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

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

  1. $n=5$
  2. $n=6$
  3. $n=7$

  4. none of these.
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $r=1,n=7$

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

In a triangle if the sum of two sides is $x$ and this product is $y ,\left(x\ge 2\sqrt{y}\right)$ such that $\left(x+z\right)\left(x-z\right)=y$ where $z$ is the third side of the triangle.
On the basis of the above information, answer the following questions:
The sides of the triangle are:

  1. $\dfrac{x\pm\sqrt{\left({x}^{2}-4y\right)}}{2},z$
  2. $\dfrac{y\pm\sqrt{\left({y}^{2}-4z\right)}}{2},z$
  3. $\dfrac{z\pm\sqrt{\left({z}^{2}-4x\right)}}{2},z$

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

$\because b+c=x,bc=y$
$\therefore b,c$ are the roots of ${t}^{2}-\left(b+c\right)t+bc=0$
or ${t}^{2}-xt+y=0$
$\therefore t=\dfrac{x\pm\sqrt{{x}^{2}-4y}}{2}$
Hence, sides are  $\dfrac{x\pm\sqrt{{x}^{2}-4y}}{2},z$ 

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

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

  1. $5$
  2. $6$
  3. $10$
  4. $18$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $n=5$

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

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

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

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

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

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

  1. $\displaystyle \frac{1}{2}\cot \frac{\pi }{2n}$
  2. $\displaystyle \cot \frac{\pi }{2n}$
  3. $\displaystyle \cot \frac{\pi }{n}$
  4. $\displaystyle \frac{1}{2}\tan \frac{\pi }{2n}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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

Multiple choice maths construction of triangle construction of triangles - i construction of triangles constructions of triangles

State true or false:

Whether it is possible to construct a triangle or not with its sides equal to $5$ cm, $7$ cm, and $4$ cm
Ans: Yes

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A triangle with three sides is possible, if sum of two smaller sides is greater than the third side.
For given sides, $5$ cm, $7$ cm and $4$ cm
$5 + 4 > 7$
$9 > 7$
Thus, a triangle is possible.

Multiple choice maths construction of triangle construction of triangles - i construction of triangles constructions of triangles

The steps for construction of $\triangle DEF$ with $DE = 4\ cm, EF=6.5\ cm$ and $DF = 8.6\ cm$ are given below in jumbled order:
1. Draw arcs of length $4\ cm$ from $4\ cm$ from $D$ and $6.5\ cm$ from $F$ and mark the intersection point as $E$.
2. Join $D-E$ and $F-E$.
3. Draw a line segment of length $DF = 8.6\ cm$.

The correct order of the steps is:

  1. $3-1-2$
  2. $1-2-3$
  3. $2-3-1$
  4. $2-1-3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Correct sequence is

Step 1: Draw a line segment of length $DF=8.6 cm$.
Step2 :Draw arcs of length $4 cm$ from $4 cm$ from $D$ and $6.5 cm$ from $F$ and mark the intersection point as $E$
Step 3: Join $D-E$ and $F-E$.
So the sequence is $3-1-2$

Multiple choice maths construction of triangle construction of triangles - i construction of triangles constructions of triangles

In $\triangle ABC$, $AB=5\ cm, BC= 6\ cm ,AC=4\ cm$. Identify the type of triangle.

  1. Right angled triangle

  2. Isosceles triangle

  3. Equilateral triangle

  4. Scalene triangle
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$5cm+6cm>4cm$
$\Rightarrow AB+BC>AC$

$6cm+4cm>5cm$
$\Rightarrow BC+AC>AB$

$4cm+5cm>6cm$
$\Rightarrow AC+AB>BC$

Sum of any two sides taken in pair is greater than the third side. So a triangle can be formed.

Now all the sides of triangle are unequal . So the triangle is Scalene triangle.

Further ${(LargestSide)}^{2}$ is not equal to $({Side1})^{2} + ({Side2})^{2}$. Hence, not right angled.

Option $D$ is correct
Multiple choice maths construction of triangle construction of triangles - i construction of triangles constructions of triangles

For construction of a $\triangle PQR$, where $\displaystyle QR=6\ cm, PR=10\ cm$ and $\angle Q=90^{\circ}$, its steps for construction is given below in jumbled form. Identify the third step from the following.

1. At point $ Q $, draw an angle of $ {90}^{\circ} $.
2. From $ R $ cut an arc of length $ PR = 10.0 \ cm $ using a compass. 
3. Name the point of intersection of the arm of the angle $ {90}^{\circ} $ and the arc drawn in step 3, as $ P $.
4. Join $P $ to $ Q $ . $ PQR $ is the required triangle. 
5. Draw the base side $ QR = 6\  cm $.

  1. $3$
  2. $4$
  3. $2$
  4. $5$
  5. $1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Step 1. Draw a line $QR=6\ \ cm$

Step 2. At point $Q$ ,draw an angle of $90^{\circ}$
Step 3. From $R$ cut an arc $PR=10\ \ cm$ using compass.
Step 4. Name the point of intersection of the arm of angle $90^{\circ}$ and the arc in step $3$ , as $P$
Step 5. Join $p$ to $Q$. $PQR$ is required triangle.
So the third step is $2$
Option $C$ is correct.