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?(a) A convex quadrilateral
(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.
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 ∆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).
(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.
Find the angle measure x in the following figures:
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.
(a) Find x + y + z
Solution:
(a) ∠a + 30° + 90° =
180° [Angle sum property of a
triangle]
⇒ ∠a + 120° = 180°
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 = 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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