Selina Concise Mathematics Class 6 ICSE Solutions Chapter 28 Polygons

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 28 Polygons

Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 28 Polygons

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../ provides step by step solutions for Selina Concise ICSE Solutions for Class 6 Mathematics. You can download the Selina Concise Mathematics ICSE Solutions for Class 6 with Free PDF download option. Selina Publishers Concise Mathematics for Class 6 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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IMPORTANT POINTS

1. Polygon : A closed plane geometrical figure, bounded by atleast three line segments, is called a

polygon.

The adjoining figure is a polygon as it is :



(i) Closed

(ii) bounded by five line segments AB, BC, CD, DE and AE.

Also, it is clear from the given polygon that:

(i) the line segments AB, BC, CD, DE and AE intersect at their end points.

(ii) two line segments, with a common vertex, are not collinear i.e. the angle at any vertex is not 180°.

A polygon is named according to the number of sides (line-segments) in it:

2. Sum of Interior Angles of a Polygon

1. Triangle : Students already know that the sum of interior angles of a triangle is always 180°.

i.e. for ∆ ABC, ∠B AC + ∠ABC + ∠ACB = 180°

⇒ ZA + ZB + ZC = 180°

2. Quadrilateral : Consider a quadrilateral ABCD as shown alongside.

If diagonal AC of the quadrilaterals drawn, the quadrilateral will be divided into two triangles ABC and ADC.

Since, the sum of interior angles of a triangle is 180°.



∴ In ∆ ABC, ∠ABC + ∠BAC +∠ACB = 180°

And, in ∆ ADC ∠DAC + ∠ADC + ∠ACD = 180°

Adding we get:

∠ABC + ∠BAC +∠ACB + ∠DAC + ∠ADC + ∠ACD = 180° +180°

⇒(∠BAC + ∠DAC) + ∠ABC + (∠ACB + ∠ACD) + ∠ADC = 360°

⇒∠BAD + ∠ABC + ∠BCD + ∠ADC = 360°

⇒ ∠A + ∠B + ∠C + ∠D = 360°

Alternative method : On drawing the diagonal AC, the given quadrilateral is divided into two triangles. And, we know the sum of the interior angles of a triangle is 180°.

∴ Sum of interior angles of the quadrilateral ABCD

= Sum of interior angles of ∆ ABC + sum of interior angles of ∆ ADC = 180°+ 180° = 360°

3. Pentagon : Consider a pentagon ABCDE as shown alongside.

On joining CA and CE, the given pentagon is divided into three triangles ABC, CDE and ACE.



Since, the sum of the interior angles of a triangle is 180°

Sum of the interior angles of the pentagon ABCDE = Sum of interior angles of (∆ ABC + ∆ CDE + ∆ACE)

= 180° + 180° + 180° = 540°

4. Hexagon :

It is clear from the given figure that the sum of the interior angles of the hexagon ABCDEF.



= Sum of inteior angles of

(∆ABC + ∆ ACF + ∆ FCE + ∆ ECD)

= 180° + 180° + 180° + 180° = 720°

3. Using Formula : The sum of interior angles of a polygon can also be obtained by using the following formula:

Note : Sum of interior angles of a polygon = (n – 2) x 180°

where, n = number of sides of the polygon.

Polygons Exercise 28A – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.

State, which of the following are polygons :



Solution:

Only figure (ii) and (iii) are polygons.

Question 2.

Find the sum of interior angles of a polygon with :

(i) 9 sides

(ii) 13 sides

(iii) 16 sides

Solution:

(i) 9 sides

No. of sides n = 9

∴Sum of interior angles of polygon = (2n – 4) x 90°

= (2 x 9 – 4) x 90°

= 14 x 90°= 1260°

(ii) 13 sides

No. of sides n = 13

∴ Sum of interior angles of polygon = (2n – 4) x 90° = (2 x 13 – 4) x 90° = 1980°

(iii) 16 sides

No. of sides n = 16

∴ Sum of interior angles of polygon = (2n – 4) x 90°

= (2 x 16 – 4) x 90°

= (32 – 4) x 90° = 28 x 90°

= 2520°

Question 3.

Find the number of sides of a polygon, if the sum of its interior angles is :

(i) 1440°

(ii) 1620°

Solution:

Question 4.

Is it possible to have a polygon, whose sum of interior angles is 1030°.

Solution:

Question 5.

(i) If all the angles of a hexagon arc equal, find the measure of each angle.

(ii) If all the angles of an octagon are equal, find the measure of each angle,

Solution:

Question 6.

One angle of a quadrilateral is 90° and all other angles are equal ; find each equal angle.

Solution:

Question 7.

If angles of quadrilateral are in the ratio 4 : 5 : 3 : 6 ; find each angle of the quadrilateral.

Solution:

Question 8.

If one angle of a pentagon is 120° and each of the remaining four angles is x°, find the magnitude of x.

Solution:

One angle of a pentagon = 120°

Let remaining four angles be x, x, x and x

Their sum = 4x + 120°

But sum of all the interior angles of a pentagon = (2n – 4) x 90°

= (2 x 5 – 4) x 90° = 540°

= 3 x 180° = 540°

∴ 4x+120o° = 540°

4x = 540° – 120°

4x = 420

x = (frac { 420 }{ 4 }) ⇒ x = 105°

∴Equal angles are 105° (Each)

Question 9.

The angles of a pentagon are in the ratio 5 : 4 : 5 : 7 : 6 ; find each angle of the pentagon.

Solution:

Question 10.

Two angles of a hexagon are 90° and 110°. If the remaining four angles arc equal, find each equal angle.

Solution:

Polygons Exercise 28B – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.

Fill in the blanks :

In case of regular polygon, with



Solution:

Question 2.

Find the number of sides in a regular polygon, if its each interior angle is :

(i) 160°

(ii) 150°

Solution:

Question 3.

Find number of sides in a regular polygon, if its each exterior angle is :

(i) 30°

(ii) 36°

Solution:

Question 4.

Is it possible to have a regular polygon whose each interior angle is :

(i) 135°

(ii) 155°

Solution:

Question 5.

Is it possible to have a regular polygon whose each exterior angle is :

(i) 100°

(ii) 36°

Solution:

Question 6.

The ratio between the interior angle and the exterior angle of a regular polygon is 2 : 1. Find :

(i) each exterior angle of this polygon.

(ii) number of sides in the polygon.

Solution: