Class 8 Maths Chapter 4 Polygons Questions and Answers Kerala Syllabus

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SCERT Class 8 Maths Chapter 4 Solutions Polygons

Class 8 Kerala Syllabus Maths Solutions Chapter 4 Polygons Questions and Answers

Polygons Class 8 Questions and Answers Kerala Syllabus

Names & Angles (Page No. 65)

Question 1.
The sum of the angles of a polygon is 1980°. What is the sum of the angles of a polygon with one side more? And for a polygon with one side less?
Answer:
When the number of sides increases by 1 angle sum increases by 180°.
Angle sum is 1980 + 180 = 2160°
When the number of sides decreases by 1 angle sum decreases by 180°.
Angle sum 1980 – 180 = 1800°

Question 2.
What is the sum of the angles of a 27-sided polygon?
Answer:
Angle sum of a polygon having n sides is 180° × (n – 2)
Angle sum 180 × 25 = 4500°

Question 3.
The sum of the angles of a polygon is 8100°. How many sides does it have?
Answer:
Sum of the angles = (n – 2) × 180°
Where n is the number of sides.
∴ (n – 2) × 180 = 8100
⇒ n – 2 = \(\frac {8100}{180}\) = 45
⇒ n = 45 + 2
⇒ n = 47

Question 4.
Is the sum of the angles of any polygon equal to 1000°? Explain.
Answer:
Inner angle sum is always a multiple of 180.
1000 is not a multiple of 180.
Angle sum cannot be 1000°.

Question 5.
A 20-sided polygon has equal angles. How much is each angle?
Answer:
Angle sum of a 20 sided polygon = 180 × (20 – 2)
= 180 × 18
= 3240°
One inner angle = \(\frac {3240}{20}\) = 162

Outer Angles (Page No. 71)

Question 1.
We have seen that in a triangle, the outer angle at any vertex is equal to the sum of the inner angles at the other two vertices; and in a quadrilateral, the sum of the outer angles at any two vertices is equal to the sum of the inner angles at the other two vertices. So, here are some possible investigations:
(i) Is there any such relation between the inner and outer angles of a pentagon?
(ii) And in a hexagon?
(iii) Is there a general relation that is true for all polygons?
Answer:
(i) Angles of a pentagon are a, b, c, d, e.
Angle sum = a + b + c + d + e = 180 × 3 = 540°.
Outer angles are 180 – a, 180 – b, 180 – c.
Sum = 540 – (a + b + c)
Sum of two inner angles = 540 – (540 – c + d)) = c + d
The sum of two inner angles is equal to the sum of the other three outer angles.

(ii) Angles are x₁, x₂, x₃, x₄, x₅, x₆
Two inner angles are x₁, x₂
Sum of four outer angles = 180 – x₃ + 180 – x₄ + 180 – x₅ + 180 – x₆ = 720 – (x₃ + x₄ + x₅ + x₆)
Sum of four outer angles = 720 – (720 – (x₁ + x₂) = x₁ + x₂
The sum of two inner angles is equal to the sum of the other outer angles.

(iii) The angles of a polygon having n sides: x₁, x₂,…, x_n
Excluding these two inner angles, there will be n – 3 outer angles.
Sum of the outer angles = 180 – x₃ + 180 – x₄ + ……. + 180 – x_n
Sum of n – 3 outer angles = 180 × (n – 2) – (x₃ + x₄ + …… + x_n)
= 180(n – 2) – (180(n – 2) – (x₁ + x₂))
= x₁ + x₂

Textbook Page No. 72

Question 1.
All inner angles of an 18-sided polygon are equal.
(i) How much is each outer angle?
(ii) How much is each inner angle?
Answer:
(i) Outer angles are equal.
One outer angle = \(\frac {360}{18}\) = 20°
(ii) One inner angle = 180 – 20 = 160°

Question 2.
(i) In which polygon is the sum of outer angles of a polygon, one from each vertex, equal to the sum of its inner angles?
(ii) In which polygon is the sum of the outer angles twice the sum of the inner angles?
(iii) In which polygon is the sum of the outer angles half the sum of the inner angles?
(iv) In which polygon is the sum of the outer angles one-third the sum of the inner angles?
Answer:
(i) Sum of the outer angles 360°.
Since the outer angle sum is 360°.
∴ 180 × (n – 2) = 360
⇒ n – 2 = 2
⇒ n = 4
Number of sides 4.
The polygon is a Quadrilateral.

(ii) Sum of the outer angles 360°.
The sum of the inner angles is 180°.
The polygon is a Triangle.

(iii) Sum of the outer angles = 2 × 360 = 720
∴ 180 × (n – 2) = 720
⇒ n – 2 = 4
⇒ n = 6
The polygon is a Hexagon.

(iv) The sum of the inner angles is three times the sum of the outer angles.
180 × (n – 2) = 3 × 360
⇒ n – 2 = \(\frac{3 \times 360}{180}\) = 6
⇒ n = 8
Number of sides = 8
The polygon is an Octagon.

Regular Polygons (Page No. 77, 78)

Question 1.
(i) Draw a hexagon of equal sides with angles different.
(ii) Draw a hexagon of equal angles with different sides.
Answer:

Question 2.
The picture shows a regular hexagon with vertices on a circle. Prove that the length of its sides is equal to the radius of the circle.

Answer:
Inner angle sum = 180 × (6 – 2)
= 180 × 4
= 720°
All angles are equal.
One angle = \(\frac {720}{6}\) = 120°
Bisectors of angles pass through the centre.
That is angle around the centre will be divided into six equal parts.
Length of side is equal to radius.

Question 3.
Draw a regular octagon (eight sides) of sides 3 centimetres.
Answer:
Angle sum of octagon = 180 × (8 – 2) = 1080
One angle of regular octagon = \(\frac {1080}{8}\) = 135°
Draw a line AB of length 3 cm.
With B as the centre, draw a 135° angle and mark C at a distance of 3 cm away.
Continue this process.

Question 4.
Draw a circle of radius 3 centimetres and draw a regular octagon with all vertices on this circle.
Answer:
Draw the circle, its diameter, and its perpendicular bisector.
Now the angle around the centre is divided into four equal parts.
Bisect the angles again and complete the octagon.

Question 5.
The picture below shows a regular hexagon and a triangle joining its alternate vertices:

Is this an equilateral triangle? Why?
Answer:

Triangles around the middle triangles are equal.
Sum of the small angles = 180 – 120 = 60°
One small angle = 30°
Angles of middle triangle are 30 + 30 = 60°
This is an equilateral triangle.

Question 6.
The picture below shows a regular hexagon and a quadrilateral joining four of its vertices:

Is this quadrilateral a rectangle? Why?
Answer:

One angle of a regular hexagon = 180 × (6 – 2) ÷ 6 = 120°
Look at the triangles on the sides of the quadrilateral.
These are isosceles triangles.
Angles are 30°, 30°, 120°
One angle of the quadrilateral = 120 – 30 = 90°
This is a rectangle.

Question 7.
Calculate the inner angle and the outer angle of a regular polygon of 15 sides.
Answer:
Measure of one outer angle = \(\frac {360}{15}\) = 24°
Measure of one inner angle = 180 – 24 = 156°

Question 8.
An outer angle of a regular polygon is 20°. How many sides does it have?
Answer:
Outer angle sum = 360°
All outer angles are equal.
One angle is \(\frac {360}{20}\) = 18°

Question 9.
An inner angle of a regular polygon is 168°. How many sides does it have?
Answer:
One outer angle = 180 – 168 = 12°
Number of sides = \(\frac {360}{12}\) = 30

Class 8 Maths Chapter 4 Kerala Syllabus Polygons Questions and Answers

Class 8 Maths Polygons Questions and Answers

Question 1.
A regular polygon has an inner and outer angle sum equal. Which of the following is that polygon?
(a) Equilateral triangle
(b) Square
(c) Regular pentagon
(d) Regular hexagon
Answer:
(b) Square

Question 2.
Two statements are given below.
p₁: Whatever the number of sides of a polygon, the outer angle sum is 360°.
p₂: inner angle sum increases by 180° if the number of sides increases by 1.
(a) p₁ and p₂ are true
(b) p₁ is true p₂ is false
(c) p₁ is false p₂ is true
(d) p₁ and p₂ are false
Answer:
(a) p₁ and p₂ are true

Question 3.
A polygon has 7 sides
(a) What is the sum of its inner angles?
(b) What is the sum of the outer angles?
Answer:
(a) Angle sum of a polygon of n sides is 180° × (n – 2)
Angle sum is 180° × 5 = 900°
(b) 360°

Question 4.
A polygon has an inner angle sum that is two times the outer angle sum.
(a) What is its inner angle sum?
(b) Find the number of sides of the polygon.
Answer:
(a) 2 × 360 = 720°
(b) 180 × (n – 2) = 720
⇒ n – 2 = 4
⇒ n = 6

Question 5.
Find the sum of the angles marked in the star.

Answer:
Diagram

In triangle BQD, the sum of the marked angles is the outer angle at Q.
In triangle PCE, the marked angle sum is the outer angle at P.
All marked angles make the sum 180°.

Question 6.
A polygon has all inner angles equal. There are 18 sides.
(a) What is the inner angle sum?
(b) Find the measure of an inner angle?
(c) What is the measure of one outer angle?
Answer:
(a) 180 × (18 – 2) = 2880°
(b) \(\frac {2880}{18}\) = 160°
(c) 180 – 160 = 20°

Question 7.
This is the section of a regular polygon.

(a) What are the angles of a triangle?
(b) What is an outer angle?
(c) How many sides does this polygon have?
Answer:
(a) 18°, 18°, 144°
(b) 180 – 144 = 36°
(c) \(\frac {360}{6}\) = 10

Question 8.
Draw a circle of radius 3 cm and construct an equilateral triangle with corners on the circle.
Answer:
Divide the angle around the centre into three equal parts by drawing radii.

Question 9.
A regular polygon has 15 sides.
(a) What is the measure of an outer angle?
(b) What is the inner angle of this polygon?
(c) Find the inner angle sum.
Answer:
(a) \(\frac {360}{15}\) = 24
(b) 180 – 24 = 156°
(c) 15 × 156 = 2340°

Question 10.
The diagram shows an equilateral triangle, a square, and one diagonal of the square. What is the value of?

Answer:
An equilateral triangle has one inner angle = 60°
Square has inner angle = 90°
A diagonal divides it into 45° each.
x = 60 + 45 = 105°

Question 11.
Squares are drawn on the sides of a regular hexagon. Find x.

Answer:
One inner angle of a regular hexagon is 120°.
Two squares in a corner make the angles 90°.
Angle around the corner is 360°.
x = 360° – (120° + 90° + 90°) = 60°

Question 12.
The measure of one outer angle of a regular polygon is 40°.
(a) What is the measure of one inner angle?
(b) Find the number of sides of this polygon?
Answer:
(a) 140°
(b) \(\frac {360}{40}\) = 9

Question 13.
The measure of an outer angle of a regular polygon is 2x, and the measure of an inner angle is 4x.
(a) Use the relationship between inner and outer angles to find x.
(b) Find the measure of one inner and outer angle.
(c) Find the number of sides in the polygon and the type of polygon.
Answer:
(a) The sum of the inner angles at a corner is 180°.
⇒ 6x = 180
⇒ x = 30
(b) Angles are 60°, 120°
(c) Regular hexagon

Question 14.
The measure of one interior angle of a regular polygon is 144°. How many sides does it have?
Answer:
One outer angle is 180 – 144 = 36°
Number of sides = \(\frac {360}{36}\) = 10

Question 15.
Five angles of a hexagon have measures 100°, 110°, 120°, 130°, 140°. What is the measure of the sixth angle?
Answer:
The inner angle sum of the hexagon is 720°.
Sixth angle = 720 – (100 + 110 + 120 + 130 + 140)
= 720 – 600
= 120°

Class 8 Maths Chapter 4 Notes Kerala Syllabus Polygons

→ Sum of the angles of a triangle is 180°. Polygons can be divided into triangles by drawing diagonals.

→ Inner angle sum of a polygon is the product of 180° and 2 less than the number of sides.

→ Using algebra for the number of sides n, then the inner angle sum is 180° × (n – 2).

→ When the number of sides increases by 1, the sum of the inner angles increases by 180°.

→ If the number of sides decreases by 1, then the angle sum decreases by 180°.

→ The sum of the inner angle and outer angle at the corner of a polygon is 180°.

→ Outer angle of a triangle when one side is extended is the sum of the inner angles of the other two corners.

→ Outer angle sum of a polygon is 360°.

→ One angle of a regular polygon of n sides is 180° × \(\frac{n-2}{n}\).

→ Outer angles are 360° × \(\frac {1}{n}\).

Polygons are closed figures with lines. The polygon having 3 sides is called a triangle. A polygon having 4 sides is a quadrilateral, a polygon with five sides pentagon.

Polygons are named based on the number of sides. The relation between sides and inner angle sum, sum of the inner angle and outer angle at a corner, sum of the outer angles, and angle around a point are discussed in this unit.

Names
A polygon is defined as a two-dimensional shape formed with straight lines. A polygon cannot have curves, and its lines should be fully connected. For example, triangles, rectangles, pentagons etc.

The polygons depend upon their sides. Triangles have three sides, quadrilaterals have four sides, and pentagons have five sides. There can be any number of sides for a polygon.

Regular polygon
If the polygon has sides of the same length and angles of the same size is called a regular polygon.

Angles
The sum of the three angles of a triangle is 180°.
The sum of all four angles of a rectangle is 360°.

In the case of a quadrilateral,

If we draw a diagonal, then two of its angles are split into two. The quadrilateral is also now split into two triangles.

If we add up all the angles of both the triangles, it amounts to adding all the angles of the quadrilateral.
Therefore, the sum of the angles of a quadrilateral = 2 × the Sum of the angles of a triangle
= 2 × 180°
= 360°
That means the sum of the angles of a quadrilateral is 360°.
The sum of the angles of a pentagon is 360° + 180° = 540°
The sum of the angles of a hexagon is 540° + 180° = 720°

The sum of the angles of a polygon is 180° multiplied by two less than the number of sides.

The sum of the angles of a polygon of ‘n’ sides is (n – 2) × 180°.

For example: Find the sum of the angles of a polygon of 20 sides.
Answer:
(20 – 2) × 180° = 18 × 180°
= (182 × 10)°
= 3240°

Worksheet – 1

Question 1.
What is the sum of the angles of a polygon with 102 sides?
Answer:
Sum of angles = (n – 2) × 180°
= (102 – 2) × 180°
= 100 × 180°
= 18000°

Question 2.
The sum of angles of a polygon is 10440°. Then
(a) The sum of angles of a polygon with one more side = 10440 + ________ = ________
(b) The sum of angles of a polygon with one side less = 10440 – ________ = ________
Answer:
(a) 10440 + 180 = 10620°
(b) 10440 – 180 = 10260°

Question 3.
The sum of angles of a polygon is 6300°. How many sides does it have?
Answer:
Sum of angles = (n – 2) × 180°
∴ (n – 2) × 180° = 6300
⇒ n – 2 = \(\frac {6300}{180}\) = 35
⇒ n = 35 + 2
⇒ n = 37

Question 4.
The number of triangles formed by drawing diagonals from one vertex of a polygon is 18. Then
(a) What is the sum of the angles of the polygon?
(b) What is the number of sides of the polygon?
Answer:
(a) 18 × 180° = 3240°
(b) Number of sides = 18 + 2 = 20

Outer Angles
An angle formed outside the triangle at a vertex by extending a side is called an outer angle of the triangle. The angle at this vertex of the triangle is called an interior angle.

The sum of the inner and outer angles of a triangle is 180°.
In any triangle, the sum of the inner and outer angles at each vertex is 180°.

In any triangle, since the sum of all three inner angles is 180°, the sum of two inner angles is equal to the third inner angle subtracted from 180°. And this equals the outer angle at the third vertex.

The same reason holds, even if we change the inner angles. That is, in any triangle, the outer angle at any vertex is equal to the sum of the inner angles at the other two vertices.

If we take one outer angle at each vertex,
The sum of the outer angles of a triangle is 360°.
The sum of the outer angles of a quadrilateral is 360°.
The sum of the outer angles of a polygon is 360°.

In general, for a polygon with n vertices,
The sum of all angles is n × 180°
Sum of inner angles is (n – 2) × 180°
The sum of the outer angles is 360°.

The sum of the outer angles at any two vertices of a quadrilateral is equal to the sum of the inner angles at the other two vertices.

Worksheet – 2

Question 1.
Compute all outer angles and inner angles of the triangle and the sum of outer angles.

Answer:
Remaining inner angle = 180° – (77° + 58°)
= 180° – 135°
= 45°
Outer angles are = 180° – 58°, 180° – 77°, 180° – 45° = 122°, 103°, 135°
Sum of outer angle is = 122° + 103° + 135° = 360°

Question 2.
Compute all outer angles and inner angles of the quadrilateral and the sum of outer angles.

Answer:
Remaining inner angle of quadrilateral is = 360° – (87° + 97° + 90°)
= 360° – 274°
= 86°
or
180° – 94° = 86°
Outer angles are 180° – 86°, 180° – 87°, 180° – 97°, 180° – 90° = 94°, 93°, 83°, 90°
Sum of outer angles = 93° + 83° + 90° + 94° = 360°

Question 3.
Find x from the figure.

Answer:
From the figure.
∠ACB = 180° – 130° = 50°
∠BAC = 180° – (41 + 50) = 89°
∴ x = 180° – 89° = 91°

Regular Polygons
Polygons with the lengths of sides equal and the sizes of angles equal are called regular polygons.
In a regular polygon of n sides
Each inner angle is \(\frac{n-2}{n}\) × 180°
Each outer angle is \(\frac {1}{n}\) × 360°

Worksheet – 3

Question 1.
In the figure, ABCD is a square. ‘O’ is the centre of the circle. Find the measure of ∠ACD, ∠AOD.

Answer:
∠ACD = ∠CAD
= \(\frac{\left(180^{\circ}-\angle \mathrm{ADC}\right)}{2}\)
= \(\frac{\left(180^{\circ}-90^{\circ}\right)}{2}\)
= \(\frac {90}{2}\)
= 45°
OA = OD
∴ ∠OAD = ∠ODA = 45°
∴ ∠AOD = 180° – (45° + 45°) = 90°

Question 2.
In the figure ABCDE is a regular pentagon ‘O’ is the centre of the circle.

Find
(i) ∠ABD
(ii) ∠ACD
(iii) ∠AOD
(iv) Is there any relation between these angles?
(v) Find ∠AED.
Answer:
(i) ABCDE is a regular pentagon.
So each angle will be \(\frac {540}{5}\) = 108°
also BC = DC means in ∠CBD = ∠CDB
= \(\frac{180^0-108^0}{2}\)
= \(\frac {72}{2}\)
= 36°
∴ ∠ABD = 108 – 36 = 72°

(ii) Considering ΔBAC, we will get
∠BAC = ∠BCA
= \(\frac{180^0-108^0}{2}\)
= \(\frac {72}{2}\)
= 36°
∴ ∠ACD = 108° – 36° = 72°

(iii) In a regular pentagon, if a circle passes through all the vertices, as from each vertex, if we join to the centre of the circle, all angles formed in the centre will be equal.

It will be \(\frac {360}{5}\) = 72°
So ∠AOD = 2 × 72° = 144°

(iv) Here ∠ABD = ∠ACD
also ∠AOD = 2 × ∠ABD = 2 × ∠ACD

(v) ∠AED = \(\frac {540}{5}\) = 108°

Question 3.
In the figure, ABCDEF is a regular hexagon. ‘O’ is the centre of the circle. Find
(i) ∠ABF
(ii) ∠ACF
(iii) ∠ADF
(iv) ∠AEF
(v) ∠AOF
(vi) Is there is any relation between these angles? Find ∠BDF and ∠BAF also.

Answer:
(i) ∠BAF = \(\frac {720}{6}\) = 120° because ABCDEF is regular hexagon.
AB = AF, so ∠ABF = ∠AFB
= \(\frac{180^{\circ}-120^{\circ}}{2}\)
= 30°
∴ ∠ABF = 30°

(ii) From the figure
∠BCF = \(\frac {120}{2}\) = 60°
∠BCA = \(\frac{\left(180^{\circ}-120^{\circ}\right)}{2}\) = 30°
∴ ∠ACF = 60° – 30° = 30°

(iii) As above ∠ADF = 30°

(iv) ∠AEF = \(\frac{\left(180^{\circ}-120^{\circ}\right)}{2}\) = 30

(v) As we know, in the case of a regular pentagon, here
∠AOF = \(\frac {360}{6}\) = 60°

(vi) ∠ABF = ∠ACF = ∠ADF
also ∠AOF = 2 × ∠ABF = 2 × ∠ACF = 2 × ∠ADF
∠BDF = 2 × 30° = 60°
∠BAF = \(\frac {720}{6}\) = 120°.