[3 marks sum]

Question 13 Marks

The angles of a pentagon are $100^\circ , 96^\circ , 74^\circ , 2x^\circ $ and $3x^\circ .$ Find the measures of the two angles $2x^\circ $ and $3x^\circ .$

Answer

A pentagon has $5$ sides
$\therefore$ Sum of interior anfles
$ =(n-2) \times 180^{\circ}$
$=(5-2) \times 180^{\circ}$
$=3 \times 180^{\circ}$
$=540^{\circ} $
Given, the angles are $100^{\circ}, 96^{\circ} 74^{\circ}, 2 x^{\circ}$ and $3 x^{\circ}$
$ \therefore 100^{\circ}+96^{\circ} 74^{\circ}+2 x ^{\circ}+3 x ^{\circ}=540^{\circ}$
$\Rightarrow 5 x ^{\circ}+270^{\circ}=540^{\circ}$
$\Rightarrow x =\frac{\left(540^{\circ}-270^{\circ}\right)}{5}$
$=54^{\circ} $
$\therefore$ The two angles $2 x^{\circ}$ and $3 x^{\circ}$ are $108^{\circ}$ and $162^{\circ}$ respectively.

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Question 23 Marks

Find the number of sides in a regular polygon, when each interior angle is $: 135^\circ$

Answer

Each interior angle of a regular polygon
$=\frac{(n-2) \times 180^{\circ}}{n}$
$\Rightarrow \frac{(n-2)^n \times 180^{\circ}}{n}=135^{\circ}$
$\Rightarrow 180^{\circ}(n-2)=135^{\circ}(n)$
$\Rightarrow 4(n-2)=3 n$
$\Rightarrow n=8 . $

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Question 33 Marks

Find the number of sides in a regular polygon, when each interior angle is $: 140^\circ$

Answer

Each interior angle of a regular polygon
$=\frac{(n-2) \times 180^{\circ}}{n} $
$\Rightarrow \frac{(n-2)^n \times 180^{\circ}}{n}=140^{\circ} $
$\Rightarrow 180^{\circ}(n-2)=140^{\circ}(n) $
$\Rightarrow 9(n-2)=7 n $
$\Rightarrow n=\frac{18}{2} $
$=9.$

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Question 43 Marks

Find the number of sides in a regular polygon, when each interior angle is $: 120^\circ$

Answer

Each interior angle of a regular polygon
$=\frac{( n -2) \times 180^{\circ}}{ n } $
$\Rightarrow \frac{( n -2)^{\circ} \times 180^{\circ}}{ n }=120^{\circ} $
$\Rightarrow 180^{\circ}( n -2)=120^{\circ}( n ) $
$\Rightarrow 3( n -2)=2 n $
$\Rightarrow n =6 .$

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Question 53 Marks

In a polygon, there are $3$ right angles and the remaining angles are equal to $165^\circ .$ Find the number of sides in the polygon.

Answer

Let the number of sides of the polygon be $n$
Number of right angles $= 3$
$\therefore $ Number of angles of $165^\circ $ each $= n - 3$
Sum of interior angles of a polygon $= (n - 2) \times 180$
$\Rightarrow 3 \times 90^\circ + (n - 3)165^\circ = 180^\circ n - 360^\circ $
$\Rightarrow 270^\circ + 165^\circ n - 495^\circ = 180^\circ n - 360^\circ $
$\Rightarrow 180^\circ n - 165^\circ k = 270^\circ - 495^\circ + 360^\circ $
$\Rightarrow 15^\circ n = 135^\circ $
$\Rightarrow n = 9$
Thus, the number of sides in the polygon is $9.$

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Question 63 Marks

Is it possible to have a polygon whose each interior angle is $105^\circ ?$

Answer

Given each interior angle $=105^{\circ}$
So, each exterior angle $=180^{\circ}-105^{\circ}=75^{\circ}$
Thus, the number of sides of the polygon
$ =\frac{360^{\circ}}{\text { Each exterior angle }}$
$=\frac{360^{\circ}}{75^{\circ}}$
$=4 \frac{4}{5}$
which is not a natural number
Therefore, no polygon is possible whose each interior angle is $105^{\circ}$.

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Question 73 Marks

Is it possible to have a polygon whose each interior angle is $124^\circ?$

Answer

Given each interior angle $=124^{\circ}$
So, each exterior angle $=180^{\circ}-124^{\circ}=56^{\circ}$
Thus, the number of sides of the polygon
$=\frac{360^{\circ}}{\text { Each exterior angle }}$
$=\frac{360^{\circ}}{56^{\circ}}$
$=6 \frac{3}{7} $
which is not a natural number
Therefore, no polygon is possible whose each interior angle is $124^{\circ}$.

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Question 83 Marks

Is it possible to have a polygon whose sum of interior angles is $7$ right angles?

Answer

Let the number of sides in the polygon be $n$.
$\therefore( n -2) \times 180^{\circ}=7 $ Right Angles
$\Rightarrow( n -2) \times 180^{\circ}=7 \times 90^{\circ} $
$\Rightarrow 180^{\circ} n -360^{\circ}=630^{\circ} $
$\Rightarrow 180^{\circ} n =990^{\circ} $
$\Rightarrow n =\frac{990^{\circ}}{180^{\circ}} $
$=\frac{11}{2} $
$=5 \frac{1}{2}$
Since the number of sides of a polygon cannot be in a fraction,
therefore the polygon is not possible.

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Question 93 Marks

Is it possible to have a polygon whose sum of interior angles is $780^\circ ?$

Answer

Let the number of sides in the polygon be $n$.
$\therefore( n -2) \times 180^{\circ}=780^{\circ} $
$\Rightarrow 180^{\circ} n -360^{\circ}=780^{\circ} $
$\Rightarrow 180^{\circ} n =1140^{\circ} $
$\Rightarrow n =\frac{1140^{\circ}}{180^{\circ}} $
$=6 \frac{1}{3}$
Since the number of sides of a polygon cannot be in a fraction,
therefore the polygon is not possible.

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Question 103 Marks

The angles of a hexagon are $(2x + 5)^\circ , (3x - 5)^\circ , (x + 40)^\circ , (2x + 20)^\circ , (2x + 25)^\circ $ and $(2x + 35)^\circ .$ Find the value of $x.$

Answer

A hexagon has $6$ sides
$\therefore $ Sum of interior angles
$= (n - 2) \times 180^\circ $
$= (6 - 2) \times 180^\circ $
$= 4 \times 180^\circ $
$= 720^\circ $
Given the angle of a hexagon are:
$(2x + 5)^\circ , (3x - 5)^\circ , (x + 40)^\circ , (2x + 20)^\circ , (2x + 25)^\circ $ and $(2x + 35)^\circ$
$\therefore (2x + 5)^\circ , + (3x - 5)^\circ , + (x + 40)^\circ , + (2x + 20)^\circ , + (2x + 25)^\circ +(2x + 35)^\circ = 720$
$\Rightarrow 12x + 120^\circ = 720^\circ $
$\Rightarrow x = 50^\circ .$

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Question 113 Marks

The angles of a pentagon are $x^\circ , (x - 10)^\circ , (x + 20)^\circ , (2x - 44)^\circ $ and $(2x - 70)^\circ .$ Find the angles.

Answer

A pentagon has $5$ sides
$\therefore$ Sum of interior angles
$=(n-2) \times 180^{\circ}$
$=(5-2) \times 180^{\circ}$
$=3 \times 180^{\circ}$
$=540^{\circ} $
Given, the angles are
$x^{\circ},(x-10)^{\circ},(x+20)^{\circ},(2 x-44)^{\circ}$ and $(2 x-70)^{\circ}$
$ \therefore x ^{\circ}+( x -10)^{\circ}+( x +20)^{\circ}+(2 x -44)^{\circ}+(2 x -70)^{\circ}=540^{\circ}$
$\Rightarrow 7 x ^{\circ}-104^{\circ}=540^{\circ}$
$\Rightarrow x =\frac{\left(540^{\circ}+104^{\circ}\right)}{7}=92^{\circ} $
$\therefore$ The interior angles of the pentagon are $92^{\circ}, 82^{\circ}, 112^{\circ}, 140^{\circ}$ and $114^{\circ}$.

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MATHEMATICS [3 marks sum]

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