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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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