Understanding Shapes

(a) Simple curve

A curve which does not cross itself at any point is called a simple curve.

Hence, figures (i), (ii), (iii), (v) and (vi) are simple curves.

(b) Simple closed curve

A curve which does not cross itself at any point and has the same beginning and end points is called a simple closed curve.

Hence, figures (iii), (v) and (vi) are simple closed curves.

(c) Polygon

A simple closed curve made up entirely of line segments is called a polygon.

Hence, figures (iii) and (vi) are polygons.

(d) Convex polygon

A polygon in which each interior angle is less than 180° is called a convex polygon.

Hence, figure (iii) is a convex polygon.

(e) Concave polygon

A polygon in which atleast one interior angle is greater than 180° is called a concave polygon.

Hence, figure (vi) is a concave polygon.

According to the properties of a polygon, if a polygon has $n$ sides, where $n$ is greater than 3, then the number of diagonals of the polygon is $\frac{n(n - 3)}{2}$.

(a) A convex quadrilateral

A quadrilateral has 4 sides, so $n$ = 4.

$$\begin{aligned}\text{Number of diagonals} = \frac{4(4 - 3)}{2} \\[1em] = \frac{4 \times 1}{2} \\[1em] = \frac{4}{2} \\[1em] = 2\end{aligned}$$

Hence, a convex quadrilateral has 2 diagonals.

(b) A regular hexagon

A hexagon has 6 sides, so $n$ = 6.

$$\begin{aligned}\text{Number of diagonals} = \frac{6(6 - 3)}{2} \\[1em] = \frac{6 \times 3}{2} \\[1em] = \frac{18}{2} \\[1em] = 9\end{aligned}$$

Hence, a regular hexagon has 9 diagonals.

According to the properties of a polygon, if a polygon has n sides, then the sum of measures of all its interior angles is (n - 2) × 180°.

(i) 8

Here, n = 8.

Sum of measures of all interior angles = (8 - 2) x 180°

= 6 x 180°

= 1080°

Hence, the sum of measures of all interior angles is 1080°.

(ii) 12

Here, n = 12.

Sum of measures of all interior angles = (12 - 2) x 180°

= 10 x 180°

= 1800°

Hence, the sum of measures of all interior angles is 1800°.

According to the properties of a polygon, if each exterior angle of a regular polygon is $x^{\circ}$, then the number of sides in the regular polygon is $\frac{360}{x}$.

(i) 24°

Number of sides = $\frac{360^{\circ}}{24^{\circ}}$

= 15

Hence, the number of sides is 15.

(ii) 60°

Number of sides = $\frac{360^{\circ}}{60^{\circ}}$

= 6

Hence, the number of sides is 6.

(iii) 72°

Number of sides = $\frac{360^{\circ}}{72^{\circ}}$

= 5

Hence, the number of sides is 5.

We know that the sum of an interior angle and its adjacent exterior angle is 180°.

Also, if each exterior angle of a regular polygon is $x^{\circ}$, then the number of sides in the regular polygon is $\frac{360}{x}$.

(i) 90°

Each exterior angle = 180° - 90° = 90°

Number of sides = $\frac{360^{\circ}}{90^{\circ}}$

= 4

A regular polygon of 4 sides is a square.

Hence, the number of sides is 4 and the polygon is a square.

(ii) 108°

Each exterior angle = 180° - 108° = 72°

Number of sides = $\frac{360^{\circ}}{72^{\circ}}$

= 5

A regular polygon of 5 sides is a pentagon.

Hence, the number of sides is 5 and the polygon is a pentagon.

(iii) 165°

Each exterior angle = 180° - 165° = 15°

Number of sides = $\frac{360^{\circ}}{15^{\circ}}$

= 24

A regular polygon of 24 sides is a 24-sided polygon.

Hence, the number of sides is 24 and the polygon is a 24-sided polygon.

According to the properties of a polygon, if a polygon has n sides, then the sum of measures of all its interior angles is (n - 2) × 180°.

(i) 1260°

⇒ (n - 2) × 180° = 1260°

⇒ n - 2 = $\frac{1260^{\circ}}{180^{\circ}}$

⇒ n - 2 = 7

⇒ n = 7 + 2

⇒ n = 9

Hence, the number of sides is 9.

(ii) 1980°

⇒ (n - 2) × 180° = 1980°

⇒ n - 2 = $\frac{1980^{\circ}}{180^{\circ}}$

⇒ n - 2 = 11

⇒ n = 11 + 2

⇒ n = 13

Hence, the number of sides is 13.

(iii) 3420°

⇒ (n - 2) × 180° = 3420°

⇒ n - 2 = $\frac{3420^{\circ}}{180^{\circ}}$

⇒ n - 2 = 19

⇒ n = 19 + 2

⇒ n = 21

Hence, the number of sides is 21.

It is given that the angles of a pentagon are in the ratio 7 : 8 : 11 : 13 : 15.

Let the common factor of the angles be a. So, the angles are 7a, 8a, 11a, 13a and 15a.

According to the properties of a polygon, if a polygon has n sides, then the sum of measures of all its interior angles is (n - 2) × 180°.

A pentagon has 5 sides, so n = 5.

Sum of interior angles = (5 - 2) x 180°

= 3 x 180°

= 540°

So,

⇒ 7a + 8a + 11a + 13a + 15a = 540°

⇒ 54a = 540°

⇒ a = $\frac{540^{\circ}}{54}$

⇒ a = 10°

Thus, the angles are:

7a = 7 x 10° = 70°

8a = 8 x 10° = 80°

11a = 11 x 10° = 110°

13a = 13 x 10° = 130°

15a = 15 x 10° = 150°

Hence, the angles of the pentagon are 70°, 80°, 110°, 130° and 150°.

It is given that the angles of a pentagon are x°, (x - 10)°, (x + 20)°, (2x - 44)° and (2x - 70)°.

According to the properties of a polygon, if a polygon has n sides, then the sum of measures of all its interior angles is (n - 2) × 180°.

A pentagon has 5 sides, so n = 5.

Sum of interior angles = (5 - 2) x 180°

= 3 x 180°

= 540°

So,

⇒ x° + (x - 10)° + (x + 20)° + (2x - 44)° + (2x - 70)° = 540°

⇒ x° + x° - 10° + x° + 20° + 2x° - 44° + 2x° - 70° = 540°

⇒ 7x° - 104° = 540°

⇒ 7x° = 540° + 104°

⇒ 7x° = 644°

⇒ x° = $\frac{644^{\circ}}{7}$

⇒ x = 92

Hence, the value of x is 92.

It is given that the exterior angles of a pentagon are in the ratio 1 : 2 : 3 : 4 : 5.

Let the common factor of the angles be a. So, the exterior angles are a, 2a, 3a, 4a and 5a.

According to the properties of a polygon, the sum of all the exterior angles of a polygon is 360°.

So,

⇒ a + 2a + 3a + 4a + 5a = 360°

⇒ 15a = 360°

⇒ a = $\frac{360^{\circ}}{15}$

⇒ a = 24°

Thus, the exterior angles are:

a = 24°

2a = 2 x 24° = 48°

3a = 3 x 24° = 72°

4a = 4 x 24° = 96°

5a = 5 x 24° = 120°

As the sum of an interior angle and its adjacent exterior angle is 180°, the interior angles are obtained by subtracting each exterior angle from 180°.

⇒ 180° - 24° = 156°

⇒ 180° - 48° = 132°

⇒ 180° - 72° = 108°

⇒ 180° - 96° = 84°

⇒ 180° - 120° = 60°

Hence, the interior angles of the pentagon are 156°, 132°, 108°, 84° and 60°.

It is given that in quadrilateral ABCD, AB ∥ DC, ∠A : ∠D = 2 : 3 and ∠B : ∠C = 7 : 8.

As AB ∥ DC and AD is a transversal, the co-interior angles ∠A and ∠D are supplementary.

⇒ ∠A + ∠D = 180°

Let the common factor of ∠A and ∠D be a. So, ∠A = 2a and ∠D = 3a.

⇒ 2a + 3a = 180°

⇒ 5a = 180°

⇒ a = $\frac{180^{\circ}}{5}$

⇒ a = 36°

Thus,

∠A = 2a = 2 x 36° = 72°

∠D = 3a = 3 x 36° = 108°

As AB ∥ DC and BC is a transversal, the co-interior angles ∠B and ∠C are supplementary.

⇒ ∠B + ∠C = 180°

Let the common factor of ∠B and ∠C be b. So, ∠B = 7b and ∠C = 8b.

⇒ 7b + 8b = 180°

⇒ 15b = 180°

⇒ b = $\frac{180^{\circ}}{15}$

⇒ b = 12°

Thus,

∠B = 7b = 7 x 12° = 84°

∠C = 8b = 8 x 12° = 96°

Hence, ∠A = 72°, ∠B = 84°, ∠C = 96° and ∠D = 108°.

(i) From the figure, in quadrilateral ABCD,

∠DAB = (3x + 4)°, ∠ABC = (50 + x)°, ∠BCD = (5x + 8)° and ∠CDA = (3x + 10)°

As we know, the sum of all the interior angles of a quadrilateral is 360°.

⇒ ∠DAB + ∠ABC + ∠BCD + ∠CDA = 360°

⇒ (3x + 4)° + (50 + x)° + (5x + 8)° + (3x + 10)° = 360°

⇒ 3x° + 4° + 50° + x° + 5x° + 8° + 3x° + 10° = 360°

⇒ 12x° + 72° = 360°

⇒ 12x° = 360° - 72°

⇒ 12x° = 288°

⇒ x° = $\frac{288^{\circ}}{12}$

⇒ x = 24

Hence, the value of x is 24.

(ii) ∠DAB = (3x + 4)°

= (3 x 24 + 4)°

= (72 + 4)°

= 76°

Hence, ∠DAB = 76°.

(iii) In ΔADB, ∠DAB = 76° and ∠ABD = 50°.

As we know, the sum of all the angles of a triangle is 180°.

⇒ ∠DAB + ∠ABD + ∠ADB = 180°

⇒ 76° + 50° + ∠ADB = 180°

⇒ 126° + ∠ADB = 180°

⇒ ∠ADB = 180° - 126°

⇒ ∠ADB = 54°

Hence, ∠ADB = 54°.

As we know, the sum of all the interior angles of a quadrilateral is 360°.

(i) From the figure, the interior angles of the quadrilateral are 40°, 140°, 100° and x.

⇒ 40° + 140° + 100° + x = 360°

⇒ 280° + x = 360°

⇒ x = 360° - 280°

⇒ x = 80°

Hence, x = 80°.

(ii) From the figure, the interior angles of the quadrilateral are 90°, x, 110° and 60°.

⇒ 90° + x + 110° + 60° = 360°

⇒ 260° + x = 360°

⇒ x = 360° - 260°

⇒ x = 100°

Hence, x = 100°.

(iii) From the figure, three interior angles of the quadrilateral are 90°, 110° and 83°.

Let the fourth interior angle be a.

⇒ 90° + 110° + 83° + a = 360°

⇒ 283° + a = 360°

⇒ a = 360° - 283°

⇒ a = 77°

As the side is produced, x and a form a linear pair.

⇒ x + a = 180°

⇒ x + 77° = 180°

⇒ x = 180° - 77°

⇒ x = 103°

Hence, x = 103°.

It is given that a heptagon has three equal angles each of 120° and four equal angles.

According to the properties of a polygon, if a polygon has n sides, then the sum of measures of all its interior angles is (n - 2) × 180°.

A heptagon has 7 sides, so n = 7.

Sum of interior angles = (7 - 2) x 180°

= 5 x 180°

= 900°

Let the size of each of the four equal angles be a.

So,

⇒ 3 × 120° + 4a = 900°

⇒ 360° + 4a = 900°

⇒ 4a = 900° - 360°

⇒ 4a = 540°

⇒ a = $\frac{540^{\circ}}{4}$

⇒ a = 135°

Hence, the size of each equal angle is 135°.

It is given that the ratio between an exterior angle and the interior angle of a regular polygon is 1 : 5.

Let the common factor be a. So, the exterior angle is a and the interior angle is 5a.

As we know, the sum of an interior angle and its adjacent exterior angle is 180°.

⇒ a + 5a = 180°

⇒ 6a = 180°

⇒ a = $\frac{180^{\circ}}{6}$

⇒ a = 30°

(i) Each exterior angle = a = 30°

Hence, the measure of each exterior angle is 30°.

(ii) Each interior angle = 5a = 5 x 30° = 150°

Hence, the measure of each interior angle is 150°.

(iii) According to the properties of a polygon, if each exterior angle of a regular polygon is $x^{\circ}$, then the number of sides in the regular polygon is $\frac{360}{x}$.

Number of sides = $\frac{360^{\circ}}{30^{\circ}}$

= 12

Hence, the number of sides in the polygon is 12.

It is given that each interior angle of a regular polygon is double of its exterior angle.

Let the exterior angle be a. So, the interior angle is 2a.

As we know, the sum of an interior angle and its adjacent exterior angle is 180°.

⇒ a + 2a = 180°

⇒ 3a = 180°

⇒ a = $\frac{180^{\circ}}{3}$

⇒ a = 60°

So, each exterior angle of the regular polygon is 60°.

According to the properties of a polygon, if each exterior angle of a regular polygon is $x^{\circ}$, then the number of sides in the regular polygon is $\frac{360}{x}$.

Number of sides = $\frac{360^{\circ}}{60^{\circ}}$

= 6

Hence, the number of sides in the polygon is 6.

From the figure, in parallelogram ABCD, AB = 9 cm, BC = 6 cm, ∠ABC = 120°, AO = 7 cm and DO = 5 cm.

(i) The opposite sides of a parallelogram are of equal length.

AD = BC = 6 cm

Hence, AD = 6 cm.

(ii) The opposite sides of a parallelogram are of equal length.

DC = AB = 9 cm

Hence, DC = 9 cm.

(iii) The adjacent angles of a parallelogram are supplementary.

⇒ ∠DCB + ∠ABC = 180°

⇒ ∠DCB + 120° = 180°

⇒ ∠DCB = 180° - 120°

⇒ ∠DCB = 60°

Hence, ∠DCB = 60°.

(iv) The opposite angles of a parallelogram are of equal measure.

∠ADC = ∠ABC = 120°

Hence, ∠ADC = 120°.

(v) The opposite angles of a parallelogram are of equal measure.

∠DAB = ∠DCB = 60°

Hence, ∠DAB = 60°.

(vi) The diagonals of a parallelogram bisect each other.

OC = AO = 7 cm

Hence, OC = 7 cm.

(vii) The diagonals of a parallelogram bisect each other.

OB = DO = 5 cm

Hence, OB = 5 cm.

(viii) The adjacent angles of a parallelogram are supplementary.

m∠DAB + m∠CDA = 60° + 120° = 180°

Hence, m∠DAB + m∠CDA = 180°.

From the figure, 120° is the exterior angle at the vertex where the side has been produced.

As the sum of an interior angle and its adjacent exterior angle is 180°, the interior angle at that vertex is:

= 180° - 120°

= 60°

As the opposite angles of a parallelogram are of equal measure,

x = 60°

As the adjacent angles of a parallelogram are supplementary,

⇒ z + 60° = 180°

⇒ z = 180° - 60°

⇒ z = 120°

As the opposite angles of a parallelogram are of equal measure,

y = z = 120°

Hence, x = 60°, y = 120° and z = 120°.

Let the parallelogram be ABCD, where A is the bottom-left vertex, B is the bottom-right vertex, C is the top-right vertex and D is the top-left vertex, with diagonals AC and BD meeting at O.

From the figure, ∠DAC = 40°, ∠CAB = 30° and ∠DOC = 100°.

As AB ∥ DC and AC is a transversal,

∠DCA = ∠CAB = 30° (alternate angles are equal)

In ΔDOC, the sum of all the angles of a triangle is 180°.

⇒ x + ∠DOC + ∠DCA = 180°

⇒ x + 100° + 30° = 180°

⇒ x + 130° = 180°

⇒ x = 180° - 130°

⇒ x = 50°

As AD ∥ BC and AC is a transversal,

z = ∠DAC = 40° (alternate angles are equal)

As ∠DOC and ∠BOC form a linear pair,

⇒ ∠BOC = 180° - 100° = 80°

In ΔBOC, the sum of all the angles of a triangle is 180°.

⇒ y + ∠BOC + z = 180°

⇒ y + 80° + 40° = 180°

⇒ y + 120° = 180°

⇒ y = 180° - 120°

⇒ y = 60°

Hence, x = 50°, y = 60° and z = 40°.

Let the parallelogram be ABCD, where A is the bottom-left vertex, B is the bottom-right vertex, C is the top-right vertex and D is the top-left vertex, with AC as the diagonal.

From the figure, ∠DAC = 35° and ∠ABC = 120°.

As the opposite angles of a parallelogram are of equal measure,

y = ∠ADC = ∠ABC = 120°

As the adjacent angles of a parallelogram are supplementary,

⇒ ∠DAB + ∠ABC = 180°

⇒ ∠DAB + 120° = 180°

⇒ ∠DAB = 180° - 120°

⇒ ∠DAB = 60°

From the figure,

⇒ ∠DAC + z = ∠DAB

⇒ 35° + z = 60°

⇒ z = 60° - 35°

⇒ z = 25°

As AB ∥ DC and AC is a transversal,

x = z = 25° (alternate angles are equal)

Hence, x = 25°, y = 120° and z = 25°.

Let the parallelogram be ABCD, where A is the bottom-left vertex, B is the bottom-right vertex, C is the top-right vertex and D is the top-left vertex, with the diagonal AC produced beyond C.

From the figure, ∠DAC = 67° and ∠ABC = 70°.

As the opposite angles of a parallelogram are of equal measure,

z = ∠ADC = ∠ABC = 70°

As the adjacent angles of a parallelogram are supplementary,

⇒ ∠DAB + ∠ABC = 180°

⇒ ∠DAB + 70° = 180°

⇒ ∠DAB = 180° - 70°

⇒ ∠DAB = 110°

From the figure,

⇒ ∠DAC + x = ∠DAB

⇒ 67° + x = 110°

⇒ x = 110° - 67°

⇒ x = 43°

As AB ∥ DC and AC is a transversal,

∠DCA = x = 43° (alternate angles are equal)

As the diagonal AC is produced, ∠DCA and y form a linear pair.

⇒ y + ∠DCA = 180°

⇒ y + 43° = 180°

⇒ y = 180° - 43°

⇒ y = 137°

Hence, x = 43°, y = 137° and z = 70°.

It is given that two adjacent sides of a parallelogram are in the ratio 5 : 7 and the perimeter of the parallelogram is 72 cm.

Let the parallelogram be ABCD and let the common factor be a. So, AB = 5a cm and BC = 7a cm.

As the opposite sides of a parallelogram are of equal length,

DC = AB = 5a cm and AD = BC = 7a cm

Perimeter of parallelogram ABCD = AB + BC + CD + DA

⇒ 72 = 5a + 7a + 5a + 7a

⇒ 72 = 24a

⇒ a = $\frac{72}{24}$

⇒ a = 3

Thus, the sides are:

5a = 5 x 3 = 15 cm

7a = 7 x 3 = 21 cm

Hence, the lengths of the sides are 15 cm and 21 cm.

It is given that the measure of two adjacent angles of a parallelogram are in the ratio 4 : 5.

Let the parallelogram be ABCD and let the common factor be a. So, ∠A = 4a and ∠B = 5a.

As the adjacent angles of a parallelogram are supplementary,

⇒ ∠A + ∠B = 180°

⇒ 4a + 5a = 180°

⇒ 9a = 180°

⇒ a = $\frac{180^{\circ}}{9}$

⇒ a = 20°

Thus,

∠A = 4a = 4 x 20° = 80°

∠B = 5a = 5 x 20° = 100°

As the opposite angles of a parallelogram are of equal measure,

∠C = ∠A = 80° and ∠D = ∠B = 100°

Hence, the angles of the parallelogram are 80°, 100°, 80° and 100°.

(i) ∠A + ∠C = 180°

In a parallelogram, the opposite angles are equal, that is ∠A = ∠C.

The given condition only tells us that ∠A and ∠C are supplementary, which does not mean that they are equal. For example, ∠A could be 100° and ∠C could be 80°.

Hence, ABCD may not be a parallelogram.

(ii) AD = BC = 6 cm, AB = 5 cm, DC = 4.5 cm

In a parallelogram, both the pairs of opposite sides are equal.

Here, AD = BC = 6 cm, but AB = 5 cm and DC = 4.5 cm, so AB ≠ DC.

Hence, ABCD is not a parallelogram as AB ≠ DC.

(iii) ∠B = 80°, ∠D = 70°

In a parallelogram, the opposite angles are equal.

Here, ∠B and ∠D are opposite angles, but ∠B ≠ ∠D.

Hence, ABCD is not a parallelogram as ∠B ≠ ∠D.

(iv) ∠B + ∠C = 180°

As ∠B and ∠C are co-interior angles, this condition shows that AB ∥ DC.

However, it gives us no information about the sides AD and BC, so we cannot say that AD ∥ BC.

Hence, ABCD may not be a parallelogram.

(i) In parallelogram HOPE, the side HO is produced and the exterior angle so formed at O is 70°.

As the sum of an interior angle and its adjacent exterior angle is 180°,

⇒ ∠HOP + 70° = 180°

⇒ ∠HOP = 180° - 70°

⇒ ∠HOP = 110°

As the opposite angles of a parallelogram are of equal measure,

x = ∠HEP = ∠HOP = 110°

As HE ∥ OP and HP is a transversal,

y = ∠HPO = ∠EHP = 40° (alternate angles are equal)

In ΔHOP, the sum of all the angles of a triangle is 180°.

⇒ z + ∠HOP + ∠HPO = 180°

⇒ z + 110° + 40° = 180°

⇒ z + 150° = 180°

⇒ z = 180° - 150°

⇒ z = 30°

Hence, x = 110°, y = 40° and z = 30°. The properties used are that the opposite angles of a parallelogram are equal, that alternate angles formed by a transversal cutting two parallel lines are equal and that the sum of the angles of a triangle is 180°.

(ii) In parallelogram ROPE, the side RO is produced and the exterior angle so formed at O is 80°.

As the sum of an interior angle and its adjacent exterior angle is 180°,

⇒ ∠ROP + 80° = 180°

⇒ ∠ROP = 180° - 80°

⇒ ∠ROP = 100°

From the figure,

⇒ x + 60° = ∠ROP

⇒ x + 60° = 100°

⇒ x = 100° - 60°

⇒ x = 40°

As the adjacent angles of a parallelogram are supplementary,

⇒ y + ∠ROP = 180°

⇒ y + 100° = 180°

⇒ y = 180° - 100°

⇒ y = 80°

As EP ∥ RO and EO is a transversal,

z = x = 40° (alternate angles are equal)

Hence, x = 40°, y = 80° and z = 40°. The properties used are that the adjacent angles of a parallelogram are supplementary and that alternate angles formed by a transversal cutting two parallel lines are equal.

(i) In parallelogram TURN,

TN = 5y - 1, NR = 28, RU = 24 and TU = 4x + 2.

As the opposite sides of a parallelogram are of equal length,

⇒ TU = NR

⇒ 4x + 2 = 28

⇒ 4x = 28 - 2

⇒ 4x = 26

⇒ x = $\frac{26}{4}$

⇒ x = 6.5

Also,

TN = RU

⇒ 5y - 1 = 24

⇒ 5y = 24 + 1

⇒ 5y = 25

⇒ y = $\frac{25}{5}$

⇒ y = 5

Hence, x = 6.5 cm and y = 5 cm.

(ii) In parallelogram BURN, the diagonals BR and NU meet at O, where NO = x + 3, OU = 18, BO = x + y and OR = 20.

As the diagonals of a parallelogram bisect each other,

NO = OU

⇒ x + 3 = 18

⇒ x = 18 - 3

⇒ x = 15

Also,

BO = OR

⇒ x + y = 20

⇒ 15 + y = 20

⇒ y = 20 - 15

⇒ y = 5

Hence, x = 15 cm and y = 5 cm.

It is given that both ABCD and PQRS are parallelograms.

Let BC and PS intersect at M.

In parallelogram ABCD, ∠DAB = 120°.

As the adjacent angles of a parallelogram are supplementary,

⇒ ∠ABC + ∠DAB = 180°

⇒ ∠ABC + 120° = 180°

⇒ ∠ABC = 180° - 120°

⇒ ∠ABC = 60°

In parallelogram PQRS, ∠SRQ = 50°.

As the opposite angles of a parallelogram are of equal measure,

∠SPQ = ∠SRQ = 50°

Now, in ΔPMB,

∠MPB = ∠SPQ = 50° and ∠MBP = ∠ABC = 60°

As the sum of all the angles of a triangle is 180°,

⇒ ∠MPB + ∠MBP + x = 180°

⇒ 50° + 60° + x = 180°

⇒ 110° + x = 180°

⇒ x = 180° - 110°

⇒ x = 70°

Hence, the value of x is 70°.

It is given that ABCD is a parallelogram whose diagonals intersect at O, ∠AOD = 68°, ∠BDC = 30° and ∠DBC = 46°.

(i) As AD ∥ BC and BD is a transversal,

∠ADB = ∠DBC = 46° (alternate angles are equal)

In ΔAOD, the sum of all the angles of a triangle is 180°.

⇒ ∠CAD + ∠AOD + ∠ADB = 180°

⇒ ∠CAD + 68° + 46° = 180°

⇒ ∠CAD + 114° = 180°

⇒ ∠CAD = 180° - 114°

⇒ ∠CAD = 66°

Hence, ∠CAD = 66°.

(ii) As ∠AOD and ∠DOC form a linear pair,

⇒ ∠DOC = 180° - 68° = 112°

In ΔDOC, the sum of all the angles of a triangle is 180°.

⇒ ∠ACD + ∠DOC + ∠BDC = 180°

⇒ ∠ACD + 112° + 30° = 180°

⇒ ∠ACD + 142° = 180°

⇒ ∠ACD = 180° - 142°

⇒ ∠ACD = 38°

Hence, ∠ACD = 38°.

(iii) From the figure,

∠ADC = ∠ADB + ∠BDC

= 46° + 30°

= 76°

Hence, ∠ADC = 76°.

It is given that ABCD is a parallelogram and DN and BP are perpendiculars drawn on the diagonal AC.

(i) In ΔDCN and ΔBAP,

∠DNC = ∠BPA = 90° (as DN ⊥ AC and BP ⊥ AC)

DC = BA (opposite sides of a parallelogram are equal)

As DC ∥ AB and AC is a transversal,

∠DCN = ∠BAP (alternate angles are equal)

Therefore, by the A.A.S. congruence rule,

ΔDCN ≅ ΔBAP

Hence, proved that ΔDCN ≅ ΔBAP.

(ii) As ΔDCN ≅ ΔBAP, the corresponding parts of congruent triangles are equal.

So, CN = AP ...............(1)

From the figure,

AN = AC - CN ...............(2)

CP = AC - AP ...............(3)

Using equation (1) in equation (3),

CP = AC - CN ...............(4)

From equations (2) and (4),

AN = CP

Hence, proved that AN = CP.

In quadrilateral ARCB,

AR ∥ BC (as the line through A is parallel to BC)

RC ∥ AB (as the line through C is parallel to AB)

As both the pairs of opposite sides are parallel, ARCB is a parallelogram.

So, AR = BC ...............(1)

In quadrilateral AQBC,

AQ ∥ BC (as the line through A is parallel to BC)

QB ∥ AC (as the line through B is parallel to CA)

As both the pairs of opposite sides are parallel, AQBC is a parallelogram.

So, AQ = BC and QB = AC ...............(2)

In quadrilateral ACPB,

CP ∥ AB (as the line through C is parallel to AB)

BP ∥ AC (as the line through B is parallel to CA)

As both the pairs of opposite sides are parallel, ACPB is a parallelogram.

So, CP = AB and BP = AC ...............(3)

Also, from parallelogram ARCB, RC = AB ...............(4)

Now, from the figure,

QR = QA + AR

Using equations (1) and (2),

⇒ QR = BC + BC = 2BC ...............(5)

RP = RC + CP

Using equations (3) and (4),

⇒ RP = AB + AB = 2AB ...............(6)

PQ = PB + BQ

Using equations (2) and (3),

⇒ PQ = AC + AC = 2AC ...............(7)

Adding equations (5), (6) and (7),

⇒ PQ + QR + RP = 2AC + 2BC + 2AB

⇒ PQ + QR + RP = 2(AB + BC + CA)

Hence, proved that 2(AB + BC + CA) = PQ + QR + RP.

(i) Four sides of equal length

A square has all its four sides equal in length and a rhombus also has all its four sides equal in length.

Hence, a square and a rhombus have four sides of equal length.

(ii) Four right angles

A square has all its four angles equal to 90° and a rectangle also has all its four angles equal to 90°.

Hence, a square and a rectangle have four right angles.

(i) A quadrilateral

A quadrilateral is a polygon bounded by four line segments.

A square is a polygon bounded by four line segments.

Hence, a square is a 4-sided polygon, so it is a quadrilateral.

(ii) A parallelogram

A parallelogram is a quadrilateral in which both the pairs of opposite sides are parallel.

In a square, both the pairs of opposite sides are parallel.

Hence, a square has its opposite sides parallel, so it is a parallelogram.

(iii) A rhombus

A rhombus is a parallelogram in which all the four sides are equal.

In a square, all the four sides are equal.

Hence, a square is a parallelogram with all the 4 sides equal, so it is a rhombus.

(iv) A rectangle

A rectangle is a parallelogram in which each angle is a right angle.

In a square, each angle is a right angle.

Hence, a square is a parallelogram with each angle a right angle, so it is a rectangle.

(i) The diagonals of a parallelogram bisect each other. As a rhombus, a rectangle and a square are all parallelograms, their diagonals also bisect each other.

Hence, the quadrilaterals are parallelogram, rhombus, square and rectangle.

(ii) The diagonals of a rhombus bisect each other at right angles. As a square is also a rhombus, its diagonals are also perpendicular bisectors of each other.

Hence, the quadrilaterals are rhombus and square.

(iii) The diagonals of a rectangle are equal in length. As a square is also a rectangle, its diagonals are also equal in length.

Hence, the quadrilaterals are rectangle and square.

It is given that one of the diagonals of a rhombus and its sides are equal.

Let ABCD be the rhombus in which the diagonal AC is equal to each of its sides.

So, AB = BC = CD = DA = AC

In ΔABC,

AB = BC = AC

As all the three sides are equal, ΔABC is an equilateral triangle.

So, ∠BAC = ∠ABC = ∠BCA = 60°

Similarly, in ΔACD,

AC = CD = DA

As all the three sides are equal, ΔACD is an equilateral triangle.

So, ∠CAD = ∠ACD = ∠ADC = 60°

Thus, the angles of the rhombus are:

⇒ ∠A = ∠BAC + ∠CAD = 60° + 60° = 120°

⇒ ∠B = ∠ABC = 60°

⇒ ∠C = ∠BCA + ∠ACD = 60° + 60° = 120°

⇒ ∠D = ∠ADC = 60°

Hence, the angles of the rhombus are 60°, 120°, 60° and 120°.

It is given that ABCD is a rhombus in which the diagonals AC and BD intersect at O, with DO = 6, OC = 8, AO = x, OB = y and AB = z.

As the diagonals of a rhombus bisect each other,

AO = OC

⇒ x = 8

Also,

OB = DO

⇒ y = 6

As the diagonals of a rhombus intersect at right angles, ∠AOB = 90°.

So, ΔAOB is a right angled triangle.

By Pythagoras theorem,

⇒ AB² = AO² + OB²

⇒ z² = 8² + 6²

⇒ z² = 64 + 36

⇒ z² = 100

⇒ z = $\sqrt{100}$

⇒ z = 10

Hence, x = 8, y = 6 and z = 10.

It is given that ABCD is a trapezium in which AB ∥ DC, ∠A : ∠D = 5 : 7, ∠B = (3x + 11)° and ∠C = (5x - 31)°.

As AB ∥ DC and AD is a transversal, the co-interior angles ∠A and ∠D are supplementary.

⇒ ∠A + ∠D = 180°

Let the common factor of ∠A and ∠D be a. So, ∠A = 5a and ∠D = 7a.

⇒ 5a + 7a = 180°

⇒ 12a = 180°

⇒ a = $\frac{180^{\circ}}{12}$

⇒ a = 15°

Thus,

∠A = 5a = 5 x 15° = 75°

∠D = 7a = 7 x 15° = 105°

As AB ∥ DC and BC is a transversal, the co-interior angles ∠B and ∠C are supplementary.

⇒ ∠B + ∠C = 180°

⇒ (3x + 11)° + (5x - 31)° = 180°

⇒ 3x° + 11° + 5x° - 31° = 180°

⇒ 8x° - 20° = 180°

⇒ 8x° = 180° + 20°

⇒ 8x° = 200°

⇒ x° = $\frac{200^{\circ}}{8}$

⇒ x = 25

Thus,

∠B = (3x + 11)° = (3 x 25 + 11)° = (75 + 11)° = 86°

∠C = (5x - 31)° = (5 x 25 - 31)° = (125 - 31)° = 94°

Hence, ∠A = 75°, ∠B = 86°, ∠C = 94° and ∠D = 105°.

It is given that ABCD is a rectangle and ∠CEB : ∠ECB = 3 : 2.

(i) As ABCD is a rectangle, each of its angles is 90°.

So, ∠ABC = 90°, which means ∠EBC = 90°.

Let the common factor be a. So, ∠CEB = 3a and ∠ECB = 2a.

In ΔEBC, the sum of all the angles of a triangle is 180°.

⇒ ∠CEB + ∠ECB + ∠EBC = 180°

⇒ 3a + 2a + 90° = 180°

⇒ 5a = 180° - 90°

⇒ 5a = 90°

⇒ a = $\frac{90^{\circ}}{5}$

⇒ a = 18°

Thus,

∠CEB = 3a = 3 x 18° = 54°

Hence, ∠CEB = 54°.

(ii) ∠ECB = 2a = 2 x 18° = 36°

As ABCD is a rectangle, ∠DCB = 90°.

From the figure,

⇒ ∠DCE = ∠DCB - ∠ECB

⇒ ∠DCE = 90° - 36°

⇒ ∠DCE = 54°

As EC is produced to F, ∠DCE and ∠DCF form a linear pair.

⇒ ∠DCF + ∠DCE = 180°

⇒ ∠DCF + 54° = 180°

⇒ ∠DCF = 180° - 54°

⇒ ∠DCF = 126°

Hence, ∠DCF = 126°.

It is given that ABCD is a rectangle whose diagonals intersect at O and ∠AOB = 118°.

As the diagonals of a rectangle are equal and bisect each other,

OA = OB = OC = OD

(i) In ΔAOB, OA = OB, so ΔAOB is an isosceles triangle.

So, ∠ABO = ∠BAO (angles opposite to equal sides are equal)

As the sum of all the angles of a triangle is 180°,

⇒ ∠ABO + ∠BAO + ∠AOB = 180°

⇒ 2∠ABO + 118° = 180°

⇒ 2∠ABO = 180° - 118°

⇒ 2∠ABO = 62°

⇒ ∠ABO = $\frac{62^{\circ}}{2}$

⇒ ∠ABO = 31°

Hence, ∠ABO = 31°.

(ii) As ∠AOB and ∠AOD form a linear pair,

⇒ ∠AOD = 180° - 118° = 62°

In ΔAOD, OA = OD, so ΔAOD is an isosceles triangle.

So, ∠ADO = ∠DAO (angles opposite to equal sides are equal)

As the sum of all the angles of a triangle is 180°,

⇒ ∠ADO + ∠DAO + ∠AOD = 180°

⇒ 2∠ADO + 62° = 180°

⇒ 2∠ADO = 180° - 62°

⇒ 2∠ADO = 118°

⇒ ∠ADO = $\frac{118^{\circ}}{2}$

⇒ ∠ADO = 59°

Hence, ∠ADO = 59°.

(iii) In ΔAOD, OA = OD, so ∠OAD = ∠ADO = 59°.

As AD ∥ BC and AC is a transversal,

∠OCB = ∠OAD (alternate angles are equal) In the adjoining isosceles trapezium ABCD, ∠C = 102°. Find all the remaining angles of the trapezium.

⇒ ∠OCB = 59°

Hence, ∠OCB = 59°.

It is given that ABCD is a rhombus and ∠ABD = 50°.

Let the diagonals AC and BD intersect at O.

As the diagonals of a rhombus intersect at right angles,

∠AOB = 90°

(i) In ΔAOB, the sum of all the angles of a triangle is 180°.

⇒ ∠CAB + ∠ABO + ∠AOB = 180°

⇒ ∠CAB + 50° + 90° = 180°

⇒ ∠CAB + 140° = 180°

⇒ ∠CAB = 180° - 140°

⇒ ∠CAB = 40°

Hence, ∠CAB = 40°.

(ii) As the diagonals of a rhombus bisect the angles of the rhombus, the diagonal BD bisects ∠ABC.

⇒ ∠ABC = 2 x ∠ABD

⇒ ∠ABC = 2 x 50°

⇒ ∠ABC = 100°

As the adjacent angles of a rhombus are supplementary,

⇒ ∠BCD + ∠ABC = 180°

⇒ ∠BCD + 100° = 180°

⇒ ∠BCD = 180° - 100°

⇒ ∠BCD = 80°

Hence, ∠BCD = 80°.

(iii) As the opposite angles of a rhombus are of equal measure,

∠ADC = ∠ABC = 100°

Hence, ∠ADC = 100°.

It is given that ABCD is an isosceles trapezium in which AB ∥ DC, AD = BC and ∠C = 102°.

As the angles on the same base of an isosceles trapezium are equal,

∠D = ∠C = 102°

As AB ∥ DC and BC is a transversal, the co-interior angles ∠B and ∠C are supplementary.

⇒ ∠B + ∠C = 180°

⇒ ∠B + 102° = 180°

⇒ ∠B = 180° - 102°

⇒ ∠B = 78°

As the angles on the same base of an isosceles trapezium are equal,

∠A = ∠B = 78°

Hence, ∠A = 78°, ∠B = 78° and ∠D = 102°.

It is given that PQRS is a kite in which ∠Q = 120°, ∠R = 50°, ∠P = y and ∠S = x.

In kite PQRS, PQ = PS and RQ = RS.

So, the angles included between the pairs of unequal sides are equal.

⇒ ∠S = ∠Q

⇒ x = 120°

As we know, the sum of all the interior angles of a quadrilateral is 360°.

⇒ ∠P + ∠Q + ∠R + ∠S = 360°

⇒ y + 120° + 50° + 120° = 360°

⇒ y + 290° = 360°

⇒ y = 360° - 290°

⇒ y = 70°

Hence, x = 120° and y = 70°.

(i) The curves which have different beginning and end points are called open curves.

(ii) A curve which does not cross itself at any point is called a simple curve.

(iii) A simple closed curve made up entirely of line segments is called a polygon.

(iv) A polygon in which each interior angle is less than 180° is called a convex polygon.

(v) 10 sided polygon is called decagon.

(vi) A polygon which has all its sides of equal length and all its angles of equal measure is called a regular polygon.

(vii) Sum of all exterior angles of a polygon is 360°.

(viii) Sum of all interior angles of a n-sided polygon is (n - 2) × 180°.

(ix) The adjacent angles of a parallelogram are supplementary.

(x) If an angle of a parallelogram is a right angle, then it is called a rectangle.

(xi) If two adjacent sides of a parallelogram are equal, then it is called a rhombus.

(xii) If two adjacent sides of a rectangle are equal, then it is called a square.

(xiii) The diagonals of a rhombus bisect each other at right angles.

(xiv) A quadrilateral in which one pair of opposite sides is parallel is called a trapezium.

(xv) A quadrilateral in which two pairs of adjacent sides are equal is called a kite.

(xvi) If two non-parallel sides of a trapezium are equal then it is called an isosceles trapezium.

(i) The curves which have same beginning and end points are called open curves.

False

The curves which have the same beginning and end points are called closed curves.

(ii) The region of the plane that lies inside the curve is called interior of curve.

True

(iii) A polygon in which atleast one interior angle is greater than 180° is called convex polygon.

False

A polygon in which atleast one interior angle is greater than 180° is called a concave polygon.

(iv) 6 sided polygon is called hexagon.

True

(v) Sum of all interior angles of a quadrilateral is 180°

False

The sum of all interior angles of a quadrilateral is 360°.

(vi) Each interior angle of a $n$-sided regular polygon is $\frac{(2n - 4) \times 90^{\circ}}{n}$

True

(vii) The diagonals of a parallelogram bisect each other at right angles.

False

The diagonals of a parallelogram bisect each other, but not necessarily at right angles.

(viii) The opposite angles of a parallelogram are equal.

True

(ix) The diagonals of a rhombus bisect the angles of rhombus.

True

(x) The diagonals of a square are not equal.

False

The diagonals of a square are equal in length.

(xi) Co-interior angles of a parallelogram are supplementary.

True

(xii) The diagonals of a kite bisect at right angles.

False

The diagonals of a kite intersect at right angles, but only one diagonal bisects the other.

(xiii) All rectangles are squares.

False

All squares are rectangles, but all rectangles are not squares.

(xiv) All rhombuses are parallelograms.

True

(xv) All squares are rhombuses and also rectangles.

True

(xvi) All squares are not parallelograms.

False

All squares are parallelograms.

(xvii) All kites are rhombuses.

False

All rhombuses are kites, but all kites are not rhombuses.

(xviii) All rhombuses are kites.

True

(xix) All parallelograms are trapeziums.

True

(xx) All squares are trapeziums.

True

According to the properties of a polygon, if a polygon has n sides, then the sum of measures of all its interior angles is (n - 2) × 180°.

Here, n = 11.

Sum of all interior angles = (11 - 2) x 180°

= 9 x 180°

= 1620°

Hence, option 1 is the correct option.

It is given that each interior angle of a regular polygon is 144°.

As the sum of an interior angle and its adjacent exterior angle is 180°,

Each exterior angle = 180° - 144° = 36°

According to the properties of a polygon, if each exterior angle of a regular polygon is $x^{\circ}$, then the number of sides in the regular polygon is $\frac{360}{x}$.

Number of sides = $\frac{360^{\circ}}{36^{\circ}}$

= 10

Hence, option 3 is the correct option.

It is given that the sum of all interior angles of a polygon is 1260°.

According to the properties of a polygon, if a polygon has n sides, then the sum of measures of all its interior angles is (n - 2) × 180°.

⇒ (n - 2) × 180° = 1260°

⇒ n - 2 = $\frac{1260^{\circ}}{180^{\circ}}$

⇒ n - 2 = 7

⇒ n = 7 + 2

⇒ n = 9

Hence, option 4 is the correct option.

According to the properties of a polygon, the sum of all the exterior angles of a polygon is always 360°.

A pentagon is a polygon of 5 sides, so the sum of all its exterior angles is also 360°.

Hence, option 2 is the correct option.

It is given that the ratio between an exterior and interior angle of a regular polygon is 1 : 5.

Let the common factor be a. So, the exterior angle is a and the interior angle is 5a.

As the sum of an interior angle and its adjacent exterior angle is 180°,

⇒ a + 5a = 180°

⇒ 6a = 180°

⇒ a = $\frac{180^{\circ}}{6}$

⇒ a = 30°

So, each exterior angle is 30°.

According to the properties of a polygon, if each exterior angle of a regular polygon is $x^{\circ}$, then the number of sides in the regular polygon is $\frac{360}{x}$.

Number of sides = $\frac{360^{\circ}}{30^{\circ}}$

= 12

Hence, option 2 is the correct option.

From the figure, the sides of the triangle have been produced in order, so 140°, 90° and x are the exterior angles of the triangle.

According to the properties of a polygon, the sum of all the exterior angles of a polygon is 360°.

⇒ 140° + 90° + x = 360°

⇒ 230° + x = 360°

⇒ x = 360° - 230°

⇒ x = 130°

Hence, option 3 is the correct option.

From the figure, the side of the quadrilateral has been produced and the exterior angle so formed is 120°.

As the sum of an interior angle and its adjacent exterior angle is 180°,

Interior angle at that vertex = 180° - 120° = 60°

As we know, the sum of all the interior angles of a quadrilateral is 360°.

⇒ 60° + x + 110° + 60° = 360°

⇒ 230° + x = 360°

⇒ x = 360° - 230°

⇒ x = 130°

Hence, option 2 is the correct option.

From the figure, the sides of the quadrilateral have been produced in order, so x, y, z and w are the exterior angles of the quadrilateral.

According to the properties of a polygon, the sum of all the exterior angles of a polygon is always 360°.

So, x + y + z + w = 360°

Hence, option 4 is the correct option.

From the figure, 110°, y, 60° and x are the interior angles of the quadrilateral.

As we know, the sum of all the interior angles of a quadrilateral is 360°.

⇒ 110° + y + 60° + x = 360°

⇒ x + y + 170° = 360°

⇒ x + y = 360° - 170°

⇒ x + y = 190°

Hence, option 2 is the correct option.

It is given that the lengths of two adjacent sides of a parallelogram are in the ratio 1 : 2 and the perimeter of the parallelogram is 60 cm.

Let the common factor be a. So, the adjacent sides are a cm and 2a cm.

As the opposite sides of a parallelogram are of equal length,

Perimeter = a + 2a + a + 2a

⇒ 60 = 6a

⇒ a = $\frac{60}{6}$

⇒ a = 10

Thus, the sides are:

a = 10 cm

2a = 2 x 10 = 20 cm

Hence, option 4 is the correct option.

From the figure, in parallelogram ABCD the diagonals AC and BD intersect at O, where AO = 6 cm, OC = 5x + 1, BO = 4 cm and OD = y + 3.

As the diagonals of a parallelogram bisect each other,

OC = AO

⇒ 5x + 1 = 6

⇒ 5x = 6 - 1

⇒ 5x = 5

⇒ x = $\frac{5}{5}$

⇒ x = 1

Also,

OD = BO

⇒ y + 3 = 4

⇒ y = 4 - 3

⇒ y = 1

Hence, option 1 is the correct option.

From the figure, in parallelogram ABCD the side AB has been produced beyond B and the exterior angle so formed is 70°.

As the sum of an interior angle and its adjacent exterior angle is 180°,

⇒ ∠ABC + 70° = 180°

⇒ ∠ABC = 180° - 70°

⇒ ∠ABC = 110°

From the figure,

⇒ 50° + x = ∠ABC

⇒ 50° + x = 110°

⇒ x = 110° - 50°

⇒ x = 60°

As AD ∥ BC and BD is a transversal,

y = ∠ADB = ∠DBC = x = 60° (alternate angles are equal)

As the opposite angles of a parallelogram are of equal measure,

z = ∠BCD = ∠DAB

As the adjacent angles of a parallelogram are supplementary,

⇒ ∠DAB = 180° - ∠ABC

⇒ ∠DAB = 180° - 110°

⇒ ∠DAB = 70°

So, z = 70°

Hence, option 1 is the correct option.

From the figure, in rectangle ABCD the diagonals intersect at O and ∠OAB = 30°.

As the diagonals of a rectangle are equal and bisect each other,

OA = OB

So, ΔAOB is an isosceles triangle.

⇒ ∠OBA = ∠OAB = 30° (angles opposite to equal sides are equal)

In ΔAOB, the sum of all the angles of a triangle is 180°.

⇒ ∠OAB + ∠OBA + ∠AOB = 180°

⇒ 30° + 30° + ∠AOB = 180°

⇒ 60° + ∠AOB = 180°

⇒ ∠AOB = 180° - 60°

⇒ ∠AOB = 120°

As x and ∠AOB are vertically opposite angles,

x = ∠AOB = 120°

Hence, option 3 is the correct option.

It is given that in rhombus ABCD, the diagonals AC and BD are respectively 8 cm and 6 cm.

Let the diagonals intersect at O.

As the diagonals of a rhombus bisect each other,

AO = $\frac{8}{2}$ = 4 cm

BO = $\frac{6}{2}$ = 3 cm

As the diagonals of a rhombus intersect at right angles, ∠AOB = 90°.

So, ΔAOB is a right angled triangle.

By Pythagoras theorem,

⇒ AB² = AO² + BO²

⇒ AB² = 4² + 3²

⇒ AB² = 16 + 9

⇒ AB² = 25

⇒ AB = $\sqrt{25}$

⇒ AB = 5 cm

Hence, option 2 is the correct option.

From the figure, in square ABCD the diagonals AC and BD are drawn and $x$ = ∠ABD.

As we know, the diagonals of a square bisect the angles of the square.

So, the diagonal BD bisects ∠ABC.

$$\begin{aligned}\Rightarrow x = \frac{1}{2} \times \angle ABC \\[1em] \Rightarrow x = \frac{1}{2} \times 90^{\circ} \\[1em] \Rightarrow x = 45^{\circ}\end{aligned}$$

Hence, option 2 is the correct option.

From the figure, in kite ABCD, AD = DC and AB = BC, the diagonals AC and BD intersect at O, ∠DCA = 50° and ∠ABD = 36°.

In ΔADC, AD = DC, so ΔADC is an isosceles triangle.

⇒ ∠DAC = ∠DCA = 50° (angles opposite to equal sides are equal)

As the diagonals of a kite intersect at right angles,

∠AOB = 90°

In ΔAOB, the sum of all the angles of a triangle is 180°.

⇒ ∠OAB + ∠ABO + ∠AOB = 180°

⇒ ∠OAB + 36° + 90° = 180°

⇒ ∠OAB + 126° = 180°

⇒ ∠OAB = 180° - 126°

⇒ ∠OAB = 54°

From the figure,

x = ∠DAB = ∠DAC + ∠OAB

⇒ x = 50° + 54°

⇒ x = 104°

Hence, option 3 is the correct option.

From the figure, ABCD is an isosceles trapezium in which AB ∥ DC, AD = BC and ∠A = 70°.

As the angles on the same base of an isosceles trapezium are equal,

z = ∠B = ∠A = 70°

As AB ∥ DC and AD is a transversal, the co-interior angles ∠A and ∠D are supplementary.

⇒ x + ∠A = 180°

⇒ x + 70° = 180°

⇒ x = 180° - 70°

⇒ x = 110°

As the angles on the same base of an isosceles trapezium are equal,

y = ∠C = ∠D = x = 110°

Hence, option 1 is the correct option.

A simple closed curve made up entirely of line segments is called a polygon.

So, Statement I is true.

In a regular polygon, all the sides are of equal length and all the interior angles are of equal measure, and each of these interior angles is less than 180°. So, every regular polygon is convex.

So, Statement II is true.

Hence, option 3 is the correct option.

A heptagon has 7 sides.

According to the properties of a polygon, if a regular polygon has $n$ sides, then each of its interior angles is $\frac{(n - 2) \times 180^{\circ}}{n}$.

$$\begin{aligned}\text{Each interior angle} = \frac{(7 - 2) \times 180^{\circ}}{7} \\[1em] = \frac{5 \times 180^{\circ}}{7} \\[1em] = \frac{900^{\circ}}{7} \\[1em] = 128\frac{4}{7}^{^{\circ}}\end{aligned}$$

As $128\frac{4}{7}^{^{\circ}}$ ≠ 120°, Statement I is false.

According to the properties of a polygon, if a regular polygon has $n$ sides, then each of its exterior angles is $\frac{360^{\circ}}{n}$.

$$\begin{aligned}\text{Each exterior angle} = \frac{360^{\circ}}{7} \\[1em] = 51\frac{3}{7}^{^{\circ}}\end{aligned}$$

As $51\frac{3}{7}^{^{\circ}}$ ≠ 60°, Statement II is false.

Hence, option 4 is the correct option.

In a rectangle, both the pairs of opposite sides are parallel and equal, so every rectangle is a parallelogram.

So, Statement I is true.

In a square, all the four sides are equal, but in a rectangle the adjacent sides need not be equal. So, every rectangle is not a square.

So, Statement II is false.

Hence, option 1 is the correct option.

The opposite sides of a parallelogram are of equal length.

So, Statement I is true.

The diagonals of a parallelogram bisect each other, but they are not equal in length. The diagonals are equal only in a rectangle and a square.

So, Statement II is false.

Hence, option 1 is the correct option.

In a rhombus, both the pairs of opposite sides are parallel and equal, so every rhombus is a parallelogram.

So, Statement I is true.

In a square, each angle is a right angle, but in a rhombus the angles need not be right angles. So, every rhombus is not a square.

So, Statement II is false.

Hence, option 1 is the correct option.

The diagonals of a rectangle bisect the angles of the rectangle only when the rectangle is a square. In a rectangle whose adjacent sides are unequal, a diagonal divides a right angle into two unequal parts.

So, Statement I is false.

A rectangle is a parallelogram and the diagonals of a parallelogram bisect each other, so the diagonals of a rectangle bisect each other.

So, Statement II is true.

Hence, option 2 is the correct option.

From the figure,

ABCD is a quadrilateral in which the angle at B is a reflex angle, ∠BCD = 53° and the side CD has been produced to E so that the exterior angle at D is 122°.

The angle marked 140° at B is the angle outside quadrilateral ABCD.

So, the reflex angle at B is:

= 360° - 140°

= 220°

As ∠ADC and ∠ADE form a linear pair,

⇒ ∠ADC + 122° = 180°

⇒ ∠ADC = 180° - 122°

⇒ ∠ADC = 58°

As we know, the sum of all the interior angles of a quadrilateral is 360°.

⇒ x + 220° + 53° + 58° = 360°

⇒ x + 331° = 360°

⇒ x = 360° - 331°

⇒ x = 29°

Hence, the value of x is 29°.

It is given that two angles of a quadrilateral are 77° and 57°, and out of the remaining two angles, one angle is 10° smaller than the other.

Let the greater of the remaining two angles be a. So, the other angle is (a - 10°).

As we know, the sum of all the interior angles of a quadrilateral is 360°.

⇒ 77° + 57° + a + (a - 10°) = 360°

⇒ 77° + 57° + a + a - 10° = 360°

⇒ 2a + 124° = 360°

⇒ 2a = 360° - 124°

⇒ 2a = 236°

⇒ a = $\frac{236^{\circ}}{2}$

⇒ a = 118°

Thus, the angles are:

a = 118°

a - 10° = 118° - 10° = 108°

Hence, the remaining two angles are 108° and 118°.

It is given that in the figure, AB ∥ DC, ∠A = 74° and ∠B : ∠C = 4 : 5.

(i) As AB ∥ DC and AD is a transversal, the co-interior angles ∠A and ∠D are supplementary.

⇒ ∠A + ∠D = 180°

⇒ 74° + ∠D = 180°

⇒ ∠D = 180° - 74°

⇒ ∠D = 106°

Hence, ∠D = 106°.

(ii) As AB ∥ DC and BC is a transversal, the co-interior angles ∠B and ∠C are supplementary.

⇒ ∠B + ∠C = 180°

Let the common factor be a. So, ∠B = 4a and ∠C = 5a.

⇒ 4a + 5a = 180°

⇒ 9a = 180°

⇒ a = $\frac{180^{\circ}}{9}$

⇒ a = 20°

Thus,

∠B = 4a = 4 x 20° = 80°

Hence, ∠B = 80°.

(iii) ∠C = 5a = 5 x 20° = 100°

Hence, ∠C = 100°.

It is given that in quadrilateral ABCD, ∠A : ∠B : ∠C : ∠D = 3 : 4 : 6 : 7.

Let the common factor of the angles be a. So, ∠A = 3a, ∠B = 4a, ∠C = 6a and ∠D = 7a.

As we know, the sum of all the interior angles of a quadrilateral is 360°.

⇒ ∠A + ∠B + ∠C + ∠D = 360°

⇒ 3a + 4a + 6a + 7a = 360°

⇒ 20a = 360°

⇒ a = $\frac{360^{\circ}}{20}$

⇒ a = 18°

Thus, the angles are:

∠A = 3a = 3 x 18° = 54°

∠B = 4a = 4 x 18° = 72°

∠C = 6a = 6 x 18° = 108°

∠D = 7a = 7 x 18° = 126°

Now, ∠A + ∠D = 54° + 126° = 180°

As ∠A and ∠D are co-interior angles formed by the transversal AD, and their sum is 180°, the lines AB and DC are parallel.

Also, ∠A + ∠B = 54° + 72° = 126°

As ∠A and ∠B are co-interior angles formed by the transversal AB, and their sum is not 180°, the lines AD and BC are not parallel.

Hence, the angles of the quadrilateral are 54°, 72°, 108° and 126°; AB and DC are parallel but BC is not parallel to AD.

It is given that one angle of a parallelogram is two-third of the other.

Let the parallelogram be ABCD and let ∠A and ∠B be its adjacent angles.

Let ∠B = $a$. So, ∠A = $\frac{2}{3}a$.

As the adjacent angles of a parallelogram are supplementary,

$$\begin{aligned}\Rightarrow \angle A + \angle B = 180^{\circ} \\[1em] \Rightarrow \frac{2}{3}a + a = 180^{\circ} \\[1em] \Rightarrow \frac{2a + 3a}{3} = 180^{\circ} \\[1em] \Rightarrow \frac{5a}{3} = 180^{\circ} \\[1em] \Rightarrow 5a = 180^{\circ} \times 3 \\[1em] \Rightarrow 5a = 540^{\circ} \\[1em] \Rightarrow a = \frac{540^{\circ}}{5} \\[1em] \Rightarrow a = 108^{\circ}\end{aligned}$$

Thus,

∠B = 108°

∠A = $\frac{2}{3}$ × 108° = 72°

As the opposite angles of a parallelogram are of equal measure,

∠C = ∠A = 72° and ∠D = ∠B = 108°

Hence, the angles of the parallelogram are 72°, 108°, 72° and 108°.

It is given that ABCD is a kite in which AD = AB and CD = CB, ∠BCD = 52° and ∠ADB = 42°.

In ΔABD, AD = AB, so ΔABD is an isosceles triangle.

⇒ x = ∠ABD = ∠ADB = 42° (angles opposite to equal sides are equal)

In ΔABD, the sum of all the angles of a triangle is 180°.

⇒ y + ∠ADB + ∠ABD = 180°

⇒ y + 42° + 42° = 180°

⇒ y + 84° = 180°

⇒ y = 180° - 84°

⇒ y = 96°

In ΔCBD, CD = CB, so ΔCBD is an isosceles triangle.

⇒ z = ∠CDB (angles opposite to equal sides are equal)

In ΔCBD, the sum of all the angles of a triangle is 180°.

⇒ z + ∠CDB + ∠BCD = 180°

⇒ 2z + 52° = 180°

⇒ 2z = 180° - 52°

⇒ 2z = 128°

⇒ z = $\frac{128^{\circ}}{2}$

⇒ z = 64°

Hence, x = 42°, y = 96° and z = 64°.

From figure,

In ΔABC and ΔBAD,

AB = BA (common side)

∠ABC = ∠BAD = 90° (each angle of a rectangle is a right angle)

BC = AD (opposite sides of a rectangle are equal)

Therefore, by the S.A.S. congruence rule,

ΔABC ≅ ΔBAD

As the corresponding parts of congruent triangles are equal,

AC = BD

Hence, proved that AC = BD.

It is given that ABCD is a rhombus, EDC is an equilateral triangle and ∠DAB = 48°.

As the opposite angles of a rhombus are of equal measure,

∠BCD = ∠DAB = 48°

As EDC is an equilateral triangle, each of its angles is 60° and all its sides are equal.

So, ∠DCE = 60° and ED = DC = CE

Also, as ABCD is a rhombus, BC = CD.

So, BC = CE ...............(1)

From the figure,

∠BCE = ∠BCD + ∠DCE

= 48° + 60°

= 108°

(i) In ΔBCE, BC = CE, so ΔBCE is an isosceles triangle.

⇒ ∠BEC = ∠EBC (angles opposite to equal sides are equal)

In ΔBCE, the sum of all the angles of a triangle is 180°.

⇒ ∠BEC + ∠EBC + ∠BCE = 180°

⇒ 2∠BEC + 108° = 180°

⇒ 2∠BEC = 180° - 108°

⇒ 2∠BEC = 72°

⇒ ∠BEC = $\frac{72^{\circ}}{2}$

⇒ ∠BEC = 36°

Hence, ∠BEC = 36°.

(ii) As EDC is an equilateral triangle, ∠DEC = 60°.

From the figure,

∠DEB = ∠DEC - ∠BEC

= 60° - 36°

= 24°

Hence, ∠DEB = 24°.

(iii) In ΔBFC,

∠FCB = ∠BCD = 48° and ∠FBC = ∠EBC = ∠BEC = 36°

As the sum of all the angles of a triangle is 180°,

⇒ ∠BFC + ∠FCB + ∠FBC = 180°

⇒ ∠BFC + 48° + 36° = 180°

⇒ ∠BFC + 84° = 180°

⇒ ∠BFC = 180° - 84°

⇒ ∠BFC = 96°

Hence, ∠BFC = 96°.

It is given that each interior angle of a regular polygon is 168°.

As the sum of an interior angle and its adjacent exterior angle is 180°,

Each exterior angle = 180° - 168° = 12°

According to the properties of a polygon, if each exterior angle of a regular polygon is $x^{\circ}$, then the number of sides in the regular polygon is $\frac{360}{x}$.

Number of sides = $\frac{360^{\circ}}{12^{\circ}}$

= 30

Hence, the number of sides is 30.

It is given that the sum of interior angles of a polygon is 3780°.

According to the properties of a polygon, if a polygon has n sides, then the sum of measures of all its interior angles is (n - 2) × 180°.

⇒ (n - 2) × 180° = 3780°

⇒ n - 2 = $\frac{3780^{\circ}}{180^{\circ}}$

⇒ n - 2 = 21

⇒ n = 21 + 2

⇒ n = 23

Hence, the number of sides is 23.

It is given that the angles of a hexagon are (2x + 5)°, (3x - 5)°, (x + 40)°, (2x + 20)°, (2x + 25)° and (2x + 35)°.

According to the properties of a polygon, if a polygon has n sides, then the sum of measures of all its interior angles is (n - 2) × 180°.

A hexagon has 6 sides, so n = 6.

Sum of interior angles = (6 - 2) x 180°

= 4 x 180°

= 720°

So,

⇒ (2x + 5)° + (3x - 5)° + (x + 40)° + (2x + 20)° + (2x + 25)° + (2x + 35)° = 720°

⇒ 2x° + 5° + 3x° - 5° + x° + 40° + 2x° + 20° + 2x° + 25° + 2x° + 35° = 720°

⇒ 12x° + 120° = 720°

⇒ 12x° = 720° - 120°

⇒ 12x° = 600°

⇒ x° = $\frac{600^{\circ}}{12}$

⇒ x = 50

Hence, the value of x is 50.

It is given that two angles of a polygon are right angles and every other angle is 120°.

Let the number of sides of the polygon be n.

So, the polygon has 2 angles of 90° each and the remaining (n - 2) angles of 120° each.

According to the properties of a polygon, if a polygon has n sides, then the sum of measures of all its interior angles is (n - 2) × 180°.

So,

⇒ 2 × 90° + (n - 2) × 120° = (n - 2) × 180°

⇒ 180° + 120° n - 240° = 180° n - 360°

⇒ 120° n - 60° = 180° n - 360°

⇒ 180° n - 120° n = 360° - 60°

⇒ 60° n = 300°

⇒ n = $\frac{300^{\circ}}{60^{\circ}}$

⇒ n = 5

Hence, the number of sides of the polygon is 5.

It is given that the sum of interior angles of a regular polygon is twice the sum of its exterior angles.

According to the properties of a polygon, the sum of all the exterior angles of a polygon is 360°.

So, the sum of its interior angles = 2 x 360° = 720°

Also, if a polygon has n sides, then the sum of measures of all its interior angles is (n - 2) × 180°.

⇒ (n - 2) × 180° = 720°

⇒ n - 2 = $\frac{720^{\circ}}{180^{\circ}}$

⇒ n - 2 = 4

⇒ n = 4 + 2

⇒ n = 6

Hence, the number of sides of the polygon is 6.

It is given that an exterior angle of a regular polygon is one-fourth of its interior angle.

Let the interior angle be $a$. So, the exterior angle is $\frac{a}{4}$.

As the sum of an interior angle and its adjacent exterior angle is 180°,

$$\begin{aligned}\Rightarrow a + \frac{a}{4} = 180^{\circ} \\[1em] \Rightarrow \frac{4a + a}{4} = 180^{\circ} \\[1em] \Rightarrow \frac{5a}{4} = 180^{\circ} \\[1em] \Rightarrow 5a = 180^{\circ} \times 4 \\[1em] \Rightarrow 5a = 720^{\circ} \\[1em] \Rightarrow a = \frac{720^{\circ}}{5} \\[1em] \Rightarrow a = 144^{\circ}\end{aligned}$$

So, the interior angle is 144° and the exterior angle is:

$= \frac{144^{\circ}}{4}$

= 36°

According to the properties of a polygon, if each exterior angle of a regular polygon is $x^{\circ}$, then the number of sides in the regular polygon is $\frac{360}{x}$.

$\text{Number of sides} = \frac{360^{\circ}}{36^{\circ}}$

= 10

Hence, the number of sides in the polygon is 10.

From the figure, in pentagon ABCDE the side AB has been produced on both the sides, so that Q, A, B and P lie on a straight line.

The exterior angle at A is 60° and the exterior angle at B is 80°.

As the sum of an interior angle and its adjacent exterior angle is 180°,

⇒ ∠EAB = 180° - 60° = 120°

⇒ ∠ABC = 180° - 80° = 100°

According to the properties of a polygon, if a polygon has n sides, then the sum of measures of all its interior angles is (n - 2) × 180°.

A pentagon has 5 sides, so n = 5.

Sum of interior angles = (5 - 2) x 180°

= 3 x 180°

= 540°

From the figure, ∠D = 40°, ∠E = x and ∠C = x.

So,

⇒ ∠EAB + ∠ABC + ∠C + ∠D + ∠E = 540°

⇒ 120° + 100° + x + 40° + x = 540°

⇒ 2x + 260° = 540°

⇒ 2x = 540° - 260°

⇒ 2x = 280°

⇒ x = $\frac{280^{\circ}}{2}$

⇒ x = 140°

Hence, the value of x is 140°.

It is given that ABCDEF is a regular hexagon, in which AE and BD meet the lines DF and AC at H and G respectively.

In a regular hexagon, the opposite sides are parallel and equal.

In quadrilateral ABDE,

AB ∥ ED and AB = ED (opposite sides of a regular hexagon)

As one pair of opposite sides is parallel and equal, ABDE is a parallelogram.

Hence, proved that quadrilateral ABDE is a parallelogram.

In quadrilateral ACDF,

AF ∥ CD and AF = CD (opposite sides of a regular hexagon)

As one pair of opposite sides is parallel and equal, ACDF is a parallelogram.

Hence, proved that quadrilateral ACDF is a parallelogram.

Now, in quadrilateral AGDH,

As ABDE is a parallelogram, AE ∥ BD.

The point H lies on AE and the point G lies on BD.

So, AH ∥ GD ...............(1)

As ACDF is a parallelogram, AC ∥ FD.

The point G lies on AC and the point H lies on FD.

So, AG ∥ HD ...............(2)

From equations (1) and (2), both the pairs of opposite sides of quadrilateral AGDH are parallel.

Hence, proved that quadrilateral AGDH is a parallelogram.

As ABCD is a parallelogram, DC ∥ AB, that is DC ∥ EF.

In ΔANE and ΔDNC,

AN = ND (as N is the mid-point of AD)

∠ANE = ∠DNC (vertically opposite angles are equal)

∠NAE = ∠NDC (alternate angles are equal)

Therefore, by the A.S.A. congruence rule,

ΔANE ≅ ΔDNC

As the corresponding parts of congruent triangles are equal,

EA = DC ...............(1)

In ΔBMF and ΔCMD,

BM = MC (as M is the mid-point of BC)

∠BMF = ∠CMD (vertically opposite angles are equal)

∠MBF = ∠MCD (alternate angles are equal)

Therefore, by the A.S.A. congruence rule,

ΔBMF ≅ ΔCMD

As the corresponding parts of congruent triangles are equal,

BF = CD ...............(2)

Also, as the opposite sides of a parallelogram are equal,

AB = DC ...............(3)

From equations (1), (2) and (3),

EA = AB = BF

Hence, proved that EA = AB = BF.

Let ABCD be a rectangle and let P, Q, R and S be the mid-points of the sides AB, BC, CD and DA respectively.

Join PQ, QR, RS and SP.

As each interior angle of a rectangle is a right angle,

∠A = ∠B = ∠C = ∠D = 90°

As the opposite sides of a rectangle are equal,

AB = CD and BC = DA

As P, Q, R and S are the mid-points of AB, BC, CD and DA respectively,

AP = PB = $\frac{1}{2}$AB

CR = RD = $\frac{1}{2}$CD

As AB = CD, we get

AP = PB = CR = RD ...............(1)

Also,

BQ = QC = $\frac{1}{2}$BC

DS = SA = $\frac{1}{2}$DA

As BC = DA, we get

BQ = QC = DS = SA ...............(2)

In ΔAPS and ΔBPQ,

AP = BP (from equation (1))

∠A = ∠B = 90° (each angle of a rectangle is a right angle)

SA = QB (from equation (2))

Therefore, by the S.A.S. congruence rule,

ΔAPS ≅ ΔBPQ

As the corresponding parts of congruent triangles are equal,

PS = PQ ...............(3)

In ΔCRQ and ΔDRS,

CR = DR (from equation (1))

∠C = ∠D = 90° (each angle of a rectangle is a right angle)

QC = SD (from equation (2))

Therefore, by the S.A.S. congruence rule,

ΔCRQ ≅ ΔDRS

As the corresponding parts of congruent triangles are equal,

RQ = RS ...............(4)

In ΔAPS and ΔCRQ,

AP = CR (from equation (1))

∠A = ∠C = 90° (each angle of a rectangle is a right angle)

SA = QC (from equation (2))

Therefore, by the S.A.S. congruence rule,

ΔAPS ≅ ΔCRQ

As the corresponding parts of congruent triangles are equal,

PS = RQ ...............(5)

From equations (3), (4) and (5),

PQ = QR = RS = SP

So, in quadrilateral PQRS,

PQ = RS and QR = SP

As both the pairs of opposite sides of quadrilateral PQRS are equal, PQRS is a parallelogram.

Hence, proved that the quadrilateral formed by joining the mid-points of the adjacent sides of a rectangle is a parallelogram.