Gujarat Board GSEB Textbook Solutions Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.3 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.3

Question 1.

Given a parallelogram ABCD. Complete each statement along with the definition or property used

(i) AD = __________

(ii) ZDCB = __________

(in) OC = __________

(iv) ∠DAB + mZCDA = __________

Solution:

(i) AD = BC [v Opposite sides are equal]

(ii) ZDCB = ZDAB [ Opposite angles are equal]

(iii) OC = OA [Y Diagonals bisect each other]

(iv) mZDAB + mZCDA = 180°

[∴ Adjacent angles are supplementary]

Question 2.

Consider the following parallelograms. Find the values of the unknowns x, y, z.

Solution:

(i) ∠y = 100°

[∴ Opposite angles of a parallelogram are equal.]

Sum of interior angles of a parallelogram = 360°

∴ x + y + z + ∠B = 360°

or x + 100° + z + 100° = 360°

or x + z = 360° – 100° – 100° = 160°

But x = z

∴ x = z = \(\frac { 160° }{ 2 }\) = 80°

Thus,

(ii) ∴ Opposite angles are equal.

∴ m ∠1 = 50

now, ∠1 + z = 180° [Linear pair]

or z = 180° – ∠1

= 180° – 50° = 130°

x + y + 50° + 50° = 360°

or x + y = 360° – 50° – 50° = 260°

But x = y

∴ x = y = \(\frac { 260 }{ 2 }\) = 130°

Thus,

(iii) ∵Vertically opposite angles are equal

∵ x = 90°

∵ Sum of the angles of a triangle = 180°

∴ 90° + 30° + y = 180°

or y = 108° – 30° – 90° = 60°

In the figure ABCD is a parallelogram

∴ AD || BC and BD is a transversal

∴ y = z [Alternate angles]

But y = 60°

∴ z = 60°

Thus, x = 90°, y = 60° and z = 60°

(iv) ABCD is a parallelogram

∴ Opposite angles are equal.

∴ y = 80°

AB || CD and BC is a transversal.

∴ x + 80° = 180°

[Interior opposite angles]

or x = 180° – 80° = 100°

Again BC || AD and CD is a transversal,

∴ z = 80° [Corresponding angles]

Thus, x = 100°, y = 80° and z = 80°

(v) ∴ In a parallelogram, opposite angles are equal.

y = 112°

In ∆ ACD, x + y + 40° = 180°

x + 112° + 40° = 180°

∴ x = 180° – 112° – 40° = 28°

∵ AD || BC and AC is a transversal.

∴ x = z [∵ Alternate angles are equal]

and z = 28°

Thus, x = 28°, y = 112° and z = 28°

Question 3.

Can a quadrilateral ABCD be a parallelogram if

(i) ∠D + ∠B = 180°?

(ii) AB = DC = 8 cm, AD = 4 cm and BC = 4.4 cm?

(in) ∠A = 70° and ∠C = 65°?

Solution:

(i) In a quadrilateral ABCD,

∠A + ∠B = Sum of adjacent angles = 180°

∴ The quadrilateral may be a parallelogram but not always

(ii) In a quadrilateral ABCD,

AB = DC = 8 cm

AD = 4 cm

BC = 4.4 cm

∵ Opposite sides AD and BC are not equal.

∴ It cannot be a parallelogram.

(iii) In a quadrilateral ABCD,

∠A = 70° and ZC = 65°

∵ Opposite angles ∠A ≠ ∠C

∴ It cannot be a parallelogram.

Question 4.

Draw a rough figure of a quadrilateral that is not a parallelogram but has exactly two opposite angles of equal measure.

Solution:

In the adjoining figure, ABCD is not a parallelogram such that opposite angles ∠B and ∠D are equal. It is a kite.

Question 5.

The measures of two adjacent angles of a parallelogram are in the ratio 3 : 2. Find the measure of each of the angles of the parallelogram.

Solution:

Let ABCD be a parallelogram in which adjacent angles ∠A and ∠B are 3x and 2x respectively, since adjacent angles are supplementary.

∴ ∠A + ∠B = 180°

∴ 3x + 2x = 180° or 5x = 180°

or x = \(\frac { 180° }{ 5 }\) = 36°

∠A = 3 x 36° = 108° and ∠B = 2 x 36° = 72°

∵ Opposite angles are equal.

∴∠D = ∠B = 72° and ∠C = ∠A = 108°

∴ ∠A = 108°, ∠B = 12°, ∠C = 108° and ∠D = 72°

Question 6.

Two adjacent angles of a parallelogram have equal measure. Find the measure of each of the angles of the parallelogram.

Solution:

Let ABCD be a parallelogram such that adjacent angles ∠A = ∠B.

Since ∠A + ∠B = 180°

∠A = ∠B = \(\frac { 180° }{ 2 }\) = 90°

Since, opposite angles of a parallelogram are equal.

∠A = ∠C = 90° and ∠B = ∠D = 90°

Thus, ∠A = 90°, ∠B = 90°, ∠C = 90° and ∠D = 90°.

Question 7.

The adjacent figure HOPE is a parallelogram. Find the angle measures x, y and z. State the properties you use to find them.

Solution:

y + z = 70° …..(1)

In a triangle, exterior angle is equal to the sum of opposite interior angles.

∴ In A ∠HOP, ∠HOP = 180° – (y + z)

= 180° – 70° = 110°

Now x = ∠HOP

[Opposite angles of a parallelogram are equal]

∴ x = 110°

EH || OP and PH is a transversal.

∴ y = 40° [Alternate angles are equal]

From (1), 40° + z = 70°

∴ z = 70° – 40° = 30°

Thus, x = 110°, y = 40° and z = 30°

Question 8.

The following figures GUNS and RUNS are parallelograms. Find x and y. (Lengths are in cm.)

(i) ∵ GUNS is a parallelogram.

Its opposite sides are equal.

GS = NU and SN = GU

or 3x = 18 and 26 = 3y – 1

Now 3x = 18 ⇒ x = \(\frac { 18 }{ 3 }\) = 6

3y – 1 = 26 ⇒ y = \(\frac { 26 + 1 }{ 3 }\) = \(\frac { 27 }{ 3 }\) = 9

Thus, x = 6 cm and y = 9 cm

(ii) RUNS is a parallelogram and thus its diagonals bisect each other.

∴ x + y = 16

and 7 + y = 20

i.e y = 20 – 7

or y = 13

∴ From x + y = 16, we have

x + 13 = 16

or x = 16 – 13 = 3

Thus, x = 3 cm and y = 13 cm

Question 9.

In the following figure, both RISK and CLUE are parallelograms. Find the value of x.

Solution:

RISK is a parallelogram.

∴ ∠R + ∠K= 180°

[∵ Adjacent angles of a parallelogram are supplementary]

or ∠R + 120° = 180°

⇒ ∠R = 180° – 120° = 60°

But ∠R and ∠S are opposite angles.

∴ ∠S = 60°

CLUE is also a parallelogram.

∴ Its opposite angles are equal.

∴ ∠E = ∠L = 70°

Now, in ∆ ESO, we have

∠E + ∠S + x = 180°

∴ 70° + 60° + x = 180°

or x = 180° – 60° – 70° ⇒ x = 50°

Question 10.

Explain how this figure is a trapezium. Which of its two sides are parallel?

Solution:

Since, 100° + 80° = 180°

i.e., ∠M and ∠L are supplementary.

NM || KL

[∵ If interior opposite angles along the transversal are supplementary]

Question 11.

Find m∠C in the adjoining figure if \(\bar { AB } \) || \(\bar { DC } \)

Solution:

∵ ABCD is a trapezium in which \(\bar { AB } \) || \(\bar { DC } \) and BC is a transversal.

∵ Interior opposite angles along BC are supplementary.

∴ m∠B + m∠C = 180° or m∠C = 180° – m∠B

∴ m∠C = 180° – 120° [∴∠B = 120°]

or m∠C = 60°

Question 12.

Find the measure of ∠P and ∠S if \(\bar { SP } \) || \(\bar { RQ } \) in figure. (If you find m∠R, is there more than one method to find m∠P?)

Solution:

PQRS is a trapezium such that \(\bar { SP } \) || \(\bar { RQ } \) and PQ is a transversal

∴ m∠p + m∠Q = 180°

[Interior opposite angles are supplementary]

or m∠p + 130° = 180°

or m∠P = 180° – 130° = 50°

Again SP || RQ and RS is a transversal

∴ m∠S + m∠R = 180°

or m∠S + 90° = 180°

∴ m∠S = 108° – 90° = 90°

Yes, using the angle sum property of a quadrilateral, we can find m∠p when m∠R is know.

∴ m∠P +m∠Q + m∠R + m∠S = 360°

or m∠P + 130° + 90° + 90° = 360°

or m∠P = 360° – 130° – 90° – 90° = 50°