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Kerala Syllabus Class 8 Mathematics - Unit 4 Polygons - Questions and Answers | Teacher Text 

Questions and Answers for Class 8 Mathematics - Unit 4 Polygons - Study Notes | Text Books Solution STD 8 - Maths: Unit 4 Polygons - Questions and Answers | ഗണിതം - ബഹുഭുജങ്ങൾ - ചോദ്യോത്തരങ്ങൾ 

എട്ടാം ക്ലാസ്സ്‌  Mathematics - Unit 4 Polygons എന്ന പാഠം ആസ്പദമാക്കി തയ്യാറാക്കിയ ചോദ്യോത്തരങ്ങള്‍. ഈ അധ്യായത്തിന്റെ Teachers Handbook ഡൗൺലോഡ് ചെയ്യാനുള്ള ലിങ്ക് ചോദ്യോത്തരങ്ങളുടെ അവസാനം നൽകിയിട്ടുണ്ട്. പുതിയ അപ്‌ഡേറ്റുകൾക്കായി ഞങ്ങളുടെ Telegram Channel ൽ ജോയിൻ ചെയ്യുക.

Kerala Syllabus STD 8 Maths: Unit 4 Polygons - Textbook Solutions

♦ Textbook Activities (Textbook Page No: 63, 64, 65)
Angles

♦ Now can you guess the sum of the angles of a hexagon?
Maybe a picture will help:
We can think of a hexagon as the figure formed by joining a pentagon and a triangle.
The sum of the angles is 540° + 180° = 720°
So what can we say in general?
• Taking the polygons in order as triangle, quadrilateral, pentagon and so on, when we move from one polygon to the next, the number of vertices and the number of sides increase by 1
• This change can be seen as adjoining a triangle
•  The sum of the angles increases by 180°
So, we can write the sum of the angles in order as shown below:
Looking at this pattern, can you say the sum of the angles of a polygon with 10 sides?
We think like this:

(i) Taking triangle as the first polygon, quadrilateral as the second, pentagon as the third and so on, what is the position of the ten-sided polygon (decagon)?
If we take the shapes in order first triangle, second Quadrilateral, third pentagon and so on the polygon with sides (decagon) would be in the 8th position.

(ii)  So, continuing the table above, how many times is 180° the sum of the angles of a decagon?
The sum of the measures of the angles of a decagon is 8 times 180°.
The sum of the angles of a 10-sided polygon is 8 × 180° = 1440°.

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°

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?
When the number of sides is increased by 1, we can draw one more triangle and the sum of the angles will be increased by 180°.
Sum of the angles
1980+ 180 = 2160°
When the number of sides is decreased by 1, the number of triangles also is decreased by 1. So the sum of the angles is reduced by 180°. Sum of the angles now = 1980-180 = 1800°

2. What is the sum of the angles of a 27-sided polygon?
Sum of angles = (n − 2) x 180°
                       = (27 − 2) x 180°
                       = 25 x 180° = 4500°

3. The sum of the angles of a polygon is 8100°. How many sides does it have?
(n-2) x 180° = 8100°, n − 2 = 8100 ÷ 180 = 45 n = 45 + 2 = 47
Number of sides = 47

4. Is the sum of the angles of any polygon equal to 1000°? Explain.
The sum of the angles of any polygon is a multiple of180°.
1000/180 = 5.55
Here, 1000 is not a multiple of 180, so the sum of the angles of a polygon cannot be 1000.

5. A 20-sided polygon has equal angles. How much is each angle?
Sum of angles = (n − 2) x 180°
                       = (20 − 2) x 180°
                       = 18 x 180°= 3240°
Sum all the angles that have the same measure; the measure of one angle = 3240 ÷ 20 = 162°
♦ Project (Textbook Page No: 71)
♦ 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 inner and outer angles of a pentagon?
(ii)  And in a hexagon?
(iii) Is there a general relation which is true for all polygons? 
Answer:
(i) In a pentagon, if the inner angles are a°, b°, c°, d°, e°, then the outer angles would be 180° − a°, 180° − b°, 180° − c°, 180° − d°, 180° − e°.
If we take the outer angles at three vertices as 180° − c°, 180° − d°, 180° − e°, then their sum would be 180° − c° + 180° − d° + 180° − e° = 540° − (c° + d° + e°) 
The sum of all inner angles in a pentagon is 540°.  
    a° + b° + c° + d° + e°= 540° 
    a° + b° = 540° − (c° + d° + e°)
    a° + b° = (180° − c°) + (180° − d°) + (180° − e°) 
That is, in any pentagon, the sum of the inner angles at two vertices is equal to the sum of the outer angles at the other three vertices.

(ii) In a hexagon, if the inner angles are a°, b°, c°, d°, e°, f°, then the outer angles would be 180° − a°, 180° − b°, 180° − c°, 180° − d°, 180° − e°, 180° − f°. 
The sum of all inner angles in a hexagon is 720°. 
      a° + b° + c° + d° + e° + f° = 720° 
      a° + b° = 720° − (c° + d° + e° + f°) 
      a° + b° = (180° − c°) + (180° − d°) + (180° − e°) + (180° − f°)
That is, in any hexagon the sum of the inner angles at two vertices is equal to the sum of the outer angles at the other four vertices.

(iii) In any polygon with n sides, the sum of the inner angles at two vertices is equal to the sum of the outer angles at the other (n − 2) vertices.
♦ Textbook Activities (Textbook Page No: 72)
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) Number of sides = 18
In a polygon with 18 sides, the sum of outer angles = 360°.
Since all inner angles of the polygon are equal, the outer angles are also equal.
Measure of each outer angle: 360/18 = 20° 

(ii) Measure of each inner angle = 180° − 20° = 160°.

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) The sum of outer angles in any polygon is 360°. 
The sum of inner angles in a polygon with n sides is 180° × (n − 2).
Given that the sum of inner angles in the polygon is equal to the sum of the outer angles.
So, 180 (n − 2) = 360°
                n − 2 = 360°/180, n − 2 = 2, n = 4
Since the number of sides is 4, the polygon obtained is a quadrilateral.

(ii) Here, since the sum of outer angles is twice the sum of inner angles
      360 = 2 x 180 (n − 2)
      360 = 360 (n − 2), n − 2 = 1, n = 3
Since the number of sides is 3, the polygon obtained is a triangle.

(iii) Since the sum of outer angles is half of the sum of inner angles,
      360 = ½ × 180 (n − 2)
      360 = 90 (n − 2), n − 2 = 4, n = 6
Since the number of sides is 6, the polygon obtained is a hexagon.

(iv) Here, since the sum of outer angles is one third of the sum of inner angles,
         360 = ⅓ × 180 (n − 2)
         360 = 60 (n − 2), n − 2 = 6, n = 8
Since the number of sides is 8, the polygon obtained is an octagon.

♦ Textbook Activities (Textbook Page No: 77, 78)
(1) (i) Draw a hexagon of equal sides with angles different.
(ii) Draw a hexagon of equal angles with different sides.
Answer:
Draw AB and ED of equal length.
Draw two arcs of radius equal to AB with centres B and D to meet at C. Draw two other arcs with the same radius, fixing centres at A and E to meet at F. Join BC, CD, EF and AF to complete the hexagon. ABCDEF is the required hexagon. 

(ii) Draw AB using any measure. Measure ∠A = 120° and mark F at any point on the line. 
Draw an angle equal to 120° at B and mark C. Draw a line parallel to AF through C and a line parallel to BC through F. On the line drawn from C, mark D at a suitable length. At D, measure 120°. This line meets the parallel line drawn from F at E. Join DE. ABCDEF is the required hexagon.

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.
In the regular hexagon, all sides are of the same length and all angles are of the same size. Here, we can draw a circle and divide it into 6 angles of 60° (360 × ⅙) each at the centre. 
That is, we can divide the hexagon into 6 equilateral triangles. Each equilateral triangle has all angles equal to 60°. Here, two sides of each equilateral triangle are the radius of the circle. Since all three sides of the triangle are equal, the third side is also equal to the radius of the circle.

3. Draw a regular octagon (eight sides) of sides 3 centimetres.
First draw a line AB of length 3 cm. Measure a 135° angle at A. Mark point 4 on this line at a distance of 3 cm from A. Measure a 135° angle at B and mark point Č at a distance of 3 cm. Again measure 135° angles at vertices H and C and mark points G and D at a distance of 3 cm. Now, draw GF parallel to BC and also draw AH parallel to DE with a length of 3 cm. Join EF. ABCDEFGH is the regular octagon required in the question. 

4. Draw a circle of radius 3 centimetres and draw a regular octagon with all vertices on this circle.
Draw a circle of radius 3 cm. 
Then draw 8 angles of 360° x ⅛ = 45° each at the centre. All the sides of this regular octagon are the same length, and the measures of the angles are 135° each.
5. The picture below shows a regular hexagon and a triangle joining its alternate vertices:
Is this an equilateral triangle? Why?
Each angle of a regular hexagon measures = 720/6 = 120°
In ∆ABC, ∠B = 120°
Since AB = BC, the angles opposite these sides are equal.
∠BCA = ∠BAC = 180° − 120°/2 = 60°/2 = 30°
In ∠CDE, ∠D = 120°
∠DCE = ∠DEC = 180° − 120°/2 = 60°/2 = 30°
In ∠AFE, ∠F = 120°
∠FAE = ∠FEA = 180° − 120°/2 = 60°/2 = 30°
In ∆ACE
∠A = 120° − (30° + 30°) = 120° − 60° = 60°
∠C = 120° − (30° + 30°) = 120° − 60° = 60°
∠E = 120° − (30° + 30°) = 120° − 60° = 60°
All 3 cm angles of ∆ACE are equal. Therefore, the sides opposite to them are also equal.
∴ ∆ACE is an equilateral triangle. 

6. The picture below shows a regular hexagon and a quadrilateral joining four of its vertices:
Is this quadrilateral a rectangle? Why?
Each angle of a regular hexagon measures = 720°/6 = 120°. 
In ∆ABC, ∠B = 120°
Since AB = BC, the angles opposite these sides are equal
∠BAC = ∠BCA = 180° − 120°/2 = 60°/2 = 30°
In ∆DEF, ∠E = 120°
Since DE = EF, the angles opposite these sides are equal.
∠EDF = ∠EFD = 180° − 120°/2 = 60°/2 = 30°
In quadrilateral ACDF, ∠A = 120° − 30° = 90° 
∠C = 120° − 30° = 90° 
∠D = 120° − 30° = 90° 
∠F = 120° − 30° = 90° 
All angles in this quadrilateral are 90° each, so ACDF is a rectangle.

7. Calculate an inner angle and an outer angle of a regular polygon of 15 sides.
For a 15-sided polygon, the sum of inner angles = 180° x (15 − 2/15)
= 180° x ¹³⁄₁₅ = 12° x 13 = 156°
For a 15-sided polygon, the measure of each outer angle = 360° x ¹⁄₁₅ = 24°

8. An outer angle of a regular polygon is 20°. How many sides does it have?
The sum of the outer angles of any polygon is 360°. 
The measure of one outer angle = 20°
360°
∴ Number of sides = 360°/ 20° = 18

9. An inner angle of a regular polygon is 168°. How many sides does it have?
The measure of one inner angle = 168° 
The measure of one outer angle = 180° − 168° = 12°
Number of sides = 360°/12° = 30