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17. Special Types of Quadrilaterals 
EXERCISE 17 
Question 1. 
In parallelogram ABCD, ?A = 3 times ?B. Find all the angles of the parallelogram. In the 
same parallelogram, if AB = 5x – 7 and CD = 3x +1 ; find the length of CD. 
Solution: 
 
Let ?B = x 
?A = 3 ?B = 3x 
AD||BC 
?A + ?B = 180° 
3x + x = 180° 
? 4x = 180° 
? x = 45° 
?B = 45° 
?A = 3x = 3 x 45 = 135° 
and ?B = ?D = 45° 
opposite angles of || gm are equal. 
?A = ?C = 135° 
opposite sides of //gm are equal. 
AB = CD 
5x – 7 = 3x + 1 
? 5x – 3x = 1+7 
? 2x = 8 
? x = 4 
CD = 3 x 4+1 = 13 
Hence 135°, 45°, 135° and 45° ; 13 
Question 2. 
In parallelogram PQRS, ?Q = (4x – 5)° and ?S = (3x + 10)°. Calculate : ?Q and ?R. 
Solution: 
In parallelogram PQRS, 
?Q = (4x – 5)° and ?S = (3x + 10)° 
Page 2


17. Special Types of Quadrilaterals 
EXERCISE 17 
Question 1. 
In parallelogram ABCD, ?A = 3 times ?B. Find all the angles of the parallelogram. In the 
same parallelogram, if AB = 5x – 7 and CD = 3x +1 ; find the length of CD. 
Solution: 
 
Let ?B = x 
?A = 3 ?B = 3x 
AD||BC 
?A + ?B = 180° 
3x + x = 180° 
? 4x = 180° 
? x = 45° 
?B = 45° 
?A = 3x = 3 x 45 = 135° 
and ?B = ?D = 45° 
opposite angles of || gm are equal. 
?A = ?C = 135° 
opposite sides of //gm are equal. 
AB = CD 
5x – 7 = 3x + 1 
? 5x – 3x = 1+7 
? 2x = 8 
? x = 4 
CD = 3 x 4+1 = 13 
Hence 135°, 45°, 135° and 45° ; 13 
Question 2. 
In parallelogram PQRS, ?Q = (4x – 5)° and ?S = (3x + 10)°. Calculate : ?Q and ?R. 
Solution: 
In parallelogram PQRS, 
?Q = (4x – 5)° and ?S = (3x + 10)° 
 
opposite ?s of //gm are equal. 
?Q = ?S 
4x – 5 = 3x + 10 
4x – 3x = 10+5 
x = 15 
?Q = 4x – 5 =4 x 15 – 5 = 55° 
Also ?Q + ?R = 180° 
55° + ?R = 180° 
?R = 180°-55° = 125° 
?Q = 55° ; ?R = 125° 
Question 3. 
In rhombus ABCD ; 
(i) if ?A = 74° ; find ?B and ?C. 
(ii) if AD = 7.5 cm ; find BC and CD. 
Solution: 
AD || BC 
?A + ?B = 180° 
74° + ?B = 180° 
?B =180° – 74°= 106° 
 
opposite angles of Rhombus are equal. 
?A = ?C = 74° 
Sides of Rhombus are equal. 
BC = CD = AD = 7.5 cm 
(i) ?B = 106° ; ?C = 74° 
(ii) BC = 7.5 cm and CD = 7.5 cm Ans. 
Question 4. 
In square PQRS : 
Page 3


17. Special Types of Quadrilaterals 
EXERCISE 17 
Question 1. 
In parallelogram ABCD, ?A = 3 times ?B. Find all the angles of the parallelogram. In the 
same parallelogram, if AB = 5x – 7 and CD = 3x +1 ; find the length of CD. 
Solution: 
 
Let ?B = x 
?A = 3 ?B = 3x 
AD||BC 
?A + ?B = 180° 
3x + x = 180° 
? 4x = 180° 
? x = 45° 
?B = 45° 
?A = 3x = 3 x 45 = 135° 
and ?B = ?D = 45° 
opposite angles of || gm are equal. 
?A = ?C = 135° 
opposite sides of //gm are equal. 
AB = CD 
5x – 7 = 3x + 1 
? 5x – 3x = 1+7 
? 2x = 8 
? x = 4 
CD = 3 x 4+1 = 13 
Hence 135°, 45°, 135° and 45° ; 13 
Question 2. 
In parallelogram PQRS, ?Q = (4x – 5)° and ?S = (3x + 10)°. Calculate : ?Q and ?R. 
Solution: 
In parallelogram PQRS, 
?Q = (4x – 5)° and ?S = (3x + 10)° 
 
opposite ?s of //gm are equal. 
?Q = ?S 
4x – 5 = 3x + 10 
4x – 3x = 10+5 
x = 15 
?Q = 4x – 5 =4 x 15 – 5 = 55° 
Also ?Q + ?R = 180° 
55° + ?R = 180° 
?R = 180°-55° = 125° 
?Q = 55° ; ?R = 125° 
Question 3. 
In rhombus ABCD ; 
(i) if ?A = 74° ; find ?B and ?C. 
(ii) if AD = 7.5 cm ; find BC and CD. 
Solution: 
AD || BC 
?A + ?B = 180° 
74° + ?B = 180° 
?B =180° – 74°= 106° 
 
opposite angles of Rhombus are equal. 
?A = ?C = 74° 
Sides of Rhombus are equal. 
BC = CD = AD = 7.5 cm 
(i) ?B = 106° ; ?C = 74° 
(ii) BC = 7.5 cm and CD = 7.5 cm Ans. 
Question 4. 
In square PQRS : 
(i) if PQ = 3x – 7 and QR = x + 3 ; find PS 
(ii) if PR = 5x and QR = 9x – 8. Find QS 
Solution: 
(i) sides of square are equal. 
 
PQ = QR 
=> 3x – 7 = x + 3 
=> 3x – x = 3 + 7 
=> 2x = 10 
x = 5 
PS=PQ = 3x – 7 = 3 x 5 – 7 =8 
(ii) PR = 5x and QS = 9x – 8 
 
As diagonals of square are equal. 
PR = QS 
5x = 9x – 8 
=> 5x – 9x = -8 
=> -4x = -8 
=> x = 2 
QS = 9x – 8 = 9 x 2 – 8 =10 
Question 5. 
ABCD is a rectangle, if ?BPC = 124° 
Calculate : (i) ?BAP (ii) ?ADP 
Page 4


17. Special Types of Quadrilaterals 
EXERCISE 17 
Question 1. 
In parallelogram ABCD, ?A = 3 times ?B. Find all the angles of the parallelogram. In the 
same parallelogram, if AB = 5x – 7 and CD = 3x +1 ; find the length of CD. 
Solution: 
 
Let ?B = x 
?A = 3 ?B = 3x 
AD||BC 
?A + ?B = 180° 
3x + x = 180° 
? 4x = 180° 
? x = 45° 
?B = 45° 
?A = 3x = 3 x 45 = 135° 
and ?B = ?D = 45° 
opposite angles of || gm are equal. 
?A = ?C = 135° 
opposite sides of //gm are equal. 
AB = CD 
5x – 7 = 3x + 1 
? 5x – 3x = 1+7 
? 2x = 8 
? x = 4 
CD = 3 x 4+1 = 13 
Hence 135°, 45°, 135° and 45° ; 13 
Question 2. 
In parallelogram PQRS, ?Q = (4x – 5)° and ?S = (3x + 10)°. Calculate : ?Q and ?R. 
Solution: 
In parallelogram PQRS, 
?Q = (4x – 5)° and ?S = (3x + 10)° 
 
opposite ?s of //gm are equal. 
?Q = ?S 
4x – 5 = 3x + 10 
4x – 3x = 10+5 
x = 15 
?Q = 4x – 5 =4 x 15 – 5 = 55° 
Also ?Q + ?R = 180° 
55° + ?R = 180° 
?R = 180°-55° = 125° 
?Q = 55° ; ?R = 125° 
Question 3. 
In rhombus ABCD ; 
(i) if ?A = 74° ; find ?B and ?C. 
(ii) if AD = 7.5 cm ; find BC and CD. 
Solution: 
AD || BC 
?A + ?B = 180° 
74° + ?B = 180° 
?B =180° – 74°= 106° 
 
opposite angles of Rhombus are equal. 
?A = ?C = 74° 
Sides of Rhombus are equal. 
BC = CD = AD = 7.5 cm 
(i) ?B = 106° ; ?C = 74° 
(ii) BC = 7.5 cm and CD = 7.5 cm Ans. 
Question 4. 
In square PQRS : 
(i) if PQ = 3x – 7 and QR = x + 3 ; find PS 
(ii) if PR = 5x and QR = 9x – 8. Find QS 
Solution: 
(i) sides of square are equal. 
 
PQ = QR 
=> 3x – 7 = x + 3 
=> 3x – x = 3 + 7 
=> 2x = 10 
x = 5 
PS=PQ = 3x – 7 = 3 x 5 – 7 =8 
(ii) PR = 5x and QS = 9x – 8 
 
As diagonals of square are equal. 
PR = QS 
5x = 9x – 8 
=> 5x – 9x = -8 
=> -4x = -8 
=> x = 2 
QS = 9x – 8 = 9 x 2 – 8 =10 
Question 5. 
ABCD is a rectangle, if ?BPC = 124° 
Calculate : (i) ?BAP (ii) ?ADP 
 
Solution: 
Diagonals of rectangle are equal and bisect each other. 
?PBC = ?PCB = x (say) 
But ?BPC + ?PBC + ?PCB = 180° 
124° + x + x = 180° 
2x = 180° – 124° 
2x = 56° 
=> x = 28° 
?PBC = 28° 
But ?PBC = ?ADP [Alternate ?s] 
?ADP = 28° 
Again ?APB = 180° – 124° = 56° 
Also PA = PB 
?BAP =  (180° – ?APB) 
=  x (180°- 56°) =  x 124° = 62° 
Hence (i) ?BAP = 62° (ii) ?ADP =28° 
Question 6. 
ABCD is a rhombus. If ?BAC = 38°, find : 
(i) ?ACB 
(ii) ?DAC 
(iii) ?ADC. 
 
Solution: 
ABCD is Rhombus (Given) 
AB = BC 
?BAC = ?ACB ( ?s opp. to equal sides) 
But ?BAC = 38° (Given) 
?ACB = 38° 
Page 5


17. Special Types of Quadrilaterals 
EXERCISE 17 
Question 1. 
In parallelogram ABCD, ?A = 3 times ?B. Find all the angles of the parallelogram. In the 
same parallelogram, if AB = 5x – 7 and CD = 3x +1 ; find the length of CD. 
Solution: 
 
Let ?B = x 
?A = 3 ?B = 3x 
AD||BC 
?A + ?B = 180° 
3x + x = 180° 
? 4x = 180° 
? x = 45° 
?B = 45° 
?A = 3x = 3 x 45 = 135° 
and ?B = ?D = 45° 
opposite angles of || gm are equal. 
?A = ?C = 135° 
opposite sides of //gm are equal. 
AB = CD 
5x – 7 = 3x + 1 
? 5x – 3x = 1+7 
? 2x = 8 
? x = 4 
CD = 3 x 4+1 = 13 
Hence 135°, 45°, 135° and 45° ; 13 
Question 2. 
In parallelogram PQRS, ?Q = (4x – 5)° and ?S = (3x + 10)°. Calculate : ?Q and ?R. 
Solution: 
In parallelogram PQRS, 
?Q = (4x – 5)° and ?S = (3x + 10)° 
 
opposite ?s of //gm are equal. 
?Q = ?S 
4x – 5 = 3x + 10 
4x – 3x = 10+5 
x = 15 
?Q = 4x – 5 =4 x 15 – 5 = 55° 
Also ?Q + ?R = 180° 
55° + ?R = 180° 
?R = 180°-55° = 125° 
?Q = 55° ; ?R = 125° 
Question 3. 
In rhombus ABCD ; 
(i) if ?A = 74° ; find ?B and ?C. 
(ii) if AD = 7.5 cm ; find BC and CD. 
Solution: 
AD || BC 
?A + ?B = 180° 
74° + ?B = 180° 
?B =180° – 74°= 106° 
 
opposite angles of Rhombus are equal. 
?A = ?C = 74° 
Sides of Rhombus are equal. 
BC = CD = AD = 7.5 cm 
(i) ?B = 106° ; ?C = 74° 
(ii) BC = 7.5 cm and CD = 7.5 cm Ans. 
Question 4. 
In square PQRS : 
(i) if PQ = 3x – 7 and QR = x + 3 ; find PS 
(ii) if PR = 5x and QR = 9x – 8. Find QS 
Solution: 
(i) sides of square are equal. 
 
PQ = QR 
=> 3x – 7 = x + 3 
=> 3x – x = 3 + 7 
=> 2x = 10 
x = 5 
PS=PQ = 3x – 7 = 3 x 5 – 7 =8 
(ii) PR = 5x and QS = 9x – 8 
 
As diagonals of square are equal. 
PR = QS 
5x = 9x – 8 
=> 5x – 9x = -8 
=> -4x = -8 
=> x = 2 
QS = 9x – 8 = 9 x 2 – 8 =10 
Question 5. 
ABCD is a rectangle, if ?BPC = 124° 
Calculate : (i) ?BAP (ii) ?ADP 
 
Solution: 
Diagonals of rectangle are equal and bisect each other. 
?PBC = ?PCB = x (say) 
But ?BPC + ?PBC + ?PCB = 180° 
124° + x + x = 180° 
2x = 180° – 124° 
2x = 56° 
=> x = 28° 
?PBC = 28° 
But ?PBC = ?ADP [Alternate ?s] 
?ADP = 28° 
Again ?APB = 180° – 124° = 56° 
Also PA = PB 
?BAP =  (180° – ?APB) 
=  x (180°- 56°) =  x 124° = 62° 
Hence (i) ?BAP = 62° (ii) ?ADP =28° 
Question 6. 
ABCD is a rhombus. If ?BAC = 38°, find : 
(i) ?ACB 
(ii) ?DAC 
(iii) ?ADC. 
 
Solution: 
ABCD is Rhombus (Given) 
AB = BC 
?BAC = ?ACB ( ?s opp. to equal sides) 
But ?BAC = 38° (Given) 
?ACB = 38° 
In ?ABC, 
?ABC + ?BAC + ?ACB = 180° 
?ABC + 38°+ 38° = 180° 
?ABC = 180° – 76° = 104° 
But ?ABC = ?ADC (opp. ?s of rhombus) 
?ADC = 104° 
?DAC = ?DCA ( AD = CD) 
?DAC =  [180° – 104°] 
?DAC =  x 76° = 38° 
Hence (i) ?ACB = 38° (ii) ?DAC = 38° (iii) ?ADC = 104° Ans. 
Question 7. 
ABCD is a rhombus. If ?BCA = 35°. find ?ADC. 
Solution: 
Given : Rhombus ABCD in which ?BCA = 35° 
 
To find : ?ADC 
Proof : AD || BC 
?DAC = ?BCA (Alternate ?s) 
But ?BCA = 35° (Given) 
?DAC = 35° 
But ?DAC = ?ACD ( AD = CD) & ?DAC + ?ACD + ?ADC = 180° 
35°+ 35° + ?ADC = 180° 
?ADC = 180° – 70° = 110° 
Hence ?ADC = 110° 
Question 8. 
PQRS is a parallelogram whose diagonals intersect at M. 
If ?PMS = 54°, ?QSR = 25° and ?SQR = 30° ; find : 
(i) ?RPS 
(ii) ?PRS 
(iii) ?PSR. 
Solution: 
Given : ||gm PQRS in which diagonals PR & QS intersect at M. 
?PMS = 54° ; ?QSR = 25° and ?SQR=30°

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