NCERT solutions for class 8 maths chapter 3 Understanding Quadrilaterals

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

 

a) A convex quadrilateral: 2.

b) A regular hexagon: 9.

c) A triangle: 0

 

 

Let ABCD be a convex quadrilateral.

From the figure, we infer that the quadrilateral ABCD is formed by two triangles,

i.e. ΔADC and ΔABC.

Since we know that sum of the interior angles of a triangle is 180°,

the sum of the measures of the angles is 180° + 180° = 360°

Let us take another quadrilateral ABCD which is not convex .

Join BC, such that it divides ABCD into two triangles ΔABC and ΔBCD. In ΔABC,

∠1 + ∠2 + ∠3 = 180° (angle sum property of triangle)

In ΔBCD,

∠4 + ∠5 + ∠6 = 180° (angle sum property of triangle)

∴, ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 = 180° + 180°

⇒ ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 = 360°

⇒ ∠A + ∠B + ∠C + ∠D = 360°

Thus, this property holds if the quadrilateral is not convex.

 

The angle sum of a polygon having side n = (n-2)×180°

a) 7

Here, n = 7

Thus, angle sum = (7-2)×180° = 5×180° = 900°

b) 8

Here, n = 8

Thus, angle sum = (8-2)×180° = 6×180° = 1080°

c) 10

Here, n = 10

Thus, angle sum = (10-2)×180° = 8×180° = 1440°

d) n

Here, n = n

Thus, angle sum = (n-2)×180°

 

Regular polygon: A polygon having sides of equal length and angles of equal measures is called a regular polygon. A regular polygon is both equilateral and equiangular.

(i) A regular polygon of 3 sides is called an equilateral triangle.

(ii) A regular polygon of 4 sides is called a square.

(iii) A regular polygon of 6 sides is called a regular hexagon.

 

a) The figure has 4 sides. Hence, it is a quadrilateral. Sum of angles of the quadrilateral = 360°

⇒ 50° + 130° + 120° + x = 360°

⇒ 300° + x = 360°

⇒ x = 360° – 300° = 60°

b) The figure has 4 sides. Hence, it is a quadrilateral. Also, one side is perpendicular forming a right angle.

Sum of angles of the quadrilateral = 360°

⇒ 90° + 70° + 60° + x = 360°

⇒ 220° + x = 360°

⇒ x = 360° – 220° = 140°

c) The figure has 5 sides. Hence, it is a pentagon.

Sum of angles of the pentagon = 540° Two angles at the bottom are a linear pair.

∴, 180° – 70° = 110°

180° – 60° = 120°

⇒ 30° + 110° + 120° + x + x = 540°

⇒ 260° + 2x = 540°

⇒ 2x = 540° – 260° = 280°

⇒ 2x = 280°

= 140°

d) The figure has 5 equal sides. Hence, it is a regular pentagon. Thus, all its angles are equal.

5x = 540°

⇒ x = 540°/5

⇒ x = 108°

 

a) Sum of all angles of triangle = 180°

One side of triangle = 180°- (90° + 30°) = 60°

x + 90° = 180° ⇒ x = 180° – 90° = 90°

y + 60° = 180° ⇒ y = 180° – 60° = 120°

z + 30° = 180° ⇒ z = 180° – 30° = 150°

x + y + z = 90° + 120° + 150° = 360°

b) Sum of all angles of quadrilateral = 360°

One side of quadrilateral = 360°- (60° + 80° + 120°) = 360° – 260° = 100°

x + 120° = 180° ⇒ x = 180° – 120° = 60°

y + 80° = 180° ⇒ y = 180° – 80° = 100°

z + 60° = 180° ⇒ z = 180° – 60° = 120°

w + 100° = 180° ⇒ w = 180° – 100° = 80°

x + y + z + w = 60° + 100° + 120° + 80° = 360°

 

a) An equilateral triangle is the regular polygon (with 3 sides) having the least possible minimum interior angle because a regular polygon can be constructed with minimum 3 sides.

Since the sum of interior angles of a triangle = 180°

Each interior angle = 180/3 = 60°

b) An equilateral triangle is the regular polygon (with 3 sides) having the maximum exterior angle because the regular polygon with the least number of sides has the maximum exterior angle possible. Maximum exterior possible = 180 – 60° = 120°

 

a)

125° + m = 180° ⇒ m = 180° – 125° = 55° (Linear pair)

125° + n = 180° ⇒ n = 180° – 125° = 55° (Linear pair)

x = m + n (The exterior angle of a triangle is equal to the sum of the two opposite interior angles)

⇒ x = 55° + 55° = 110°

b)

Two interior angles are right angles = 90°

70° + m = 180° ⇒ m = 180° – 70° = 110° (Linear pair)

60° + n = 180° ⇒ n = 180° – 60° = 120° (Linear pair) The figure is having five sides and is a pentagon.

Thus, sum of the angles of a pentagon = 540°

⇒ 90° + 90° + 110° + 120° + y = 540°

⇒ 410° + y = 540° ⇒ y = 540° – 410° = 130°

x + y = 180° (Linear pair)

⇒ x + 130° = 180°

⇒ x = 180° – 130° = 50°

 

Sum of the angles of a regular polygon having side n = (n-2)×180°

(i) Sum of the angles of a regular polygon having 9 sides = (9-2)×180°= 7×180° = 1260°

Each interior angle=1260/9 = 140°

Each exterior angle = 180° – 140° = 40°

Or,

Each exterior angle = Sum of exterior angles/Number of angles = 360/9 = 40°

(ii) Sum of angles of a regular polygon having side 15 = (15-2)×180°

= 13×180° = 2340°

Each interior angle = 2340/15 = 156°

Each exterior angle = 180° – 156° = 24°

Or,

Each exterior angle = sum of exterior angles/Number of angles = 360/15 = 24°

 

Each exterior angle = sum of exterior angles/Number of angles

24°= 360/ Number of sides

⇒ Number of sides = 360/24 = 15

Thus, the regular polygon has 15 sides.

 

Interior angle = 165°

Exterior angle = 180° – 165° = 15°

Number of sides = sum of exterior angles/exterior angles

⇒ Number of sides = 360/15 = 24

Thus, the regular polygon has 24 sides.

 

a) Exterior angle = 22°

Number of sides = sum of exterior angles/ exterior angle

⇒ Number of sides = 360/22 = 16.36

No, we can’t have a regular polygon with each exterior angle as 22° as it is not a divisor of 360.

b) Interior angle = 22°

Exterior angle = 180° – 22°= 158°

No, we can’t have a regular polygon with each exterior angle as 158° as it is not a divisor of 360.

 

ABCD is a figure of quadrilateral that is not a parallelogram but has exactly two opposite angles, that is, ∠B = ∠D of equal measure. It is not a parallelogram because ∠A ≠ ∠C.

 

Let the measures of two adjacent angles ∠A and ∠B be 3x and 2x, respectively in

parallelogram ABCD.

∠A + ∠B = 180°

⇒ 3x + 2x = 180°

⇒ 5x = 180°

⇒ x = 36°

We know that opposite sides of a parallelogram are equal.

∠A = ∠C = 3x = 3 × 36° = 108°

∠B = ∠D = 2x = 2 × 36° = 72°

 

(i) AD = BC (Opposite sides of a parallelogram are equal)

(ii) ∠DCB = ∠DAB (Opposite angles of a parallelogram are equal)

(iii) OC = OA (Diagonals of a parallelogram are equal)

(iv) m ∠DAB + m ∠CDA = 180°

 

(i)

y = 100° (opposite angles of a parallelogram)

x + 100° = 180° (adjacent angles of a parallelogram)

⇒ x = 180° – 100° = 80°

x = z = 80° (opposite angles of a parallelogram)

∴, x = 80°, y = 100° and z = 80°

(ii)

50° + x = 180° ⇒ x = 180° – 50° = 130° (adjacent angles of a parallelogram) x = y = 130° (opposite angles of a parallelogram)

x = z = 130° (corresponding angle)

(iii)

x = 90° (vertical opposite angles)

x + y + 30° = 180° (angle sum property of a triangle)

⇒ 90° + y + 30° = 180°

⇒ y = 180° – 120° = 60°

also, y = z = 60° (alternate angles)

(iv)

z = 80° (corresponding angle) z = y = 80° (alternate angles) x + y = 180° (adjacent angles)

⇒ x + 80° = 180° ⇒ x = 180° – 80° = 100°

(v)

x=28o

y = 112o z = 28o

 

(i) Yes, a quadrilateral ABCD can be a parallelogram if ∠D + ∠B = 180° but it should also fulfil some conditions, which are:

(a) The sum of the adjacent angles should be 180°.

(b) Opposite angles must be equal.

(ii) No, opposite sides should be of the same length. Here, AD ≠ BC

(iii) No, opposite angles should be of the same measures. ∠A ≠ ∠C

 

Let ABCD be a parallelogram.

Sum of adjacent angles of a parallelogram = 180°

∠A + ∠B = 180°

⇒ 2∠A = 180°

⇒ ∠A = 90°

also, 90° + ∠B = 180°

⇒ ∠B = 180° – 90° = 90°

∠A = ∠C = 90°

∠B = ∠D = 90

 

y = 40° (alternate interior angle)

∠P = 70° (alternate interior angle)

∠P = ∠H = 70° (opposite angles of a parallelogram)

z = ∠H – 40°= 70° – 40° = 30°

∠H + x = 180°

⇒ 70° + x = 180°

⇒ x = 180° – 70° = 110°

 

(i) SG = NU and SN = GU (opposite sides of a parallelogram are equal) 3x = 18

x = 18/3

⇒ x =6

3y – 1 = 26

⇒ 3y = 26 + 1

⇒ y = 27/3=9

x = 6 and y = 9

(ii) 20 = y + 7 and 16 = x + y (diagonals of a parallelogram bisect each other) y + 7 = 20

⇒ y = 20 – 7 = 13 and,

x + y = 16

⇒ x + 13 = 16

⇒ x = 16 – 13 = 3

x = 3 and y = 13

 

∠K + ∠R = 180° (adjacent angles of a parallelogram are supplementary)

⇒ 120° + ∠R = 180°

⇒ ∠R = 180° – 120° = 60°

also, ∠R = ∠SIL (corresponding angles)

⇒ ∠SIL = 60°

also, ∠ECR = ∠L = 70° (corresponding angles) x + 60° + 70° = 180° (angle sum of a triangle)

⇒ x + 130° = 180°

⇒ x = 180° – 130° = 50°

 

When a transversal line intersects two lines in such a way that the sum of the adjacent angles on the same side of transversal is 180°, then the lines are parallel to each other. Here, ∠M + ∠L = 100° + 80° = 180°

Thus, MN || LK

As the quadrilateral KLMN has one pair of parallel lines, it is a trapezium. MN and LK are parallel lines.

 

m∠C + m∠B = 180° (angles on the same side of transversal)

⇒ m∠C + 120° = 180°

⇒ m∠C = 180°- 120° = 60°

 

∠P + ∠Q = 180° (angles on the same side of transversal)

⇒ ∠P + 130° = 180°

⇒ ∠P = 180° – 130° = 50°

also, ∠R + ∠S = 180° (angles on the same side of transversal)

⇒ 90° + ∠S = 180°

⇒ ∠S = 180° – 90° = 90°

Thus, ∠P = 50° and ∠S = 90°

Yes, there are more than one method to find m∠P.

PQRS is a quadrilateral. Sum of measures of all angles is 360°.

Since, we know the measurement of ∠Q, ∠R and ∠S.

∠Q = 130°, ∠R = 90° and ∠S = 90°

∠P + 130° + 90° + 90° = 360°

⇒ ∠P + 310° = 360°

⇒ ∠P = 360° – 310° = 50°