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Understanding Quadrilaterals : Exercise 3.1 (Mathematics NCERT Class 8th)


Q.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
Sol. (a) Simple Curve – 1, 2, 5, 6, 7
(b) Simple closed curve – 1, 2, 5, 6, 7
(c) Polygon – 1, 2
(d) Convex polygon – 2
(e) Concave polygon – 1

Q.2 How many diagonals does each of the following have?
(a) A convex quadrilateral
(b) A regular hexagon
(c) A triangle
Sol. (a) A convex quadrilateral has two diagonals.
(b) A regular hexagon has 9 diagonals.
(c) A triangle has no diagonal.

Q.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?
Sol. The sum of the measures of the angles of a convex quadrilateral is 360ᵒ.
Yes. This property will hold true even if the quadrilateral is not convex. Since, any quadrilateral can be divided into two triangles.

Q.4 Examine the table. (Each figure is divided into triangles and the sum of the angles deduced from that.)

Figure

Side

3

4

5

6

Angle sum 180ᵒ 2 ×180ᵒ

= (4 – 2) × 180ᵒ

3 ×180ᵒ

= (5 – 2) × 180ᵒ

4 ×180ᵒ

= (6 – 2) × 180ᵒ

What can you say about the angle sum of a convex polygon with number of sides?
(a) 7 (b) 8 (c) 10 (d) n
Sol. From the table, it can be seen that the angle sum for any convex polygon of ‘n’ sides is (n – 2) × 180ᵒ. Hence, the angle sum of a convex polygon with given number of sides will be as follows:
(a) (7 – 2) × 180ᵒ = 900ᵒ
(b) (8 – 2) × 180ᵒ = 1080ᵒ
(c) (10 – 2) × 180ᵒ = 1440ᵒ
(d) (n – 2) × 180ᵒ

Q.5 What is a regular polygon?
State the name of a regular polygon of
(i) 3 sides (ii) 4 sides (iii) 6 sides
Sol. A polygon which is both equiangular and equilateral is called a regular polygon.
(i) A regular polygon of 3 sides is named as triangle.
(ii) A regular polygon of 4 sides is named as quadrilateral.
(iii) A regular polygon of 6 sides is named as hexagon.

Q.6 Find the angle measure x in the following figures.Sol. (a) Using the angle sum property of a quadrilateral, we know that the sum of the measure of all interior angles of a quadrilateral is 360ᵒ.
Hence, 50ᵒ + 130ᵒ + 120ᵒ + x = 360ᵒ
300ᵒ + x = 360ᵒ
x = 360ᵒ - 300ᵒ
x = 60ᵒ
(b) Using the angle sum property of a quadrilateral,
90ᵒ + 60ᵒ + 70ᵒ + x = 360ᵒ
220ᵒ + x = 360ᵒ
x = 360ᵒ - 220ᵒ
x = 140ᵒ
(c) From the figure, it can be seen that,
70ᵒ + a = 180ᵒ (Linear Pair)
a = 110ᵒ
60ᵒ + b = 180ᵒ (Linear Pair)
b = 120ᵒ
We know that, the sum of measure of interior angles of a pentagon is 540ᵒ
Hence, 120ᵒ + 110ᵒ + 30ᵒ + x + x = 540ᵒ
260ᵒ + 2x = 540ᵒ
2x = 280ᵒ
x = 140ᵒ
(d) We know that, the sum of measure of interior angles of a pentagon is 540ᵒ
5x = 280ᵒ
x = 108ᵒ

Q.7 (a) Find x + y + z (b) Find x + y + z + w
Sol. (a) We know that, sum of linear pair of angles is 180ᵒ
Therefore, 90ᵒ + x = 180ᵒ
x = 180ᵒ - 90ᵒ
x = 90ᵒ
Similarly, z + 30ᵒ = 180ᵒ
z = 180ᵒ - 30ᵒ
z = 150ᵒ
Using the exterior angle property, we can write,
Also, y = 90ᵒ + 30ᵒ = 120ᵒ
Now, x + y + z = 90ᵒ + 120ᵒ +150ᵒ = 360ᵒ

(b)

Using the angle sum property of a quadrilateral, we can write,
60ᵒ + 80ᵒ +120ᵒ+n = 360ᵒ
260ᵒ+n = 360ᵒ
n = 360ᵒ - 260ᵒ
n = 100ᵒ
We know that, sum of linear pair of angles is 180ᵒ
Hence, w + 100ᵒ= 180ᵒ
w = 80ᵒ
x+ 120ᵒ= 180ᵒ
x= 60ᵒ
y+ 80ᵒ= 180ᵒ
y = 100ᵒ
z + 60ᵒ= 180ᵒ
z = 120ᵒ
Therefore, x + y + z + w = 60ᵒ + 100ᵒ + 120ᵒ + 80ᵒ
x + y + z + w = 360ᵒ

 



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