# GSEB Solutions Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.3

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) āµVertically opposite angles are equal

āµ x = 90Ā°
āµ 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°
āµ AD || BC and AC is a transversal.
ā“ x = z [āµ 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
āµ 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°
āµ 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°
āµ 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) āµ 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°
[āµ 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
[āµ If interior opposite angles along the transversal are supplementary]

Question 11.
Find mā C in the adjoining figure if $$\bar { AB }$$ || $$\bar { DC }$$

Solution:
āµ ABCD is a trapezium in which $$\bar { AB }$$Ā  || $$\bar { DC }$$ and BC is a transversal.
āµ 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°