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

Answer:

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.

Answer:

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)

Answer:2018-12-21_17-53-19

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 .

Answer:2018-12-21_17-55-58

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?

Answer:

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

Answer:

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

Answer:

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.

Answer:

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 .

Answer:

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.

Answer:

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.

Answer:

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 .

Answer:

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

Answer:

Regular hexagon: n = 6 2018-12-21_17-59-03

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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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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