NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals
Get here NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals.
More Detail
Latest News
Finished viewing? Save it for later —
Download NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals (PDF · 18 pages)Downloaded 138 times
About NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals
NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals is available here for free download. Published by NCERT for Class 8, this solution can be viewed online or downloaded as a PDF (18 pages). Candidates preparing for Class 8 can use NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals to understand the exam pattern, the type of questions asked, and the overall difficulty level.
Frequently Asked Questions
How can I download NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals?
Open this page and click the Download button to save NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals as a PDF. It is completely free on AglaSem Docs.
Is NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals free to download?
Yes. NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals can be viewed online and downloaded as a PDF free of cost on AglaSem Docs.
How many pages does NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals have?
NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals contains 18 pages, which you can read online or download together as a single PDF.
Where can I find more Class 8 study material?
You can find more Class 8 question papers, sample papers, syllabus, and answer keys on AglaSem Docs.
NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals – Text
Read the full text of this solution below — useful to quickly search, copy and reference the content online without downloading the PDF.
📄 View text version (18 pages)
Page 1
NCERT
SOLUTIONS
CLASS - 8TH
aglase .co
SOLUTIONS
CLASS - 8TH
aglase .co
Page 2
Book : Mathematics Ncert Solutions | Chapter-3 Maths
Class : 8th
Subject : Maths
Chapter : 3
Chapter Name : Understanding Quadrilaterals
Exercise 3.1
Q1 Given here are some gures.
Classify each of them on the basis of the following.
(a) Simple curve (b) Simple closed curve (c) Polygon (d) Convex polygon (e) Concave polygon.
Answer. (a) 1, 2, 5, 6, 7
(b) 1, 2, 5, 6, 7
(c) 1, 2
(d) 2
(e) 1
Page : 41 , Block Name : Exercise 3.1
Q2 How many diagonals does each of the following have?
(a) A convex quadrilateral (b) A regular hexagon (c) A triangle
Answer.
Page 1 of 17 Aglasem Schools
Class : 8th
Subject : Maths
Chapter : 3
Chapter Name : Understanding Quadrilaterals
Exercise 3.1
Q1 Given here are some gures.
Classify each of them on the basis of the following.
(a) Simple curve (b) Simple closed curve (c) Polygon (d) Convex polygon (e) Concave polygon.
Answer. (a) 1, 2, 5, 6, 7
(b) 1, 2, 5, 6, 7
(c) 1, 2
(d) 2
(e) 1
Page : 41 , Block Name : Exercise 3.1
Q2 How many diagonals does each of the following have?
(a) A convex quadrilateral (b) A regular hexagon (c) A triangle
Answer.
Page 1 of 17 Aglasem Schools
Page 3
Book : Mathematics Ncert Solutions | Chapter-3 Maths
Page : 41 , Block Name : Exercise 3.1
Q3 What is the sum of the measures of the angles of a convex quadrilateral? Will this property
hold if the quadrilateral is not convex? (Make a non-convex quadrilateral and try!)
Answer. The sum of the measures of the angles of a convex quadrilateral is 3600 as a convex
quadrilateral is made of two triangles.
Page 2 of 17 Aglasem Schools
Page : 41 , Block Name : Exercise 3.1
Q3 What is the sum of the measures of the angles of a convex quadrilateral? Will this property
hold if the quadrilateral is not convex? (Make a non-convex quadrilateral and try!)
Answer. The sum of the measures of the angles of a convex quadrilateral is 3600 as a convex
quadrilateral is made of two triangles.
Page 2 of 17 Aglasem Schools
Page 4
Book : Mathematics Ncert Solutions | Chapter-3 Maths
Here, ABCD is a convex quadrilateral, made Of two triangles △ABD and △BCD.Therefore, the
sum of all the interior angles of this quadrilateral will be same as the sum of all the interior angles
of these two triangles i.e., 180 + 180 = 360
∘ ∘ ∘
Yes, this property also holds true for a quadrilateral which is not convex. This is
because any quadrilateral can be divided into two triangles.
Here again, ABCD is a concave quadrilateral, made of two triangles △ABD and △BCD .
Therefore, sum of all the interior angles of this quadrilateral will also be 180 + 180 = 360 .
∘ ∘ ∘
Page : 41 , Block Name : Exercise 3.1
Q4 Examine the table. (Each gure is divided into triangles and the sum of the angles deduced
from that.)
What can you say about the angle sum of a convex polygon with number of sides? (a) 7
(b) 8
(c) 10
(d) n
Answer. From the table, it can be observed that the angle sum of a convex polygon of n sides is (n
—2) x 180 . Hence, the angle sum of the convex polygons having number of sides as above will be
∘
as follows.
(a) (7 - 2) x 180 = 900
∘ ∘
(b) (8 - 2) x 180 = 1800
∘ ∘
(c) (10 - 2) x 180 = 180
∘ ∘
(d) (n - 2) x 180
∘
Page 3 of 17 Aglasem Schools
Here, ABCD is a convex quadrilateral, made Of two triangles △ABD and △BCD.Therefore, the
sum of all the interior angles of this quadrilateral will be same as the sum of all the interior angles
of these two triangles i.e., 180 + 180 = 360
∘ ∘ ∘
Yes, this property also holds true for a quadrilateral which is not convex. This is
because any quadrilateral can be divided into two triangles.
Here again, ABCD is a concave quadrilateral, made of two triangles △ABD and △BCD .
Therefore, sum of all the interior angles of this quadrilateral will also be 180 + 180 = 360 .
∘ ∘ ∘
Page : 41 , Block Name : Exercise 3.1
Q4 Examine the table. (Each gure is divided into triangles and the sum of the angles deduced
from that.)
What can you say about the angle sum of a convex polygon with number of sides? (a) 7
(b) 8
(c) 10
(d) n
Answer. From the table, it can be observed that the angle sum of a convex polygon of n sides is (n
—2) x 180 . Hence, the angle sum of the convex polygons having number of sides as above will be
∘
as follows.
(a) (7 - 2) x 180 = 900
∘ ∘
(b) (8 - 2) x 180 = 1800
∘ ∘
(c) (10 - 2) x 180 = 180
∘ ∘
(d) (n - 2) x 180
∘
Page 3 of 17 Aglasem Schools
Page 5
Book : Mathematics Ncert Solutions | Chapter-3 Maths
Page : 41 , Block Name : Exercise 3.1
Q5 What is a regular polygon?
State the name of a regular polygon of
(i) 3 sides
(ii) 4 sides
(iii) 6 sides
Answer. A polygon with equal sides and equal angles is called a regular polygon.
Page : 42 , Block Name : Exercise 3.1
Q6 Find the angle measure x in the following gures.
Page 4 of 17 Aglasem Schools
Page : 41 , Block Name : Exercise 3.1
Q5 What is a regular polygon?
State the name of a regular polygon of
(i) 3 sides
(ii) 4 sides
(iii) 6 sides
Answer. A polygon with equal sides and equal angles is called a regular polygon.
Page : 42 , Block Name : Exercise 3.1
Q6 Find the angle measure x in the following gures.
Page 4 of 17 Aglasem Schools
Page 6
Book : Mathematics Ncert Solutions | Chapter-3 Maths
Answer. (a) Sum of the measures of all interior angles of a quadrilateral is 360 Therefore, in the
∘
given quadrilateral,
50 + 130 + 120 + x = 360
∘ ∘ ∘ ∘
300 + x = 360
∘ ∘
x = 60
∘
(b)
From the gure, it can be concluded that,
90 + a = 180 (Linear pair)
∘ ∘
a = 180 - 90 = 90
∘ ∘ ∘
Sum Of the measures Of all interior angles of a quadrilateral is 360 . Therefore, in the given
∘
quadrilateral,
60 + 70 + x + 90 = 360
∘ ∘ ∘ ∘
220 + x = 360
∘ ∘
x = 140
∘
(c)
Page 5 of 17 Aglasem Schools
Answer. (a) Sum of the measures of all interior angles of a quadrilateral is 360 Therefore, in the
∘
given quadrilateral,
50 + 130 + 120 + x = 360
∘ ∘ ∘ ∘
300 + x = 360
∘ ∘
x = 60
∘
(b)
From the gure, it can be concluded that,
90 + a = 180 (Linear pair)
∘ ∘
a = 180 - 90 = 90
∘ ∘ ∘
Sum Of the measures Of all interior angles of a quadrilateral is 360 . Therefore, in the given
∘
quadrilateral,
60 + 70 + x + 90 = 360
∘ ∘ ∘ ∘
220 + x = 360
∘ ∘
x = 140
∘
(c)
Page 5 of 17 Aglasem Schools
Page 7
Book : Mathematics Ncert Solutions | Chapter-3 Maths
From the gure, it can be concluded that,
70 + a = 180 (Linear pair) ∘
a = 110
∘
60 + b = 180 (Linear pair)
∘ ∘
b = 120 ∘
Sum Of the measures Of all interior angles Of a pentagon is 540 ,
∘
Therefore, in the given pentagon,
120 + 110 + 30 + x + x = 540
∘ ∘ ∘ ∘
260 + 2x = 540
∘ ∘
2x = 280 ∘
X = 140
∘
(d) Sum of the measures of all interior angles of a pentagon is 540 ∘
5x = 540
∘
x = 108 ∘
Page : 42 , Block Name : Exercise 3.1
Q7
Answer. (a) x + 90 = 180 (Linear pair) ∘ ∘
x = 90 ∘
z + 30 = 180 (Linear pair)
∘ ∘
z = 150 ∘
y = 90 + 30 (Exterior angle theorem)\)
∘ ∘
y = 120 ∘
x + y + z = 90 + 120 + 150 = 360
∘ ∘ ∘ ∘
Page 6 of 17 Aglasem Schools
From the gure, it can be concluded that,
70 + a = 180 (Linear pair) ∘
a = 110
∘
60 + b = 180 (Linear pair)
∘ ∘
b = 120 ∘
Sum Of the measures Of all interior angles Of a pentagon is 540 ,
∘
Therefore, in the given pentagon,
120 + 110 + 30 + x + x = 540
∘ ∘ ∘ ∘
260 + 2x = 540
∘ ∘
2x = 280 ∘
X = 140
∘
(d) Sum of the measures of all interior angles of a pentagon is 540 ∘
5x = 540
∘
x = 108 ∘
Page : 42 , Block Name : Exercise 3.1
Q7
Answer. (a) x + 90 = 180 (Linear pair) ∘ ∘
x = 90 ∘
z + 30 = 180 (Linear pair)
∘ ∘
z = 150 ∘
y = 90 + 30 (Exterior angle theorem)\)
∘ ∘
y = 120 ∘
x + y + z = 90 + 120 + 150 = 360
∘ ∘ ∘ ∘
Page 6 of 17 Aglasem Schools
Page 8
Book : Mathematics Ncert Solutions | Chapter-3 Maths
(b)
Sum of the measures of all interior angles of a quadrilateral is 360 . Therefore, in the given
∘
quadrilateral,
a + 60 + 80 + 120 = 360
∘ ∘ ∘ ∘
a + 260 =360
∘ ∘
a = 100 ∘
x + 120 = 180 ( Linear pair )
∘ ∘
x = 60
∘
y + 80 = 180 (Linear pair)
∘ ∘
y = 100 ∘
z + 60 = 180
∘ ∘
z = 120 ∘
w + 100 = 180
∘ ∘
w = 80 ∘
Sum of the measures of all interior angles = x + y + z + w
= 60 + 100 + 120 + 80
∘ ∘ ∘ ∘
= 360 ∘
Page : 42 , Block Name : Exercise 3.1
Exercise 3.2
Q1 Find x in the following gures.
Answer. We know that the sum of all exterior angles of any polygon is 360 .
∘
(a) 125 + 125 + x = 360
∘ ∘ ∘
250 + x = 360
∘ ∘
Page 7 of 17 Aglasem Schools
(b)
Sum of the measures of all interior angles of a quadrilateral is 360 . Therefore, in the given
∘
quadrilateral,
a + 60 + 80 + 120 = 360
∘ ∘ ∘ ∘
a + 260 =360
∘ ∘
a = 100 ∘
x + 120 = 180 ( Linear pair )
∘ ∘
x = 60
∘
y + 80 = 180 (Linear pair)
∘ ∘
y = 100 ∘
z + 60 = 180
∘ ∘
z = 120 ∘
w + 100 = 180
∘ ∘
w = 80 ∘
Sum of the measures of all interior angles = x + y + z + w
= 60 + 100 + 120 + 80
∘ ∘ ∘ ∘
= 360 ∘
Page : 42 , Block Name : Exercise 3.1
Exercise 3.2
Q1 Find x in the following gures.
Answer. We know that the sum of all exterior angles of any polygon is 360 .
∘
(a) 125 + 125 + x = 360
∘ ∘ ∘
250 + x = 360
∘ ∘
Page 7 of 17 Aglasem Schools
Page 9
Book : Mathematics Ncert Solutions | Chapter-3 Maths
x = 110
∘
(b)
60
∘
+ 90 +70 + x + 90 = 360
∘ ∘ ∘ ∘
310 + x = 360
∘ ∘
x = 50
∘
Page : 44 , Block Name : Exercise 3.2
Q2 Find the measure of each exterior angle of a regular polygon of
(i) 9 sides
(ii) 15 sides
Answer. (i)Sum of all exterior angles of the given polygon = 360 ∘
Each exterior angle of a regular polygon has the same measure.
∘
Thus, measure of each exterior angle of a regular polygon of 9 sides = 360
9
= 40
∘
(ii) Sum Of all exterior angles Of the given polygon = 360
∘
Each exterior angle Of a regular polygon has the same measure.
∘
Thus, measure of each exterior angle Of a regular polygon of 15 sides = 360
15
= 24
∘
Page : 44 , Block Name : Exercise 3.2
Q3 How many sides does a regular polygon have if the measure of an exterior angle is 24°?
Answer. Sum of all exterior angles of the given polygon = 360 ∘
Measure of each exterior angle = 24 ∘
∘
Thus, number of sides of the regular polygon = =
360 ∘
= 15
24
Page : 44 , Block Name : Exercise 3.2
Q4 How many sides does a regular polygon have if each of its interior angles is 165°?
Answer. Measure Of each interior angle = 165 ∘
Measure of each exterior angle = 180 - 165 = 15
∘ ∘ ∘
Page 8 of 17 Aglasem Schools
x = 110
∘
(b)
60
∘
+ 90 +70 + x + 90 = 360
∘ ∘ ∘ ∘
310 + x = 360
∘ ∘
x = 50
∘
Page : 44 , Block Name : Exercise 3.2
Q2 Find the measure of each exterior angle of a regular polygon of
(i) 9 sides
(ii) 15 sides
Answer. (i)Sum of all exterior angles of the given polygon = 360 ∘
Each exterior angle of a regular polygon has the same measure.
∘
Thus, measure of each exterior angle of a regular polygon of 9 sides = 360
9
= 40
∘
(ii) Sum Of all exterior angles Of the given polygon = 360
∘
Each exterior angle Of a regular polygon has the same measure.
∘
Thus, measure of each exterior angle Of a regular polygon of 15 sides = 360
15
= 24
∘
Page : 44 , Block Name : Exercise 3.2
Q3 How many sides does a regular polygon have if the measure of an exterior angle is 24°?
Answer. Sum of all exterior angles of the given polygon = 360 ∘
Measure of each exterior angle = 24 ∘
∘
Thus, number of sides of the regular polygon = =
360 ∘
= 15
24
Page : 44 , Block Name : Exercise 3.2
Q4 How many sides does a regular polygon have if each of its interior angles is 165°?
Answer. Measure Of each interior angle = 165 ∘
Measure of each exterior angle = 180 - 165 = 15
∘ ∘ ∘
Page 8 of 17 Aglasem Schools
Page 10
Book : Mathematics Ncert Solutions | Chapter-3 Maths
The sum Of all exterior angles Of any polygon is 360 . ∘
∘
Thus, number of sides of the polygon = 360
15
= 24
∘
Page : 44 , Block Name : Exercise 3.2
Q5 (a) Is it possible to have a regular polygon with measure of each exterior angle as 22°?
(b) Can it be an interior angle of a regular polygon? Why?
Answer. The sum of all exterior angles of all polygons is 360 . Also, in a regular polygon, each
∘
exterior angle is Of the same measure. Hence, if 360 is a perfect multiple Of the given exterior
∘
angle, then the given polygon will be possible.
(a) Exterior angle = 22
∘
360 is not a perfect multiple of 22 . Hence, such polygon is not possible.
∘ ∘
(b) Interior angle = 22
∘
Such a polygon is not possible as 360 is not a perfect multiple of 158
∘ ∘
Page : 44 , Block Name : Exercise 3.2
Q6 (a) What is the minimum interior angle possible for a regular polygon? Why?
(b) What is the maximum exterior angle possible for a regular polygon?
Answer. Consider a regular polygon having the lowest possible number Of sides (i.e., an
equilateral triangle). The exterior angle of this triangle will be the maximum exterior angle
possible for any regular polygon.
∘
Exterior angle of an equilateral triangle = 360
= 120
3
∘
Hence, maximum possible measure of exterior angle for any polygon is 120 . Also, we know that
∘
an exterior angle and an interior angle are always in a linear pair.
Hence, minimum interior angle = 180 - 120 = 60
∘ ∘ ∘
Page : 44 , Block Name : Exercise 3.2
Exercise 3.3
Q1 Given a parallelogram ABCD . Complete each statement along with
the de nition or property used.
(i) AD = ......
(ii) ∠ DCB = ......
(iii) OC = ......
(iv) m ∠DAB + m ∠CDA = ......
Answer. (i) In a parallelogram, opposite sides are equal in length.
AD = BC
Page 9 of 17 Aglasem Schools
The sum Of all exterior angles Of any polygon is 360 . ∘
∘
Thus, number of sides of the polygon = 360
15
= 24
∘
Page : 44 , Block Name : Exercise 3.2
Q5 (a) Is it possible to have a regular polygon with measure of each exterior angle as 22°?
(b) Can it be an interior angle of a regular polygon? Why?
Answer. The sum of all exterior angles of all polygons is 360 . Also, in a regular polygon, each
∘
exterior angle is Of the same measure. Hence, if 360 is a perfect multiple Of the given exterior
∘
angle, then the given polygon will be possible.
(a) Exterior angle = 22
∘
360 is not a perfect multiple of 22 . Hence, such polygon is not possible.
∘ ∘
(b) Interior angle = 22
∘
Such a polygon is not possible as 360 is not a perfect multiple of 158
∘ ∘
Page : 44 , Block Name : Exercise 3.2
Q6 (a) What is the minimum interior angle possible for a regular polygon? Why?
(b) What is the maximum exterior angle possible for a regular polygon?
Answer. Consider a regular polygon having the lowest possible number Of sides (i.e., an
equilateral triangle). The exterior angle of this triangle will be the maximum exterior angle
possible for any regular polygon.
∘
Exterior angle of an equilateral triangle = 360
= 120
3
∘
Hence, maximum possible measure of exterior angle for any polygon is 120 . Also, we know that
∘
an exterior angle and an interior angle are always in a linear pair.
Hence, minimum interior angle = 180 - 120 = 60
∘ ∘ ∘
Page : 44 , Block Name : Exercise 3.2
Exercise 3.3
Q1 Given a parallelogram ABCD . Complete each statement along with
the de nition or property used.
(i) AD = ......
(ii) ∠ DCB = ......
(iii) OC = ......
(iv) m ∠DAB + m ∠CDA = ......
Answer. (i) In a parallelogram, opposite sides are equal in length.
AD = BC
Page 9 of 17 Aglasem Schools
Page 11
Book : Mathematics Ncert Solutions | Chapter-3 Maths
(ii) In a parallelogram, opposite angles are equal in measure.
∠DAB
(iii) In a parallelogram, diagonals bisect each other.
Hence, OC = OA
(iv) In a parallelogram, adjacent angles are supplementary to each other.
Hence, m∠DAB + m∠CDA = 180 ∘
Page : 50 , Block Name : Exercise 3.3
Q2 Consider the following parallelograms. Find the values of the unknowns x, y, z.
Answer. (i) x + 100 = 180 (Adjacent angles are supplementary)
∘ ∘
X = 80 ∘
z = x = 80 ( Opposite angles are equal )
∘
y = 100 ( Opposite angles are equal )
∘
(ii) 50 + y = 180 ( Adjacent angles are supplementary )
∘ ∘
y = 130
∘
x = y = 130 ( Opposite angles are equal )
∘
z = x = 130 ( Corresponding angles )
∘
(iii) x = 90 (Vertically Opposite angles )
∘
x + y + 30 = 180 (Angle sum property of triangles )
∘ ∘
120 + y = 180
∘ ∘
y = 60 ∘
z = y = 60 ( Alternate interior angles )
∘
(iv) z = 80 ( Corresponding angles )
∘
y = 80 ( Opposite angles are equal )
∘
x + y = 180 (Adjacent angles are supplementary )
∘
x = 180 - 80 = 100
∘ ∘ ∘
(v) y = 112 (Opposite angles are equal)
∘
x + y + 40 = 180 ( Angle sum property of Triangles )
∘ ∘
x + 112 + 40 = 180
∘ ∘ ∘
x + 152 = 180
∘ ∘
x = 28
∘
z = x = 28 (Alternate Interior angles )
∘
Page : 51 , Block Name : Exercise 3.3
Page 10 of 17 Aglasem Schools
(ii) In a parallelogram, opposite angles are equal in measure.
∠DAB
(iii) In a parallelogram, diagonals bisect each other.
Hence, OC = OA
(iv) In a parallelogram, adjacent angles are supplementary to each other.
Hence, m∠DAB + m∠CDA = 180 ∘
Page : 50 , Block Name : Exercise 3.3
Q2 Consider the following parallelograms. Find the values of the unknowns x, y, z.
Answer. (i) x + 100 = 180 (Adjacent angles are supplementary)
∘ ∘
X = 80 ∘
z = x = 80 ( Opposite angles are equal )
∘
y = 100 ( Opposite angles are equal )
∘
(ii) 50 + y = 180 ( Adjacent angles are supplementary )
∘ ∘
y = 130
∘
x = y = 130 ( Opposite angles are equal )
∘
z = x = 130 ( Corresponding angles )
∘
(iii) x = 90 (Vertically Opposite angles )
∘
x + y + 30 = 180 (Angle sum property of triangles )
∘ ∘
120 + y = 180
∘ ∘
y = 60 ∘
z = y = 60 ( Alternate interior angles )
∘
(iv) z = 80 ( Corresponding angles )
∘
y = 80 ( Opposite angles are equal )
∘
x + y = 180 (Adjacent angles are supplementary )
∘
x = 180 - 80 = 100
∘ ∘ ∘
(v) y = 112 (Opposite angles are equal)
∘
x + y + 40 = 180 ( Angle sum property of Triangles )
∘ ∘
x + 112 + 40 = 180
∘ ∘ ∘
x + 152 = 180
∘ ∘
x = 28
∘
z = x = 28 (Alternate Interior angles )
∘
Page : 51 , Block Name : Exercise 3.3
Page 10 of 17 Aglasem Schools
Page 12
Book : Mathematics Ncert Solutions | Chapter-3 Maths
Q3 Can a quadrilateral ABCD be a parallelogram if
(i) ∠D + ∠B = 180°?
(ii) AB = DC = 8 cm, AD = 4 cm and BC = 4.4 cm?
(iii) ∠A = 70° and ∠C = 65°?
Answer. (i) For ∠D + ∠B = 180 , quadrilateral ABCD may or may not be a parallelogram.
∘
Along with this condition, the following conditions should also be ful lled.
The sum of the measures of adjacent angles should be 180 ,∘
Opposite angles should also be of same measures.
(ii) No. Opposite sides AD and BC are of different lengths.
(iii) No. Opposite angles A and C have different measures.
Page : 51 , Block Name : Exercise 3.3
Q4 Draw a rough gure of a quadrilateral that is not a parallelogram but has exactly two opposite
angles of equal measure.
Answer. Here, quadrilateral ABCD (kite) has two of its interior angles,∠B and ∠D, of same
measures. However, still the quadrilateral ABCD is not a parallelogram as the measures Of the
remaining pair Of opposite angles,∠A and ∠C , are not equal.
Page : 51 , Block Name : Exercise 3.3
Q5 The measures of two adjacent angles of a parallelogram are in the ratio 3 : 2. Find the measure
of each of the angles of the parallelogram.
Answer. Let the measures Of two adjacent angles, ∠A and ∠B, of parallelogram ABCD are in the
ratio of 3:2. Let ∠A = 3x and ∠B = 2x
We know that the sum Of the measures Of adjacent angles is 180 for a
∘
parallelogram.
∘
∠A + ∠B = 180
∘
3x + 2x = 180
∘
5x = 180
∘
180 ∘
x = = 36
5
∘
□A = □C = 3x = 108 ( Opposite angles )
∘
□B = □D = 2x = 72 ( Opposite angles )
Page 11 of 17 Aglasem Schools
Q3 Can a quadrilateral ABCD be a parallelogram if
(i) ∠D + ∠B = 180°?
(ii) AB = DC = 8 cm, AD = 4 cm and BC = 4.4 cm?
(iii) ∠A = 70° and ∠C = 65°?
Answer. (i) For ∠D + ∠B = 180 , quadrilateral ABCD may or may not be a parallelogram.
∘
Along with this condition, the following conditions should also be ful lled.
The sum of the measures of adjacent angles should be 180 ,∘
Opposite angles should also be of same measures.
(ii) No. Opposite sides AD and BC are of different lengths.
(iii) No. Opposite angles A and C have different measures.
Page : 51 , Block Name : Exercise 3.3
Q4 Draw a rough gure of a quadrilateral that is not a parallelogram but has exactly two opposite
angles of equal measure.
Answer. Here, quadrilateral ABCD (kite) has two of its interior angles,∠B and ∠D, of same
measures. However, still the quadrilateral ABCD is not a parallelogram as the measures Of the
remaining pair Of opposite angles,∠A and ∠C , are not equal.
Page : 51 , Block Name : Exercise 3.3
Q5 The measures of two adjacent angles of a parallelogram are in the ratio 3 : 2. Find the measure
of each of the angles of the parallelogram.
Answer. Let the measures Of two adjacent angles, ∠A and ∠B, of parallelogram ABCD are in the
ratio of 3:2. Let ∠A = 3x and ∠B = 2x
We know that the sum Of the measures Of adjacent angles is 180 for a
∘
parallelogram.
∘
∠A + ∠B = 180
∘
3x + 2x = 180
∘
5x = 180
∘
180 ∘
x = = 36
5
∘
□A = □C = 3x = 108 ( Opposite angles )
∘
□B = □D = 2x = 72 ( Opposite angles )
Page 11 of 17 Aglasem Schools
Page 13
Book : Mathematics Ncert Solutions | Chapter-3 Maths
Thus, the measures of the angles Of the parallelogram are 108 , 72 , 108 , and 72 .
∘ ∘ ∘ ∘
Page : 51 , Block Name : Exercise 3.3
Q6 Two adjacent angles of a parallelogram have equal measure. Find the measure of each of the
angles of the parallelogram.
Answer. Sum of adjacent angles = 180 ∘
∘
□A + □B = 180
∘
2□A = 180 (□A = □B)
∘
□A = 90
∘
□B = □A = 90
∘
□C = □A = 90 ( Opposite angles )
∘
□D = □B = 90 ( Opposite angles )
Thus, each angle of the parallelogram measures 90 . ∘
Page : 51 , Block Name : Exercise 3.3
Q7 The adjacent gure HOPE is a parallelogram. Find the angle measures x, y and z. State the
properties you use to nd them.
Answer.
∘
y = 40 (Alternate interior angles)
∘ ∘
70 = z + 40 (Corresponding angles)
∘ ∘
70 − 40 = z
∘
z = 30
∘ ∘
x + (z + 40 ) = 180 ( Adjacent pair of angles)
∘ ∘
x + 70 = 180
∘
x = 110
Page : 51 , Block Name : Exercise 3.3
Q8 The following gures GUNS and RUNS are parallelograms. Find x and y. (Lengths are in cm)
Page 12 of 17 Aglasem Schools
Thus, the measures of the angles Of the parallelogram are 108 , 72 , 108 , and 72 .
∘ ∘ ∘ ∘
Page : 51 , Block Name : Exercise 3.3
Q6 Two adjacent angles of a parallelogram have equal measure. Find the measure of each of the
angles of the parallelogram.
Answer. Sum of adjacent angles = 180 ∘
∘
□A + □B = 180
∘
2□A = 180 (□A = □B)
∘
□A = 90
∘
□B = □A = 90
∘
□C = □A = 90 ( Opposite angles )
∘
□D = □B = 90 ( Opposite angles )
Thus, each angle of the parallelogram measures 90 . ∘
Page : 51 , Block Name : Exercise 3.3
Q7 The adjacent gure HOPE is a parallelogram. Find the angle measures x, y and z. State the
properties you use to nd them.
Answer.
∘
y = 40 (Alternate interior angles)
∘ ∘
70 = z + 40 (Corresponding angles)
∘ ∘
70 − 40 = z
∘
z = 30
∘ ∘
x + (z + 40 ) = 180 ( Adjacent pair of angles)
∘ ∘
x + 70 = 180
∘
x = 110
Page : 51 , Block Name : Exercise 3.3
Q8 The following gures GUNS and RUNS are parallelograms. Find x and y. (Lengths are in cm)
Page 12 of 17 Aglasem Schools
Page 14
Book : Mathematics Ncert Solutions | Chapter-3 Maths
Answer. (i)We know that the lengths of opposite sides of a parallelogram are equal to each other.
GU = SN
3y − 1 = 26
3y = 27
y = 9
SG = NU
3x = 18
x = 6
Hence, the measures of x and y are 6 cm and 9 cm respectively.
(ii)We know that the diagonals of a parallelogram bisect each other.
y + 7 = 20
y = 13
x + y = 16
x + 13 = 16
x = 3
Hence, the measures Of x and y are 3 cm and 13 cm respectively.
Page : 51 , Block Name : Exercise 3.3
Q9
In the above gure both RISK and CLUE are parallelograms. Find the value of x.
Answer. Adjacent angles of a parallelogram are supplementary.
In parallelogram RISK, □RKS + □I SK = 180 ∘
∘ ∘
120 + □I SK = 180
∘
□I SK = 60
Also, opposite angles of a parallelogram are equal.
In parallelogram CLUE, □ ∪ LC = □CEU = 70 ∘
The sum of the measures of all the interior angles of a triangle is 180 ∘
∘ ∘ ∘
x + 60 + 70 = 180
∘
x = 50
Page 13 of 17 Aglasem Schools
Answer. (i)We know that the lengths of opposite sides of a parallelogram are equal to each other.
GU = SN
3y − 1 = 26
3y = 27
y = 9
SG = NU
3x = 18
x = 6
Hence, the measures of x and y are 6 cm and 9 cm respectively.
(ii)We know that the diagonals of a parallelogram bisect each other.
y + 7 = 20
y = 13
x + y = 16
x + 13 = 16
x = 3
Hence, the measures Of x and y are 3 cm and 13 cm respectively.
Page : 51 , Block Name : Exercise 3.3
Q9
In the above gure both RISK and CLUE are parallelograms. Find the value of x.
Answer. Adjacent angles of a parallelogram are supplementary.
In parallelogram RISK, □RKS + □I SK = 180 ∘
∘ ∘
120 + □I SK = 180
∘
□I SK = 60
Also, opposite angles of a parallelogram are equal.
In parallelogram CLUE, □ ∪ LC = □CEU = 70 ∘
The sum of the measures of all the interior angles of a triangle is 180 ∘
∘ ∘ ∘
x + 60 + 70 = 180
∘
x = 50
Page 13 of 17 Aglasem Schools
Page 15
Book : Mathematics Ncert Solutions | Chapter-3 Maths
Page : 51 , Block Name : Exercise 3.3
Q10 Explain how this gure is a trapezium. Which of its two sides are parallel?
Answer. If a transversal line is intersecting two given lines such that the sum of the measures of
the angles on the same side of transversal is 180 , then the given two lines will be parallel to each
∘
other.
Here, □NML + □MLK = 180
∘
Page : 52 , Block Name : Exercise 3.3
Q11 Find m∠C if AIIDC.
¯
¯¯¯ ¯¯
¯¯¯
¯¯¯
Answer.Given that, AB∥DC
¯¯¯¯¯¯¯¯ ¯
¯¯¯
¯¯¯
¯¯
□B + □C = 180
∘
(Angles on the same side of transversal)
∘ ∘
120 + □C = 180
∘
□C = 60
Page : 52 , Block Name : Exercise 3.3
Q12 Find the measure of ∠P and ∠S if ABIIDC. (If you nd m∠R, is there more than one method
¯¯
¯¯¯
¯¯¯ ¯¯
¯¯¯
¯¯¯
to nd m∠P?)
Answer.
Page 14 of 17 Aglasem Schools
Page : 51 , Block Name : Exercise 3.3
Q10 Explain how this gure is a trapezium. Which of its two sides are parallel?
Answer. If a transversal line is intersecting two given lines such that the sum of the measures of
the angles on the same side of transversal is 180 , then the given two lines will be parallel to each
∘
other.
Here, □NML + □MLK = 180
∘
Page : 52 , Block Name : Exercise 3.3
Q11 Find m∠C if AIIDC.
¯
¯¯¯ ¯¯
¯¯¯
¯¯¯
Answer.Given that, AB∥DC
¯¯¯¯¯¯¯¯ ¯
¯¯¯
¯¯¯
¯¯
□B + □C = 180
∘
(Angles on the same side of transversal)
∘ ∘
120 + □C = 180
∘
□C = 60
Page : 52 , Block Name : Exercise 3.3
Q12 Find the measure of ∠P and ∠S if ABIIDC. (If you nd m∠R, is there more than one method
¯¯
¯¯¯
¯¯¯ ¯¯
¯¯¯
¯¯¯
to nd m∠P?)
Answer.
Page 14 of 17 Aglasem Schools
Page 16
Book : Mathematics Ncert Solutions | Chapter-3 Maths
∘
□P + □Q = 180 (Angles on the same side of transversal)
∘ ∘
□P + 130 = 180
∘
□P = 50
∘
□R + □S = 180 ( Angles on the same side of transversal)
∘ ∘
90 + □R = 180
∘
□S = 90
Yes. There is one more method to nd the measure of m□P .
m□R and m□Q are given. After nding m□S , the angle sum property of a quadrilateral can be
applied to nd m□P .
Page : 52 , Block Name : Exercise 3.3
Exercise 3.4
Q1 State whether True or False.
(a) All rectangles are squares
(b) All rhombuses are parallelograms
(c) All squares are rhombuses and also rectangles
(d) All squares are not parallelograms.
(e) All kites are rhombuses.
(f) All rhombuses are kites.
(g) All parallelograms are trapeziums.
(h) All squares are trapeziums.
Answer. (a) False. All squares are rectangles but all rectangles are not squares.
(b) True. Opposite sides of a rhombus are equal and parallel to each other.
(c) True. All squares are rhombuses as all sides of a square are of equal lengths. All squares are also
rectangles as each internal angle measures 900.
(d) False. All squares are parallelograms as opposite sides are equal and parallel.
(e) False. A kite does not have all sides of the same length.
(f) True. A rhombus also has two distinct consecutive pairs of sides of equal length.
(g) True. All parallelograms have a pair of parallel sides.
(h) True. All squares have a pair of parallel sides.
Page : 55 , Block Name : Exercise 3.4
Q2 Identify all the quadrilaterals that have.
(a) four sides of equal length
(b) four right angles
Answer. (a) Rhombus and Square are the quadrilaterals that have 4 sides of equal length.
(b) Square and rectangle are the quadrilaterals that have 4 right angles.
Page : 55 , Block Name : Exercise 3.4
Page 15 of 17 Aglasem Schools
∘
□P + □Q = 180 (Angles on the same side of transversal)
∘ ∘
□P + 130 = 180
∘
□P = 50
∘
□R + □S = 180 ( Angles on the same side of transversal)
∘ ∘
90 + □R = 180
∘
□S = 90
Yes. There is one more method to nd the measure of m□P .
m□R and m□Q are given. After nding m□S , the angle sum property of a quadrilateral can be
applied to nd m□P .
Page : 52 , Block Name : Exercise 3.3
Exercise 3.4
Q1 State whether True or False.
(a) All rectangles are squares
(b) All rhombuses are parallelograms
(c) All squares are rhombuses and also rectangles
(d) All squares are not parallelograms.
(e) All kites are rhombuses.
(f) All rhombuses are kites.
(g) All parallelograms are trapeziums.
(h) All squares are trapeziums.
Answer. (a) False. All squares are rectangles but all rectangles are not squares.
(b) True. Opposite sides of a rhombus are equal and parallel to each other.
(c) True. All squares are rhombuses as all sides of a square are of equal lengths. All squares are also
rectangles as each internal angle measures 900.
(d) False. All squares are parallelograms as opposite sides are equal and parallel.
(e) False. A kite does not have all sides of the same length.
(f) True. A rhombus also has two distinct consecutive pairs of sides of equal length.
(g) True. All parallelograms have a pair of parallel sides.
(h) True. All squares have a pair of parallel sides.
Page : 55 , Block Name : Exercise 3.4
Q2 Identify all the quadrilaterals that have.
(a) four sides of equal length
(b) four right angles
Answer. (a) Rhombus and Square are the quadrilaterals that have 4 sides of equal length.
(b) Square and rectangle are the quadrilaterals that have 4 right angles.
Page : 55 , Block Name : Exercise 3.4
Page 15 of 17 Aglasem Schools
Page 17
Book : Mathematics Ncert Solutions | Chapter-3 Maths
Q3 Explain how a square is.
(i) a quadrilateral
(ii) a parallelogram
(iii) a rhombus
(iv) a rectangle
Answer. (i) A square is a quadrilateral since it has four sides.
(ii) A square is a parallelogram since its opposite sides are parallel to each other.
(iii) A square is a rhombus since its four sides are of the same length.
(iv) A square is a rectangle since each interior angle measures 900.
Page : 55 , Block Name : Exercise 3.4
Q4 Name the quadrilaterals whose diagonals.
(i) bisect each other
(ii) are perpendicular bisectors of each other
(iii) are equal
Answer. (i) The diagonals of a parallelogram, rhombus, square, and rectangle bisect each other.
(ii) The diagonals of a rhombus and square act as perpendicular bisectors.
(iii) The diagonals of a rectangle and square are equal.
Page : 55 , Block Name : Exercise 3.4
Q5 Explain why a rectangle is a convex quadrilateral.
Answer. In a rectangle, there are two diagonals, both lying in the interior Of the rectangle.
Hence, it is a convex quadrilateral.
Page : 55 , Block Name : Exercise 3.4
Q6 ABC is a right-angled triangle and O is the mid point of the side opposite to the right angle.
Explain why O is equidistant from A, B and C. (The dotted lines are drawn additionally to help
you).
Answer. Draw lines AD and DC such that ADIIBC, ABIIDC
AD = DC, AB = DC
ABCD is a rectangle as opposite sides are equal and parallel to each other and all the interior
angles are of 90
∘
In a rectangle, diagonals are of equal length and also these bisect each other.
Hence, AO = OC = BO = OD
Thus, O is equidistant from A, B, and C.
Page 16 of 17 Aglasem Schools
Q3 Explain how a square is.
(i) a quadrilateral
(ii) a parallelogram
(iii) a rhombus
(iv) a rectangle
Answer. (i) A square is a quadrilateral since it has four sides.
(ii) A square is a parallelogram since its opposite sides are parallel to each other.
(iii) A square is a rhombus since its four sides are of the same length.
(iv) A square is a rectangle since each interior angle measures 900.
Page : 55 , Block Name : Exercise 3.4
Q4 Name the quadrilaterals whose diagonals.
(i) bisect each other
(ii) are perpendicular bisectors of each other
(iii) are equal
Answer. (i) The diagonals of a parallelogram, rhombus, square, and rectangle bisect each other.
(ii) The diagonals of a rhombus and square act as perpendicular bisectors.
(iii) The diagonals of a rectangle and square are equal.
Page : 55 , Block Name : Exercise 3.4
Q5 Explain why a rectangle is a convex quadrilateral.
Answer. In a rectangle, there are two diagonals, both lying in the interior Of the rectangle.
Hence, it is a convex quadrilateral.
Page : 55 , Block Name : Exercise 3.4
Q6 ABC is a right-angled triangle and O is the mid point of the side opposite to the right angle.
Explain why O is equidistant from A, B and C. (The dotted lines are drawn additionally to help
you).
Answer. Draw lines AD and DC such that ADIIBC, ABIIDC
AD = DC, AB = DC
ABCD is a rectangle as opposite sides are equal and parallel to each other and all the interior
angles are of 90
∘
In a rectangle, diagonals are of equal length and also these bisect each other.
Hence, AO = OC = BO = OD
Thus, O is equidistant from A, B, and C.
Page 16 of 17 Aglasem Schools
Page 18
Book : Mathematics Ncert Solutions | Chapter-3 Maths
Page : 55 , Block Name : Exercise 3.4
Page 17 of 17 Aglasem Schools
Page : 55 , Block Name : Exercise 3.4
Page 17 of 17 Aglasem Schools
Document Details
| Board / Org | NCERT |
| Exam | Class 8 |
| Type | Solution |
| Pages | 18 |
| Language | english |
| Updated | 30 Apr 2026 |