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

Answer:

Given: Interior angle = 8 \times Exterior angle

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

\Rightarrow 2n-4 = 32

\Rightarrow 2n = 36

\Rightarrow n = 18

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

Answer:

Let the number of sides of the first polygon = n

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

\Rightarrow 20n - 40 = 16n

\Rightarrow 4n = 40

\Rightarrow n = 10

For the second polygon: n = 20

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

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

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

= 162^o

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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 7:2  .

Answer:

Given: 2 \times Interior angle =  7 \times Exterior angle

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

\Rightarrow 4n-8 = 28

\Rightarrow 4n = 36

\Rightarrow n = 9

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

Answer:2018-12-21_18-06-02

To prove: AC = AD

Consider \triangle ABC and \triangle ADE

Since ABCDE is a regular pentagon,

AB = AE (given)

BC = DE (given)

and \angle ABC = \angle AED (given)

Therefore \triangle ABC \cong \triangle ADE (by S.A.S criterion)

Therefore AC = AD   (corresponding sides of congruent triangles are equal)

Hence the diagonals are equal from any vertex.

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

Answer:

Polygon 1: Sides = n_1

Polygon 2: Sides = n_2

Given: \frac{n_1}{n_2} = \frac{1}{2}

\Rightarrow 2n_1 = n_2 … … … … … i)

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

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} 

\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

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

\Rightarrow 16n_1 - 32 = 12n_1 - 12

\Rightarrow 4n_1 = 20

\Rightarrow n_1 = 5

Therefore n_2 = 10

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

Answer:

Polygon 1: Sides = n_1

Polygon 2: Sides = n_2

Given: \frac{n_1}{n_2} = \frac{3}{4}

\Rightarrow 4n_1 = 3n_2 … … … … … i)

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

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

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

Substituting i) in ii) we get

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

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

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

\Rightarrow n_1 = 6

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

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

Answer:

Polygon 1: Sides = n

Polygon 2: Sides = n+1

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

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

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

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

\Rightarrow 5n^2 + 5n - 360 = 0

\Rightarrow n^2 + n - 72 = 0

\Rightarrow (n+9)(n-8) = 0

\Rightarrow  n = -9, 8

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

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

Answer:

Heptagon: n = 7

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

Therefore (2 \times 7 - 4) \times 90^o = 4 \times 132^o + 3x

\Rightarrow 900^o = 528^o + 3x

\Rightarrow 3x = 372^o

\Rightarrow x = 124^o

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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.

Answer:

Let the number of sides = n

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

Therefore (2n-4)\times 90^o = 6 \times 360^o

\Rightarrow 2n-4 = 24

\Rightarrow 2n = 28

\Rightarrow n = 14

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

Answer:

Given: n = 5

Let the angles be 4x, 5x, 6x, 7x , and 5x

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

Therefore 4x + 5x + 6x + 7x + 5x = 540^o

\Rightarrow 27x = 540^o

\Rightarrow x = 20^o

Therefore the angles are 80^o, 100^o, 120^o, 140^o, 100^o

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

Answer:

Angles of a hexagon are (2x+ 5)^o, (3x -5)^o, (x + 40)^o, (2x +20)^o, (2x + 25)^o and (2x + 35)^o

Number of sides: n = 6

Sum of interior angles = (2n-4) \times 90^o = 8 \times 90^o = 720^o

Therefore 720^o = (2x+ 5)^o+ (3x -5)^o+ (x + 40)^o+ (2x +20)^o+ (2x + 25)^o + (2x+35)^o

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

\Rightarrow 12x = 600^o

\Rightarrow x = 50^o

Therefore the angles are 105^o, 145^o, 90^o, 120^o, 125^o and 135^o

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

Answer:

Pentagon: n = 5

Sum of interior angles = (2n-4) \times 90^o = 68 \times 90^o = 540^o

Therefore x^o + (x-10)^o + (x+20)^o + (2x-44)^o + (2x -70)^o = 540^o

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

\Rightarrow 7x = 644^o

\Rightarrow x = 92^o

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

Answer:

Hexagon: n = 6

Sum of the exterior angles = 360^o

Therefore 40^o + 51^o + 86^o + 3x = 360^o

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

\Rightarrow x = 61^o

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

Answer:2018-12-21_18-09-20

Regular pentagon: n = 5

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

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

Since AE = AB (sides of a regular pentagon)

\angle AEB = \angle ABE = x

Therefore 2x + 84^o = 180^o \Rightarrow x = 48^o

Therefore \angle BED = 84^o - 48^o = 36^o

Hence y = 180^o - 36^o - 96^o = 48^o

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

Answer:2018-12-21_18-12-26

To prove: \triangle ACE is equilateral triangle

Consider \triangle ABC and \triangle DCE

CB = DC

AB = ED

\angle ABC = \angle CDE

Therefore \triangle ABC \cong \triangle DCE (By S.A.S criterion)

\Rightarrow AC = EC

Similarly, \triangle AEF \cong \triangle DEC

\Rightarrow AE = EC

Therefore AC = EC = AE

\Rightarrow \triangle ACE is equilateral

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

Answer:2018-12-21_18-13-57

To prove: AP \parallel EC

n = 5

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

Noe in \triangle CDE     CD = ED

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

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

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

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

Therefore \angle BAE + \angle AEC = 108^o + 72^o = 180^o

Therefore AB \parallel CE since the sum of the interior alternate angles is 180^o

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Question 34: If in a pentagon ABCDE , we have

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

Answer:

Regular pentagon: n = 5

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

Therefore 180^o + 153^o + 3x = 540^o

\Rightarrow 3x = 207^o

\Rightarrow x = 69^o

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

Answer:2018-12-21_18-27-03

\angle BCD = 108

Therefore 110^o + 142^o + 128^o + 2x = (2 \times 5 -4) \times 90

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

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

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

Answer:2018-12-21_18-29-51

Consider \triangle ABC and \triangle AED

AB = AE

BC = ED

\angle ABC = \angle AED

Therefore \triangle ABC \cong \triangle AED (By S.A.S criterion)

Therefore AC = AD

Therefore \angle ACD = \angle ADC

Also since AB = BC= AE = AD

\angle BAC = \angle BCA = \angle EAD = \angle EDA

Therefore \angle BCA + \angle ACD =\angle EDA + \angle ADC

\Rightarrow \angle BCD = \angle EDC

(iv) BC and ED are produced to meet at X , prove that BX = EX

Answer:2018-12-21_18-33-22

Assuming that ABCDE is a regular pentagon

To prove: BX = EX

\angle BCD = \angle EDC

\Rightarrow XCD = \angle XDC

\Rightarrow CX = DX

Therefore BC + CX = ED + DX

\Rightarrow BX = EX

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

Answer:

n = 5

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

Therefore 540^o = 140^o + 180^o + 5x + 6x

\Rightarrow 11x = 220^o

\Rightarrow x = 20^o

Therefore \angle C = 100^o, \ \ \ \angle D = 120^o

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

Answer:2018-12-21_18-37-26

Regular pentagon: n = 5

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

In \triangle AED

x + x + 108^o = 180^o

\Rightarrow 2x = 72^o

\Rightarrow x = 36^o

Therefore y = 108^o - 36^o = 72^o

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

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

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