[2 Mark Question Answer]

Question 12 Marks

Find the number of sides in a regular polygon, when each exterior angle is $: 72^\circ$

Answer

Each exterior angle
$ =\frac{360^{\circ}}{ n }$
$\Rightarrow \frac{360^{\circ}}{ n }=72^{\circ}$
$\Rightarrow n =5 .$

Question 22 Marks

Find the number of sides in a regular polygon, when each exterior angle is $: 60^\circ$

Answer

Each exterior angle
$ =\frac{360^{\circ}}{360^{\circ}}$
$\Rightarrow \frac{30^{\circ}}{n}=60^{\circ}$
$\Rightarrow n=6 . $

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Question 32 Marks

Find the number of sides in a regular polygon, when each exterior angle is $: 20^\circ$

Answer

Each exterior angle
$ =\frac{360^{\circ}}{3 \frac{n}{6}}$
$\Rightarrow \frac{360^{\circ}}{n}=20^{\circ}$
$\Rightarrow n=18 . $

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Question 42 Marks

Find each exterior angle of a regular polygon of $: 18$ sides

Answer

When $n=18$
$\therefore$ Each exterior angle of a regular polygon
$=\frac{360^{\circ}}{360^{\circ}}$
$=\frac{18}{18}$
$=20^{\circ} . $

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Question 52 Marks

Find each exterior angle of a regular polygon of $: 15$ sides

Answer

When $n=15$
$ \therefore$ Each exterior angle of a regular polygon
$=\frac{360^{\circ}}{15}$
$=24^{\circ}$

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Question 62 Marks

Find each exterior angle of a regular polygon of $: 9$ sides

Answer

When $n=9$
$ \therefore$ Each exterior angle of a regular polygon
$=\frac{360^{\circ}}{360^{\circ}}$
$=\frac{3}{9}$
$=40^{\circ} . $

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Question 72 Marks

Find the measure of each interior angle of a regular polygon of $: 15$ sides

Answer

When $n=15$
$\therefore$ Each interior angle of a regular polygon
$=\frac{(n-2) \times 180^{\circ}}{n}$
$=\frac{(15-2) \times 180^{\circ}}{15}$
$=156^{\circ} . $

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Question 82 Marks

Find the measure of each interior angle of a regular polygon of $: 10$ sides

Answer

When $n=10$
$\therefore$ Each interior angle of a regular polygon
$=\frac{(n-2) \times 180^{\circ}}{n}$
$=\frac{(10-2) \times 180^{\circ}}{10}$
$=144^{\circ} .$

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Question 92 Marks

Find the measure of each interior angle of a regular polygon of $: 6$ sides

Answer

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

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Question 102 Marks

Calculate the measure of each angle of a regular polygon of $20$ sides.

Answer

Here $n=20$
$\therefore$ Each interior angle of the regular polygon
$=\frac{( n -2) \times 180^{\circ}}{ n }$
$=\frac{(20-2) \times 180^{\circ}}{20}$
$=162^{\circ} . $

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Question 112 Marks

Calculate the measure of each angle of a nonagon.

Answer

A nonagon has $9$ sides.
$\therefore$ Each interior angle of a regular polygon
$ =\frac{(n-2) \times 180^{\circ}}{n}$
$=\frac{(9-2)^n \times 180^{\circ}}{9}$
$=140^{\circ} . $

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Question 122 Marks

Find the sum of the interior angles of a polygon of$: 9$ sides

Answer

When $n = 9$
$\therefore $ Sum of interior angles
$= (n - 2) \times 180^\circ $
$= (9 - 2) \times 180^\circ$
$= 7 \times 180^\circ $
$= 1260^\circ .$

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Question 132 Marks

Find the sum of the interior angles of a polygon of$: 12$ sides

Answer

When $n = 12$
$\therefore $ Sum of interior angles
$= (n - 2) \times 180^\circ $
$= (12 - 2) \times 180^\circ $
$= 10 \times 180^\circ $
$= 1800^\circ .$

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Question 142 Marks

Find the sum of the interior angles of a polygon of$: 7$ sides

Answer

When $n = 7$
$\therefore $ Sum of interior angles
$= (n - 2) \times 180^\circ $
$= (7 - 2) \times 180^\circ $
$= 5 \times 180^\circ $
$= 900^\circ .$

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MATHEMATICS [2 Mark Question Answer]

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