Mathematics Part I Solutions Solutions for Class 9 Math Chapter 1 Polygons are provided here with simple step-by-step explanations. These solutions for Polygons are extremely popular among Class 9 students for Math Polygons Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part I Solutions Book of Class 9 Math Chapter 1 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Part I Solutions Solutions. All Mathematics Part I Solutions Solutions for class Class 9 Math are prepared by experts and are 100% accurate.

Page No 10:

Question 1:

The sum of the angles of a polygon is 2160°. What is the sum of the angles of a polygon with one more side? What about a polygon with one side less?

Answer:

Sum of angles of the given polygon = 2160°

We know that the sum of the angles of polygon with n sides = (n 2) × 180°

(n 2) × 180° = 2160°

n 2 =

n 2 = 12

n = 14

Number of sides in the given polygon = 14

Number of sides in the second polygon = 14 + 1 = 15

Sum of the angles of the second polygon = (n 2) × 180°

= (15 2) × 180°

= 13 × 180°

= 2340°

Thus, sum of the angles of a polygon with 15 sides is 2340°.

Number of sides in the third polygon = 14 1 = 13

Sum of the angles of the third polygon = (n 2) × 180°

= (13 2) × 180°

= 11 × 180°

= 1980°

Thus, the sum of the angles of a polygon with 13 sides is 1980°.

Page No 10:

Question 2:

What is the sum of the angles of a polygon with 102 sides?

Answer:

Number of sides of the given polygon = 102

We know that the sum of the angles of a polygon with n sides = (n 2) × 180°

Sum of the angles of the given polygon = (102 2) × 180°

= 100 × 180°

= 18000°

Thus, the sum of the angles of a polygon with 102 sides is 18000°.

Page No 10:

Question 3:

The sum of the angles of a polygon is 2700°. How many sides does it have?

Answer:

Sum of angles of the given polygon = 2700°

We know that the sum of the angles of a polygon with n sides = (n 2) × 180°

(n 2) × 180° = 2700°

n 2 =

n 2 = 15

n = 17

Thus, the given polygon has 17 sides. 

Page No 10:

Question 4:

Does any polygon have the sum of its angles equal to 1000°? How about 900°?

Answer:

We know that the sum of the angles of a polygon with n sides = (n 2) × 180°

For the sum of the angles of a polygon to be 1000°, we must have:

(n 2) × 180° = 1000°

n 2 =

n 2 = 5.555…

n is not a natural number. 

Thus, there is no polygon whose sum of all its angles is 1000°.

For the sum of the angles of a polygon to be 900°, we must have:

(n 2) × 180° = 900°

n 2 =

n 2 = 5

n = 7

Thus, a polygon with seven sides has the sum of its angles as 900°.

Page No 10:

Question 5:

A 10 sided polygon has all its angles equal. How much is each angle?

Answer:

Number of sides of the given polygon = 10

We know that the sum of the angles of a polygon with n sides = (n 2) × 180°

Sum of the angles of the given polygon = (10 2) × 180°

= 8 × 180°

= 1440°

Given: All the angles of the given polygon are equal in measure.

Let the measure of each angle of the given polygon be x°.

10 × x = 1440°

x = 144°

Thus, the measure of each angle of a polygon with 10 sides is 144°.

Page No 10:

Question 6:

Each angle of a polygon is 150°. How many sides does it have?

Answer:

Measure of each angle of the given polygon = 150°

We know that the sum of the angles of a polygon with n sides = (n 2) × 180°

Sum of the angles of the given polygon = 150° × n

150° × n = (n 2) × 180°

5° × n = (n 2) × 6°

Thus, the given polygon has 12 sides.



Page No 13:

Question 1:

The angles of a triangle are 30°, 40° and 110°. Find the measures of its external angles.

Answer:

Given: ΔABC with A = 30°, B = 40° and C = 110°

The measures of the external angles of a triangle can be calculated using the property of linear pair.

We know that the sum of the measures of angles forming a linear pair is 180°.

⇒ ∠ACS + ACB = 180°

⇒ ∠ACS = 180° 110°

= 70°

Similarly, CBU + CBA = 180°

⇒ ∠CBU + 40° = 180°

⇒ ∠CBU = 180°− 40°

= 140°

CAT + CAB = 180°

⇒ ∠CAT = 180° 30°

= 150°

Thus, the measures of the external angles of the given triangle are 150°, 140° and 70°.

Page No 13:

Question 2:

Three angles of a quadrilateral are 60°, 75° and 100°. Find the fourth angle. Also, find all the four external angles.

Answer:

Given: Quadrilateral PQRS with SPQ = 60°, PQR = 75° and QRS = 100°

We know that the sum of the angles of a quadrilateral is 360°.

⇒ ∠SPQ + PQR + QRS + RSP = 360°

60° + 75° + 100° + RSP = 360°

235° + RSP = 360°

⇒∠ RSP = 360° 235°

= 125°

The measures of the external angles of a quadrilateral can be calculated using the property of linear pair.

We know that the sum of the measures of angles forming a linear pair is 180°.

⇒ ∠QRS + QRX = 180°

⇒ ∠QRS = 180° 100°

= 80°

Similarly, RQP + PQW = 180°

⇒ ∠PQW = 180° 75°

= 105°

SPQ + SPZ = 180°

⇒ ∠SPZ = 180° 60°

= 120°

RSY + RSP = 180°

⇒∠ RSY = 180° 125°

= 55°

Thus, the measures of the external angles of the given quadrilateral are 80°, 105°, 120° and 55°.



Page No 15:

Question 1:

Prove that in any triangle, cash external angle is equal to the sum of the internal angles at the other two vertices.

Answer:

Consider ΔPQR:

In ΔPQR, by angle sum property, we have:

QPR + PQR +QRP = 180°

⇒ ∠QRP = 180° PQR QPR ... (1)

Also, PRT and PRQ are forming a linear pair.

⇒ ∠PRT + PRQ = 180°

⇒∠PRT + 180° PQR QPR = 180° (Using equation (1))

⇒∠PRT = PQR + QPR

Hence, in any triangle, each external angle is equal to the sum of the internal angles at the other two vertices. 

Page No 15:

Question 2:

Prove that in any quadrilateral, the sum of the external angles at two vertices is equal to the sum of the internal angles at the other two vertices.

Answer:

Consider quadrilateral PQRS:

In quadrilateral PQRS, by angle sum property, we have:

SPQ + PQR + QRS +RSP = 360°

⇒ ∠PQR + QRS = 360° SPQ RSP ... (1)

Also, QRS and QRU are forming a linear pair.

⇒∠ QRS + QRU = 180° ... (2)

Similarly, PQT and PQR are forming a linear pair.

⇒∠ PQT + PQR = 180° ... (3)

Adding equation (2) and equation (3):

⇒∠QRS + QRU + PQT + PQR = 360°

⇒∠QRS + PQR = 360° QRU PQT …(4)

From equation (1) and equation (4):

360° QRU −∠PQT = 360° SPQ RSP 

⇒ ∠QRU + PQT = SPQ + PSR 

Hence, the sum of the external angles at the two vertices of a quadrilateral is equal to the sum of the internal angles at the other two vertices.

Page No 15:

Question 3:

In a 12-sided polygon, the external angles are all equal. How much is each external angle? And each angle of the polygon.

Answer:

Total number of sides = 12

We know that in a polygon, the sum of all the external angles is 360°.

Given: All the external angles of the given polygon are equal in measure.

Let the measure of each external angle of the given polygon be x°.

12x = 360°

x = 30°

Thus, each external angle measures 30°.

We know that for a polygon with n sides, the sum of the angles = (n 2) × 180°

Sum of the angles of the given polygon = (12 2) × 180°

= 10 × 180°

= 1800°

Since each external angle of the given polygon is equal in measure, each internal angle should also be equal.

Let the measure of each internal angle of the given polygon be y°.

12y = 1800°

y = 150°

Thus, the measure of each internal angle of the given polygon is 150°.

Page No 15:

Question 4:

A 10-sided polygon has all its angles equal. How much is each external angle?

Answer:

Total number of sides of the given polygon = 10

We know that in a polygon, the sum of all the external angles = 360°

Given: All the external angles of the given polygon are equal in measure.

Let the measure of each external angle of the given polygon be x°.

10x = 360°

x = 36°

Thus, the measure of each external angle of the given polygon is 36°.

Page No 15:

Question 5:

The sum of the angles of a polygon and the sum of its external angles are equal. How many sides does it have?

Answer:

Given: Sum of the external angles of a polygon is equal to the sum of its internal angles. 

We know that in a polygon, the sum of all the external angles = 360°

Also, the sum of the angles of a polygon with n sides = (n 2) × 180°

360° = (n 2) × 180°

n 2 = 2

n = 4

Thus, a polygon with four sides has the sum of the internal angles equal to the sum of its external angles.

Page No 15:

Question 6:

Each external angle of a polygon is 20°. How many sides does it have?

Answer:

Measure of each external angle of the given polygon = 20°

We know that in a polygon, the sum of all the external angles = 360°

Let the given polygon has n external angles.

20 × n = 360°

n = 18

Thus, the given polygon has 18 sides. 



Page No 18:

Question 1:

Draw a hexagon with all angles equal, but not all sides equal.

Answer:

A hexagon with all angles equal should have each side of equal length.

So, it is not possible to draw a hexagon with all angles equal but not all sides equal.

Page No 18:

Question 2:

Draw a hexagon with all sides equal, but not all angles equal.

Answer:

A hexagon with all sides equal in length should have all angles of equal measure.

So, it is not possible to draw a hexagon with all sides equal but not all angles equal.

Page No 18:

Question 3:

How much is each angle of a 12-sided regular polygon? How much is each of its external angle?

Answer:

Total number of sides of the given regular polygon = 12

We know that sum of the angles of a polygon with n sides = (n 2) × 180°

Sum of all the angles of the given regular polygon = (n 2) × 180°

= (12 2) × 180°

= 10 × 180°

=1800°

Let the measure of each angle of the given regular polygon be x°.

12x = 1800°

x = 150°

The external angle forms a linear pair with the adjacent internal angle

150° + External angle = 180°

External angle = 180° 150° = 30°

Thus, the measure of each external angle is 30°.

Page No 18:

Question 4:

Prove that in a regular pentagon, the perpendicular from any vertex to the opposite side bisects that side.

Answer:

Consider the regular pentagon, ABCDE:

Construction: Draw DF perpendicular to AB. Join AD and DB.

In ΔAFD and ΔBFD:

AD = BD (Diagonals of a regular pentagon are equal in length)

DF = DF (Common)

AFD = BFD = 90°

∴ ΔAFD ΔBFD (RHS congruence criterion)

AF = FB (By c.p.c.t.)

Thus, in a regular pentagon, the perpendicular from any vertex to the opposite side bisects that side.

Page No 18:

Question 5:

In the figure below, ABCDEF is a regular hexagon.

Prove that ABCDE is a rectangle.

Answer:

The given figure is a regular hexagon, so all its angles are equal in measure.

Number of sides of a regular hexagon = 6

We know that the sum of the angles of a polygon with n sides = (n 2) × 180°

Sum of the angles of the given regular hexagon = (6 2) × 180°

= 4 × 180° 

= 720°

Let the measure of each angle of the given regular hexagon be x°.

6x = 720°

x = 120°

In ΔFEA, by angle sum property, we have:

AFE +FEA + EAF = 180° …(1)

FE = FA (All sides of a regular hexagon are equal in length)

⇒∠FEA = FAE (Angles opposite to equal sides are equal in measure)

Putting in equation (1):

AFE + 2FEA = 180°

120° + 2FEA = 180°

2FEA = 60°

⇒∠FEA = 30°

Similarly, CDB = CBD = FAE = 30°

As FED = 120° 

⇒ ∠FEA + AED = 120°

30° + AED = 120°

⇒∠AED = 90°

Similarly, EAB = ABD = BDE = 90°

In quadrilateral ABDE, EAB = ABD = BDE = AED = 90°.

Also, ED = AB

Thus, quadrilateral ABDE is a rectangle.

Page No 18:

Question 6:

In the figure below, ABCDEF is regular hexagon.

Prove that ACE is an equilateral triangle.

Answer:

The given figure is a regular hexagon, so all its angles are equal in measure.

Number of sides of a regular hexagon = 6

We know that the sum of the angles of a polygon with n sides = (n 2) × 180°

Sum of the angles of the regular hexagon = (6 2) × 180°

= 4 × 180° 

= 720°

Let the measure of each angle of the given regular hexagon be x°.

6x = 720°

x = 120°

In ΔFEA, by angle sum property, we have:

AFE +FEA + EAF = 180° …(1)

FE = FA (All sides of a regular hexagon are equal in length)

⇒∠FEA = FAE (Angles opposite to equal sides are equal in measure)

Putting in equation (1):

AFE + 2FEA = 180°

120° + 2FEA = 180°

2FEA = 60°

⇒∠FEA = 30°

Similarly, DEC = DCE = BCA = BAC = FAE = 30°

As FED = 120° 

⇒ ∠FEA + AEC + CED = 120°

30° + AEC + 30° = 120°

60° + AEC = 120°

⇒∠AEC = 60°

Similarly, ACE = EAC = 60°

Thus, ΔAEC is an equilateral triangle.

Page No 18:

Question 7:

How much is an angle of a 36-sided regular polygon?

Answer:

Total number of sides of the given regular polygon = 36

We know that the sum of the angles of a polygon with n sides = (n 2) × 180°

Sum of the angles of the given regular polygon = (36 2) × 180°

= 34 × 180°

= 6120°

Let the measure of each angle of the given regular polygon be x°.

36x = 6120°

x = 170°

Thus, the measure of each angle of the given regular polygon is 170°.

Page No 18:

Question 8:

One angle of a regular polygon is 144°. How many sides does it have?

Answer:

Measure of one angle of a regular polygon = 144°

We know that the measure of each angle of a regular polygon is equal.

Measure of each angle of the regular polygon = 144°

Let the number of sides of the given regular polygon be n.

We know that the sum of the angles of a polygon with n sides = (n 2) × 180°

144° × n = (n 2) × 180°

Thus, the given regular polygon has 10 sides.

Page No 18:

Question 9:

Can you draw a regular polygon with each of its angles equal to 47°?

Answer:

Measure of each angle of the regular polygon = 47°

Let the number of sides of the given regular polygon be n.

We know that the sum of the angles of a polygon with n sides = (n 2) × 180°

47° × n = (n 2) × 180°

As n is not a natural number, therefore, such a regular polygon does not exist. 

Hence, we cannot make a regular polygon with each angle equal to 47°.



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