The DAV Maths Class 8 Solutions and **DAV Class 8 Maths Chapter 11 Worksheet 1** Solutions of Understanding Quadrilaterals offer comprehensive answers to textbook questions.

## DAV Class 8 Maths Ch 11 WS 1 Solutions

Question 1.

Find the sum of interior angles of hexagon ABCDEF by dividing it into triangles.

Solution:

Given: ABCDEF is a hexagon.

Joining AC, CF, and DF, we get four triangles.

In ΔABC, ∠1 + ∠2 + ∠3 = 180° ………(i)

In ΔACF, ∠4 + ∠5 + ∠6 = 180° ………(ii)

In ΔCDF, ∠7 + ∠8 + ∠9 = 180° ……….(iii)

In ΔDEF, ∠10 + ∠11 + ∠12 = 180° ……….(iv)

Adding (i), (ii), (iii) and (iv), we get

∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 + ∠9 + ∠10 + ∠11 + ∠12 = 180° + 180° + 180° + 180°

⇒ (∠1 + ∠5) + ∠2 + (∠3 + ∠4 + ∠7) + (∠6 + ∠9 + ∠12) + (∠8 + ∠10) + ∠11 = 720°

⇒ ∠A + ∠B + ∠C + ∠F + ∠D + ∠E = 720°

⇒ ∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 720°

Question 2.

Find the sum of interior angles of a polygon of a given number of sides by using the formula (n – 2) × 180°.

(i) 12 sides

(ii) 9 sides

(iii) 22 sides

Solution:

(i) Given: n = 12

Then, the sum of interior angles of a polygon having 12 sides = (12 – 2) × 180°

= 10 × 180°

= 1800°

(ii) Given: n = 9

Then, the sum of the interior angles of a polygon having 9 sides = (9 – 2) × 180°

= 7 × 180°

= 1260°

(iii) Given: n = 22

Then, the sum of the interior angles of a polygon having 22 sides = (22 – 2) × 180°

= 20 × 180°

= 3600°

Question 3.

Find the measure of each angle of a regular octagon.

Solution:

Number of sides of a regular octagon, n = 8

Measure of each angle of a regular octagon = \(\frac{(n-2) \times 180^{\circ}}{n}\)

= \(\frac{(8-2) \times 180^{\circ}}{8}\)

= \(\frac{6 \times 180^{\circ}}{8}\)

= 3 × 45°

= 135°

Question 4.

Four angles of a pentagon are 100°, 175°, 85° and 75°. Find the fifth angle.

Solution:

The number of angles of a pentagon = 5

Sum of five angles = (5 – 2) × 180°

= 3 × 180°

= 540°

Let the fifth angle of the pentagon be x.

Then, 100° + 175° + 85° + 75° + x = 540°

⇒ 435° + x = 540°

⇒ x = 540° – 435°

⇒ x = 105°

Question 5.

Find the measure of x in the given figure.

Solution:

We know that the sum of measures of exterior angles of a pentagon is 360°.

∴ 100° + 45° + x + 40° + 65° = 360°

⇒ x + 250° = 360°

⇒ x = 360° – 250°

⇒ x = 110°

### DAV Class 8 Maths Chapter 11 Worksheet 1 Notes

- Quadrilateral is a polygon enclosed by four sides.
- The sum of all the interior angles of a quadrilateral is 360°.
- The sum of all interior angles of a pentagon is 540°.
- The sum of all exterior angles of any polygon is 360°.
- Trapezium is a quadrilateral in which one pair of opposite sides is parallel.
- Parallelogram is a quadrilateral in which (i) opposite sides are equal and parallel to each other (ii)
- opposite angles are equal to each other.
- Diagonals of a parallelogram bisect each other.
- Rhombus is a quadrilateral whose all sides are equal and its diagonals bisect each other at right angle.
- Rectangle is a quadrilateral whose each angle is 90°, opposite sides are equal and
- parallel to each other. Its diagonals are equal and bisect each other.
- Square is a quadrilateral whose all sides are equal. Each angle is 90°. Its diagonals are equal and bisect each other at right angles.

Example 1.

In the given figure find the measure of ∠x.

Solution:

Sum of all the exterior angles of a polygon is 360°.

∴ ∠x + ∠50° + ∠70° + ∠130° = 360°

⇒ ∠x + 250° = 360°

⇒ ∠x = 360 – 250

∴ ∠x = 110°.

Example 2.

PQRS is a parallelogram in which ¿P = 60°, find the remaining angles of the parallelogram.

Solution:

∠P + ∠Q = 180° [Sum of interior angles]

⇒ 60°+ ∠Q = 180°

∴∠Q = 180° – 60° = 120°

∠P = ∠R

[Opposite angles of a parallelogram]

∴ ∠R = 60°

Similarly ∠Q = ∠S

∴ ∠S = 120°.

Example 3.

The adjacent sides of a parallelogram are in the ratio 3 : 4. If its perimeter is 140 cm, find the sides.

Solution:

Let the two adjacent sides be 3x and 4x cm.

∴ Perimeter = 2 (3x + 4x) = 14x cm

14x = 140

∴ x = 10

Hence the sides are 3 × 10 = 30 cm and 4 × 10 = 40 cm.

Example 4.

The diagonals of a rhombus are in the ratio 3 : 4. If its perimeter is 40 cm, find the length of sides and the diagonals of the rhombus.

Solution:

Perimeter = 40 cm

4 × Side = 40 cm

∴ Side = 10 cm

AC : BD = 4 : 3

As the diagonals of the rhombus bisect at 90°,

∴ In rt.ΔAOB

OA = \(\frac{1}{2}\) AC

= \(\frac{1}{2}\) × 4x = 2x

OB = \(\frac{1}{2}\) BD

= \(\frac{1}{2}\) × 3x

= \(\frac{3}{2}\) x

From Pythagorous Theorem,

OA^{2} + OB^{2} = AB^{2}

(2x)^{2} + (\(\frac{3}{2}\)x)^{2} = (10)^{2}

4x^{2} + \(\frac{9 x^2}{4}\) = 100

16x^{2} + 9x^{2} = 400

25x^{2} = 400

x^{2} = \(\frac{400}{25}\) = 16

∴ x = 4.

Hence AB = BC = CD = DA = 10 cm

AC = 4 × 4 = 16 cm

and BD = 3 × 4 = 12 cm.