NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.1

# NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.1

## NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.1

NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.1 are the part of NCERT Solutions for Class 8 Maths. Here you can find the NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.1.

### Ex 3.1 Class 8 Maths Question 1.

Given here are some figures.

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

Solution:
(a) Simple curve: (1), (2), (5), (6) and (7) are simple curves.
(b) Simple closed curve: (1), (2), (5), (6) and (7) are simple curves.
(c) Polygon: (1) and (2) are polygons.
(d) Convex polygon: (2) is convex polygon.
(e) Concave polygon: (1) is concave polygon.

### Ex 3.1 Class 8 Maths Question 2.

How many diagonals does each of the following have?
(b) A regular hexagon
(c) A triangle

Solution:
(a) In the following figure, ABCD is a convex quadrilateral which has two diagonals AC and BD.

(b) In the following figure, ABCDEF is a regular hexagon which has nine diagonals AE, AD, AC, BF, BE, BD, CF, CE and DF.

### Ex 3.1 Class 8 Maths Question 3.

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 verify)

Solution:
In the following figure, we have a quadrilateral ABCD. Join the diagonal AC, which divides the quadrilateral into two triangles ABC and ACD.

In ∆ABC, 3 + 4 + 6 = 180°        …(i)         (angle sum property)
In ∆ACD,
1 + 2 + 5 = 180°        …(ii)        (angle sum property)
Adding equations (i) and (ii), we get
1 + 3 + 2 + 4 + 5 + 6 = 180° + 180°
A + C + D + B = 360°
Hence, the sum of the measures of the angles of a convex quadrilateral is 360°.
Let us draw a non-convex quadrilateral.

Join the diagonal BD and prove as above in triangles BCD and ABD.
Thus, this property also holds true for a non-convex quadrilateral. ### Ex 3.1 Class 8 Maths Question 4.

Examine the table. (Each figure 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

Solution:
From the given table, we can conclude that the sum of the angles of a convex polygon with side ‘n’ = (n – 2) × 180°
(a) If the number of sides = 7
Then the angle sum = (7 – 2) × 180° = 5 × 180° = 900°
(b) If the number of sides = 8
Then the angle sum = (8 – 2) × 180° = 6 × 180° = 1080°
(c) If the number of sides = 10

Then the angle sum = (10 – 2) × 180° = 8 × 180° = 1440°
(d) If the number of sides = n
Then the angle sum = (n – 2) × 180°

### Ex 3.1 Class 8 Maths Question 5.

What is a regular polygon? State the name of a regular polygon of
(i) 3 sides             (ii) 4 sides              (iii) 6 sides

Solution:
A polygon, whose all sides and all angles are equal, is called a regular polygon.
(i) A polygon with 3 sides is called an equilateral triangle. (ii) A polygon with 4 sides is called a square. (iii) A polygon with 6 sides is called a regular hexagon.

### Ex 3.1 Class 8 Maths Question 6.

Solution:
(a) Sum of the angles of a quadrilateral = 360°
50° + 130° + 120° + x = 360°
300° + x = 360°
x = 360° – 300°
x = 60°

(b) Sum of the angles of a quadrilateral = 360°
x + 70° + 60° + 90° = 360°        [ 180° – 90° = 90°]
x + 220° = 360°
x = 360° – 220°
x = 140°

(c) Sum of the angles of a pentagon = 540°
30° + x + 110° + 120° + x = 540°

[ 180° – 70° = 110°; 180° – 60° = 120°]
2x + 260° = 540°
2x = 540° – 260°
2x = 280°
x = 140°
(d) Sum of the angles of a regular pentagon = 540°
x + x + x + x + x = 540°

[All the angles of a regular pentagon are equal]
5x = 540°
x = 108°

### Ex 3.1 Class 8 Maths Question 7. Solution:
(a) a + 30° + 90° = 180°      [Angle sum property of a triangle]
a + 120° = 180°

a = 180° – 120° = 60°
Now, y = 180° – a     [Linear pair]
y = 180° – 60°
y = 120°
and, z + 30° = 180°     [Linear pair]
z = 180° – 30° = 150°
also, x + 90° = 180°     [Linear pair]
x = 180° – 90° = 90°
Thus, x + y + z = 90° + 120° + 150° = 360°
(b)
r + 120° + 80° + 60° = 360°     [Angle sum property of a quadrilateral]
r + 260° = 360°
r = 360° – 260° = 100°
Now, x + 120° = 180°     (Linear pair)
x = 180° – 120° = 60°
and y + 80° = 180°          (Linear pair)
y = 180° – 80° = 100°
again, z + 60° = 180°       (Linear pair)
z = 180° – 60° = 120°
also, w = 180° –
r = 180° – 100° = 80°    (Linear pair)
Therefore, x + y + z + w = 60° + 100° + 120° + 80° = 360°.

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