The DAV Maths Class 8 Solutions and **DAV Class 8 Maths Chapter 11 Worksheet 2** Solutions of Understanding Quadrilaterals offer comprehensive answers to textbook questions.

## DAV Class 8 Maths Ch 11 WS 2 Solutions

Question 1.

PQRS is a trapezium with PQ || SR. If ∠P = 30°, ∠Q = 50°, find ∠R and ∠S.

Solution:

In trapezium PQRS

PQ || SR

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

30° + ∠S = 180°

∴ ∠S = 180° – 30° = 150°

Similarly ∠Q + ∠R = 180°

50° + ∠R = 180°

∴ ∠R = 180° – 50° = 130°.

Question 2.

ABCD is a quadrilateral with ∠A = 80°, ∠B = 40°, ∠C = 140°, ∠D = 100°.

(i) Is ABCD a trapezium?

(ii) Is ABCD a parallelogram?

Justify your answer.

Solution:

(i) ∠A + ∠D = 80° +100° = 180°

and ∠B + ∠C = 40° + 140° = 180°

The sum of the interior angles is 180°

∴ AB || DC

Hence it is a trapezium.

(ii) ∠A + ∠B = 80° + 40° = 120°.

The sum of the interior angle is not 180°.

∴ AD is not parallel to BC.

Hence it is not a parallelogram.

Question 3.

One of the angles of a parallelogram is 75°. Find the measures of the of the remaining angles parallelogram.

Solution:

ABCD is a parallelogram in which ∠A = 75°

∴ ∠A = ∠C [Opposite angles of the parallelogram]

∴ ∠C = 75°

∠A + ∠B = 180° [Adjacent angles]

75° + ∠B = 180°

∴ ∠B = 180° – 75° = 105°

∠B = ∠D [Opposite angles of the parallelogram]

105° = ∠D

∴ ∠D = 105°.

Question 4.

Two adjacent angles of a parallelogram are in the ratio 1 : 5. Find all the angles of the parallelogram.

Solution:

Let the two adjacent angles be x° and 5x°.

∴ x° + 5x° = 180° [Sum of adjacent angles]

⇒ 6x = 180

⇒ x = 30

Hence the required adjacent angles are 1 × 30 = 30° and 5 × 30 = 150°.

Question 5.

The exterior angle of a parallelogram is 110°. Find the angles of the parallelogram.

Solution:

∠A + ∠DAE = 180° [Linear pairs]

∠A + 110° = 180°

∴ ∠A = 180° – 110° = 70°

∠C = ∠A = 70° [Opposite angles of a || gram]

∠A + ∠B = 180° [Adjacent angles]

70° + ∠B = 180°

∴ ∠B = 180° – 70° = 110°

∠B = ∠D [Opposite angles of the parallelogram]

110° = ∠D

Hence ∠A = 70°, ∠B = 110°, ∠C = 70° and ∠D = 110°.

Question 6.

Two adjacent sides of a parallelogram are in the ratio 3 : 8 and its perimeter is 110 cm. Find the sides of the parallelogram.

Solution:

Let AB be 8x cm and AD be 3x cm

∴ Perimeter = 2[AB + AD]

⇒ 110 = 2[8x + 3x]

⇒ 110 = 2 × 11x

⇒ 22x = 110

⇒ x = \(\frac{110}{22}\) = 5

∴ AB = 8 × 5 = 40 cm

BC = AD = 3 × 5 = 15 cm

DC = AB = 40 cm.

Question 7.

One side of a parallelogram is \(\frac{3}{4}\) times its adjacent side. If the perimeter of the parallelogram is 70 cm, find the sides of the parallelogram.

Solution:

Let one side of the parallelogram be x cm

∴ The adjacent side = \(\frac{3}{4}\)x cm

∴ Perimeter = 2[l + b]

⇒ 70 = 2[x + \(\frac{3}{4}\)x]

⇒ 70 = 2 × \(\frac{7}{4}\)x

⇒ 70 = \(\frac{7}{2}\)x

⇒ x = 70 × \(\frac{2}{7}\)

⇒ x = 20

∴ The required sides are 20 cm = \(\frac{3}{4}\) × 20 = 15 cm, 20 cm and 15 cm.

Question 8.

ABCD is a parallelogram whose diagonals intersect each other at right angles. If the length of the diagonals is 8 cm and 8 cm, find the lengths of all the sides of the parallelogram.

Solution:

AC = 8 cm

∴ AO = 4 cm

BD = 6 cm

∴ BO = 3 cm

[The diagonals of a parallelogram bisect each other]

∠AOB = 90° [Given]

∴ In rt. ∆AOB

AB^{2} = AO^{2} + BO^{2} [From Pythagoras Theorem]

⇒ AB^{2} = (4)^{2} + (3)^{2}

⇒ AB^{2} = 16 + 9

⇒ AB^{2} = 25

⇒ AB = 5 cm

∴ ABCD is a rhombus.

[∵ If the diagonals of a quadrilateral bisect each other at 90° then it is a rhombus]

Hence AB = BC = CD = DA = 5 cm.

Question 9.

In the given figure, one pair of adjacent sides of a parallelogram is in the ratio 3 : 4. If one of its angles, ∠A is a right angle and diagonal \(\overline{\mathrm{BD}}\) = 10 cm, find the

(i) lengths of the sides of the parallelogram

(ii) perimeter of the parallelogram.

Solution:

∠A = 90° [Given]

In rt. ∆ABD,

BD^{2} = AB^{2} + AD^{2} [From Pythagoras Theorem]

⇒ (10)^{2} = (4x)^{2} + (3x)^{2}

⇒ 100 = 16x^{2} + 9x^{2}

⇒ 100 = 25x^{2}

⇒ x^{2} = \(\frac{100}{25}\) = 4

∴ x = 2

Hence AB = 4 × 2 = 8 cm

AD = 3 × 2 = 6 cm

BC = AD = 6 cm

CD = AB = 8 cm

Question 10.

ABCD is a quadrilateral in which \(\overline{\mathrm{AB}}=\overline{\mathrm{CD}}\) and \(\overline{\mathrm{AD}}=\overline{\mathrm{BC}}\). Show that it is a parallelogram.

[Hint: Draw one of the diagonals]

Solution:

Draw diagonal AC

In ∆ABC and ∆ADC

\(\overline{\mathrm{AB}}=\overline{\mathrm{CD}}\)

\(\overline{\mathrm{AD}}=\overline{\mathrm{BC}}\) [Given]

\(\overline{\mathrm{AC}}=\overline{\mathrm{AC}}\) [Common]

∴ ∆ABC = ∆ADC [SSS]

∴ ∠1 = ∠2 [C.P.C.T.]

But these are alternate angles

∴ \(\overline{\mathrm{AB}} \| \overline{\mathrm{DC}}\)

Similarly \(\overline{\mathrm{AD}} \| \overline{\mathrm{BC}}\)

Hence ABCD is a parallelogram.