RBSE Solutions for Class 8 Maths Chapter 6 Polygons Additional Questions

RBSE Solutions for Class 8 Maths Chapter 6 Polygons Additional Questions is part of RBSE Solutions for Class 8 Maths. Here we have given Rajasthan Board RBSE Class 8 Maths Chapter 6 Polygons Additional Questions.

Board RBSE
Textbook SIERT, Rajasthan
Class Class 8
Subject Maths
Chapter Chapter 6
Chapter Name Polygons
Exercise Additional Questions
Number of Questions 34
Category RBSE Solutions

Rajasthan Board RBSE Class 8 Maths Chapter 6 Polygons Additional Questions

I. Objective Type Questions

Question 1.
Which of the following closed curve has four sides
(a) diagonal
(b) quadrilaterals
(c) triangle
(d) circle

Question 2.
Which of the quadrilaterals has all angles has right angles, opposite side equal and diagonals bisect each other,
(a) rectangle
(b) rhombus
(c) square
(d) none of these

Question 3.
Which of the following quadrilaterals is a regular quadrilateral?
(a) rhombus
(b) square
(c) kite
(d) rectangle

RBSE Solutions

Question 4.
The angle of a square are
(a) 90°, 120°, 120°, 30°
(b) 90°, 90°, 90°, 90°
(c) 90°, 50°, 100°, 120°
(d) 90°, 80°, 70°, 60°

Question 5.
What are the measurements of(RBSESolutions.com)sides of given parallelogram?
RBSE Solutions for Class 8 Maths Chapter 6 Polygons Additional Questions img-1
(a) 9 cm and 2 cm
(b) 3 cm and 6 cm
(c) 4 cm and 5 cm
(d) 7 cm and 7.7 cm

RBSE Solutions

Question 6.
The sum of the measures of the external angles of any polygon is
(a) 90°
(b) 180°
(c) 240°
(d) 360°

Question 7.
A quadrilateral whose diagonals bisect each other at right angles is ….
(a) Rectangle
(b) Square
(c) Parallelogram
(d) Trapezium

Question 8.
Sum of all exterior angles(RBSESolutions.com)of n sided regular polygon is ……
(a) 360°
(b) 720°
(c) 540°
(d) 900°

RBSE Solutions

Question 9.
The formula for sum of interior angles of n-sided polygon is ……..
(a) (n – 4) x 180°
(b) (n – 2) x 90°
(c) (n – 3) x 180°
(d) (n – 2) x 180°

Question 10.
Sum of all the exterior angles(RBSESolutions.com)formed by increasing the sides of a convex polygon in same order is
(a) 360°
(b) 270°
(c) 180°
(d) 90°

Answers
1. (b)
2. (a)
3. (b)
4. (b)
5. (c)
6. (d)
7. (b)
8. (a)
9. (d)
10. (a).

RBSE Solutions

II. Fill in the blanks

Fill in the blanks choosing the right option.

Question 1.
A quadrilateral which has two pairs of equal adjacent sides is called ……(parallelogram/kite)

Question 2.
Those polygons whose all(RBSESolutions.com)diagonals are in the interior, are called ……. (convex/concave)

Question 3.
Sum of all exterior angles of a polygon is ……….. (180°/360°)

Question 4.
Diagonals are perpendicular in ………… (parallelogram/rhombus)

Answer
1. kite
2. convex
3. 360°
4. rhombus.

RBSE Solutions

III. True/False Type Questions

(1) All rectangles are squares
(2) All rhombuses are parallelograms
(3) All squares are rhombuses and also rectangles
(4) All squares(RBSESolutions.com)are not parallelograms
(5) All kites are rhombuses
(6) All rhombuses are kites
(7) All parallelograms are trapeziums
(8) All squares are trapeziums.

Solution
(1) False
(2) True
(3) True
(4) False
(5) False
(6) True
(7) True
(8) True.

RBSE Solutions

IV. Very Short Answer Type Questions

Question 1.
Write the value of(RBSESolutions.com)each angle in a rectangle.
Answer
90°

Question 2.
Diagonals of a rhombus are………
Answer
angle bisectors

Question 3.
Which sides of a(RBSESolutions.com)parallelogram are equal?
Answer
opposite sides.

RBSE Solutions

Question 4.
Write two necessary points to make a rhombus.
Answer

  • Measurement of a side and(RBSESolutions.com)angle between two adjacent sides.
  • diagonals.

Question 5.
Define rectangle.
Answer
A rectangle is a parallelogram with equal angles.

RBSE Solutions

Short Answer Type Questions

Question 1.
The following figure RING is a parallelogram. If ∠R = 70° then final out remaining all angels of this parallelogram.
RBSE Solutions for Class 8 Maths Chapter 6 Polygons Additional Questions img-2
Solution
By showing given(RBSESolutions.com)parallelogram RING,
∵∠R = 70° (given that)
∴∠N = 70°
∵∠N, ∠R are opposite angles
we know that the sum of any two adjacent angles of a parallelogram is 180°, therefore ∠I = 180° – 70° = 110°
so we can write ∠G = 110° ∵ ∠I, ∠G are opposite angles.

Question 2.
The following figure PQRS in a parallelogram. Find x and y (length is in cm).
RBSE Solutions for Class 8 Maths Chapter 6 Polygons Additional Questions img-3
Solution
In parallelogram PQ = SR
so 3x – 1 = 29
or 3x = 29 + 1
or 3x = 30
or x = \(\frac { 30 }{ 3 }\)
= 10 cm
In parallelogram PS = QR
so 4y = 24
or y = \(\frac { 24 }{ 4 }\)
= 6 cm
so we can write x = 10 cm and y = 6 cm

RBSE Solutions

Question 3.
A mason has made a concrete slab. He needs it to be rectangular. In what different ways can he make sure that it is rectangular?
Solution
To make a concrete(RBSESolutions.com)rectangular slab the man son should ensure

  • its opposite sides equal;
  • its diagonals equal;
  • each angle measure 90°.

Question 4.
Find out value of x and y in the given parallelogram.
RBSE Solutions for Class 8 Maths Chapter 6 Polygons Additional Questions img-4
Solution
We know that the diagonals(RBSESolutions.com)of parallelogram bisect each other
∴ x + 3 = 20 ….(1)
y = 16 ……(2)
From equation (1)
x = 20 – 3 = 17
y = 16

Question 5.
A square was defined as a(RBSESolutions.com)rectangle with all sides equal. Can we define it as rhombus with equal angles? Explore this idea.
Solution
Square can not be defined as a rhombus with equal angles as its diagonal will not be equal unless its each angle is 90°.

RBSE Solutions

Question 6.
Can a trapezium have all angles equal? Can it have all sides equal? Explain.
Solution
A trapezium cannot have all angles equal as its opposite sides become parallel. But trapezium is a quadrilateral with one pair of parallel sides. A trapezium cannot have all sides equal as again its(RBSESolutions.com)opposite sides become parallel. But trapezium is a quadrilateral with one pair of parallel sides.

Question 7.
The following figures GUNS and RUNS are parallelograms. Find x and y. (Lengths are in cm)
RBSE Solutions for Class 8 Maths Chapter 6 Polygons Additional Questions img-5
Solution
(i) Since GUNS is a parallelogram, therefore, its opposite sides are equal,
i. e., GS = UN and GU = SN
or 3x = 18, => x = 6
and 3y – 1 = 26
or 3y = 26 + 1 = 27
or y = 9
∴x = 6 cm and y = 9 cm Ans.

(ii) Diagonals of a parallelogram(RBSESolutions.com)bisect each other, therefore,
OR = ON i.e. 16 = x + y ….(1)
and OU = OS i.e. y + 7 = 20 ….(2)
From (2), y = 20 – 7 = 13 cm
putting y = 13 in (1), we get
16 = x + 13
or x = 16 – 13 = 3 cm
Hence x = 3 cm and y = 13 cm

RBSE Solutions

Question 8.
Two adjacent angles of a parallelogram have equal measure. Find the measure of each of the angles of the parallelogram.
Solution
Let x be the measure of two adjacent angles A and B.
We know that the adjacent angles(RBSESolutions.com)of a parallelogram are supplementary.
∴ ∠P + ∠Q = 180°
or X + x = 180°
or 2x = 180°
or x = 90°
∴ ∠P = 90° and ∠Q = 90°
We know that the opposite angles are equal in a parallelogram
∴ ∠P = ∠Q = ∠R = ∠S = 90°

Question 9.
Find the values of x and y from given parallelogram ABCD.
RBSE Solutions for Class 8 Maths Chapter 6 Polygons Additional Questions img-6
Solution
Since opposite sides of a parallelogram(RBSESolutions.com)are equal in lengths, therefore
3x + 1 = 16 ⇒ 3x = 16 – 1
⇒ 3x = 15 ⇒ x = 5 cm.
5y – 6 = 24 ⇒ 5y = 24 + 6
⇒ 5y = 30 ⇒ y = 6 cm. Ans.

RBSE Solutions

Question 10.
Find the values of unknown angles x, y, z and w in the following figure
RBSE Solutions for Class 8 Maths Chapter 6 Polygons Additional Questions img-7
Solution
Here x + 60° = 180° (Linear pair)
⇒ x = 180° – 60° = 120°
Again, y + 100° = 180°
⇒ y = 180° – 100 = 80°
Similarly, z + 130° = 180°
⇒ z = 180° – 130° = 50°
We know that the sum of all interior angles of an n sided polygon is (2n – 4) right angle.
∴The sum of all interior(RBSESolutions.com)angles of 5 sided polygon
= (2 x 5 – 4) x 90°
= 6 x 90°
= 540°
Let the remaining two angles of pentagon be a and b.
RBSE Solutions for Class 8 Maths Chapter 6 Polygons Additional Questions img-8
Then a + b + x + 100° + 130° = 540°
⇒ a + (180° – 85°) + 120° + 100° + 130° = 540°
⇒ a + 95° + 120° + 100° + 130° = 540°
⇒ a + 445° = 540°
⇒ a – 540° – 445° = 95°
∴w = 180° – a = 180° – 95°
⇒ w = 85°

RBSE Solutions

Question 11.
Find the value of x and y in given parallelogram PQRS.
RBSE Solutions for Class 8 Maths Chapter 6 Polygons Additional Questions img-9
Also write any two properties used to find the value of angles.
Solution
PQRS is a parallelogram
∴PS || RQ and PR is transversal
∴y = 40° (alternate angle)
When PS || RQ and PQ is transversal
∠SPQ = 70° (corresponding angle)
40° + x° = 70°
x = 70° – 40°
x = 30°.
Two properties :

  • Alternate angles are equal.
  • Corresponding(RBSESolutions.com)angles are equal.

We hope the given RBSE Solutions for Class 8 Maths Chapter 6 Polygons Additional Questions will help you. If you have any query regarding Rajasthan Board RBSE Class 8 Maths Chapter 6 Polygons Additional Questions, drop a comment below and we will get back to you at the earliest.

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