Tag: regular polygons

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Questions Related to regular polygons

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

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

  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. $\left|\cos{A}-\cos{B}\right|=\dfrac{1}{2\sqrt{3}}$
Reveal answer Fill a bubble to check yourself
B,C Correct answer
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)

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
A polygon has $n$ sides. If all the sides and all the angles are same then this polygon is called a regular polygon. Let ${A} _{1},{A} _{2},{A} _{3},...{A} _{n}$ be a regular polygon of $n$ sides. Let $R$ be the radius of the circumscribed circle of a regular polygon and $r$ be the radius of the inscribed circle of a regular polygon.
If ${A} _{1}{A} _{2}={A} _{2}{A} _{3}={A} _{3}{A} _{4}=...={A} _{n}{A} _{1}=a$

Based on the above information, answer the question:

The area of a regular polygon of $n$ sides is

  1. $\dfrac{n{R}^{2}}{2}\sin{\left(\dfrac{2\pi}{n}\right)}$
  2. $n{R}^{2}\tan{\left(\dfrac{\pi}{n}\right)}$
  3. $\dfrac{n{r}^{2}}{2}\sin{\left(\dfrac{2\pi}{n}\right)}$
  4. $n{r}^{2}\tan{\left(\dfrac{\pi}{n}\right)}$
Reveal answer Fill a bubble to check yourself
A,D Correct answer
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

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

  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. $\displaystyle nR^{2}\tan(\frac {\pi}{n})$
Reveal answer Fill a bubble to check yourself
A,B,C Correct answer
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.

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