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

- Class 8 Maths Understanding Quadrilaterals Exercise 3.1
- Class 8 Maths Understanding Quadrilaterals Exercise 3.2
- Class 8 Maths Understanding Quadrilaterals Exercise 3.3
- Class 8 Maths Understanding Quadrilaterals Exercise 3.4
- Understanding Quadrilaterals Class 8 Extra Questions

**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°.

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