Mathematics · Quantitative Aptitude

Circle and Arc Properties

115 Questions

Circle and arc properties involve calculating arc lengths, understanding radius relationships, and solving geometric proofs. These geometry concepts are crucial for quantitative aptitude tests. Review these questions to improve your spatial reasoning and accuracy.

Arc length calculationsCircle theoremsRadius and diameterInscribed polygonsCentral angles

Circle and Arc Properties Questions

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 angles in our surroundings measuring and drawing angles angles in a clock checking angles between hands of a clock

A circular paper is divided into $4$ equal parts by cutting it through two diameters. Then the central angle of each part is equal to:

  1. $45^\circ$
  2. $90^\circ$
  3. $60^\circ$
  4. $30^\circ$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Central angle $=360^{\circ}$

Now circular piece of paper is divided in four equal parts .
Let measure of central angle $=x$
$\ \Rightarrow x+x+x+x={ 360 }^{ \circ  }\ \Rightarrow 4x={ 360 }^{ \circ  }\ \Rightarrow x=\dfrac { { 360 }^{ \circ  } }{ 4 } \ \Rightarrow x={ 90 }^{ \circ  }$
So option $B$ is correct.

Multiple choice maths angles in our surroundings measuring and drawing angles angles in a clock checking angles between hands of a clock

The central angle of a part of a circle which is divided into $6$ equal parts is:

  1. $60^\circ$
  2. $30^\circ$
  3. $45^\circ$
  4. $90^\circ$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Central angle $=360^{\circ}$

Now park is divided in four equal parts.

Let measure of central angle $=x$
$\ \Rightarrow x+x+x+x+x+x={ 360 }^{ \circ  }\ \Rightarrow 6x={ 360 }^{ \circ  }\ \Rightarrow x=\dfrac { { 360 }^{ \circ  } }{ 6 } \ \Rightarrow x={ 60 }^{ \circ  }$

So option $A$ is correct.

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 construction of line segment and circle of given radius construction related to lines

Let C be the circle with centre at $(1, 1)$ and radius $=1$. If T is the circle centred at $(0, y)$, passing through origin and touching the circle C externally, then the radius of T is equal to?

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

Let circle C have center (1, 1) and radius 1. Circle T has center (0, y), passes through the origin (0, 0), so its radius is r = y. Circle T touches C externally, meaning the distance between their centers equals the sum of their radii: sqrt((0 - 1)^2 + (y - 1)^2) = r + 1 = y + 1. Squaring both sides: 1 + y^2 - 2y + 1 = y^2 + 2y + 1, which simplifies to 2 - 2y = 2y + 1, giving 4y = 1, so y = 1/4. Thus, the radius is 1/4.

Multiple choice maths construction circumscribing and inscribing a circle on a regular hexagon construction of tangent to a circle construction of tangents construction of line segment and circle of given radius construction related to lines

The sides of a triangle are $25,39$ and $40$. The diameter of the circumscribed circle is: 

  1. $\cfrac { 133 }{ 3 } $
  2. $\cfrac { 125 }{ 3 } $
  3. $42$
  4. $41$
  5. $40$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Circum radius formula

$R$ $=\cfrac { abc }{ \sqrt { (a+b+c)(b+c-a)(c+a-b)(a+b-c) }  }$ .
Where  $a, b, c$  are sides of triangle 
$\Rightarrow$ $R$ $=\cfrac { 25\times 39\times 40\quad  }{ \sqrt { (140\quad \times (54)\times (26)\quad \times (240) }  } $
$=\cfrac { 25\times 39\times 40\quad  }{ \sqrt { { 2 }^{ 3 } } \times 13\times 2\times { 3 }^{ 3 }\times 2\times 13\times { 2 }^{ 3 }\times 3 } $
$=\cfrac { 25\times 39\times 40\quad  }{ \sqrt { { 2 }^{ 8 } } \times { 3 }^{ 4 }\times { 13 }^{ 2 } } $.
$=\cfrac { 25 \times \ 39 \times 40  }{ { 2 }^{ 4 }\times { 3 }^{ 2 }\times { 13 } } =\quad \cfrac { 25 \times 39 \times40\quad  }{ 16\times 9\times { 13 } }$ 
$=\cfrac { 125 }{ 6 }$ 
$\therefore$   Diameter $=\cfrac { 125\times \ 2 }{ 6 } = \cfrac { 125 }{ 3 } $

$\therefore$ B) Answer.

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

Two consecutive vertices of a regular hexagon $A _1A _2A _3A _4A _5A _6$ are $A _1\equiv (1, 0), A _2\equiv (3, 0)$. If the centre of hexagon lies above the x-axis, then equation of the circumcircle of the hexagon is?

  1. $x^2+y^2-4x-2\sqrt{3}y+\dfrac{17}{3}=0$
  2. $x^2+y^2-4x-2\sqrt{3}y+\dfrac{25}{3}=0$
  3. $x^2+y^2-4x-2\sqrt{3}y+3=0$

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

The side length is 2. The center of the hexagon is found by rotating the vector (A2-A1) by 60 degrees. With A1=(1,0) and A2=(3,0), the center is (2, sqrt(3)). The radius is 2. The circle equation is (x-2)^2 + (y-sqrt(3))^2 = 4, which simplifies to x^2 + y^2 - 4x - 2sqrt(3)y + 17/3 = 0.

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

Let ${A} _{0}{A} _{1}{A} _{2}{A} _{3}{A} _{4}{A} _{5}$ be a regular hexagon inscribed in a circle of unit radius.Then the product of the length of  ${A} _{0}{A} _{1}.{A} _{0}{A} _{2}.{A} _{0}{A} _{4}$ is

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

Given ${A} _{0}{A} _{1}{A} _{2}{A} _{3}{A} _{4}{A} _{5}$ is  a regular hexagon inscribed in a circle of unit radius
$\Rightarrow {A} _{0}{A} _{1}=1$
$\Rightarrow {A} _{0}{A} _{2}=2\sin{{60}^{0}}=2\times\dfrac{\sqrt{3}}{2}=\sqrt{3}$
$\Rightarrow {A} _{0}{A} _{4}=2\sin{{60}^{0}}=2\times\dfrac{\sqrt{3}}{2}=\sqrt{3}$
$\therefore {A} _{0}{A} _{1}.{A} _{0}{A} _{2}.{A} _{0}{A} _{4}=1\times \sqrt{3}\times\sqrt{3}=3$

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 the inscribed circle of a regular polygon of $n$ sides, then $r$ is equal to?

  1. $\dfrac{a}{2} cot (\dfrac{\pi}{2n})$
  2. $\dfrac{a}{2} cot (\dfrac{\pi}{n})$
  3. $\dfrac{a}{2} tan (\dfrac{\pi}{n})$
  4. $\dfrac{a}{2} cos (\dfrac{\pi}{n})$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

in $\Delta{ABL}$, $AL$ is the radius of the inscribed circle. 


$ BL=\cfrac{BC}{2}=\cfrac{a}{2}$

$\cot(\cfrac{\pi}{n})=\cfrac{AL}{BL}=\cfrac{r}{\dfrac{a}{2}}$

Hence $r=\cfrac{a(\cot(\dfrac{\pi}{n}))}{2}$

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

Let ${A} _{0}{A} _{1}{A} _{2}{A} _{3}{A} _{4}{A} _{5}$ be a regular hexagon inscribed in a circle of unit radius.The product of the length of the line segments ${A} _{0}{A} _{1},{A} _{0}{A} _{2}$ and ${A} _{0}{A} _{4}$ is

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

${A} _{0}{A} _{1}=2.1.\cos{\dfrac{\pi}{3}}=2.1.\dfrac{1}{2}=1={A} _{1}{A} _{2}$
$\cos{\dfrac{2\pi}{3}}=\dfrac{{\left({A} _{0}{A} _{1}\right)}^{2}+{\left({A} _{1}{A} _{2}\right)}^{2}-{\left({A} _{0}{A} _{2}\right)}^{2}}{2{A} _{0}{A} _{1}.{A} _{1}{A} _{2}}$
$\Rightarrow \cos{\left(\pi-\dfrac{\pi}{3}\right)}=\dfrac{1+1-{\left({A} _{0}{A} _{2}\right)}^{2}}{2.1.1}$
$\Rightarrow -\cos{\dfrac{\pi}{3}}=\dfrac{2-{\left({A} _{0}{A} _{2}\right)}^{2}}{2}$
$\Rightarrow \dfrac{-1}{2}=\dfrac{2-{\left({A} _{0}{A} _{2}\right)}^{2}}{2}$
$\Rightarrow -1=2-{\left({A} _{0}{A} _{2}\right)}^{2}$
$\Rightarrow {\left({A} _{0}{A} _{2}\right)}^{2}=3$
$\therefore {A} _{0}{A} _{2}=\sqrt{3}={A} _{0}{A} _{4}$
$\therefore {A} _{0}{A} _{1}\times {A} _{0}{A} _{2}\times{A} _{0}{A} _{4}=1\times\sqrt{3}\times\sqrt{3}=3$

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 $R=r$)
  $=8{r}^{2}\sin{\dfrac{\pi}{8}}\cos{\dfrac{\pi}{8}}$ 
$\therefore \dfrac{{A} _{2}}{{A} _{1}}=\dfrac{8{r}^{2}\sin{\dfrac{\pi}{8}}\cos{\dfrac{\pi}{8}}}{8{r}^{2}\tan{\dfrac{\pi}{8}}}$
$=\dfrac{\sin{\dfrac{\pi}{8}}\cos{\dfrac{\pi}{8}}}{\dfrac{\sin{\dfrac{\pi}{8}}}{\cos{\dfrac{\pi}{8}}}}$
$={\cos}^{2}{\dfrac{\pi}{8}}$