ICSE Class 8 Maths Selina Solutions for Chapter 16 Understanding Shapes

Chapter 16 Understanding shapes explains about properties of quadrilaterals, angle sum property, properties of parallelograms, kinds of quadrilaterals and theorems related to them. Knowing this chapter is important because it has applications in everyday life for eg: thinking about home projects, and in various careers, like architecture. For the better understanding of the concepts related to shapes, we have provided ICSE Class 8 maths Selina Solutions Chapter 16 – Understanding Shapes. These solutions are prepared by our subject experts. The questions are solved in the easiest way so that students can understand it easily.

Students of Class 8 ICSE board can download the ICSE Class 8 maths Selina Solutions for Chapter 16 Understanding Shapes from the downloadable link provided below:

Download ICSE Class 8 Maths Selina Solutions Chapter 16 Understanding Shapes

Chapter 16 of ICSE Class 8 Maths Selina Solutions is based on the topic Understanding Shapes. This chapter consists of a total number of 15 questions in one exercise. All the solutions are solved in a proper step by step method for easy understanding of the students.

CHAPTER 16 – UNDERSTANDING SHAPES

Question 1. 

 State which of the following are polygons: 

Class 8 Maths Selina Solutions for Chapter 16 Understanding Shapes-1

If the given figure is a polygon, name it as convex or concave. 

 Solution: 

In given Fig. (ii), (iii) and (v) are polygons.

Fig. (ii) and (iii) are concave polygons while

Fig. (v) is convex.

Question 2. 

Calculate the sum of angles of a polygon with: 

 (i) 10 sides 

Solution:-

No. of sides n=10

Sum of angles of polygon \(=(n-2) \times 180° \) \(=(10-2) \times 180°=1440°\)

(ii) 12 sides 

Solution:-

No. of sides n=12

Sum of angles \(=(n-2) \times 180°\) \(=(12-2) \times 180°=10 \times 180°=1800°\)

(iii) 20 sides 

Solution:-

  n =20

Sum of angles of Polygon \(=(n-2) \times 180° \) \(=(20-2) \times 180°=3240°\)

(iv) 25 sides 

Solution:-

(iv) n = 25

Sum of angles of polygon \(=(n-2) \times 180° \) \(=(25-2) \times 180°=4140°\)

Question 3. 

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

 (i) 900°

Solution:-

Let no. of sides = n

Sum of angles of polygon= 900°

\( (n-2) \times 180°=900\) \(n-2=\frac{900}{180}\)

n-2=5

n=5+2

n=7

(ii) 1620°

Solution:-

Let no. of sides = n

Sum of angles of polygon = 1620°

\( (n-2) \times 180°=1620°\) \(n-2=\frac{1620}{180}\) \(n-2=\frac{1620}{180}\)

n-2=9

n=9+2

n=11

(iii) 16 right-angles 

Solution:-

Let no. of sides =n

Sum of angles of polygon \(=16 \text { right angles }=16 \times 90=1440° \) \( (\mathrm{n}-2) \times 180°=1440°\) \(\mathrm{n}-2=\frac{1440}{180}\)

n-2=8

n=8+2

n=10

(iv) 32 right-angles. 

Solution:-

Let no. of sides =n

Sum of angles of polygon =32 right angles \(=32 \times 90=2880°  \) \( (n-2) \times 180°=2880\) \(n-2=\frac{2880}{180}\)

n-2=16

n=16+2

n=18

Question 4. 

 Is it possible to have a polygon; whose sum of interior angles is? 

 (i) 870°

Solution:-

 (i) Let no. of sides = n

Sum of angles =870°

\( (\mathrm{n}-2) \times 180°=870°\) \(n-2=\frac{870}{180}\) \(\mathrm{n}-2=\frac{20}{6}\) \(n=\frac{29}{6}+2\) \(\mathrm{n}=\frac{41}{6}\)

Which is not a whole number.

Hence it is not possible to have a polygon, the sum of whose interior angles is 870°

(ii) 2340°

Solution:

Let no. of sides =n

Sum of angles =2340°

\( (n-2) \times 180°=2340°\) \(n-2=\frac{2340}{180}\)

n-2=13

n=13+2=15

Which is a whole number.

Hence it is possible to have a polygon, the sum of whose interior angles is 2340°.

(iii) 7 right-angles 

Solution:-

Let no. of sides =n

Sum of angles\(=7 right angles =7 \times 90=630° \) \( (n-2) \times 180°=630° \) \(\mathrm{n}-2=\frac{630}{180}\) \(\mathrm{n}-2=\frac{7}{2}\) \(\mathrm{n}=\frac{7}{2}+2\) \(\mathrm{n}=\frac{11}{2}\)

Which is not a whole number. Hence it is not possible to have a polygon, the sum of whose interior angles is 7 right-angles.

 (iv) 4500°

Solution:-

 Let no. of sides = n

\( (\mathrm{n}-2) \times 180°=4500°\) \(n-2=\frac{4500}{180}\)

n-2=25

n=25+2

n=27

Which is a whole number.

Hence it is possible to have a polygon, the sum of whose interior angles is 4500°.

Question 5.

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

Solution:-

No. of sides of hexagon, n=6

Let each angle be =x°

Sum of angles =6x°

\( (n-2) \times 180°\)= Sum of angles

\( (6-2) \times 180°=6 x°\) \(4 \times 180=6 x\) \(x=\frac{4 \times 180}{6}\)

x=120°

∴ Each angle of hexagon =120°

(ii) If all the angles of a 14 – sided figure are equal; find the measure of each angle.

Solution:-

No. of sides of polygon, n=14

Let each angle =x°

Sum of angles =14x°

\(∵(n-2) \times 180°\)=Sum of angles of polygon

\(∵(14-2) \times 180°=14 x\) \(12 \times 180°=14 x\) \(x=\frac{12 \times 180}{14}\) \(x=\frac{1080}{7}\) \(x=\left(154 \frac{2}{7}\right)°\)

Question 6. 

Find the sum of exterior angles obtained on producing, in order, the sides of a polygon with:

(i) 7 sides

(ii) 10 sides

(iii) 250 sides.

(i) Solution:

No. of sides n=7 

Sum of interior exterior angles at one vertex =180°

Sum of all interior exterior angles \(=7 \times 180°\)

=1260°

Sum of interior angles \(=(n-2) \times 180°\) \(=(7-2) \times 180°\) \(=(7-2) \times 180°\)

=900°

∴Sum of exterior angles =1260°-900°

=360°

(ii)Solution

No. of sides n=10

Sum of interior and exterior angles \(=10° \times 180° \)

=1800°

But sum of interior angles \(=(n-2) \times 180°\) \(=(10-2) \times 180°\)

=1440°

∴Sum of exterior angles  = 1800 – 1440

Sum of exterior angles = 360°

(iii)  Solution:

No. of side n=250

Sum of all interior and exterior angles

\(=250 \times 180°\)

=45000°

But sum of interior angles \(=(n-2) \times 180°\) \(=(250-2) \times 180°\) \(=248 \times 180°\)

=44640°

∴ Sum of exterior angles =45000-44640

=360°

Question 7 :

The sides of a hexagon are produced in order. If the measures of exterior angles so obtained are

(6x-1)° , (10 x+2)°,(8 x+2)°( 9 x-3)°,(5 x+4)° and  (12 x+6)°;Find each exterior angle.

Solution:-

Sum of exterior angles of hexagon formed by producing sides of order =360°

∴(6x-1)°+(10x+2)°+(8x+2)°+(9x-3)° +(5x+4)°+(12x+6)°=360°

50x+10°=360°

50x=360°-10°

50x=350°

\(x=\frac{350}{50}\)

x=7

∴ Angles are (6x-1)°:(10x+2)°:(8x+2)°;(9x-3)° (5x+4)° and (12x+6)°

i.e., (6×7-1)° ; (10×7+2)° ; (8×7+2)° ;  (9×7-3)° ; (5×7+4)°; (12×7+6)°)

41° ; 72° ; 58° ; 60° ; 39°  and  90°

 Question 8. 

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

Solution:-

Let the interior angles of the pentagon be 4 x, 5 x, 6 x, 7 x, 5 x

Their sum = 4 x+5 x+6 x+7 x+5 x=27 x

Sum of interior angles of polygon \(=(n-2) \times 180°=(5-2) \times 180°=540°\) \(27 x=540 x=\frac{540}{27}=20°\)

∴ Angles are  \(4 \times 20°=80°\) \(5 \times 20°=100°\) \(6 \times 20°=120°\) \(7 \times 20°=140°\) \(5 \times 20°=100°\)

Question 9

Two angles of a hexagon are 120°and 160°. If the remaining four angles are equal, find each equal angle.

Solution:-

Two angles of a hexagon are 120°, 160°

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

Their sum = 4 x+280°

But sum of all the interior angles of a hexagon \(=(6-2) \times 180°=4 \times 180°=720°\)

∴ 4x+280°=720°

⇒ 4x=720°-280°=440° ⇒ x=110°

∴Equal angles are 110° (each)

Question 10

The figure, given below, shows a pentagon ABCDE with sides AB and ED parallel to each other, and

<B : <C : <D:5:6:7.

(i) Using formula, find the sum of interior angles of the pentagon.

(ii) Write the value of ∠A+∠E

(iii) Find angles B, C and D .

Solution:-

(i) Sum of interior angles of the pentagon \(=(5-2) \times 180°\) \(=3 \times 180°=540° (∵sum for a polygon of x sides =(x-2) \times 180°\)

(ii) Since AB || ED

∴∠A+∠E=180°

(iii) Let ∠B=5 x ∠C=6 x ∠D=7x

∴5x+6x+7x+180°=540°

∠A+ ∠E=180° Proved in (ii)

18x=540°-180°

⇒  18x=360°  ⇒  x=20°

\(∵ \angle B=5 \times 20°=100°, \angle C=6 \times 20=120°\) \(\angle D=7 \times 20=140°\)

Question 11.

Two angles of a polygon are right angles and the remaining are 120° each. Find the number of sides in it.

Solution:-

 Let number of sides = n

Sum of interior angles \(=(n-2) \times 180° \)

=180n-360°

Sum of 2 right angles \(=2 \times 90°\)

=180°

∴   Sum of other angles =180n-360°-180°

=180 n-540

No. of vertices at which these angles are formed

=n-2

∴   Each interior angle \(=\frac{180 n-540}{n-2}\) \(∵ \quad \frac{180 n-540}{n-2}=120°\)

180 n-540=120 n-240

180 n-120 n=-240+540

60 n=300

\(n=\frac{300}{60}\)

n=5

Question 12.

In a hexagon ABCDEF, side AB is parallel to side FE and ∠B:∠C:∠D:∠E=6:4:2:3.find ∠B and ∠D.

Solution:-

Class 8 Maths Selina Solutions for Chapter 16 Understanding Shapes-2

Given: Hexagon ABCDEF in which AB II EF

and ∠B:∠C:∠D:∠E=6:4:2:3

To find : ∠B and ∠D

Proof: No of sides n=6

∴ Sum of interior angles \(=(n-2) \times 180°\) \(=(6-2) \times 180°=720°\)

∵AB‖EF (Given)

\(∵ \mathrm{A}+\angle \mathrm{F}=180°\)

But \(\angle A+\angle B+\angle C+\angle D+\angle E+\angle F=720°\)

(proved)

\(\angle B+\angle C+\angle D+\angle E+\angle 180°=720°\) \(∵ \angle B+\angle C+\angle D+\angle E=720°-180°=540°\)

Ratio =6:4:2:3

Sum of parts =6+4+2+3=15

\(∵ \angle B=\frac{6}{15} \times 540=216°\) \(\angle \mathrm{D}=\frac{2}{15} \times 540°=72°\)

Hence \(\angle B=216° : \angle D=72° \)

 Question 13.

the angles of a hexagon are x+10°,2x+20°,2x-20°,3x-50°,x+40° and x+20°. Find x.

Solution:-

Angles of a hexagon are x+10°, 2x+20°,

2x-20°, 3x-50°,x+40° and x+20°

∴ But sum of angles of a hexagon \(=(x-2) \times 180°\) \(=(6-2) \times 180°=4 \times 180°=720°\)

But sum=x+10+2x+20°+2x-20°+3x-50°+x+40+x+20

=10x+90-70=10x+20

∴10x+20=720°⇒10x=720-20=700

\(\Rightarrow x=\frac{700°}{10}=70°\)

∴x=70°

 Question 14.

In a pentagon, two angles are 40° and 60° and the rest are in the ratio 1:3:7. Find the biggest angle of the pentagon.

Solution:-

 In a pentagon, two angles are 40° and 60° Sum of remaining 3 angles \(=3 \times 180°\)

=540°-40°-60°=540°-100°=440°

Ratio in these 3 angles =1:3:7

Sum of ratios =1+3+7=11

Biggest angle \(=\frac{440 \times 7}{11}=280°\)

Question 15

 Fill in the blanks: 

In case of regular polygon, with:

No. of sides Each exterior angle Each interior angle
(i)…..8…..

(ii)….12….

(iii)………..

(iv)……

(v)…….

(vi)……….

………..

………..

……72°…..

……45°….

………………

…………..

……………

……………

…………….

……………….

……150°…..

……140°…..

Solution:-

No. Of sides Each exterior angle Each interior angle
(i)8

(ii)12

(iii)5

(iv)8

(v)12

(vi)9

45°

30°

72°

45°

30°

40°

135°

150°

108°

135°

150°

140°

Explanation:

(i) Each exterior angle \(=\frac{360°}{8}=45°\)

Each interior angle=180°-45°-135°

(ii) Each exterior angle \(=\frac{360°}{12}=30°\)

Each interior angle =180°-30°=150°

(iii) Since each exterior =72°

∴ Number of sides \(=\frac{360°}{72°}=5\)

Also interior angle =180°-72°=108°

(iv) Since each exterior angle =45°

∴ Number of sides \(=\frac{360°}{45°}=8\)

Interior angle =180°-45°=135°

(v) Since interior angle =150°

∴ Exterior angle =180°-150°=30°

∴ Number of sides \(=\frac{360°}{30°}=12\)

(vi) Since interior angle =140°

∴ Exterior angle =180°-140°=40°

∴ Number of sides \(=\frac{360°}{40°}=9\)

ICSE Class 8 Maths Selina Solutions Chapter 16 – Understanding Shapes

The concepts discussed in Chapter 16 – Understanding Shapes will help students to solve the problems related to the topic by using different types of theorems explained in the chapter. The topic comes under the unit of Geometry and it is important from exam point of view. This topic helps in understanding the world around us through different patterns, areas, volumes, lengths and angles.

For all the other subjects like Physics, Chemistry and Biology students can get the answers by clicking on ICSE Class 8 Selina Solutions.

We hope that the information provided here for ICSE Class 8 Maths Selina Solutions Chapter 16 Understanding Shapes will help students while preparing for their Class 8 final exam.

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