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

\\

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

\\

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)

\\

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

\\

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

\\

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

\\

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

\\

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

\\

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

\\

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

\\

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

\\

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

\\

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

\\

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

\\

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

\\

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

\\

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

\\

Advertisements