
For $n=8$ we have
${A} _{2}=\dfrac{8{R}^{2}}{2}\sin{\dfrac{2\pi}{8}}$
  $=\dfrac{8{R}^{2}}{2}\sin{\dfrac{\pi}{4}}$
  $=\dfrac{8{r}^{2}}{2}2\sin{\dfrac{\pi}{8}}\cos{\dfrac{\pi}{8}}$ (for $R=r$)
  $=8{r}^{2}\sin{\dfrac{\pi}{8}}\cos{\dfrac{\pi}{8}}$ 
$\therefore \dfrac{{A} _{2}}{{A} _{1}}=\dfrac{8{r}^{2}\sin{\dfrac{\pi}{8}}\cos{\dfrac{\pi}{8}}}{8{r}^{2}\tan{\dfrac{\pi}{8}}}$
$=\dfrac{\sin{\dfrac{\pi}{8}}\cos{\dfrac{\pi}{8}}}{\dfrac{\sin{\dfrac{\pi}{8}}}{\cos{\dfrac{\pi}{8}}}}$
$={\cos}^{2}{\dfrac{\pi}{8}}$

Multiple choice mathematics and statistics coordinates, points and lines what is meant by the equation of a straight line or of a curve introduction to slope introduction to straight lines

A circle of radius 2 is concentric with ellipse $\frac{x^{2}}{7}+\frac{y^{2}}{3}=1$ then inclination of common tangent with x-axis - 

  1. $\frac{\pi }{2}$
  2. $\frac{\pi }{4}$
  3. $\frac{\pi }{3}$
  4. $\frac{\pi }{6}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For an ellipse x^2/a^2 + y^2/b^2 = 1, the tangent with slope m is y = mx +/- sqrt(a^2m^2 + b^2). For a concentric circle x^2 + y^2 = r^2, the tangent is y = mx +/- r*sqrt(1+m^2). Equating these leads to the vertical tangent case.