Class 8 Exam > Class 8 Notes > Mathematics (Maths) Class 8 > Understanding Shapes – III (Special Types of Quadrilaterals) - EXERCISE 17.1
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Page 1
RD Sharma Solutions for Class 8 Maths Chapter 17 –
Understanding Shapes – III (Special Types of Quadrilaterals)
EXERCISE 17. 1 2 . 7 1 : O N S E G A P 1
1. Given below is a parallelogram ABCD. Complete each statement along with the
definition or property used.
(i) AD =
(ii) ?DCB =
(iii) OC =
(iv) ?DAB + ?CDA =
Solution:
(i) AD = BC. Because, diagonals bisect each other in a parallelogram.
(ii) ?DCB = ?BAD. Because, alternate interior angles are equal.
(iii) OC = OA. Because, diagonals bisect each other in a parallelogram.
(iv) ?DAB+ ?CDA = 180°. Because sum of adjacent angles in a parallelogram is 180°.
2. The following figures are parallelograms. Find the degree values of the
unknowns x, y, z.
Page 2
RD Sharma Solutions for Class 8 Maths Chapter 17 –
Understanding Shapes – III (Special Types of Quadrilaterals)
EXERCISE 17. 1 2 . 7 1 : O N S E G A P 1
1. Given below is a parallelogram ABCD. Complete each statement along with the
definition or property used.
(i) AD =
(ii) ?DCB =
(iii) OC =
(iv) ?DAB + ?CDA =
Solution:
(i) AD = BC. Because, diagonals bisect each other in a parallelogram.
(ii) ?DCB = ?BAD. Because, alternate interior angles are equal.
(iii) OC = OA. Because, diagonals bisect each other in a parallelogram.
(iv) ?DAB+ ?CDA = 180°. Because sum of adjacent angles in a parallelogram is 180°.
2. The following figures are parallelograms. Find the degree values of the
unknowns x, y, z.
RD Sharma Solutions for Class 8 Maths Chapter 17 –
Understanding Shapes – III (Special Types of Quadrilaterals)
Solution:
(i) ?ABC = ?y = 100
o
(opposite angles are equal in a parallelogram)
?x + ?y = 180
o
(sum of adjacent angles is = 180
o
in a parallelogram)
?x + 100
o
= 180
o
?x = 180
o
- 100
o
= 80
o
? ?x = 80
o
?y = 100
o
?z = 80
o
(opposite angles are equal in a parallelogram)
(ii) ?RSP + ?y = 180
o
(sum of adjacent angles is = 180
o
in a parallelogram)
?y + 50
o
= 180
o
?y = 180
o
- 50
o
= 130
o
? ?x = ?y = 130
o
(opposite angles are equal in a parallelogram)
?RSP = ?RQP = 50
o
(opposite angles are equal in a parallelogram)
?RQP + ?z = 180
o
(linear pair)
50
o
+ ?z = 180
o
?z = 180
o
– 50
o
= 130
o
? ?x = 130
o
?y = 130
o
?z = 130
o
(iii) In ?PMN
?NPM + ?NMP + ?MNP = 180° [Sum of all the angles of a triangle is 180°]
Page 3
RD Sharma Solutions for Class 8 Maths Chapter 17 –
Understanding Shapes – III (Special Types of Quadrilaterals)
EXERCISE 17. 1 2 . 7 1 : O N S E G A P 1
1. Given below is a parallelogram ABCD. Complete each statement along with the
definition or property used.
(i) AD =
(ii) ?DCB =
(iii) OC =
(iv) ?DAB + ?CDA =
Solution:
(i) AD = BC. Because, diagonals bisect each other in a parallelogram.
(ii) ?DCB = ?BAD. Because, alternate interior angles are equal.
(iii) OC = OA. Because, diagonals bisect each other in a parallelogram.
(iv) ?DAB+ ?CDA = 180°. Because sum of adjacent angles in a parallelogram is 180°.
2. The following figures are parallelograms. Find the degree values of the
unknowns x, y, z.
RD Sharma Solutions for Class 8 Maths Chapter 17 –
Understanding Shapes – III (Special Types of Quadrilaterals)
Solution:
(i) ?ABC = ?y = 100
o
(opposite angles are equal in a parallelogram)
?x + ?y = 180
o
(sum of adjacent angles is = 180
o
in a parallelogram)
?x + 100
o
= 180
o
?x = 180
o
- 100
o
= 80
o
? ?x = 80
o
?y = 100
o
?z = 80
o
(opposite angles are equal in a parallelogram)
(ii) ?RSP + ?y = 180
o
(sum of adjacent angles is = 180
o
in a parallelogram)
?y + 50
o
= 180
o
?y = 180
o
- 50
o
= 130
o
? ?x = ?y = 130
o
(opposite angles are equal in a parallelogram)
?RSP = ?RQP = 50
o
(opposite angles are equal in a parallelogram)
?RQP + ?z = 180
o
(linear pair)
50
o
+ ?z = 180
o
?z = 180
o
– 50
o
= 130
o
? ?x = 130
o
?y = 130
o
?z = 130
o
(iii) In ?PMN
?NPM + ?NMP + ?MNP = 180° [Sum of all the angles of a triangle is 180°]
RD Sharma Solutions for Class 8 Maths Chapter 17 –
Understanding Shapes – III (Special Types of Quadrilaterals)
30° + 90° + ?z = 180°
?z = 180°-120°
= 60°
?y = ?z = 60° [opposite angles are equal in a parallelogram]
?z = 180°-120° [sum of the adjacent angles is equal to 180° in a parallelogram]
?z = 60°
?z + ?LMN = 180° [sum of the adjacent angles is equal to 180° in a parallelogram]
60° + 90°+ ?x = 180°
?x = 180°-150°
?x = 30°
? ?x = 30
o
?y = 60
o
?z = 60
o
(iv) ?x = 90° [vertically opposite angles are equal]
In ?DOC
?x + ?y + 30° = 180° [Sum of all the angles of a triangle is 180°]
90° + 30° + ?y = 180°
?y = 180°-120°
?y = 60°
?y = ?z = 60° [alternate interior angles are equal]
? ?x = 90
o
?y = 60
o
?z = 60
o
(v) ?x + ?POR = 180° [sum of the adjacent angles is equal to 180° in a parallelogram]
?x + 80° = 180°
?x = 180°-80°
?x = 100°
?y = 80° [opposite angles are equal in a parallelogram]
?SRQ = ?x = 100°
?SRQ + ?z = 180° [Linear pair]
100° + ?z = 180°
?z = 180°-100°
?z = 80°
? ?x = 100
o
?y = 80
o
?z = 80
o
(vi) ?y = 112° [In a parallelogram opposite angles are equal]
?y + ?VUT = 180° [In a parallelogram sum of the adjacent angles is equal to 180°]
?z + 40° + 112° = 180°
?z = 180°-152°
?z = 28°
?z = ?x = 28° [alternate interior angles are equal]
Page 4
RD Sharma Solutions for Class 8 Maths Chapter 17 –
Understanding Shapes – III (Special Types of Quadrilaterals)
EXERCISE 17. 1 2 . 7 1 : O N S E G A P 1
1. Given below is a parallelogram ABCD. Complete each statement along with the
definition or property used.
(i) AD =
(ii) ?DCB =
(iii) OC =
(iv) ?DAB + ?CDA =
Solution:
(i) AD = BC. Because, diagonals bisect each other in a parallelogram.
(ii) ?DCB = ?BAD. Because, alternate interior angles are equal.
(iii) OC = OA. Because, diagonals bisect each other in a parallelogram.
(iv) ?DAB+ ?CDA = 180°. Because sum of adjacent angles in a parallelogram is 180°.
2. The following figures are parallelograms. Find the degree values of the
unknowns x, y, z.
RD Sharma Solutions for Class 8 Maths Chapter 17 –
Understanding Shapes – III (Special Types of Quadrilaterals)
Solution:
(i) ?ABC = ?y = 100
o
(opposite angles are equal in a parallelogram)
?x + ?y = 180
o
(sum of adjacent angles is = 180
o
in a parallelogram)
?x + 100
o
= 180
o
?x = 180
o
- 100
o
= 80
o
? ?x = 80
o
?y = 100
o
?z = 80
o
(opposite angles are equal in a parallelogram)
(ii) ?RSP + ?y = 180
o
(sum of adjacent angles is = 180
o
in a parallelogram)
?y + 50
o
= 180
o
?y = 180
o
- 50
o
= 130
o
? ?x = ?y = 130
o
(opposite angles are equal in a parallelogram)
?RSP = ?RQP = 50
o
(opposite angles are equal in a parallelogram)
?RQP + ?z = 180
o
(linear pair)
50
o
+ ?z = 180
o
?z = 180
o
– 50
o
= 130
o
? ?x = 130
o
?y = 130
o
?z = 130
o
(iii) In ?PMN
?NPM + ?NMP + ?MNP = 180° [Sum of all the angles of a triangle is 180°]
RD Sharma Solutions for Class 8 Maths Chapter 17 –
Understanding Shapes – III (Special Types of Quadrilaterals)
30° + 90° + ?z = 180°
?z = 180°-120°
= 60°
?y = ?z = 60° [opposite angles are equal in a parallelogram]
?z = 180°-120° [sum of the adjacent angles is equal to 180° in a parallelogram]
?z = 60°
?z + ?LMN = 180° [sum of the adjacent angles is equal to 180° in a parallelogram]
60° + 90°+ ?x = 180°
?x = 180°-150°
?x = 30°
? ?x = 30
o
?y = 60
o
?z = 60
o
(iv) ?x = 90° [vertically opposite angles are equal]
In ?DOC
?x + ?y + 30° = 180° [Sum of all the angles of a triangle is 180°]
90° + 30° + ?y = 180°
?y = 180°-120°
?y = 60°
?y = ?z = 60° [alternate interior angles are equal]
? ?x = 90
o
?y = 60
o
?z = 60
o
(v) ?x + ?POR = 180° [sum of the adjacent angles is equal to 180° in a parallelogram]
?x + 80° = 180°
?x = 180°-80°
?x = 100°
?y = 80° [opposite angles are equal in a parallelogram]
?SRQ = ?x = 100°
?SRQ + ?z = 180° [Linear pair]
100° + ?z = 180°
?z = 180°-100°
?z = 80°
? ?x = 100
o
?y = 80
o
?z = 80
o
(vi) ?y = 112° [In a parallelogram opposite angles are equal]
?y + ?VUT = 180° [In a parallelogram sum of the adjacent angles is equal to 180°]
?z + 40° + 112° = 180°
?z = 180°-152°
?z = 28°
?z = ?x = 28° [alternate interior angles are equal]
RD Sharma Solutions for Class 8 Maths Chapter 17 –
Understanding Shapes – III (Special Types of Quadrilaterals)
? ?x = 28
o
?y = 112
o
?z = 28
o
3. Can the following figures be parallelograms? Justify your answer.
Solution:
(i) No, opposite angles are equal in a parallelogram.
(ii) Yes, opposite sides are equal and parallel in a parallelogram.
(iii) No, diagonals bisect each other in a parallelogram.
4. In the adjacent figure HOPE is a parallelogram. Find the angle measures x,
y and z. State the geometrical truths you use to find them.
Solution:
We know that
?POH + 70° = 180° [Linear pair]
?POH = 180°-70°
?POH = 110°
Page 5
RD Sharma Solutions for Class 8 Maths Chapter 17 –
Understanding Shapes – III (Special Types of Quadrilaterals)
EXERCISE 17. 1 2 . 7 1 : O N S E G A P 1
1. Given below is a parallelogram ABCD. Complete each statement along with the
definition or property used.
(i) AD =
(ii) ?DCB =
(iii) OC =
(iv) ?DAB + ?CDA =
Solution:
(i) AD = BC. Because, diagonals bisect each other in a parallelogram.
(ii) ?DCB = ?BAD. Because, alternate interior angles are equal.
(iii) OC = OA. Because, diagonals bisect each other in a parallelogram.
(iv) ?DAB+ ?CDA = 180°. Because sum of adjacent angles in a parallelogram is 180°.
2. The following figures are parallelograms. Find the degree values of the
unknowns x, y, z.
RD Sharma Solutions for Class 8 Maths Chapter 17 –
Understanding Shapes – III (Special Types of Quadrilaterals)
Solution:
(i) ?ABC = ?y = 100
o
(opposite angles are equal in a parallelogram)
?x + ?y = 180
o
(sum of adjacent angles is = 180
o
in a parallelogram)
?x + 100
o
= 180
o
?x = 180
o
- 100
o
= 80
o
? ?x = 80
o
?y = 100
o
?z = 80
o
(opposite angles are equal in a parallelogram)
(ii) ?RSP + ?y = 180
o
(sum of adjacent angles is = 180
o
in a parallelogram)
?y + 50
o
= 180
o
?y = 180
o
- 50
o
= 130
o
? ?x = ?y = 130
o
(opposite angles are equal in a parallelogram)
?RSP = ?RQP = 50
o
(opposite angles are equal in a parallelogram)
?RQP + ?z = 180
o
(linear pair)
50
o
+ ?z = 180
o
?z = 180
o
– 50
o
= 130
o
? ?x = 130
o
?y = 130
o
?z = 130
o
(iii) In ?PMN
?NPM + ?NMP + ?MNP = 180° [Sum of all the angles of a triangle is 180°]
RD Sharma Solutions for Class 8 Maths Chapter 17 –
Understanding Shapes – III (Special Types of Quadrilaterals)
30° + 90° + ?z = 180°
?z = 180°-120°
= 60°
?y = ?z = 60° [opposite angles are equal in a parallelogram]
?z = 180°-120° [sum of the adjacent angles is equal to 180° in a parallelogram]
?z = 60°
?z + ?LMN = 180° [sum of the adjacent angles is equal to 180° in a parallelogram]
60° + 90°+ ?x = 180°
?x = 180°-150°
?x = 30°
? ?x = 30
o
?y = 60
o
?z = 60
o
(iv) ?x = 90° [vertically opposite angles are equal]
In ?DOC
?x + ?y + 30° = 180° [Sum of all the angles of a triangle is 180°]
90° + 30° + ?y = 180°
?y = 180°-120°
?y = 60°
?y = ?z = 60° [alternate interior angles are equal]
? ?x = 90
o
?y = 60
o
?z = 60
o
(v) ?x + ?POR = 180° [sum of the adjacent angles is equal to 180° in a parallelogram]
?x + 80° = 180°
?x = 180°-80°
?x = 100°
?y = 80° [opposite angles are equal in a parallelogram]
?SRQ = ?x = 100°
?SRQ + ?z = 180° [Linear pair]
100° + ?z = 180°
?z = 180°-100°
?z = 80°
? ?x = 100
o
?y = 80
o
?z = 80
o
(vi) ?y = 112° [In a parallelogram opposite angles are equal]
?y + ?VUT = 180° [In a parallelogram sum of the adjacent angles is equal to 180°]
?z + 40° + 112° = 180°
?z = 180°-152°
?z = 28°
?z = ?x = 28° [alternate interior angles are equal]
RD Sharma Solutions for Class 8 Maths Chapter 17 –
Understanding Shapes – III (Special Types of Quadrilaterals)
? ?x = 28
o
?y = 112
o
?z = 28
o
3. Can the following figures be parallelograms? Justify your answer.
Solution:
(i) No, opposite angles are equal in a parallelogram.
(ii) Yes, opposite sides are equal and parallel in a parallelogram.
(iii) No, diagonals bisect each other in a parallelogram.
4. In the adjacent figure HOPE is a parallelogram. Find the angle measures x,
y and z. State the geometrical truths you use to find them.
Solution:
We know that
?POH + 70° = 180° [Linear pair]
?POH = 180°-70°
?POH = 110°
RD Sharma Solutions for Class 8 Maths Chapter 17 –
Understanding Shapes – III (Special Types of Quadrilaterals)
?POH = ?x = 110° [opposite angles are equal in a parallelogram]
?x + ?z + 40° = 180° [sum of the adjacent angles is equal to 180° in a parallelogram]
110° + ?z + 40° = 180°
?z = 180° - 150°
?z = 30°
?z + ?y = 70°
?y + 30° = 70°
?y = 70°- 30°
?y = 40°
5. In the following figures GUNS and RUNS are parallelograms. Find x and y.
Solution:
(i) 3y – 1 = 26 [opposite sides are of equal length in a parallelogram]
3y = 26 + 1
y = 27/3
y = 9
3x = 18 [opposite sides are of equal length in a parallelogram]
x = 18/3
x = 6
? x = 6 and y = 9
(ii) y – 7 = 20 [diagonals bisect each other in a parallelogram]
y = 20 + 7
y = 27
x - y = 16 [diagonals bisect each other in a parallelogram]
x -27 = 16
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