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

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

Ex 3.1 Class 8 Maths Question 1.
Given here are some figures.

Classify each of the above figure 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)
(b) Simple closed curve: (1), (2), (5), (6) and (7)
(c) Polygon: (1) and (2)
(d) Convex polygon: (2)
(e) Concave polygon: (1)

Ex 3.1 Class 8 MathsÂ Question 2.
How many diagonals does each of the following have?
(a) A convex quadrilateral
(b) A regular hexagon
(c) A triangle
Solution:
(a) In Fig. (i) ABCD is a convex quadrilateral which has two diagonals AC and BD.

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

(c) In Fig. (iii) ABC is a triangle which has no diagonal.

Ex 3.1 Class 8 MathsÂ 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 given figure, we have a quadrilateral ABCD. Join AC diagonal which divides the quadrilateral into two triangles ABC and ADC.

In âˆ†ABC, âˆ 3 + âˆ 4 + âˆ 6 = 180Â°…(i) (angle sum property)
In âˆ†ADC, âˆ 1 + âˆ 2 + âˆ 5 = 180Â° …(ii) (angle sum property)
Adding, (i) and (ii)
âˆ 1 + âˆ 3 + âˆ 2 + âˆ 4 + âˆ 5 + âˆ 6 = 180Â° + 180Â°
â‡’ âˆ A + âˆ C + âˆ D + âˆ B = 360Â°
Hence, the sum of all the angles of a convex quadrilateral = 360Â°.
Let us draw a non-convex quadrilateral.
Yes, this property also holds true for a non-convex quadrilateral.

Ex 3.1 Class 8 MathsÂ Question 4.
Examine the table. (Each figure is divided into triangles and the sum of the angles reduced 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 above table, we conclude that the sum of all the angles of a polygon of side â€˜nâ€™
= (n – 2) Ã— 180Â°
(a) Number of sides = 7
Angles sum = (7 – 2) Ã— 180Â° = 5 Ã— 180Â° = 900Â°
(b) Number of sides = 8
Angle sum = (8 – 2) Ã— 180Â° = 6 Ã— 180Â° = 1080Â°
(c) Number of sides = 10 Angle sum = (10 – 2) Ã— 180Â° = 8 Ã— 180Â° = 1440Â°
(d) Number of sides = n
Angle sum = (n – 2) Ã— 180Â°

Ex 3.1 Class 8 MathsÂ 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 with equal sides and equal angles is called a regular polygon.
(i) Equilateral triangle

(ii) Square

(iii) Regular Hexagon

Ex 3.1 Class 8 MathsÂ Question 6.
Find the angle measure x in the following figures:

Solution:
(a) Angle sum of a quadrilateral = 360Â°
â‡’ 50Â° + 130Â° + 120Â° + x = 360Â°
â‡’ 300Â° + x = 360Â°
â‡’ x = 360Â° – 300Â° = 60Â°
(b) Angle sum of a quadrilateral = 360Â°
â‡’ x + 70Â° + 60Â° + 90Â° = 360Â° [âˆµ 180Â° – 90Â° = 90Â°]
â‡’ x + 220Â° = 360Â°
â‡’ x = 360Â° – 220Â° = 140Â°
(c) Angle sum of a pentagon = 540Â°
â‡’ 30Â° + x + 110Â° + 120Â° + x = 540Â° [âˆµ 180Â° – 70Â° = 110Â°; 180Â° – 60Â° = 120Â°]
â‡’ 2x + 260Â° = 540Â°
â‡’ 2x = 540Â° – 260Â°
â‡’ 2x = 280Â°
â‡’ x = 140Â°
(d) Angle sum of a regular pentagon = 540Â°
â‡’ x + x + x + x + x = 540Â° [All angles of a regular pentagon are equal]
â‡’ 5x = 540Â°
â‡’ x = 108Â°

Ex 3.1 Class 8 MathsÂ Question 7.
(a) Find x + y + z

(b) Find x + y + z + w

Solution:
(a) âˆ a + 30Â° + 90Â° = 180Â° [Angle sum property]
â‡’ âˆ a + 120Â° = 180Â°

â‡’ âˆ a = 180Â° – 120Â° = 60Â°
Now, y = 180Â° – a (Linear pair)
â‡’ y = 180Â° – 60Â°
â‡’ y = 120Â°
and, z + 30Â° = 180Â° [Linear pair]
â‡’ z = 180Â° – 30Â° = 150Â°
also, x + 90Â° = 180Â° [Linear pair]
â‡’ x = 180Â° – 90Â° = 90Â°
Thus x + y + z = 90Â° + 120Â° + 150Â° = 360Â°
(b) âˆ r + 120Â° + 80Â° + 60Â° = 360Â° [Angle sum property of a quadrilateral]

âˆ r + 260Â° = 360Â°
âˆ r = 360Â° – 260Â° = 100Â°
Now x + 120Â° = 180Â° (Linear pair)
x = 180Â° – 120Â° = 60Â°
y + 80Â° = 180Â° (Linear pair)
â‡’ y = 180Â° – 80Â° = 100Â°
z + 60Â° = 180Â° (Linear pair)
â‡’ z = 180Â° – 60Â° = 120Â°
w = 180Â° – âˆ r = 180Â° – 100Â° = 80Â° (Linear pair)
x + y + z + w = 60Â° + 100Â° + 120Â° + 80Â° = 360Â°.