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

$\displaystyle \text{Number of sides } (n) = 4$

Given angles: $\displaystyle \text{Given angles: } 110^o, 50^o \text{ and } 40^o$

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

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

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

$\displaystyle \text{Given: } \angle A : \angle B: \angle C : \angle D = 1:2:4:5$

$\displaystyle \text{Number of sides } (n) = 4$

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

$\displaystyle \text{Therefore } x + 2x + 4x + 5x = 360^o$

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

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

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

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

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

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

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

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

Substituting in i)

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

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

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

$\displaystyle \\$

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

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

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

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

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

$\displaystyle = 180^o + 180^o + 180^o = 540^o$

$\displaystyle \\$

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

$\displaystyle \text{Regular Octagon: } (n) = 8$

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

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

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

$\displaystyle \text{Exterior Angle } = 180^o - 135^o = 45^o$

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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) $\displaystyle 160^o$ ii) $\displaystyle 135^o$ iii) $\displaystyle 175^o$ iv) $\displaystyle 162^o$ v) $\displaystyle 150^o$

i) $\displaystyle \text{Let the Number of sides } = n$

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

$\displaystyle \text{Given Interior Angle } = 160^o$

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

$\displaystyle \Rightarrow 18n - 36 = 16n$

$\displaystyle \Rightarrow 2n = 36 \Rightarrow 18$

ii) $\displaystyle \text{Let the Number of sides } = n$

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

$\displaystyle \text{Given Interior Angle } = 135^o$

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

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

$\displaystyle \Rightarrow 4n - 8 = 3n$

$\displaystyle \Rightarrow n = 8$

iii) $\displaystyle \text{Let the Number of sides } = n$

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

$\displaystyle \text{Given Interior Angle } = 175^o$

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

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

$\displaystyle \Rightarrow 36n - 72 = 35n$

$\displaystyle \Rightarrow n = 72$

iv) $\displaystyle \text{Let the Number of sides } = n$

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

$\displaystyle \text{Given Interior Angle } = 162^o$

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

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

$\displaystyle \Rightarrow 10n - 20 = 9n$

$\displaystyle \Rightarrow n = 20$

v) $\displaystyle \text{Let the Number of sides } = n$

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

$\displaystyle \text{Given Interior Angle } = 150^o$

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

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

$\displaystyle \Rightarrow 6n - 12 = 5n$

$\displaystyle \Rightarrow n = 12$

$\displaystyle \\$

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

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

$\displaystyle \text{Therefore } 1440^o = (2n-4) \times 90^o$

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

$\displaystyle \Rightarrow 144 = 18n - 36$

$\displaystyle \Rightarrow 18n = 180$

$\displaystyle \Rightarrow n = 10$

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

$\displaystyle \text{Therefore } 28 \times 90^o = (2n-4) \times 90^o$

$\displaystyle \Rightarrow 28 = 2n-4$

$\displaystyle \Rightarrow 2n = 32$

$\displaystyle \Rightarrow n = 16$

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

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

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

$\displaystyle \Rightarrow 38 = 2n - 4$

$\displaystyle \Rightarrow 2n = 42$

$\displaystyle \Rightarrow n = 21$

$\displaystyle \Rightarrow n = 10$

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

$\displaystyle \text{Regular pentagon: } n = 5$

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

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

$\displaystyle \\$

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

$\displaystyle \text{Regular hexagon: } n = 6$

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

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

$\displaystyle = 720^0$

$\displaystyle \text{Therefore } x^o + (x-5)^o + (x-5)^o + (2x-5)^o + (2x-5)^o + (2x+20)^o = 720^o$

$\displaystyle \Rightarrow 9x = 720^o$

$\displaystyle \Rightarrow x = 80^o$

$\displaystyle \text{Hence the Exterior Angle } = 180^o - 80^o = 100^o$

$\displaystyle \\$

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: $\displaystyle n = 6$

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

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

$\displaystyle = 720^0$

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

Hence Sum of all interior angles $\displaystyle = 2 \times$ 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 $\displaystyle = 3 \times$ sum of all exterior angles

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

$\displaystyle \Rightarrow 2n-4 = 12$

$\displaystyle \Rightarrow n = 8$

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

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

$\displaystyle \text{Exterior Angle } = \frac{360^o}{n}$

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

$\displaystyle \Rightarrow 2n - 4 = 20$

$\displaystyle \Rightarrow 2n = 24$

$\displaystyle \Rightarrow n = 12$

$\displaystyle \\$

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

$\displaystyle \text{Regular hexagon: } n = 6$

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

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

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

Since $\displaystyle TU = UP$

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

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

$\displaystyle \text{Therefore } 120^o + \angle TQP = 180^o$

$\displaystyle \Rightarrow \angle TQP = 60^o$

$\displaystyle \angle UPQ = 120^o$

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

$\displaystyle \Rightarrow \angle TPQ = 90^o$

$\displaystyle \text{Therefore } \angle PTQ = 180^o - 90^o - 60^o = 30^o$

$\displaystyle \\$

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

i) Given: $\displaystyle \text{Interior Angle } = 120^o$

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

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

$\displaystyle \Rightarrow 6n - 12 = 4n$

$\displaystyle \Rightarrow 2n = 12$

$\displaystyle \Rightarrow n = 6$

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

ii) Given: $\displaystyle \text{Interior Angle } = 110^o$

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

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

$\displaystyle \Rightarrow 18n - 36 = 11n$

$\displaystyle \Rightarrow 7n = 36$

$\displaystyle \Rightarrow n = 5.142$

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

iii) Given: $\displaystyle \text{Interior Angle } = 150^o$

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

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

$\displaystyle \Rightarrow 6n - 12 = 5n$

$\displaystyle \Rightarrow n = 12$

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

$\displaystyle \\$

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

i) $\displaystyle \text{Exterior Angle } = 24^o$

We know $\displaystyle \text{Exterior Angle } = \frac{360^o}{n}$

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

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

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

ii) $\displaystyle \text{Exterior Angle } = 50^o$

We know $\displaystyle \text{Exterior Angle } = \frac{360^o}{n}$

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

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

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

iii) $\displaystyle \text{Exterior Angle } = 70^o$

We know $\displaystyle \text{Exterior Angle } = \frac{360^o}{n}$

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

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

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

$\displaystyle \\$

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

i) $\displaystyle \text{Sum of interior angles } = 520^o$

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

$\displaystyle \text{Therefore } (2n-4) \times 90^o = 520^o$

$\displaystyle \Rightarrow 18n - 36 = 52$

$\displaystyle \Rightarrow 18n = 88$

$\displaystyle \Rightarrow n = 4.69$

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

ii) $\displaystyle \text{Sum of interior angles } = 1260^o$

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

$\displaystyle \text{Therefore } (2n-4) \times 90^o = 1260^o$

$\displaystyle \Rightarrow 2n - 4 = 14$

$\displaystyle \Rightarrow 2n = 18$

$\displaystyle \Rightarrow n = 9$

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

iii) $\displaystyle \text{Sum of interior angles } = 9 \times 90^o = 810^o$

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

$\displaystyle \text{Therefore } (2n-4) \times 90^o = 810^o$

$\displaystyle \Rightarrow 2n - 4 = 9$

$\displaystyle \Rightarrow 2n = 13$

$\displaystyle \Rightarrow n = 6.5$

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

$\displaystyle \\$

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: $\displaystyle \text{Interior Angle } = 2 \times$ Exterior angle

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

$\displaystyle \Rightarrow 2n-4 = 8$

$\displaystyle \Rightarrow 2n = 12$

$\displaystyle \Rightarrow n = 6$

$\displaystyle \\$

Question 18: Find the number of sides of a regular polygon if it is given that the ratio of an interior angle and an exterior angle is $\displaystyle 8:1$.

Given: $\displaystyle \text{Interior Angle } = 8 \times$ Exterior angle

$\displaystyle \Rightarrow \Big( \frac{2n-4}{n} \Big) \times 90^o = 8 \times \Big( \frac{360^o}{n} \Big)$

$\displaystyle \Rightarrow 2n-4 = 32$

$\displaystyle \Rightarrow 2n = 36$

$\displaystyle \Rightarrow n = 18$

$\displaystyle \\$

Question 19: The measure of each interior angle of a regular polygon is $\displaystyle 144^o$. Determine the interior angle of another regular polygon the number of whose sides is twice of the first polygon.

Let the number of sides of the first polygon $\displaystyle = n$

$\displaystyle \text{Therefore } \Big( \frac{2n-4}{n} \Big) \times 90^o = 144^o$

$\displaystyle \Rightarrow 20n - 40 = 16n$

$\displaystyle \Rightarrow 4n = 40$

$\displaystyle \Rightarrow n = 10$

For the second polygon: $\displaystyle n = 20$

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

$\displaystyle = \Big( \frac{2 \times 20-4}{20} \Big) \times 90^o$

$\displaystyle = \frac{36}{20} \times 90^o$

$\displaystyle = 162^o$

$\displaystyle \\$

Question 20: Find the number of sides of the regular polygon if it is given that an interior angle and an exterior angle are in the ratio of $\displaystyle 7:2$.

$\displaystyle \text{Given: } 2 \times$ $\displaystyle \text{Interior Angle } = 7 \times$ Exterior angle

$\displaystyle \Rightarrow 2 \times \Big( \frac{2n-4}{n} \Big) \times 90^o = 7 \times \Big( \frac{360^o}{n} \Big)$

$\displaystyle \Rightarrow 4n-8 = 28$

$\displaystyle \Rightarrow 4n = 36$

$\displaystyle \Rightarrow n = 9$

$\displaystyle \\$

Question 21: Show that the diagonals of a regular pentagon are equal.

$\displaystyle \text{To prove: } AC = AD$

Consider $\displaystyle \triangle ABC \text{ and } \triangle ADE$

Since $\displaystyle ABCDE$ is a regular pentagon,

$\displaystyle AB = AE$ (given)

$\displaystyle BC = DE$ (given)

and $\displaystyle \angle ABC = \angle AED$ (given)

$\displaystyle \text{Therefore } \triangle ABC \cong \triangle ADE$ (by S.A.S criterion)

$\displaystyle \text{Therefore } AC = AD$ (corresponding sides of congruent triangles are equal)

Hence the diagonals are equal from any vertex.

$\displaystyle \\$

Question 22: The number of sides of two regular polygons are in the ratio of $\displaystyle 1:2$ and their interior angles are in the ratio of $\displaystyle 3:4$, find the number of sides of the polygon.

Polygon 1: Sides $\displaystyle = n_1$

Polygon 2: Sides $\displaystyle = n_2$

$\displaystyle \text{Given: } \frac{n_1}{n_2} = \frac{1}{2}$

$\displaystyle \Rightarrow 2n_1 = n_2$ … … … … … i)

Also, their interior angles are in the ratio of $\displaystyle 3:4$

$\displaystyle \text{Therefore } \frac{\Big( \frac{2n_1-4}{n_1} \Big) \times 90^o}{\Big( \frac{2n_2-4}{n_2} \Big) \times 90^o} = \frac{3}{4}$

$\displaystyle \Rightarrow 4 \Big( \frac{2n_1-4}{n_1} \Big) = 3 \Big( \frac{2n_2-4}{n_2} \Big)$ … … … … … ii)

Substituting i) in ii) we get

$\displaystyle 4 \Big( \frac{2n_1-4}{n_1} \Big) = 3 \Big( \frac{4n_1-4}{2n_1} \Big)$

$\displaystyle \Rightarrow 16n_1 - 32 = 12n_1 - 12$

$\displaystyle \Rightarrow 4n_1 = 20$

$\displaystyle \Rightarrow n_1 = 5$

$\displaystyle \text{Therefore } n_2 = 10$

$\displaystyle \\$

Question 23: The number of sides of two regular polygons are in the ratio $\displaystyle 3 : 4$ and their sums of their interior angles are in the ratio $\displaystyle 2 : 3$. Find the number of sides of each polygon.

Polygon 1: Sides $\displaystyle = n_1$

Polygon 2: Sides $\displaystyle = n_2$

$\displaystyle \text{Given: } \frac{n_1}{n_2} = \frac{3}{4}$

$\displaystyle \Rightarrow 4n_1 = 3n_2$ … … … … … i)

Also, their sum interior angles are in the ratio of $\displaystyle 2:3$

$\displaystyle \text{Therefore } \frac{(2n_1 - 4) \times 90^o}{(2n_2-4) \times 90^o} = \frac{2}{3}$

$\displaystyle \Rightarrow 3(2n_1-4) = 2(2n_2-4)$ … … … … … ii)

Substituting i) in ii) we get

$\displaystyle 3(2n_1 - 4) = 2 (2 \times \frac{4}{3} n_1 - 4)$

$\displaystyle \Rightarrow 6n_1 - 12 = \frac{16}{3} n_1 - 8$

$\displaystyle \Rightarrow \frac{2}{3} n_1 = 4$

$\displaystyle \Rightarrow n_1 = 6$

$\displaystyle \text{Therefore } n_2 = \frac{4}{3} \times 6 = 8$

$\displaystyle \\$

Question 24: The difference between the exterior angles of two regular polygons is $\displaystyle 5^o$. If the number of sides of a polygonal is one more than the other, find the number of sides of each Polygon.

Polygon 1: Sides $\displaystyle = n$

Polygon 2: Sides $\displaystyle = n+1$

Exterior angle of polygon 1 $\displaystyle = \frac{360}{n}$

Exterior angle of polygon 2 $\displaystyle = \frac{360}{n+1}$

$\displaystyle \text{Therefore } \frac{360}{n} - \frac{360}{n+1} = 5$

$\displaystyle \Rightarrow 360 n + 360 - 360n = 5n(n+1)$

$\displaystyle \Rightarrow 5n^2 + 5n - 360 = 0$

$\displaystyle \Rightarrow n^2 + n - 72 = 0$

$\displaystyle \Rightarrow (n+9)(n-8) = 0$

$\displaystyle \Rightarrow n = -9, 8$

Now n cannot be a negative number. Hence $\displaystyle n = 8$. This implies that polygon 2 has $\displaystyle 9$ sides.

$\displaystyle \\$

Question 25: A heptagon has $\displaystyle 4$ equal angles each of $\displaystyle 132^o$ and three equal angles. Find the measure of equal angles.

Heptagon: $\displaystyle n = 7$

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

$\displaystyle \text{Therefore } (2 \times 7 - 4) \times 90^o = 4 \times 132^o + 3x$

$\displaystyle \Rightarrow 900^o = 528^o + 3x$

$\displaystyle \Rightarrow 3x = 372^o$

$\displaystyle \Rightarrow x = 124^o$

$\displaystyle \\$

Question 26: Find the number of sides of a polygon if the sum of interior angles is six times the sum of its exterior angles.

$\displaystyle \text{Let the Number of sides } = n$

Given sum of interior angles is six times the sum of its exterior angles

$\displaystyle \text{Therefore } (2n-4)\times 90^o = 6 \times 360^o$

$\displaystyle \Rightarrow 2n-4 = 24$

$\displaystyle \Rightarrow 2n = 28$

$\displaystyle \Rightarrow n = 14$

$\displaystyle \\$

Question 27: If the sum of interior angles of a pentagon are in the ratio $\displaystyle 4 : 5 : 6 : 7 : 5$, find the angles.

$\displaystyle \text{Given: } n = 5$

$\displaystyle \text{Let the angles be } 4x, 5x, 6x, 7x \text{ and } 5x$

$\displaystyle \text{Sum of interior angles } = (2n-4) \times 90^o = 6 \times 90^o = 540^o$

$\displaystyle \text{Therefore } 4x + 5x + 6x + 7x + 5x = 540^o$

$\displaystyle \Rightarrow 27x = 540^o$

$\displaystyle \Rightarrow x = 20^o$

Therefore the angles are $\displaystyle 80^o, 100^o, 120^o, 140^o, 100^o$

$\displaystyle \\$

Question 28: If the angles of a hexagon are $\displaystyle (2x+ 5)^o, (3x -5)^o, (x + 40)^o, (2x +20)^o, (2x + 25)^o \text{ and } (2x + 35)^o$, find the value of $\displaystyle x$.

Angles of a hexagon are $\displaystyle (2x+ 5)^o, (3x -5)^o, (x + 40)^o, (2x +20)^o, (2x + 25)^o \text{ and } (2x + 35)^o$

Number of sides: $\displaystyle n = 6$

$\displaystyle \text{Sum of interior angles } = (2n-4) \times 90^o = 8 \times 90^o = 720^o$

$\displaystyle \text{Therefore } 720^o = (2x+ 5)^o+ (3x -5)^o+ (x + 40)^o+ (2x +20)^o+ (2x + 25)^o + (2x+35)^o$

$\displaystyle \Rightarrow 720^o = 12x + 120^o$

$\displaystyle \Rightarrow 12x = 600^o$

$\displaystyle \Rightarrow x = 50^o$

Therefore the angles are $\displaystyle 105^o, 145^o, 90^o, 120^o, 125^o \text{ and } 135^o$

$\displaystyle \\$

Question 29: The angles of a pentagon are $\displaystyle x^o, (x-10)^o,(x+20)^o, (2x-44)^o \text{ and } (2x -70)^o$ Find the value of $\displaystyle x$.

Pentagon: $\displaystyle n = 5$

$\displaystyle \text{Sum of interior angles } = (2n-4) \times 90^o = 68 \times 90^o = 540^o$

$\displaystyle \text{Therefore } x^o + (x-10)^o + (x+20)^o + (2x-44)^o + (2x -70)^o = 540^o$

$\displaystyle \Rightarrow 7x - 104^o = 540^o$

$\displaystyle \Rightarrow 7x = 644^o$

$\displaystyle \Rightarrow x = 92^o$

$\displaystyle \\$

Question 30: The measures of three exterior angles of a hexagon are $\displaystyle 40^o, 51^o \text{ and } 86^o$, if each of the remaining exterior angles is $\displaystyle x^o$, find the value of $\displaystyle x^o$.

Hexagon: $\displaystyle n = 6$

Sum of the exterior angles $\displaystyle = 360^o$

$\displaystyle \text{Therefore } 40^o + 51^o + 86^o + 3x = 360^o$

$\displaystyle \Rightarrow 3x = 360^o - 177^o = 183^o$

$\displaystyle \Rightarrow x = 61^o$

$\displaystyle \\$

Question 31: In the adjoining figure $\displaystyle ABCDE$ is a regular pentagon. Find the measures of the angles marked $\displaystyle x, y, z$.

$\displaystyle \text{Regular pentagon: } n = 5$

$\displaystyle \text{Interior Angle } = \Big( \frac{2n-4}{n} \Big) \times 90^o = \frac{6}{5} \times 90^o = 84^o$

In the quadrilateral $\displaystyle ABDE: \angle A + \angle z = 180^o \Rightarrow \angle z = 180^o -84^o = 96^o$

Since $\displaystyle AE = AB$ (sides of a regular pentagon)

$\displaystyle \angle AEB = \angle ABE = x$

$\displaystyle \text{Therefore } 2x + 84^o = 180^o \Rightarrow x = 48^o$

$\displaystyle \text{Therefore } \angle BED = 84^o - 48^o = 36^o$

Hence $\displaystyle y = 180^o - 36^o - 96^o = 48^o$

$\displaystyle \\$

Question 32: In a regular hexagon $\displaystyle ABCDEF$, prove that $\displaystyle \triangle ACE$ is an equilateral triangle.

$\displaystyle \text{To prove: } \triangle ACE$ is equilateral triangle

Consider $\displaystyle \triangle ABC \text{ and } \triangle DCE$

$\displaystyle CB = DC$

$\displaystyle AB = ED$

$\displaystyle \angle ABC = \angle CDE$

$\displaystyle \text{Therefore } \triangle ABC \cong \triangle DCE$ (By S.A.S criterion)

$\displaystyle \Rightarrow AC = EC$

Similarly, $\displaystyle \triangle AEF \cong \triangle DEC$

$\displaystyle \Rightarrow AE = EC$

$\displaystyle \text{Therefore } AC = EC = AE$

$\displaystyle \Rightarrow \triangle ACE$ is equilateral

$\displaystyle \\$

Question 33: In a regular pentagon $\displaystyle ABCDE$, show that $\displaystyle AB$ is parallel to $\displaystyle EC$

$\displaystyle \text{To prove: } AP \parallel EC$

$\displaystyle n = 5$

$\displaystyle \text{Interior Angle } = \Big( \frac{2n - 4}{n} \Big) \times 90^o = \frac{6}{5} \times 90^o = 6 \times 18 = 108^o$

Noe in $\displaystyle \triangle CDE$ $\displaystyle CD = ED$

$\displaystyle \Rightarrow \angle DCE = \angle DEC = 36^o$

$\displaystyle \Rightarrow \angle BCE = 108^o - 36^o = 72^o$

$\displaystyle \Rightarrow \angle ABC + \angle BCE = 108^o + 72^o = 180^o$

Similarly, $\displaystyle \angle CEA = 108^o - 36^o = 72^o$

$\displaystyle \text{Therefore } \angle BAE + \angle AEC = 108^o + 72^o = 180^o$

$\displaystyle \text{Therefore } AB \parallel CE$ since the sum of the interior alternate angles is $\displaystyle 180^o$

$\displaystyle \\$

Question 34: If in a pentagon $\displaystyle ABCDE$, we have

(i) $\displaystyle AE \parallel BC, \angle C = 153, \angle D = x^o \text{ and } \angle E = 2x^o$, find the value of $\displaystyle x$.

(ii) $\displaystyle \angle A = 110^o, \angle B = 142^o , \angle D = \angle E$ and sides $\displaystyle AB \text{ and } DC$ when produced meet at right angles, find $\displaystyle \angle BCD \text{ and } \angle E$

(iii) $\displaystyle AB = AE,BC=ED \text{ and } \angle ABC = \angle AED$ prove that $\displaystyle AC = AD \text{ and } \angle BCD = \angle EDC$

(iv) $\displaystyle BC \text{ and } ED$ are produced to meet at $\displaystyle X$, prove that $\displaystyle BX = EX$

(v) $\displaystyle AB \parallel ED, \angle B = 140^o \text{ and } \angle C : \angle D = 5: 6$, find $\displaystyle \angle C \text{ and } \angle D$

(i) $\displaystyle AE \parallel BC, \angle C = 153, \angle D = x^o \text{ and } \angle E = 2x^o$, find the value of $\displaystyle x$.

$\displaystyle \text{Regular pentagon: } n = 5$

$\displaystyle \text{Sum of interior angles } = (2n-4) \times 90^o = 6 \times 90^o = 540^o$

$\displaystyle \text{Therefore } 180^o + 153^o + 3x = 540^o$

$\displaystyle \Rightarrow 3x = 207^o$

$\displaystyle \Rightarrow x = 69^o$

(ii) $\displaystyle \angle A = 110^o, \angle B = 142^o , \angle D = \angle E$ and sides $\displaystyle AB \text{ and } DC$ when produced meet at right angles, find $\displaystyle \angle BCD \text{ and } \angle E$

$\displaystyle \angle BCD = 108$

$\displaystyle \text{Therefore } 110^o + 142^o + 128^o + 2x = (2 \times 5 -4) \times 90$

$\displaystyle \Rightarrow 400^o + 2x = 540^o$

$\displaystyle \Rightarrow x = 80^o$. Hence $\displaystyle \angle E = 80^o$

(iii) $\displaystyle AB = AE,BC=ED \text{ and } \angle ABC = \angle AED$ prove that $\displaystyle AC = AD \text{ and } \angle BCD = \angle EDC$

Consider $\displaystyle \triangle ABC \text{ and } \triangle AED$

$\displaystyle AB = AE$

$\displaystyle BC = ED$

$\displaystyle \angle ABC = \angle AED$

$\displaystyle \text{Therefore } \triangle ABC \cong \triangle AED$ (By S.A.S criterion)

$\displaystyle \text{Therefore } AC = AD$

$\displaystyle \text{Therefore } \angle ACD = \angle ADC$

Also since $\displaystyle AB = BC= AE = AD$

$\displaystyle \angle BAC = \angle BCA = \angle EAD = \angle EDA$

$\displaystyle \text{Therefore } \angle BCA + \angle ACD =\angle EDA + \angle ADC$

$\displaystyle \Rightarrow \angle BCD = \angle EDC$

(iv) $\displaystyle BC \text{ and } ED$ are produced to meet at $\displaystyle X$, prove that $\displaystyle BX = EX$

Assuming that $\displaystyle ABCDE$ is a regular pentagon

$\displaystyle \text{To prove: } BX = EX$

$\displaystyle \angle BCD = \angle EDC$

$\displaystyle \Rightarrow XCD = \angle XDC$

$\displaystyle \Rightarrow CX = DX$

$\displaystyle \text{Therefore } BC + CX = ED + DX$

$\displaystyle \Rightarrow BX = EX$

(v) $\displaystyle AB \parallel ED, \angle B = 140^o \text{ and } \angle C : \angle D = 5: 6$, find $\displaystyle \angle C \text{ and } \angle D$

$\displaystyle n = 5$

Sum of internal angles $\displaystyle = (2n-4) \times 90^o = 6 \times 90^o = 540^o$

$\displaystyle \text{Therefore } 540^o = 140^o + 180^o + 5x + 6x$

$\displaystyle \Rightarrow 11x = 220^o$

$\displaystyle \Rightarrow x = 20^o$

$\displaystyle \text{Therefore } \angle C = 100^o, \ \ \ \angle D = 120^o$

$\displaystyle \\$

Question 35: $\displaystyle ABCDE$ is a regular pentagon such that diagonal $\displaystyle AD$ divides $\displaystyle \angle CDE$ into two parts. Find the ratio $\displaystyle \frac{ \angle ADE}{ \angle ADC}$

$\displaystyle \text{Regular pentagon: } n = 5$

Internal angle $\displaystyle = \frac{2n-4}{n} \times 90^o = \frac{6}{5} \times 90^o = 108^o$

In $\displaystyle \triangle AED$

$\displaystyle x + x + 108^o = 180^o$

$\displaystyle \Rightarrow 2x = 72^o$

$\displaystyle \Rightarrow x = 36^o$

$\displaystyle \text{Therefore } y = 108^o - 36^o = 72^o$

$\displaystyle \text{Therefore } \frac{\angle ADE}{\angle ADC} = \frac{36^o}{72^o} = \frac{1}{2}$

Therefore the ratio of $\displaystyle \angle ADE : \angle ADC = 1:2$