Question 1: Three angles of a quadrilateral are respectively equal to its fourth $110^o$, $50^o$ and $40^o$. Find angle.

Number of sides $(n) = 4$

Given angles: $110^o, 50^o$ and $40^o$

The sum of the interior angles of  a quadrilateral  $= (2n-4) \times 90^o = (8-4) \times 90^o = 360^o$

Therefore the 4th angle $= 360^o - 110^o - 50^o - 40^o = 160^o$

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Question 2: In a quadrilateral $ABCD$, the angles $A , B, C$ and $D$ are in the ratio $1 : 2 :4 :5$. Find the measure of each angle of the quadrilateral.

Given: $\angle A : \angle B: \angle C : \angle D = 1:2:4:5$

Number of sides $(n) = 4$

The sum of the interior angles of  a quadrilateral  $= (2n-4) \times 90^o = (8-4) \times 90^o = 360^o$

Therefore $x + 2x + 4x + 5x = 360^o$

$\Rightarrow 12x = 360^o \Rightarrow x = 30^o$

Hence the angles are: $\angle A = 30^o, \angle B = 60^o, \angle C = 120^o$ and $\angle D = 150^o$

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Question 3: ln a quadrilateral $ABCD, CO$ and. $DO$ are the bisectors of $\angle C$ and $\angle D$ respectively. Prove that $\angle COD =$ $\frac{1}{2}$ $(\angle A + \angle B)$

To prove: $\angle COD =$ $\frac{1}{2}$ $(\angle A + \angle B)$

$\angle COD +$ $\frac{1}{2}$ $\angle C +$ $\frac{1}{2}$ $\angle D = 180^o$

$\Rightarrow \angle COD = 180^o -$ $\frac{1}{2}$ $(\angle C + \angle D)$ … … … … … i)

$\angle C + \angle A = 180^o \Rightarrow \angle C = 180^o - \angle A$

$\angle D + \angle B = 180^o \Rightarrow \angle D = 180^o - \angle B$

Substituting in i)

$\angle COD = 180^o -$ $\frac{1}{2}$ $(180^o - \angle A + 180 - \angle B)$

$\Rightarrow \angle COD = 180^o - 90^o - 90^o +$ $\frac{1}{2}$ $(\angle A + \angle B)$

$\Rightarrow \angle COD =$ $\frac{1}{2}$ $(\angle A + \angle B)$

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Question 4: Prove that the sum of all the interior angles of a pentagon is $540^o$.

The pentagon $ABCDE$ comprises of three triangles $\triangle ABC, \triangle BEC$ and $\triangle CED$ (please refer to the adjoining diagram)

Sum of the internal angles of $ABCDE = \angle A + \angle B + \angle C + \angle D + \angle E$

$= \angle A + \angle ABE + \angle EBC + \angle BCE + \angle ECD + \angle D + \angle AEB + \angle BEC + \angle CED$

$= (\angle A + \angle ABE + \angle AEB) + (\angle EBC + \angle BCE + \angle BEC) + (\angle ECD + \angle CED + \angle D)$

$= 180^o + 180^o + 180^o = 540^o$

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Question 5: What is the measure of each angle of a regular octagon?

Regular Octagon $(n) = 8$

Interior Angle $= \Big($ $\frac{2n-4}{n}$ $\Big) \times 90^o$

$= \Big($ $\frac{2\times 8-4}{8}$ $\Big) \times 90^o$

$=$ $\frac{12}{8}$ $\times 90^o = 135^0$

Exterior Angle $= 180^o - 135^o = 45^o$

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Question 6: Find the number of sides of a regular polygon, when each of its angles has a measure of i) $160^o$ ii) $135^o$ iii) $175^o$ iv) $162^o$ v) $150^o$

i)    Let the number of sides $= n$

We know Interior Angle $= \Big($ $\frac{2n-4}{n}$ $\Big) \times 90^o$

Given interior angle $= 160^o$

Therefore $\Big($ $\frac{2n-4}{n}$ $\Big) \times 90^o = 160^o$

$\Rightarrow 18n - 36 = 16n$

$\Rightarrow 2n = 36 \Rightarrow 18$

ii)   Let the number of sides $= n$

We know Interior Angle $= \Big($ $\frac{2n-4}{n}$ $\Big) \times 90^o$

Given interior angle $= 135^o$

Therefore $\Big($ $\frac{2n-4}{n}$ $\Big) \times 90^o = 135^o$

$\Rightarrow$ $\Big($ $\frac{2n-4}{n}$ $\Big) \times 2 = 3$

$\Rightarrow 4n - 8 = 3n$

$\Rightarrow n = 8$

iii)   Let the number of sides $= n$

We know Interior Angle $= \Big($ $\frac{2n-4}{n}$ $\Big) \times 90^o$

Given interior angle $= 175^o$

Therefore $\Big($ $\frac{2n-4}{n}$ $\Big) \times 90^o = 175^o$

$\Rightarrow$ $\Big($ $\frac{2n-4}{n}$ $\Big) \times 18 = 35$

$\Rightarrow 36n - 72 = 35n$

$\Rightarrow n = 72$

iv)   Let the number of sides $= n$

We know Interior Angle $= \Big($ $\frac{2n-4}{n}$ $\Big) \times 90^o$

Given interior angle $= 162^o$

Therefore $\Big($ $\frac{2n-4}{n}$ $\Big) \times 90^o = 162^o$

$\Rightarrow$ $\Big($ $\frac{2n-4}{n}$ $\Big) \times 5 = 9$

$\Rightarrow 10n - 20 = 9n$

$\Rightarrow n = 20$

v)   Let the number of sides $= n$

We know Interior Angle $= \Big($ $\frac{2n-4}{n}$ $\Big) \times 90^o$

Given interior angle $= 150^o$

Therefore $\Big($ $\frac{2n-4}{n}$ $\Big) \times 90^o = 150^o$

$\Rightarrow$ $\Big($ $\frac{2n-4}{n}$ $\Big) \times 3 = 5$

$\Rightarrow 6n - 12 = 5n$

$\Rightarrow n = 12$

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Question 7: Find the number of sides of a polygon the sum of whose interior angles is i) $1440^o$ ii) $28$ right angles iii) $19$ straight angles

i)    Sum of all interior angles of a polygon $= (2n-4) \times 90^o$

Therefore $1440^o = (2n-4) \times 90^o$

$\Rightarrow 1440 = (2n-4) \times 9$

$\Rightarrow 144 = 18n - 36$

$\Rightarrow 18n = 180$

$\Rightarrow n = 10$

ii)   Sum of all interior angles of a polygon $= (2n-4) \times 90^o$

Therefore $28 \times 90^o = (2n-4) \times 90^o$

$\Rightarrow 28 = 2n-4$

$\Rightarrow 2n = 32$

$\Rightarrow n = 16$

iii)  Sum of all interior angles of a polygon $= (2n-4) \times 90^o$

Therefore $19 \times 180^o = (2n-4) \times 90^o$

$\Rightarrow 19 \times 2 = 2n-4$

$\Rightarrow 38 = 2n - 4$

$\Rightarrow 2n = 42$

$\Rightarrow n = 21$

$\Rightarrow n = 10$

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Question 8: Find the number of degrees in each exterior angle of regular pentagon.

Regular pentagon: $n = 5$

Exterior angle of a regular polygon of $n$ sides $=$ $\frac{360^o}{n}$

Therefore exterior angle of pentagon $=$ $\frac{360^o}{5}$ $= 72^o$

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Question 9: The measure of angles of hexagon are $x^o, (x-5)^o, (x-5)^o, (2x-5)^o, (2x-5)^o, (2x+20)^o$. Find the value of $x$.

Regular hexagon: $n = 6$

Sum of interior angles $= (2n - 4) \times 90^o$

$= (12 - 4) \times 90^o$

$= 720^0$

Therefore $x^o + (x-5)^o + (x-5)^o + (2x-5)^o + (2x-5)^o + (2x+20)^o = 720^o$

$\Rightarrow 9x = 720^o$

$\Rightarrow x = 80^o$

Hence the Exterior angle $= 180^o - 80^o = 100^o$

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Question 10: In a convex hexagon, prove that the sum of all interior angles is equal to twice the sum of exterior angles formed by producing the sides in the same order.

Convex hexagon: $n = 6$

Sum of interior angles $= (2n - 4) \times 90^o$

$= (12 - 4) \times 90^o$

$= 720^0$

Sum of exterior angles of a hexagon $= \Big($ $\frac{360^o}{n}$ $\Big) \times n = 360^o$

Hence Sum of all interior angles $= 2 \times$ sum of all exterior angles

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Question 11: The sum of the interior angles of a polygon is three times the sum of its exterior angles. Determine the number of sides of the polygon.

Given: Sum of all interior angles $= 3 \times$ sum of all exterior angles

Therefore $(2n-4) \times 90^o = 3 \times \Big($ $\frac{360^o}{n}$ $\Big) \times n$

$\Rightarrow 2n-4 = 12$

$\Rightarrow n = 8$

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Question 12: Determine the number of sides of a polygon whose exterior and interior angles are in the ratio of $1:5$.

Interior Angle $= \Big($ $\frac{2n-4}{n}$ $\Big) \times 90^o$

Exterior angle $=$ $\frac{360^o}{n}$

Given: $\Big($ $\frac{2n-4}{n}$ $\Big) \times 90^o = 5 \times$ $\frac{360^o}{n}$

$\Rightarrow 2n - 4 = 20$

$\Rightarrow 2n = 24$

$\Rightarrow n = 12$

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Question 13: $PQRSIU$ is a regular hexagon. Determine each angle of $\triangle PQT$

Regular hexagon: $n = 6$

Therefore Interior Angle $= \Big($ $\frac{2n-4}{n}$ $\Big) \times 90^o$

$= \Big($ $\frac{2 \times 6-4}{6}$ $\Big) \times 90^o$

$=$ $\frac{8}{6}$ $\times 90^o = 120^o$

Since $TU = UP$

$\angle UTP = \angle UPT = 30^o$

In a quadrilateral sum of opposite angles $= 180^o$

Therefore $120^o + \angle TQP = 180^o$

$\Rightarrow \angle TQP = 60^o$

$\angle UPQ = 120^o$

$\Rightarrow 30 + \angle TPQ = 120^o$

$\Rightarrow \angle TPQ = 90^o$

Therefore $\angle PTQ = 180^o - 90^o - 60^o = 30^o$

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Question 14: Is it possible to construct a regular polygon the measure of whose each interior angle is i) $120^o$ ii) $110^o$ and iii) $150^o$

i)    Given: Interior angle $= 120^o$

Interior Angle $= \Big($ $\frac{2n-4}{n}$ $\Big) \times 90^o$

Therefore $\Big($ $\frac{2n-4}{n}$ $\Big) \times 90^o = 120^o$

$\Rightarrow 6n - 12 = 4n$

$\Rightarrow 2n = 12$

$\Rightarrow n = 6$

Therefore it is possible to construct a regular polygon with an interior angle of $120^o$

ii)   Given: Interior angle $= 110^o$

Interior Angle $= \Big($ $\frac{2n-4}{n}$ $\Big) \times 90^o$

Therefore $\Big($ $\frac{2n-4}{n}$ $\Big) \times 90^o = 110^o$

$\Rightarrow 18n - 36 = 11n$

$\Rightarrow 7n = 36$

$\Rightarrow n = 5.142$

Therefore it is not possible to construct a regular polygon with an interior angle of $110^o$ since $n$ should be a positive integer.

iii)  Given: Interior angle $= 150^o$

Interior Angle $= \Big($ $\frac{2n-4}{n}$ $\Big) \times 90^o$

Therefore $\Big($ $\frac{2n-4}{n}$ $\Big) \times 90^o = 150^o$

$\Rightarrow 6n - 12 = 5n$

$\Rightarrow n = 12$

Therefore it is possible to construct a regular polygon with an interior angle of $120^o$

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Question 15: Is it possible to construct a regular polygon the measure of whose each exterior angle is i) $24^o$ ii) $50^o$ iii) $70^o$

i)     Exterior angle $= 24^o$

We know exterior angle $=$ $\frac{360^o}{n}$

Therefore $\frac{360^o}{n}$ $= 24^o$

$\Rightarrow n =$ $\frac{360^o}{24^o}$ $= 15$

Therefore it is possible to construct a regular polygon with an exterior angle of $24^o$

ii)   Exterior angle $= 50^o$

We know exterior angle $=$ $\frac{360^o}{n}$

Therefore $\frac{360^o}{n}$ $= 50^o$

$\Rightarrow n =$ $\frac{360^o}{50^o}$ $= 7.2$

Therefore it is not possible to construct a regular polygon with an exterior angle of $50^o$

iii)  Exterior angle $= 70^o$

We know exterior angle $=$ $\frac{360^o}{n}$

Therefore $\frac{360^o}{n}$ $= 70^o$

$\Rightarrow n =$ $\frac{360^o}{70^o}$ $= 5.15$

Therefore it is not possible to construct a regular polygon with an exterior angle of $70^o$

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Question 16: Can a regular polygon be described the sum of whose interior angles is i) $520^o$ ii) $1260^o$ iii) $9$ right angles

i)    Sum of interior angles $= 520^o$

We know that sum of interior angles of a polygon $= (2n-4) \times 90^o$

Therefore $(2n-4) \times 90^o = 520^o$

$\Rightarrow 18n - 36 = 52$

$\Rightarrow 18n = 88$

$\Rightarrow n = 4.69$

Hence it is not possible to construct a polygon where the sum of interior angles is $520^o$

ii)   Sum of interior angles $= 1260^o$

We know that sum of interior angles of a polygon $= (2n-4) \times 90^o$

Therefore $(2n-4) \times 90^o = 1260^o$

$\Rightarrow 2n - 4 = 14$

$\Rightarrow 2n = 18$

$\Rightarrow n = 9$

Hence it is possible to construct a polygon where the sum of interior angles is $1260^o$

iii)  Sum of interior angles $= 9 \times 90^o = 810^o$

We know that sum of interior angles of a polygon $= (2n-4) \times 90^o$

Therefore $(2n-4) \times 90^o = 810^o$

$\Rightarrow 2n - 4 = 9$

$\Rightarrow 2n = 13$

$\Rightarrow n = 6.5$

Hence it is not possible to construct a polygon where the sum of interior angles is $810^o$

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Question 17: Determine the number of sides of regular, polygon the measure of whose each interior angle is double that of the exterior angle

Given: Interior angle $= 2 \times$ Exterior angle

$\Rightarrow \Big($ $\frac{2n-4}{n}$ $\Big) \times 90^o = 2 \times \Big($ $\frac{360^o}{n}$ $\Big)$

$\Rightarrow 2n-4 = 8$

$\Rightarrow 2n = 12$

$\Rightarrow n = 6$

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