• Defining regular polygons - class-IX
Back
0

Defining regular polygons - class-IX

Description: defining regular polygons
Number of Questions: 32
Created by:
Tags: shapes and geometric reasoning polygon introduction to trigonometry polygons similar triangles geometric proof understanding shapes understanding elementary shapes rectilinear figures understanding 3d and 2d shapes maths
Start the Quiz
Attempted 0/29 Correct 0 Score 0
‹
›

If $A+B=\dfrac{\pi}{3}$ and $\cos{A}+\cos{B}=1$, then which of the following is true

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

  1. $\cos{\left(A-B\right)}=\dfrac{1}{3}$

  2. $\left|\cos{A}-\cos{B}\right|=\sqrt{\dfrac{2}{3}}$

  3. $\cos{\left(A-B\right)}=-\dfrac{1}{3}$

  4. $\left|\cos{A}-\cos{B}\right|=\dfrac{1}{2\sqrt{3}}$

Show answer
Correct Option: B,C
Explanation:

$\cos{A}+\cos{B}=1$


$\Rightarrow 2\cos{\left(\dfrac{A+B}{2}\right)}\cos{\left(\dfrac{A-B}{2}\right)}=1$

Since $A+B=\dfrac{\pi}{3}\Rightarrow \dfrac{A+B}{2}=\dfrac{\pi}{6}$
Hence $\cos{\left(\dfrac{A+B}{3}\right)}=\cos{\left(\dfrac{\pi}{6}\right)}=\dfrac{\sqrt{3}}{2}$

$\Rightarrow 2\cos{\left(\dfrac{A-B}{2}\right)}=\dfrac{1}{\dfrac{\sqrt{3}}{2}}$

$\Rightarrow \cos{\left(\dfrac{A-B}{2}\right)}=\dfrac{1}{\sqrt{3}}$

Squaring both sides, we get

${\cos}^{2}{\left(\dfrac{A-B}{2}\right)}=\dfrac{1}{3}$

$\Rightarrow 2{\cos}^{2}{\left(\dfrac{A-B}{2}\right)}=\dfrac{2}{3}$

$\Rightarrow 2{\cos}^{2}{\left(\dfrac{A-B}{2}\right)}-1=\dfrac{2}{3}-1=\cos {(A-B)}=\dfrac{-1}{3}$

$\left|\cos{A}-\cos{B}\right|=2\sin{\left(\dfrac{A+B}{2}\right)}\sin{\left(\dfrac{B-A}{2}\right)}$

                    $=2\times\dfrac{1}{2}\sqrt{1-\dfrac{1}{3}}$

                    $=\sqrt{\dfrac{2}{3}}$ (on simplification)

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

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

  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. $\dfrac{a}{2} cosec (\dfrac{\pi}{2n})$

Show answer
Correct Option: C
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 $C$

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?

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

  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

Show answer
Correct Option: A

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

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

  1. $\dfrac{3}{4}$

  2. $3\sqrt{3}$

  3. $3$

  4. $\dfrac{3\sqrt{3}}{2}$

Show answer
Correct Option: C
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
$\Rightarrow {A} _{0}{A} _{1}=1$
$\Rightarrow {A} _{0}{A} _{2}=2\sin{{60}^{0}}=2\times\dfrac{\sqrt{3}}{2}=\sqrt{3}$
$\Rightarrow {A} _{0}{A} _{4}=2\sin{{60}^{0}}=2\times\dfrac{\sqrt{3}}{2}=\sqrt{3}$
$\therefore {A} _{0}{A} _{1}.{A} _{0}{A} _{2}.{A} _{0}{A} _{4}=1\times \sqrt{3}\times\sqrt{3}=3$

In the given regular hexagon of side $8\ cm$, six circles of equal radius are inscribe as shown in figure. The area of the unshaded region is $(in\ cm^{2})$

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

  1. $99\sqrt{3} -144 \sqrt{3} (\pi -2)$

  2. $99\sqrt{3} -144 \sqrt{3} (2-\sqrt{3})$

  3. $96\sqrt{3}-144 \pi$

  4. $99\sqrt{3} -144 \sqrt{3} (\pi -\sqrt{3})$

Show answer
Correct Option: A

The area of a regular polygon of n sides is (where r is inradius, R is circumradius, and a is side of the triangle)

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

  1. $\displaystyle \frac{nR^{2}}{2}\sin \left ( \frac{2\pi }{n} \right )$

  2. $\displaystyle nr^{2}\tan \left( \frac{\pi }{n} \right )$

  3. $\displaystyle \frac{na^{2}}{4}\cot \frac{\pi }{n} $

  4. $\displaystyle nR^{2}\tan(\frac {\pi}{n})$

Show answer
Correct Option: A,B,C
Explanation:

Area of the regular polygon will be 
$=\dfrac{nR^{2}}{2}sin(\dfrac{2\pi}{n})$.
Now 
$R=\dfrac{s}{2sin(\dfrac{\pi}{n})}$
Hence
$A=\dfrac{ns^{2}}{8sin^{2}\dfrac{\pi}{n}}.2sin(\dfrac{\pi}{n}).cos(\dfrac{\pi}{n})$

$=\dfrac{ns^{2}}{4}.cot(\dfrac{\pi}{n})$. where s is the side of the polygon.

$=nr^{2}.tan(\dfrac{\pi}{n})$ where r is the incentre.

If the area of the pentagon $ABCDE$ be $\dfrac{45}{2}$ where $A = (1, 3), B = (-2, 5), C = (-3, -1), D = (0, -2)$ and $E = (2, t)$, then $t$ is:

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

  1. $-1$

  2. $99$

  3. $-1, 99$

  4. $-1, \dfrac{1}{99}$

Show answer
Correct Option: A

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

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

  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. $\dfrac{a}{2} cos (\dfrac{\pi}{n})$

Show answer
Correct Option: B
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 $r=\cfrac{a(\cot(\dfrac{\pi}{n}))}{2}$

Area of the regular hexagon each of whose sides measures $1 \,cm$ is:

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

  1. $2.598 \,cm^2$

  2. $25.98 \,cm^2$

  3. $259.8 \,cm^2$

  4. None of these

Show answer
Correct Option: A

if $\frac { 1 }{ { a } _{ x }+1 } are\quad 8$ vertices of a rectengular octagon where ${ a } _{ k }\epsilon$ R, K =1,2,3,.....,8(where $ i =\sqrt { -1 } )$then area of the regular octagon is

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

  1. $1$

  2. $\sqrt { 2 } $

  3. $\frac { 1 }{ \sqrt { 2 } } $

  4. none

Show answer
Correct Option: A

in the given figure,BD is a side a regular hexagon,DC is a side of a regular pentagon and AD is a diameter calculate

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

  1. $\angle ADC$

  2. $\angle BDA$

  3. $\angle ABC$

  4. $\angle AEC$

Show answer
Correct Option: A

What is the solid angle subtended by a hemisphere at its center? 

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

  1. $2\pi$ steradian

  2. $\pi$ steradian

  3. $3\pi$ steradian

  4. $4\pi$ steradian

Show answer
Correct Option: A
Explanation:
Solid angle $(\Omega)=\dfrac{A}{r^2}$

for a hemisphere, $A=2\pi r^2$

so, $\Omega =\dfrac{2\pi r^2}{r^2}=2\pi$ steradians

Ans is (A).

The area of a regular polygon of $2n$ sides inscribed in a circle is given by?

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

  1. The geometric mean of the areas of the inscribed and circumscribed polygons of $n$ sides.

  2. The arithmetic mean of the areas of the inscribed and circumscribed polygons of $n$ sides.

  3. The harmonic mean of the areas of the inscribed and circumscribed polygons of $n$ sides.

  4. None of the above

Show answer
Correct Option: A
Explanation:

Let $a$ be the radius of the circle 


Then,$\displaystyle s _{1}= $ Area of regular polygon of n sides inscribed in the circle $\displaystyle =\frac{1}{2}na^{2}\sin\left ( \frac{2\pi }{n} \right )$

$\displaystyle s _{2}= $  Area of regular polygon of n sides circumscribing in the circle $\displaystyle  = na^{2}\tan \frac{\pi }{n}$

$\displaystyle s _{3}= $ Area of regular polygon of 2n sides inscribed in the circle $\displaystyle  = na^{2}\tan \frac{\pi }{n}$ 

[replacing $n$ by $2n$ is $\displaystyle {(S _{1}}$]

$\displaystyle \therefore $ Geometric mean of $\displaystyle {S _{1}}$ and 

$\displaystyle {S _{2}}$ $\displaystyle = \sqrt{(S _{1}S _{2})}= na^{2}\sin\left ( \frac{\pi }{n}\right ) = S _{3}$

If A B C D E F is a regular hexagon with A B = a and B C = b, then CE equals

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

  1. b-a

  2. -b

  3. b-2a

  4. None of these

Show answer
Correct Option: A

If A B C D E F  is a regular hexagon with A B = a and B C = b , then CE equals

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

  1. b-a

  2. -b

  3. b-2a

  4. None of these

Show answer
Correct Option: A

If  $\alpha$  is the angle which each side of a regular polygon of  $n$  sides subtends at its centre, then  $1 + \cos \alpha + \cos 2 \alpha + \cos 3 \alpha \ldots + \cos ( n - 1 ) \alpha$  is equal to

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

  1. $n$

  2. $0$

  3. $1$

  4. None of these

Show answer
Correct Option: A

Relation between circumradius and number of sides is given by-

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

  1. $Area=\dfrac{r^2n\sin(\dfrac{360}{n})}{3}$

  2. $Area=\dfrac{r^2n\sin(\dfrac{360}{n})}{2}$

  3. $Area=\dfrac{r^2n\cos(\dfrac{360}{n})}{2}$

  4. None of the above

Show answer
Correct Option: B

The sum of the radii of inscribed and circumscribed circles of an n sided regular polygon of side $'a'$ is

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

  1. $=\frac{a}{2} \left ( \frac{1}{\sin \pi/2x} + \cot \frac{\pi}{x} \right )$

  2. $=\frac{a}{2} \left ( \frac{1}{\sin \pi/x} + \cot \frac{\pi}{2x} \right )$

  3. $=\frac{a}{2} \left ( \frac{1}{\sin \pi/x} + \cot \frac{\pi}{x} \right )$

  4. None of these

Show answer
Correct Option: C
Explanation:

$R\sin \theta  = \frac{a}{2}$
$R = \frac{a}{2\sin \theta }$
$\tan \theta = \frac{a/2}{r}$
$r = \frac{a}{2\tan \theta }                                   \theta = \frac{2\pi}{n} \times\frac{1}{2}$
$R+r = \frac{a}{2} \left ( \frac{1}{\sin \theta}+\frac{\sin \theta}{\cos \theta } \right )             = \frac{\pi}{x}$
    $= \frac{a}{2} \left ( \frac{1}{\sin \pi/x} + \cot \frac{\pi}{x} \right )$

In $\Delta ABC$, there are 35 lines drawn parallel to the base BC such that each line divides the other side into, equal parts. 
If BC =1.8 m find the length of $P _7 Q _7$.

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

  1. 1.8 m

  2. 3.5 m

  3. 0.35 m

  4. 0.18 m

Show answer
Correct Option: C

For a regular hexagon with apothem $5m$, the side length is about $5.77m$. The area of the regular hexagon is (in $m^2$).

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

  1. $75.5$

  2. $85.5$

  3. $76.5$

  4. $86.5$

Show answer
Correct Option: D
Explanation:

For a regular hexagon with apothem $5 m$ the side length is about $5.77m$.
Use the formula
$A = \frac{1}{2}pa$
to find the area of the hexagon.
The perimeter of the hexagon is about $6(5.77)$ or $34.62m$.
Now substitute the values.
$A = \frac{1}{2} (34.62)(5)$
Simplify.
$A = \frac{1}{2}(173.1)$
= $86.5$
$A = \frac{1}{2}(173.1)$
= $86.5$
Therefore, the area of the regular hexagon is about $86.5 m^2$.

If $D$ is the midpoint of side $BC$ of a triangle $ABC$ and $AD$ is perpendicular to $AC$ then

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

  1. $3{a}^{2}={b}^{2}-3{c}^{2}$

  2. $3{b}^{2}={a}^{2}-{c}^{2}$

  3. ${b}^{2}={a}^{2}-{c}^{2}$

  4. ${a}^{2}+{b}^{2}=5{c}^{2}$

Show answer
Correct Option: B
Explanation:

In $\triangle ACD, \cos{C}=\dfrac{b}{\left(\dfrac{a}{2}\right)}$
$\Rightarrow \dfrac{{a}^{2}+{b}^{2}-{c}^{2}}{2ab}=\dfrac{2b}{a}$
$\Rightarrow \dfrac{{a}^{2}+{b}^{2}-{c}^{2}}{2b}=2b$
$\Rightarrow {a}^{2}+{b}^{2}-{c}^{2}=4{b}^{2}$
$\Rightarrow {a}^{2}-{c}^{2}=3{b}^{2}$

If the angles of a triangle are in the ratio $2:3:7,$ then the sides opposite to these angles are in the ratio

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

  1. $\sqrt{2}:2:\left(\sqrt{3}+1\right)$

  2. $2:\sqrt{2}:\left(\sqrt{3}+1\right)$

  3. $1:\sqrt{2}:\dfrac{\sqrt{2}}{\left(\sqrt{3}-1\right)}$

  4. $\dfrac{1}{\sqrt{2}}:1:\left(\dfrac{\sqrt{3}+1}{2}\right)$

Show answer
Correct Option: A,C,D
Explanation:

Let $A=2\alpha, B=3\alpha, C=7\alpha$
$\therefore A+B+C={180}^{0}$
$\Rightarrow 2\alpha+3\alpha+7\alpha={180}^{0}$
$\Rightarrow 12\alpha={180}^{0}$
$\Rightarrow \alpha={15}^{0}$
$\therefore A=2\alpha=2\times{15}^{0}={30}^{0}$
$B=3\alpha=3\times{15}^{0}={45}^{0}$
$C=7\alpha=7\times{15}^{0}={105}^{0}$
$a:b:c=\sin{{30}^{0}}:\sin{{45}^{0}}:\sin{{105}^{0}}$
       $=\dfrac{1}{2}:\dfrac{1}{\sqrt{2}}:\dfrac{\sqrt{3}+1}{2\sqrt{2}}$
$a:b:c=\sqrt{2}:2:\sqrt{3}+1$
        $=\dfrac{1}{\sqrt{2}}:1:\dfrac{\sqrt{3}+1}{2}$
or $1:\sqrt{2}:\dfrac{\sqrt{2}\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}{2\left(\sqrt{3}-1\right)}$
or $1:\sqrt{2}:\dfrac{2\sqrt{2}}{2\left(\sqrt{3}-1\right)}$
$\therefore a:b:c= 1:\sqrt{2}:\dfrac{\sqrt{2}}{\sqrt{3}-1}$

In a triangle $ABC, \cos{A}+\cos{B}+\cos{C}=\dfrac{3}{2}$ then the triangle is

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

  1. isosceles

  2. right-angled

  3. equilateral

  4. none of these.

Show answer
Correct Option: C
Explanation:

$\cos{A}+\cos{B}+\cos{C}=\dfrac{3}{2}$
Using transformation angle formula, we have
$\left(\cos{A}+\cos{B}\right)+\cos{C}=\dfrac{3}{2}$
$\Rightarrow 2\cos{\left(\dfrac{A+B}{2}\right)}\cos{\left(\dfrac{A-B}{2}\right)}+\cos{C}=\dfrac{3}{2}$
Using sub-multiple angle formula to $\cos{C}=1-2{\sin}^{2}{\dfrac{C}{2}}$ we get
$\Rightarrow  2\cos{\left(\dfrac{A+B}{2}\right)}\cos{\left(\dfrac{A-B}{2}\right)}+1-2{\sin}^{2}{\dfrac{C}{2}}=\dfrac{3}{2}$ 
Since $A+B+C=\pi\Rightarrow \dfrac{A+B}{2}=\dfrac{\pi}{2}-\dfrac{C}{2}$
$\Rightarrow  2\cos{\left(\dfrac{\pi}{2}-\dfrac{C}{2}\right)}\cos{\left(\dfrac{A-B}{2}\right)}-2{\sin}^{2}{\dfrac{C}{2}}=\dfrac{3}{2}-1$
$\Rightarrow 2\sin{\left(\dfrac{C}{2}\right)}\cos{\left(\dfrac{A-B}{2}\right)}-2{\sin}^{2}{\dfrac{C}{2}}=\dfrac{1}{2}$
$\Rightarrow 2\sin{\left(\dfrac{C}{2}\right)}\left[\cos{\left(\dfrac{A-B}{2}\right)}+{\sin\left(\dfrac{C}{2}\right)}\right]=\dfrac{1}{2}$
Again
$\Rightarrow 2\sin{\left(\dfrac{C}{2}\right)}\left[\cos{\left(\dfrac{A-B}{2}\right)}+{\sin\left(\dfrac{\pi}{2}-\dfrac{A+B}{2}\right)}\right]=\dfrac{1}{2}$
$\Rightarrow 2\sin{\left(\dfrac{C}{2}\right)}\left[\cos{\left(\dfrac{A-B}{2}\right)}+{\cos\left(\dfrac{A+B}{2}\right)}\right]=\dfrac{1}{2}$
Using transformation angle formula, we get
$\Rightarrow \sin{\left(\dfrac{C}{2}\right)}\left[2\sin{\left(\dfrac{B}{2}\right)}\sin{\left(\dfrac{A}{2}\right)}\right]=\dfrac{1}{4}$
$\Rightarrow \sin{\left(\dfrac{A}{2}\right)}\sin{\left(\dfrac{B}{2}\right)}\sin{\left(\dfrac{C}{2}\right)}=\dfrac{1}{8}$
$\therefore \sin{\left(\dfrac{A}{2}\right)}=\dfrac{1}{2}, \sin{\left(\dfrac{B}{2}\right)}=\dfrac{1}{2},\sin{\left(\dfrac{C}{2}\right)}=\dfrac{1}{2}$
$\Rightarrow \sin{\left(\dfrac{A}{2}\right)}=\sin{\dfrac{\pi}{6}}, \sin{\left(\dfrac{B}{2}\right)}=\sin{\dfrac{\pi}{6}},\sin{\left(\dfrac{C}{2}\right)}=\sin{\dfrac{\pi}{6}}$
$\therefore {\left(\dfrac{A}{2}\right)}={\left(\dfrac{B}{2}\right)}={\left(\dfrac{C}{2}\right)}=\dfrac{\pi}{6}$
$\Rightarrow \angle{A}=\angle{B}=\angle{C}=2\times\dfrac{\pi}{6}=\dfrac{\pi}{3}$
Hence, the triangle is equilateral.

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

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

  1. $\dfrac{3}{4}$

  2. $3\sqrt{3}$

  3. $3$

  4. $\dfrac{3\sqrt{3}}{2}$

Show answer
Correct Option: C
Explanation:

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

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

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

  1. $\cos{\dfrac{\pi}{8}}$

  2. ${\sin}^{2}{\dfrac{\pi}{8}}$

  3. ${\cos}^{2}{\dfrac{\pi}{8}}$

  4. ${\tan}^{2}{\dfrac{\pi}{8}}$

Show answer
Correct Option: C
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 $R=r$)
  $=8{r}^{2}\sin{\dfrac{\pi}{8}}\cos{\dfrac{\pi}{8}}$ 
$\therefore \dfrac{{A} _{2}}{{A} _{1}}=\dfrac{8{r}^{2}\sin{\dfrac{\pi}{8}}\cos{\dfrac{\pi}{8}}}{8{r}^{2}\tan{\dfrac{\pi}{8}}}$
$=\dfrac{\sin{\dfrac{\pi}{8}}\cos{\dfrac{\pi}{8}}}{\dfrac{\sin{\dfrac{\pi}{8}}}{\cos{\dfrac{\pi}{8}}}}$
$={\cos}^{2}{\dfrac{\pi}{8}}$

If ${A} _{1}{A} _{2}{A} _{3}...{A} _{n}$ be a regular polygon of $n$ sides and 
$\dfrac{1}{{A} _{1}{A} _{2}}=\dfrac{1}{{A} _{1}{A} _{3}}+\dfrac{1}{{A} _{1}{A} _{4}},$then

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

  1. $n=5$

  2. $n=6$

  3. $n=7$

  4. none of these.

Show answer
Correct Option: C
Explanation:

If radius of circle is $r$ then 
${A} _{1}{A} _{2}=2r\sin{\left(\dfrac{\pi}{n}\right)}$
${A} _{1}{A} _{3}=2r\sin{\left(\dfrac{2\pi}{n}\right)}$
${A} _{1}{A} _{4}=2r\sin{\left(\dfrac{3\pi}{n}\right)}$
$\because \dfrac{1}{{A} _{1}{A} _{2}}=\dfrac{1}{{A} _{1}{A} _{3}}+\dfrac{1}{{A} _{1}{A} _{4}}$
$\Rightarrow \dfrac{1}{2r\sin{\left(\dfrac{\pi}{n}\right)}}=\dfrac{1}{2r\sin{\left(\dfrac{2\pi}{n}\right)}}+\dfrac{1}{2r\sin{\left(\dfrac{3\pi}{n}\right)}}$
$\Rightarrow \sin{\left(\dfrac{2\pi}{n}\right)}\sin{\left(\dfrac{3\pi}{n}\right)}=\sin{\left(\dfrac{3\pi}{n}\right)}\sin{\left(\dfrac{\pi}{n}\right)}+\sin{\left(\dfrac{2\pi}{n}\right)}\sin{\left(\dfrac{\pi}{n}\right)}$
$\Rightarrow \sin{\left(\dfrac{2\pi}{n}\right)}\left[\sin{\left(\dfrac{3\pi}{n}\right)}-\sin{\left(\dfrac{\pi}{n}\right)}\right]=\sin{\left(\dfrac{3\pi}{n}\right)}\sin{\left(\dfrac{\pi}{n}\right)}$
Using transformation angle formula, we get
$\Rightarrow \sin{\left(\dfrac{2\pi}{n}\right)}.2\cos{\left(\dfrac{2\pi}{n}\right)}\sin{\left(\dfrac{\pi}{n}\right)}=\sin{\left(\dfrac{3\pi}{n}\right)}\sin{\left(\dfrac{\pi}{n}\right)}$
$\Rightarrow 2\sin{\left(\dfrac{2\pi}{n}\right)}\cos{\left(\dfrac{2\pi}{n}\right)}=\sin{\left(\dfrac{3\pi}{n}\right)}$
Using multiple angle formula, $2\sin{A}\cos{A}=\sin{2A}$ we get
$\sin{\left(\dfrac{4\pi}{n}\right)}=\sin{\left(\dfrac{3\pi}{n}\right)}$
$\therefore \dfrac{4\pi}{n}=r+{\left(-1\right)}^{r}\dfrac{3}{n}$ for $r=1,n=7$

If $r$ and $R$ are respectively the radii of the inscribed and circumscribed circles of a regular polygon of $n$ sides such that $\dfrac{R}{r}=\sqrt{5}-1$, then $n$ is equal to

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

  1. $5$

  2. $6$

  3. $10$

  4. $18$

Show answer
Correct Option: A
Explanation:

Let $a$ be the length of side of regular polygon then
$R=\dfrac{a}{2}\csc{\left(\dfrac{\pi}{n}\right)}$ and $r=\dfrac{a}{2}\cot{\left(\dfrac{\pi}{n}\right)}$
$\therefore \dfrac{R}{r}=\dfrac{\csc{\left(\dfrac{\pi}{n}\right)}}{\cot{\left(\dfrac{\pi}{n}\right)}}=\dfrac{\dfrac{1}{\sin\left(\frac{\pi}{n}\right)}}{\dfrac{\cos{\frac{\pi}{n}}}{\sin{\frac{\pi}{n}}}}=\dfrac{1}{\cos{\left(\frac{\pi}{n}\right)}}$
$\therefore \cos{\left(\dfrac{\pi}{n}\right)}=\dfrac{1}{\sqrt{5}-1}$
On rationalising the denominator, we get 
$ \cos{\left(\dfrac{\pi}{n}\right)}=\dfrac{1}{\sqrt{5}-1}=\dfrac{1}{\sqrt{5}-1}\times\dfrac{\sqrt{5}-1}{\sqrt{5}+1}=\dfrac{\sqrt{5}+1}{4}=\cos{\left(\dfrac{\pi}{5}\right)}$
On comparing $\cos{\left(\dfrac{\pi}{n}\right)}=\cos{\left(\dfrac{\pi}{5}\right)}$ we get $n=5$

The sum of inradius and circumradius of incircle and circumcircle of a regular polygon of side $n$ is

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

  1. $\dfrac {a}{4}\cot \dfrac {\pi}{2n}$

  2. $a\cot \dfrac {\pi}{n}$

  3. $\dfrac {a}{2} \cot \dfrac {\pi}{2n}$

  4. $a\cot \dfrac {\pi}{2n}$

Show answer
Correct Option: C
Explanation:

$r + R = \dfrac {a}{2}\cot \dfrac {\pi}{n} + \dfrac {a}{2}cosec \dfrac {\pi}{n}$
$= \dfrac {a}{2} \left (\dfrac {1 + \cos \frac{\pi}{n}}{\sin \frac{\pi}{n}}\right ) = \dfrac {a}{2} \dfrac {2\cos^{2} \dfrac {\pi}{2n}}{2\sin \dfrac {\pi}{2n}\cdot \cos \dfrac {\pi}{2n}}$
$= \dfrac {a}{2} \cot \dfrac {\pi}{2n}$.

The sum of the radii of inscribed and circumscribed circles of an $n$ -sided regular polygon with side equal to one unit is?

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

  1. $\displaystyle \frac{1}{2}\cot \frac{\pi }{2n}$

  2. $\displaystyle \cot \frac{\pi }{2n}$

  3. $\displaystyle \cot \frac{\pi }{n}$

  4. $\displaystyle \frac{1}{2}\tan \frac{\pi }{2n}$

Show answer
Correct Option: A
Explanation:

From the figure:
Side of polygon $(AB)=1$
$AO=\dfrac { 1 }{ 2 } $
$\angle O=\dfrac { \pi  }{ 2n } $

In right angled $\triangle COA$ :
$\sin { O } =\dfrac { AC }{ AO } $
$\Rightarrow \sin { \dfrac { \pi  }{ n }  } =\dfrac { 1 }{ 2R } $       ..(1)

$\tan { O } =\dfrac { AC }{ CO } $
$\Rightarrow \tan { \dfrac { \pi  }{ n }  } =\dfrac { 1 }{ 2r } $       ...(2)

From (1) and (2)
$R+r=\dfrac { 1 }{ 2 } \left( \dfrac { 1 }{ \sin { \dfrac { \pi  }{ n }  }  } +\dfrac { 1 }{ \tan { \dfrac { \pi  }{ n }  }  }  \right) $

$\Rightarrow R+r=\dfrac { 1 }{ 2 } \left( \dfrac { 1+\cos { \dfrac { \pi  }{ n }  }  }{ \sin { \dfrac { \pi  }{ n }  }  }  \right) =\dfrac { 1 }{ 2 } \left( \dfrac { 2\cos ^{ 2 }{ \dfrac { \pi  }{ 2n }  }  }{ 2\cos { \dfrac { \pi  }{ 2n }  } \sin { \dfrac { \pi  }{ 2n }  }  }  \right) $

$\Rightarrow R+r=\dfrac { 1 }{ 2 } \cot { \dfrac { \pi  }{ 2n }  } $

Ans: A

- Hide questions
ADVERTISEMENT

Find more quizzes from top tags

general knowledge
3615 Quizzes
119858 Questions
 technology
307 Quizzes
36705 Questions
 sports
963 Quizzes
19148 Questions
 history
1187 Quizzes
13475 Questions
 physics
598 Quizzes
13441 Questions
 biology
1160 Quizzes
12174 Questions
 science & technology
551 Quizzes
11015 Questions
 chemistry
490 Quizzes
9892 Questions
 maths
518 Quizzes
9324 Questions
 softskills
0 Quizzes
7131 Questions