RBSE Maths Class 8 Chapter 6: Important Questions and Solutions

RBSE Maths Chapter 6 – Polygons Class 8 Important questions and solutions can be accessed here. The important questions and solutions of Chapter 6, available at BYJU’S, contain detailed step by step explanations. All the questions solved on this page are based on the new pattern prescribed by the RBSE. Students can also get the syllabus, textbooks and additional questions on RBSE Class 8 solutions.

Chapter 6 of the RBSE Class 8 Maths will help the students to solve problems related to diagonal of a polygon, concave and convex polygon, regular and irregular polygons, the sum of the measure of exterior angles of a polygon, characteristic of a quadrilateral, different types of quadrilaterals, special cases of a parallelogram.

RBSE Maths Chapter 6: Exercise 6.1 Textbook Important Questions and Solutions

Question 1: Draw the diagonal for the following figures and answer the questions given below.

RBSE Class 8 Maths Solutions Chapter 6 Exercise 6.1 Question Number 1
[vii] In which of the above shapes, the diagonal is inside the polygon?

[viii] In which of the above shapes, the diagonal is outside the polygon?

[ix] Check whether the given polygons are concave or convex.

Solution:

RBSE Class 8 Maths Solutions Chapter 6 Exercise 6.1 Question Number 1: Answers
[i] In the figure (i), the diagonals are AC, BE, AD, BD, CE.

[ii] In the figure (ii), the diagonals are AC and BD.

[iii] In the figure (iii), the diagonals are AC and BD.

[iv] In the figure (iv), the diagonals are AC and BD.

[v] In the figure (v), the diagonals are AC, AE, AD, AF, AG, BE, BD, BF, BG, BH, CE, CF, CG, CH, DF, DG, DH, EG, EH, FH.

[vi] In the figure (vi), the diagonals are AC, AD, BD, BE, CE.

[vii] In the figures (iii), (iv), (v) and (vi), the diagonals are inside the polygon.

[viii] In the figures, (i) and (ii), the diagonals are outside the polygon.

[ix] Concave figures: (i), (ii)

Convex figures: (iii), (iv), (v), (vi).

Question 2: Write the

[i] interior angles and

[ii] exterior angles of the given polygon ABCDE.

RBSE Class 8 Maths Solutions Chapter 6 Exercise 6.1 Question Number 2
Solution:

[i] The interior angles are ∠a, ∠b, ∠c, ∠d, ∠e.

[ii] The exterior angles are ∠p, ∠q, ∠r, ∠s, ∠t.

Question 3: Define a regular polygon. Write the names of the regular polygons with

[i] 5 sides

[ii] 6 sides

[iii] 8 sides

Solution:

A polygon is said to be a regular polygon if all the angles and sides are equal [otherwise it is an irregular polygon].

[i] A polygon with 5 sides is a pentagon.

[ii] A polygon with 6 sides is a hexagon.

[iii] A polygon with 8 sides is an octagon.

Question 4: Solve for unknown angles in the figures given below.

RBSE Class 8 Maths Solutions Chapter 6 Exercise 6.1 Question Number 4
Solution:

[i] Sum of the interior angles of a quadrilateral = 360o.

The given angles are

xo + 50o + 120o + 130o = 360o

xo + 300o = 360o

xo = 360o – 300o

xo = 60o

[ii] Sum of the interior angles of a quadrilateral = 360o.

The given angles are

xo + 70o + 60o + 90o = 360o

xo + 220o = 360o

xo = 360o – 220o

xo = 140o

[iii] The given figure is a pentagon. Hence,

The sum of the interior angles of a pentagon = 180o * [5 – 2] = 180o x 3 = 540o

The given angles are

2xo + 30o + [180o – 70o] + [180o – 60o] = 540o

2xo + 30o + 110o + 120o = 540o

2xo + 260o = 540o

2xo = 540o – 260o

xo = [280o] / 2

xo = 140o

[iv] The given figure is a triangle. Hence,

The sum of the angles of a triangle = 180o

30o + 90o + [180o – y] = 180o

120o + 180o – y = 180o

300o – y = 180o

y = 120o

x + 90o = 180o

x = 90o

z + 30o = 180o

z = 150o

[v] x + 120o = 180o

x = 180o – 120o

x = 60o

y + 80o = 180o

y = 180o – 80o

y = 100o

z + 60o = 180o

z = 120o

The sum of the interior angles of a quadrilateral = 360o

180o – w + 120o + 60o + 80o = 360o

440o – w = 360o

w = 440o – 360o

w = 80o

Question 5: What is the number of sides of a regular polygon whose measure of each exterior angle is 45o?

Solution:

Let the number of sides be x.

The sum of all exterior angles of a ‘x’ sided regular polygon = 360o

Each exterior angle = [360o] / n

45o = [360o] / n

n = [360o] / [45o]

n = 8

Hence, the number of sides is 8.

Question 6: What is the number of sides of a regular polygon whose measure of each interior angle is 165o?

Solution:

Let the number of sides be x.

Each interior angle = 165o

So, each exterior angle = 180o – 165o = 15o

The sum of all exterior angles of a ‘x’ sided regular polygon = 360o

Each exterior angle = [360o] / n

15o = [360o] / n

n = [360o] / [15o]

n = 24

Hence, the number of sides is 24.

Question 7: What is the number of sides of a regular polygon whose measure of each exterior angle is 24o?

Solution:

Let the number of sides be x.

The sum of all exterior angles of a ‘x’ sided regular polygon = 360o

Each exterior angle = [360o] / n

24o = [360o] / n

n = [360o] / [24o]

n = 15

Hence, the number of sides is 15.

Question 8: Find the value of each interior angle of a regular polygon of 10 sides.

Solution:

n = 10 [number of sides]

Each interior angle of ‘n’ sided regular polygon

= {[n – 2] * 180o} / [n]

= {[10 – 2] * 180o} / [10]

= {[8] * 180o} / [10]

= 1440 / 10

= 144o

The value of each interior angle of a regular polygon of 10 sides is 144o.

Question 9: If the interior angle of any polygon is 115o, then will it be a regular polygon?

Solution:

Each interior angle of ‘n’ sided regular polygon = {[n – 2] * 180o} / [n]

115o = {[n – 2] * 180o} / [n]

115o x n = {[n – 2] * 180o}

115o x n = 180o (n – 2)

115o x n = 180o x n – 360o

115o x n = 180o x n – 360o

360o = (180o – 115o) n

360o / 65o = n

n = 5.54

It is not a whole number.

Hence, it will not be a regular polygon.

Question 10: One interior angle of a hexagon is 165o and the measure of the remaining interior angle is xo. Find the measure of all the angles.

Solution:

The sum of interior angles of a hexagon = (n – 2) * 180o = (6 – 2) * 180o = 4 * 180o = 720o

165o + xo + xo + xo + xo + xo = 720o

165o + 5xo = 720o

5xo = 720o – 165o

5xo = 555o

xo = 555o / 5

xo = 111o

Question 11: By increasing the sides of a triangle in a single direction, obtained exterior angles are 110o, 115o and xo. Find the value of x.

Solution:

The sum of all exterior angles of a regular polygon = 360o

110o + 115o + xo = 360o

225o + xo = 360o

xo = 360o – 225o

x = 135o

Question 12: Find the sum of all interior angles of a regular heptagon.

Solution:

The sum of interior angles of a heptagon

= (n – 2) * 180o

= (7 – 2) * 180o

= 5 * 180o

= 900o

RBSE Maths Chapter 6: Exercise 6.2 Textbook Important Questions and Solutions

Question 1: Fill in the blanks choosing the right option.

(i) Adjacent angles of a parallelogram are ……… (equal/supplement).

(ii) Diagonals of a rectangle are ……… (equal/perpendicular bisector).

(iii) In any trapezium AB || CD, If A = 100° then the value of D will be ……….. (1000/80o).

(iv) If in any quadrilateral, diagonals bisect each other on a right angle then, it is called ……… (parallelogram/rhombus).

(v) All squares are…….. (congruent/similar).

Solution:

(i) supplementary (ii) equal (iii) 80° (iv) rhombus (v) similar

Question 2: In the figure below, BEST is a parallelogram. Find the values of x, y and z.

RBSE Class 8 Maths Solutions Chapter 6 Exercise 6.2 Question Number 2
Solution:

BEST is a parallelogram.

∠B + xo = 180o

In a parallelogram, adjacent angles are supplementary.

110o + xo = 180o

xo = 180o – 110o

xo = 70o

In a parallelogram, opposite angles are equal.

xo = yo

yo = 70o

110o = zo

z = 110o

Question 3: Find the values of x, y and z in the following parallelograms.

RBSE Class 8 Maths Solutions Chapter 6 Exercise 6.2 Question Number 3
Solution:

[i]

RBSE Class 8 Maths Solutions Chapter 6 Exercise 6.2 Question Number 3 : Answer
xo = 90o

In △OAD, OA = AD

∠OAD = ∠ODA = yo

∠AOD + ∠OAD + ∠ODA = 180o [Because sum of the angles of a triangle is 180o.]

90o + yo + yo = 180o

2yo = 90o

yo = 45o

Since the diagonals of a parallelogram bisect each other,

OA = AC

OA = AD

OC = OD

∠DOC = 90o

zo = 45o

[ii] Since, the opposite angles of a parallelogram are equal, yo = 112o

40° + zo + y° = 180° [ As the adjacent angles of a parallelogram are supplementary]

40° + zo + 112° = 180°

zo = 180° – 152o

z = 28o

xo + yo + 40° = 180o [As the sum of angles of a triangle is 180o.]

xo + 112° + 40o = 180°

xo + 152o = 180°

xo = 180o – 152°

xo = 28o

Question 4: In any parallelogram, the ratio of two adjacent angles is 1:5. Find the value of all the angles of a parallelogram.

Solution:

According to the question, let ABCD be a parallelogram.

∠A : ∠B = 1:5

Since the sum of two adjacent angles of a parallelogram is 180°

∠A + ∠B = 180o

Sum of Ratio = 1 + 5 = 6

Hence, ∠A = [⅙] * 180o = 30°

∠B = [⅚] * 180o = 150o

Since, the opposite angles of a parallelogram are equal,

∠C = ∠A = 30o and ∠D = ∠B= 150°

Angles of a parallelogram are 30°,150°, 30° and 150°.

Question 5: Given below are two parallelograms RISK and STEW. Find the values of x and y [length in cm].

RBSE Class 8 Maths Solutions Chapter 6 Exercise 6.2 Question Number 5
Solution:

(i) The opposite sides of a parallelogram are equal.

3x = 18 and 3y – 1 = 26

x = 6 and 3y = 26 + 1

3y = 27

y = 9

So. x = 6 cm and y = 9 cm.

(ii) The diagonals of a parallelogram bisect each other

x + y = 16 …………(1) and y + 7 = 20 ……… (2)

From equation (2), y = 20 – 7 = 13

Puting y = 13 in equation (1) x + 13 = 16

So. x = 3 and y = 13 cm.

Question 6: The rectangle HOPE is given below. Its diagonals intersect each other at the point S. Find the value of x if SH = 2x + 4 and SE = 3x + 1.

RBSE Class 8 Maths Solutions Chapter 6 Exercise 6.2 Question Number 6
Solution:

We know that the diagonals of a rectangle bisect each other.

SE = SO and SH = SP

EO = 2SE = 2 (3x + 1) and HP = 2 SH = 2 (2x + 4)

Since the diagonals of a rectangle are equal,

EO = HP

2 (3x + 1) = 2 (2x + 4)

3x + 1 = 2x + 4

3x – 2x = 4 -1

x = 3

Question 7: Given PEAR is a rhombus. Find the values of x, y and z. Also, write the causes.

RBSE Class 8 Maths Solutions Chapter 6 Exercise 6.2 Question Number 7
Solution:

In a rhombus, the diagonals bisect each other.

OA = OP

Given that OP = 5 cm, OE = OR and OR = 12cm.

x = 5 cm

y = 12 cm

As the sides of a rhombus are equal, EP = PR

z = 13 cm.

Question 8: In the given trapezium PQRS, PQ || SR. Find the angles x and y.

 

RBSE Class 8 Maths Solutions Chapter 6 Exercise 6.2 Question Number 8
Solution:

Given that PQ || SR

x + 90o = 180o

x = 90o

Therefore PQRS is a quadrilateral.

x + y + 130o + 90o = 360o

The sum of all angles of a quadrilateral = 3600

90o + y + 130°+ 90o = 360o

310o + y =360o

y = 360o – 310

y = 50o

RBSE Maths Chapter 6: Additional Questions and Solutions

Question 1: A concrete slab was made by a mason and he wants it to be rectangular in shape. In what different ways can it be made a rectangular?

Solution:

The concrete slab will be rectangular in shape if the following is ensured:

• The opposite sides should be equal.

• The diagonals should be equal.

• Each angle should measure 90°.

Question 2: A square is defined as a rectangle with equal sides. Is it possible that a rhombus of equal angles be defined? Explain.

Solution:

It is not possible to define a square as a rhombus with equal angles as the diagonals will not be equal unless every angle is a right angle [90o].

Question 3: Is it possible that a trapezium will have all sides and angles equal? Explain.

Solution:

In the case of a trapezium, it is not possible to have all angles equal as its opposite sides become parallel. A trapezium is a quadrilateral in which a pair of sides are parallel. It cannot have all sides equal as its opposite sides become parallel.

Question 4: Find the measure of each of the angles of a parallelogram in which two adjacent angles of a parallelogram are equal.

Solution:

Let the measures of the 2 adjacent angles be P and Q.

The adjacent angles of a parallelogram are supplementary.

∠P + ∠Q = 180o

x + x =180o

2x = 180o

x = 90o

Therefore, ∠P = ∠Q = 90o

Since, the opposite angles of a parallelogram are equal, ∠P = ∠Q = ∠R = ∠S = 90o.

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