Gujarat BoardĀ 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) ā 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 = 180Ā°

[ā“ Adjacent angles are supplementary]

Question 2.

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

Solution:

(i) ā y = 100Ā°

[ā“ Opposite angles of a parallelogram are equal.]

Sum of interior angles of a parallelogram = 360Ā°

ā“ x + y + z + ā B = 360Ā°

or x + 100Ā° + z + 100Ā° = 360Ā°

or x + z = 360Ā° – 100Ā° – 100Ā° = 160Ā°

But x = z

ā“ x = z = \(\frac { 160Ā° }{ 2 }\) = 80Ā°

Thus,

(ii) ā“ Opposite angles are equal.

ā“ m ā 1 = 50

now, ā 1 + z = 180Ā° [Linear pair]

or z = 180Ā° – ā 1

= 180Ā° – 50Ā° = 130Ā°

x + y + 50Ā° + 50Ā° = 360Ā°

or x + y = 360Ā° – 50Ā° – 50Ā° = 260Ā°

But x = y

ā“ x = y = \(\frac { 260 }{ 2 }\) = 130Ā°

Thus,

(iii) āµVertically opposite angles are equal

āµ x = 90Ā°

āµ Sum of the angles of a triangle = 180Ā°

ā“ 90Ā° + 30Ā° + y = 180Ā°

or y = 108Ā° – 30Ā° – 90Ā° = 60Ā°

In the figure ABCD is a parallelogram

ā“ AD || BC and BD is a transversal

ā“ y = z [Alternate angles]

But y = 60Ā°

ā“ z = 60Ā°

Thus, x = 90Ā°, y = 60Ā° and z = 60Ā°

(iv) ABCD is a parallelogram

ā“ Opposite angles are equal.

ā“ y = 80Ā°

AB || CD and BC is a transversal.

ā“ x + 80Ā° = 180Ā°

[Interior opposite angles]

or x = 180Ā° – 80Ā° = 100Ā°

Again BC || AD and CD is a transversal,

ā“ z = 80Ā° [Corresponding angles]

Thus, x = 100Ā°, y = 80Ā° and z = 80Ā°

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

y = 112Ā°

In ā ACD, x + y + 40Ā° = 180Ā°

x + 112Ā° + 40Ā° = 180Ā°

ā“ x = 180Ā° – 112Ā° – 40Ā° = 28Ā°

āµ AD || BC and AC is a transversal.

ā“ x = z [āµ Alternate angles are equal]

and z = 28Ā°

Thus, x = 28Ā°, y = 112Ā° and z = 28Ā°

Question 3.

Can a quadrilateral ABCD be a parallelogram if

(i) ā D + ā B = 180Ā°?

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

(in) ā A = 70Ā° and ā C = 65Ā°?

Solution:

(i) In a quadrilateral ABCD,

ā A + ā B = Sum of adjacent angles = 180Ā°

ā“ 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

āµ Opposite sides AD and BC are not equal.

ā“ It cannot be a parallelogram.

(iii) In a quadrilateral ABCD,

ā A = 70Ā° and ZC = 65Ā°

āµ Opposite angles ā A ā ā C

ā“ 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 ā B and ā 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 ā A and ā B are 3x and 2x respectively, since adjacent angles are supplementary.

ā“ ā A + ā B = 180Ā°

ā“ 3x + 2x = 180Ā° or 5x = 180Ā°

or x = \(\frac { 180Ā° }{ 5 }\) = 36Ā°

ā A = 3 x 36Ā° = 108Ā° and ā B = 2 x 36Ā° = 72Ā°

āµ Opposite angles are equal.

ā“ā D = ā B = 72Ā° and ā C = ā A = 108Ā°

ā“ ā A = 108Ā°, ā B = 12Ā°, ā C = 108Ā° and ā D = 72Ā°

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 ā A = ā B.

Since ā A + ā B = 180Ā°

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

Since, opposite angles of a parallelogram are equal.

ā A = ā C = 90Ā° and ā B = ā D = 90Ā°

Thus, ā A = 90Ā°, ā B = 90Ā°, ā C = 90Ā° and ā D = 90Ā°.

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 = 70Ā° …..(1)

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

ā“ In A ā HOP, ā HOP = 180Ā° – (y + z)

= 180Ā° – 70Ā° = 110Ā°

Now x = ā HOP

[Opposite angles of a parallelogram are equal]

ā“ x = 110Ā°

EH || OP and PH is a transversal.

ā“ y = 40Ā° [Alternate angles are equal]

From (1), 40Ā° + z = 70Ā°

ā“ z = 70Ā° – 40Ā° = 30Ā°

Thus, x = 110Ā°, y = 40Ā° and z = 30Ā°

Question 8.

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

(i) āµ GUNS is a parallelogram.

Its opposite sides are equal.

GS = NU and SN = GU

or 3x = 18 and 26 = 3y – 1

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

3y – 1 = 26 ā 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.

ā“ x + y = 16

and 7 + y = 20

i.e y = 20 – 7

or y = 13

ā“ 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.

ā“ ā R + ā K= 180Ā°

[āµ Adjacent angles of a parallelogram are supplementary]

or ā R + 120Ā° = 180Ā°

ā ā R = 180Ā° – 120Ā° = 60Ā°

But ā R and ā S are opposite angles.

ā“ ā S = 60Ā°

CLUE is also a parallelogram.

ā“ Its opposite angles are equal.

ā“ ā E = ā L = 70Ā°

Now, in ā ESO, we have

ā E + ā S + x = 180Ā°

ā“ 70Ā° + 60Ā° + x = 180Ā°

or x = 180Ā° – 60Ā° – 70Ā° ā x = 50Ā°

Question 10.

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

Solution:

Since, 100Ā° + 80Ā° = 180Ā°

i.e., ā M and ā L are supplementary.

NM || KL

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

Question 11.

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

Solution:

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

āµ Interior opposite angles along BC are supplementary.

ā“ mā B + mā C = 180Ā° or mā C = 180Ā° – mā B

ā“ mā C = 180Ā° – 120Ā° [ā“ā B = 120Ā°]

or mā C = 60Ā°

Question 12.

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

Solution:

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

ā“ mā p + mā Q = 180Ā°

[Interior opposite angles are supplementary]

or mā p + 130Ā° = 180Ā°

or mā P = 180Ā° – 130Ā° = 50Ā°

Again SP || RQ and RS is a transversal

ā“ mā S + mā R = 180Ā°

or mā S + 90Ā° = 180Ā°

ā“ mā S = 108Ā° – 90Ā° = 90Ā°

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

ā“ mā P +mā Q + mā R + mā S = 360Ā°

or mā P + 130Ā° + 90Ā° + 90Ā° = 360Ā°

or mā P = 360Ā° – 130Ā° – 90Ā° – 90Ā° = 50Ā°