Area of a Trapezium and a Polygon | Mathematics (Maths) Class 8 PDF Download

 Page 1


Q u e s t i o n : 1
Find the area of a trapezium whose parallel sides are 24 cm and 20 cm and the distance between them is 15 cm.
S o l u t i o n :
Area of a trapezium =
1
2
×(Sum of parallel sides)×(Distance between them)
                      =
1
2
×(24 +20)×15 cm
2
=
1
2
×44 ×15 cm
2
= (22 ×15) cm
2
= 330 cm
2
Hence, the area of the trapezium is 330 cm
2
.
Q u e s t i o n : 2
Find the area of a trapezium whose parallel sides are 38.7 cm and 22.3 cm, and the distance between them is 16 cm.
S o l u t i o n :
Area of a trapezium =
1
2
×(Sum of parallel sides)×(Distance between them)
                      =
1
2
×(38. 7 +22. 3)×16 cm
2
=
1
2
×61 ×16 cm
2
= (61 ×8) cm
2
= 488 cm
2
Hence, the area of the trapezium is 488 cm
2
.
Q u e s t i o n : 3
The shape of the top surface of a table is trapezium. Its parallel sides are 1 m and 1.4 m and the perpendicular distance between them is 0.9 m. Find its area.
S o l u t i o n :
Area of a trapezium =
1
2
×(Sum of parallel sides)×(Distance between them)
                       =
1
2
×(1 +1. 4)×0. 9 m
2
=
1
2
×2. 4 ×0. 9 m
2
= (1. 2 ×0. 9) m
2
= 1. 08 m
2
Hence, the area of the top surface of the table is 1. 08 m
2
.
Q u e s t i o n : 4
The area of a trapezium is 1080 cm
2
. If the lengths of its parallel sides be 55 cm and 35 cm, find the distance between them.
S o l u t i o n :
Let the distance between the parallel sides be x. Now, Area of trapezium =
1
2
×(55 +35)×x cm
2
                    =
1
2
×90 ×x cm
2
= 45x cm
2
Area of the trapezium = 1080 cm
2 
(Given) ? 45x = 1080 ? x =
1080
45
? x = 24 cmHence, the distance between the parallel sides is 24 cm.
Q u e s t i o n : 5
A field is in the form of a trapezium. Its area is 1586 m
2
 and the distance between its parallel sides is 26 m. If one of the parallel sides is 84 m, find the other.
S o l u t i o n :
Let the length of the required side be x cm. Now, Area of trapezium =
1
2
×(84 +x)×26 m
2
                        = (1092 +13x) m
2
Area of trapezium = 1586 m
2
  (Given) ? 1092 +13x = 1586 ? 13x = (1586 -1092) ? 13x = 494 ? x =
494
13
? x = 38 mHence, the length of the other side is 38 m.
Q u e s t i o n : 6
The area of a trapezium is 405 cm
2
. Its parallel sides are in the ratio 4 : 5 and the distance between them is 18 cm. Find the length of each of the parallel sides.
S o l u t i o n :
Let the lengths of the parallel sides of the trapezium be 4x cm and 5x cm, respectively. Now, Area of trapezium =
1
2
×(4x +5x)×18 cm
2
 
                                =
1
2
×9x ×18 cm
2
= 81x cm
2
{ } ( )
{ } ( )
{ } ( )
{ }
( )
{ }
{ }
( )
Page 2


Q u e s t i o n : 1
Find the area of a trapezium whose parallel sides are 24 cm and 20 cm and the distance between them is 15 cm.
S o l u t i o n :
Area of a trapezium =
1
2
×(Sum of parallel sides)×(Distance between them)
                      =
1
2
×(24 +20)×15 cm
2
=
1
2
×44 ×15 cm
2
= (22 ×15) cm
2
= 330 cm
2
Hence, the area of the trapezium is 330 cm
2
.
Q u e s t i o n : 2
Find the area of a trapezium whose parallel sides are 38.7 cm and 22.3 cm, and the distance between them is 16 cm.
S o l u t i o n :
Area of a trapezium =
1
2
×(Sum of parallel sides)×(Distance between them)
                      =
1
2
×(38. 7 +22. 3)×16 cm
2
=
1
2
×61 ×16 cm
2
= (61 ×8) cm
2
= 488 cm
2
Hence, the area of the trapezium is 488 cm
2
.
Q u e s t i o n : 3
The shape of the top surface of a table is trapezium. Its parallel sides are 1 m and 1.4 m and the perpendicular distance between them is 0.9 m. Find its area.
S o l u t i o n :
Area of a trapezium =
1
2
×(Sum of parallel sides)×(Distance between them)
                       =
1
2
×(1 +1. 4)×0. 9 m
2
=
1
2
×2. 4 ×0. 9 m
2
= (1. 2 ×0. 9) m
2
= 1. 08 m
2
Hence, the area of the top surface of the table is 1. 08 m
2
.
Q u e s t i o n : 4
The area of a trapezium is 1080 cm
2
. If the lengths of its parallel sides be 55 cm and 35 cm, find the distance between them.
S o l u t i o n :
Let the distance between the parallel sides be x. Now, Area of trapezium =
1
2
×(55 +35)×x cm
2
                    =
1
2
×90 ×x cm
2
= 45x cm
2
Area of the trapezium = 1080 cm
2 
(Given) ? 45x = 1080 ? x =
1080
45
? x = 24 cmHence, the distance between the parallel sides is 24 cm.
Q u e s t i o n : 5
A field is in the form of a trapezium. Its area is 1586 m
2
 and the distance between its parallel sides is 26 m. If one of the parallel sides is 84 m, find the other.
S o l u t i o n :
Let the length of the required side be x cm. Now, Area of trapezium =
1
2
×(84 +x)×26 m
2
                        = (1092 +13x) m
2
Area of trapezium = 1586 m
2
  (Given) ? 1092 +13x = 1586 ? 13x = (1586 -1092) ? 13x = 494 ? x =
494
13
? x = 38 mHence, the length of the other side is 38 m.
Q u e s t i o n : 6
The area of a trapezium is 405 cm
2
. Its parallel sides are in the ratio 4 : 5 and the distance between them is 18 cm. Find the length of each of the parallel sides.
S o l u t i o n :
Let the lengths of the parallel sides of the trapezium be 4x cm and 5x cm, respectively. Now, Area of trapezium =
1
2
×(4x +5x)×18 cm
2
 
                                =
1
2
×9x ×18 cm
2
= 81x cm
2
{ } ( )
{ } ( )
{ } ( )
{ }
( )
{ }
{ }
( )
Area of trapezium = 405 cm
2
  (Given) ? 81x = 405 ? x =
405
81
? x = 5 cmLength of one side = (4 ×5) cm = 20 cm Length of the other side = (5 ×5) cm = 25 cm
Q u e s t i o n : 7
The area of a trapezium is 180 cm
2
 and its height is 9 cm. If one of the parallel sides is longer than the other by 6 cm, find the two parallel sides.
S o l u t i o n :
Let the lengths of the parallel sides be x cm and (x +6) cm. Now, Area of trapezium =
1
2
×(x +x +6)×9 cm
2
                     =
1
2
×(2x +6)×9 cm
2
= 4. 5(2x +6) cm
2
= (9x +27) cm
2
Area of trapezium = 180 cm
2
 (Given) ? 9x +27 = 180 ? 9x = (180 -27) ? 9x = 153 ? x =
153
9
? x = 17Hence, the lengths of the parallel sides are 17 cm and 23 cm, that is, (17 +6) cm. 
Q u e s t i o n : 8
In a trapezium-shaped field, one of the parallel sides is twice the other. If the area of the field is 9450 m
2
 and the perpendicular distance between the two parallel sides is 84 m, find the
length of the longer of the parallel sides.
S o l u t i o n :
Let the lengths of the parallel sides be x cm and 2x cm. Area of trapezium =
1
2
×(x +2x)×84 m
2
                    =
1
2
×3x ×84 m
2
= (42 ×3x) m
2
= 126x m
2
Area of the trapezium = 9450 m
2
 (Given) ? 126x = 9450 ? x =
9450
126
? x = 75Thus, the length of the parallel sides are 75 m and 150 m, that is, (2 ×75) m, and the length of the longer side is
Q u e s t i o n : 9
The length of the fence of a trapezium-shaped field ABCD is 130 m and side AB is perpendicular to each of the parallel sides AD and BC. If BC = 54 m, CD = 19 m and AD = 42 m, find
the area of the field.
S o l u t i o n :
Length of the side AB = (130 -(54 +19 +42)) m
                         = 15 m
Area of the trapezium -shaped field =
1
2
×(AD +BC)×AB
                                        =
1
2
×(42 +54)×15 m
2
=
1
2
×96 ×15 m
2
= (48 ×15) m
2
= 720 m
2
Hence, the area of the field is 720 m
2
.
Q u e s t i o n : 1 0
In the given figure, ABCD is a trapezium in which AD||BC, ?ABC = 90°, AD = 16 cm, AC = 41 cm and BC = 40 cm. Find the area of the trapezium.
S o l u t i o n :
?ABC = 90°From the right ? ABC, we have: AB
2
= AC
2
-BC
2
? AB
2
= 41
2
- 40
2
? AB
2
= (1681 -1600) ? AB
2
= 81 ? AB = v 81 ? AB = 9 cm ? Length AB = 9 cmNow, Area of the 
                         =
1
2
×(16 +40)×9 cm
2
=
1
2
×56 ×9 cm
2
= (28 ×9) cm
2
= 252 cm
2
Hence, the area of the trapezium is 252 cm
2
.
Q u e s t i o n : 1 1
The parallel sides of a trapezium are 20 cm and 10 cm. Its nonparallel sides are both equal, each being 13 cm. Find the area of the trapezium.
S o l u t i o n :
Let ABCD be the given trapezium in which AB ? DC, AB = 20 cm, DC = 10 cm and AD = BC = 13 cm. Draw CL ? AB and CM ? DA meeting AB at L and M, respectively. Clearly, AMCD is
{ }
( )
{ }
( )
{ }
{ } ( )
( ) { ( ) ( )}
( ) ( )
Page 3


Q u e s t i o n : 1
Find the area of a trapezium whose parallel sides are 24 cm and 20 cm and the distance between them is 15 cm.
S o l u t i o n :
Area of a trapezium =
1
2
×(Sum of parallel sides)×(Distance between them)
                      =
1
2
×(24 +20)×15 cm
2
=
1
2
×44 ×15 cm
2
= (22 ×15) cm
2
= 330 cm
2
Hence, the area of the trapezium is 330 cm
2
.
Q u e s t i o n : 2
Find the area of a trapezium whose parallel sides are 38.7 cm and 22.3 cm, and the distance between them is 16 cm.
S o l u t i o n :
Area of a trapezium =
1
2
×(Sum of parallel sides)×(Distance between them)
                      =
1
2
×(38. 7 +22. 3)×16 cm
2
=
1
2
×61 ×16 cm
2
= (61 ×8) cm
2
= 488 cm
2
Hence, the area of the trapezium is 488 cm
2
.
Q u e s t i o n : 3
The shape of the top surface of a table is trapezium. Its parallel sides are 1 m and 1.4 m and the perpendicular distance between them is 0.9 m. Find its area.
S o l u t i o n :
Area of a trapezium =
1
2
×(Sum of parallel sides)×(Distance between them)
                       =
1
2
×(1 +1. 4)×0. 9 m
2
=
1
2
×2. 4 ×0. 9 m
2
= (1. 2 ×0. 9) m
2
= 1. 08 m
2
Hence, the area of the top surface of the table is 1. 08 m
2
.
Q u e s t i o n : 4
The area of a trapezium is 1080 cm
2
. If the lengths of its parallel sides be 55 cm and 35 cm, find the distance between them.
S o l u t i o n :
Let the distance between the parallel sides be x. Now, Area of trapezium =
1
2
×(55 +35)×x cm
2
                    =
1
2
×90 ×x cm
2
= 45x cm
2
Area of the trapezium = 1080 cm
2 
(Given) ? 45x = 1080 ? x =
1080
45
? x = 24 cmHence, the distance between the parallel sides is 24 cm.
Q u e s t i o n : 5
A field is in the form of a trapezium. Its area is 1586 m
2
 and the distance between its parallel sides is 26 m. If one of the parallel sides is 84 m, find the other.
S o l u t i o n :
Let the length of the required side be x cm. Now, Area of trapezium =
1
2
×(84 +x)×26 m
2
                        = (1092 +13x) m
2
Area of trapezium = 1586 m
2
  (Given) ? 1092 +13x = 1586 ? 13x = (1586 -1092) ? 13x = 494 ? x =
494
13
? x = 38 mHence, the length of the other side is 38 m.
Q u e s t i o n : 6
The area of a trapezium is 405 cm
2
. Its parallel sides are in the ratio 4 : 5 and the distance between them is 18 cm. Find the length of each of the parallel sides.
S o l u t i o n :
Let the lengths of the parallel sides of the trapezium be 4x cm and 5x cm, respectively. Now, Area of trapezium =
1
2
×(4x +5x)×18 cm
2
 
                                =
1
2
×9x ×18 cm
2
= 81x cm
2
{ } ( )
{ } ( )
{ } ( )
{ }
( )
{ }
{ }
( )
Area of trapezium = 405 cm
2
  (Given) ? 81x = 405 ? x =
405
81
? x = 5 cmLength of one side = (4 ×5) cm = 20 cm Length of the other side = (5 ×5) cm = 25 cm
Q u e s t i o n : 7
The area of a trapezium is 180 cm
2
 and its height is 9 cm. If one of the parallel sides is longer than the other by 6 cm, find the two parallel sides.
S o l u t i o n :
Let the lengths of the parallel sides be x cm and (x +6) cm. Now, Area of trapezium =
1
2
×(x +x +6)×9 cm
2
                     =
1
2
×(2x +6)×9 cm
2
= 4. 5(2x +6) cm
2
= (9x +27) cm
2
Area of trapezium = 180 cm
2
 (Given) ? 9x +27 = 180 ? 9x = (180 -27) ? 9x = 153 ? x =
153
9
? x = 17Hence, the lengths of the parallel sides are 17 cm and 23 cm, that is, (17 +6) cm. 
Q u e s t i o n : 8
In a trapezium-shaped field, one of the parallel sides is twice the other. If the area of the field is 9450 m
2
 and the perpendicular distance between the two parallel sides is 84 m, find the
length of the longer of the parallel sides.
S o l u t i o n :
Let the lengths of the parallel sides be x cm and 2x cm. Area of trapezium =
1
2
×(x +2x)×84 m
2
                    =
1
2
×3x ×84 m
2
= (42 ×3x) m
2
= 126x m
2
Area of the trapezium = 9450 m
2
 (Given) ? 126x = 9450 ? x =
9450
126
? x = 75Thus, the length of the parallel sides are 75 m and 150 m, that is, (2 ×75) m, and the length of the longer side is
Q u e s t i o n : 9
The length of the fence of a trapezium-shaped field ABCD is 130 m and side AB is perpendicular to each of the parallel sides AD and BC. If BC = 54 m, CD = 19 m and AD = 42 m, find
the area of the field.
S o l u t i o n :
Length of the side AB = (130 -(54 +19 +42)) m
                         = 15 m
Area of the trapezium -shaped field =
1
2
×(AD +BC)×AB
                                        =
1
2
×(42 +54)×15 m
2
=
1
2
×96 ×15 m
2
= (48 ×15) m
2
= 720 m
2
Hence, the area of the field is 720 m
2
.
Q u e s t i o n : 1 0
In the given figure, ABCD is a trapezium in which AD||BC, ?ABC = 90°, AD = 16 cm, AC = 41 cm and BC = 40 cm. Find the area of the trapezium.
S o l u t i o n :
?ABC = 90°From the right ? ABC, we have: AB
2
= AC
2
-BC
2
? AB
2
= 41
2
- 40
2
? AB
2
= (1681 -1600) ? AB
2
= 81 ? AB = v 81 ? AB = 9 cm ? Length AB = 9 cmNow, Area of the 
                         =
1
2
×(16 +40)×9 cm
2
=
1
2
×56 ×9 cm
2
= (28 ×9) cm
2
= 252 cm
2
Hence, the area of the trapezium is 252 cm
2
.
Q u e s t i o n : 1 1
The parallel sides of a trapezium are 20 cm and 10 cm. Its nonparallel sides are both equal, each being 13 cm. Find the area of the trapezium.
S o l u t i o n :
Let ABCD be the given trapezium in which AB ? DC, AB = 20 cm, DC = 10 cm and AD = BC = 13 cm. Draw CL ? AB and CM ? DA meeting AB at L and M, respectively. Clearly, AMCD is
{ }
( )
{ }
( )
{ }
{ } ( )
( ) { ( ) ( )}
( ) ( )
              =
1
2
×10 cm = 5 cm
From right ? CLM, we have: CL
2
= CM
2
-ML
2
 cm
2
? CL
2
= (13)
2
-(5)
2
 cm
2
? CL
2
= (109 -25) cm
2
? CL
2
= 144 cm
2
? CL = v 144 cm ? CL = 12 cm ? Length of CL = 12 cmArea of the trapezium
                         =
1
2
×(20 +10)×12 cm
2
=
1
2
×30 ×12 cm
2
= (15 ×12) cm
2
= 180 cm
2
Hence, the area of the trapezium is 180 cm
2
.
Q u e s t i o n : 1 2
The parallel sides of a trapezium are 25 cm and 11 cm, while its nonparallel sides are 15 cm and 13 cm. Find the area of the trapezium.
S o l u t i o n :
Let ABCD be the trapezium in which AB ? DC, AB = 25 cm, CD = 11 cm, AD = 13 cm and BC = 15 cm. Draw CL ? AB and CM ? DA meeting AB at L and M, respectively. Clearly, AMCD
          = (25 -11) cm = 14 cm
Thus, in ? CMB, we have: CM = 13 cmMB = 14 cm BC = 15 cm ? s =
1
2
(13 +14 +15) cm       =
1
2
42 cm       = 21 cm(s -a) = (21 -13) cm          = 8 cm(s -b) = (21 -14) cm          = 7 cm(s -c) =
                      = v 21 ×8 ×7 ×6  cm
2
= 84  cm
2
? 
1
2
×MB ×CL = 84 cm
2
?
1
2
×14 ×CL = 84 cm
2
 ? CL =
84
7
? CL = 12 cm
Area of the trapezium =
1
2
×(AB +DC)×CL
                         =
1
2
×(25 +11)×12 cm
2
=
1
2
×36 ×12 cm
2
= (18 ×12) cm
2
= 216 cm
2
Hence, the area of the trapezium is 216 cm
2
.
Q u e s t i o n : 1 3
In the given figure, ABCD is a quadrilateral in which AC = 24 cm, BL ? AC and DM ? AC such that BL = 8 cm and DM = 7 cm. Find the area of quad. ABCD.
S o l u t i o n :
Area of quadrilateral ABCD = (Area of ? ADC)+(Area of ? ACB)
                                             =
1
2
×AC ×DM +
1
2
×AC ×BL =
1
2
×24 ×7 +
1
2
×24 ×8 cm
2
= (84 +96) cm
2
= 180 cm
2
Hence, the area of the quadrilateral is 180 cm
2
.
Q u e s t i o n : 1 4
In the given figure, ABCD is a quadrilateral-shaped field in which diagonal BD is 36 m, AL ? BD and CM ? BD such that AL = 19 m and CM = 11 m. Find the area of the field.
S o l u t i o n :
Area of quadrilateral ABCD = (Area of ? ABD)+(Area of ? BCD)
                                         =
1
2
×BD ×AL +
1
2
×BD ×CM
                                         =
1
2
×36 ×19 +
1
2
×36 ×11 m
2
= (342 +198) m
2
= 540 m
2
( )
( ) { }
{ } ( )
{ }
{ } ( )
( ) ( ) [ ( ) ( )]
( ) ( )
[ ( ) ( )]
Page 4


Q u e s t i o n : 1
Find the area of a trapezium whose parallel sides are 24 cm and 20 cm and the distance between them is 15 cm.
S o l u t i o n :
Area of a trapezium =
1
2
×(Sum of parallel sides)×(Distance between them)
                      =
1
2
×(24 +20)×15 cm
2
=
1
2
×44 ×15 cm
2
= (22 ×15) cm
2
= 330 cm
2
Hence, the area of the trapezium is 330 cm
2
.
Q u e s t i o n : 2
Find the area of a trapezium whose parallel sides are 38.7 cm and 22.3 cm, and the distance between them is 16 cm.
S o l u t i o n :
Area of a trapezium =
1
2
×(Sum of parallel sides)×(Distance between them)
                      =
1
2
×(38. 7 +22. 3)×16 cm
2
=
1
2
×61 ×16 cm
2
= (61 ×8) cm
2
= 488 cm
2
Hence, the area of the trapezium is 488 cm
2
.
Q u e s t i o n : 3
The shape of the top surface of a table is trapezium. Its parallel sides are 1 m and 1.4 m and the perpendicular distance between them is 0.9 m. Find its area.
S o l u t i o n :
Area of a trapezium =
1
2
×(Sum of parallel sides)×(Distance between them)
                       =
1
2
×(1 +1. 4)×0. 9 m
2
=
1
2
×2. 4 ×0. 9 m
2
= (1. 2 ×0. 9) m
2
= 1. 08 m
2
Hence, the area of the top surface of the table is 1. 08 m
2
.
Q u e s t i o n : 4
The area of a trapezium is 1080 cm
2
. If the lengths of its parallel sides be 55 cm and 35 cm, find the distance between them.
S o l u t i o n :
Let the distance between the parallel sides be x. Now, Area of trapezium =
1
2
×(55 +35)×x cm
2
                    =
1
2
×90 ×x cm
2
= 45x cm
2
Area of the trapezium = 1080 cm
2 
(Given) ? 45x = 1080 ? x =
1080
45
? x = 24 cmHence, the distance between the parallel sides is 24 cm.
Q u e s t i o n : 5
A field is in the form of a trapezium. Its area is 1586 m
2
 and the distance between its parallel sides is 26 m. If one of the parallel sides is 84 m, find the other.
S o l u t i o n :
Let the length of the required side be x cm. Now, Area of trapezium =
1
2
×(84 +x)×26 m
2
                        = (1092 +13x) m
2
Area of trapezium = 1586 m
2
  (Given) ? 1092 +13x = 1586 ? 13x = (1586 -1092) ? 13x = 494 ? x =
494
13
? x = 38 mHence, the length of the other side is 38 m.
Q u e s t i o n : 6
The area of a trapezium is 405 cm
2
. Its parallel sides are in the ratio 4 : 5 and the distance between them is 18 cm. Find the length of each of the parallel sides.
S o l u t i o n :
Let the lengths of the parallel sides of the trapezium be 4x cm and 5x cm, respectively. Now, Area of trapezium =
1
2
×(4x +5x)×18 cm
2
 
                                =
1
2
×9x ×18 cm
2
= 81x cm
2
{ } ( )
{ } ( )
{ } ( )
{ }
( )
{ }
{ }
( )
Area of trapezium = 405 cm
2
  (Given) ? 81x = 405 ? x =
405
81
? x = 5 cmLength of one side = (4 ×5) cm = 20 cm Length of the other side = (5 ×5) cm = 25 cm
Q u e s t i o n : 7
The area of a trapezium is 180 cm
2
 and its height is 9 cm. If one of the parallel sides is longer than the other by 6 cm, find the two parallel sides.
S o l u t i o n :
Let the lengths of the parallel sides be x cm and (x +6) cm. Now, Area of trapezium =
1
2
×(x +x +6)×9 cm
2
                     =
1
2
×(2x +6)×9 cm
2
= 4. 5(2x +6) cm
2
= (9x +27) cm
2
Area of trapezium = 180 cm
2
 (Given) ? 9x +27 = 180 ? 9x = (180 -27) ? 9x = 153 ? x =
153
9
? x = 17Hence, the lengths of the parallel sides are 17 cm and 23 cm, that is, (17 +6) cm. 
Q u e s t i o n : 8
In a trapezium-shaped field, one of the parallel sides is twice the other. If the area of the field is 9450 m
2
 and the perpendicular distance between the two parallel sides is 84 m, find the
length of the longer of the parallel sides.
S o l u t i o n :
Let the lengths of the parallel sides be x cm and 2x cm. Area of trapezium =
1
2
×(x +2x)×84 m
2
                    =
1
2
×3x ×84 m
2
= (42 ×3x) m
2
= 126x m
2
Area of the trapezium = 9450 m
2
 (Given) ? 126x = 9450 ? x =
9450
126
? x = 75Thus, the length of the parallel sides are 75 m and 150 m, that is, (2 ×75) m, and the length of the longer side is
Q u e s t i o n : 9
The length of the fence of a trapezium-shaped field ABCD is 130 m and side AB is perpendicular to each of the parallel sides AD and BC. If BC = 54 m, CD = 19 m and AD = 42 m, find
the area of the field.
S o l u t i o n :
Length of the side AB = (130 -(54 +19 +42)) m
                         = 15 m
Area of the trapezium -shaped field =
1
2
×(AD +BC)×AB
                                        =
1
2
×(42 +54)×15 m
2
=
1
2
×96 ×15 m
2
= (48 ×15) m
2
= 720 m
2
Hence, the area of the field is 720 m
2
.
Q u e s t i o n : 1 0
In the given figure, ABCD is a trapezium in which AD||BC, ?ABC = 90°, AD = 16 cm, AC = 41 cm and BC = 40 cm. Find the area of the trapezium.
S o l u t i o n :
?ABC = 90°From the right ? ABC, we have: AB
2
= AC
2
-BC
2
? AB
2
= 41
2
- 40
2
? AB
2
= (1681 -1600) ? AB
2
= 81 ? AB = v 81 ? AB = 9 cm ? Length AB = 9 cmNow, Area of the 
                         =
1
2
×(16 +40)×9 cm
2
=
1
2
×56 ×9 cm
2
= (28 ×9) cm
2
= 252 cm
2
Hence, the area of the trapezium is 252 cm
2
.
Q u e s t i o n : 1 1
The parallel sides of a trapezium are 20 cm and 10 cm. Its nonparallel sides are both equal, each being 13 cm. Find the area of the trapezium.
S o l u t i o n :
Let ABCD be the given trapezium in which AB ? DC, AB = 20 cm, DC = 10 cm and AD = BC = 13 cm. Draw CL ? AB and CM ? DA meeting AB at L and M, respectively. Clearly, AMCD is
{ }
( )
{ }
( )
{ }
{ } ( )
( ) { ( ) ( )}
( ) ( )
              =
1
2
×10 cm = 5 cm
From right ? CLM, we have: CL
2
= CM
2
-ML
2
 cm
2
? CL
2
= (13)
2
-(5)
2
 cm
2
? CL
2
= (109 -25) cm
2
? CL
2
= 144 cm
2
? CL = v 144 cm ? CL = 12 cm ? Length of CL = 12 cmArea of the trapezium
                         =
1
2
×(20 +10)×12 cm
2
=
1
2
×30 ×12 cm
2
= (15 ×12) cm
2
= 180 cm
2
Hence, the area of the trapezium is 180 cm
2
.
Q u e s t i o n : 1 2
The parallel sides of a trapezium are 25 cm and 11 cm, while its nonparallel sides are 15 cm and 13 cm. Find the area of the trapezium.
S o l u t i o n :
Let ABCD be the trapezium in which AB ? DC, AB = 25 cm, CD = 11 cm, AD = 13 cm and BC = 15 cm. Draw CL ? AB and CM ? DA meeting AB at L and M, respectively. Clearly, AMCD
          = (25 -11) cm = 14 cm
Thus, in ? CMB, we have: CM = 13 cmMB = 14 cm BC = 15 cm ? s =
1
2
(13 +14 +15) cm       =
1
2
42 cm       = 21 cm(s -a) = (21 -13) cm          = 8 cm(s -b) = (21 -14) cm          = 7 cm(s -c) =
                      = v 21 ×8 ×7 ×6  cm
2
= 84  cm
2
? 
1
2
×MB ×CL = 84 cm
2
?
1
2
×14 ×CL = 84 cm
2
 ? CL =
84
7
? CL = 12 cm
Area of the trapezium =
1
2
×(AB +DC)×CL
                         =
1
2
×(25 +11)×12 cm
2
=
1
2
×36 ×12 cm
2
= (18 ×12) cm
2
= 216 cm
2
Hence, the area of the trapezium is 216 cm
2
.
Q u e s t i o n : 1 3
In the given figure, ABCD is a quadrilateral in which AC = 24 cm, BL ? AC and DM ? AC such that BL = 8 cm and DM = 7 cm. Find the area of quad. ABCD.
S o l u t i o n :
Area of quadrilateral ABCD = (Area of ? ADC)+(Area of ? ACB)
                                             =
1
2
×AC ×DM +
1
2
×AC ×BL =
1
2
×24 ×7 +
1
2
×24 ×8 cm
2
= (84 +96) cm
2
= 180 cm
2
Hence, the area of the quadrilateral is 180 cm
2
.
Q u e s t i o n : 1 4
In the given figure, ABCD is a quadrilateral-shaped field in which diagonal BD is 36 m, AL ? BD and CM ? BD such that AL = 19 m and CM = 11 m. Find the area of the field.
S o l u t i o n :
Area of quadrilateral ABCD = (Area of ? ABD)+(Area of ? BCD)
                                         =
1
2
×BD ×AL +
1
2
×BD ×CM
                                         =
1
2
×36 ×19 +
1
2
×36 ×11 m
2
= (342 +198) m
2
= 540 m
2
( )
( ) { }
{ } ( )
{ }
{ } ( )
( ) ( ) [ ( ) ( )]
( ) ( )
[ ( ) ( )]
Hence, the area of the field is 540 m
2
.
Q u e s t i o n : 1 5
Find the area of pentagon ABCDE in which BL ? AC, DM ? AC and EN ? AC such that AC = 18 cm, AM = 14 cm, AN = 6 cm, BL = 4 cm, DM = 12 cm and EN = 9 cm.
S o l u t i o n :
Area of pentagon ABCDE = (Area of ? AEN)+(Area of trapezium EDMN)+(Area of ? DMC)+(Area of ? ACB)
                                      
=
1
2
×AN ×EN +
1
2
×(EN +DM)×NM +
1
2
×MC ×DM +
1
2
×AC ×BL =
1
2
×AN ×EN +
1
2
×(EN +DM)×(AM -AN) +
1
2
×(AC -AM)×DM +
1
2
×AC ×BL =
1
2
×6 ×9 +
1
2
×(9 +12)×(14
Hence, the area of the given pentagon is 171 cm
2
.
Q u e s t i o n : 1 6
Find the area of hexagon ABCDEF in which BL ? AD, CM ? AD, EN ? AD and FP ? AD such that AP = 6 cm, PL = 2 cm, LN = 8 cm, NM = 2 cm, MD = 3 cm, FP = 8 cm, EN = 12 cm, BL
= 8 cm and CM = 6 cm.
S o l u t i o n :
Area of hexagon ABCDEF = (Area of ? AFP)+(Area of trapezium FENP)+(Area of ? ALB)
=
1
2
×AP ×FP +
1
2
×(FP +EN)×PN +
1
2
×ND ×EN +
1
2
×MD ×CM +
1
2
×(CM +BL)×LM +
1
2
×AL ×BL =
1
2
×AP ×FP +
1
2
×(FP +EN)×(PL +LN) +
1
2
×(NM +MD)×CM +
1
2
×MD ×CM
Hence, the area of the hexagon is 265 cm
2
.
Q u e s t i o n : 1 7
Find the area of pentagon ABCDE in which BL ? AC, CM ? AD and EN ? AD such that AC = 10 cm, AD = 12 cm, BL = 3 cm, CM = 7 cm and EN = 5 cm.
S o l u t i o n :
Area  of pentagon ABCDE = (Area of ? ABC)+(Area of ? ACD)+(Area of ? ADE)
                                             =
1
2
×AC ×BL +
1
2
×AD ×CM +
1
2
×AD ×EM =
1
2
×10 ×3 +
1
2
×12 ×7 +
1
2
×12 ×5 cm
2
= (15 +42 +30) cm
2
= 87 cm
2
Hence, the area of the pentagon is 87 cm
2
.
Q u e s t i o n : 1 8
Find the area enclosed by the given figure ABCDEF as per dimensions given herewith.
S o l u t i o n :
Area enclosed by the given figure = (Area of trapezium FEDC)+(Area of square ABCF)
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) [ ( ) (
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) (
( ) ( ) ( ) [ ( ) ( ) ( )]
[{ } ]
Page 5


Q u e s t i o n : 1
Find the area of a trapezium whose parallel sides are 24 cm and 20 cm and the distance between them is 15 cm.
S o l u t i o n :
Area of a trapezium =
1
2
×(Sum of parallel sides)×(Distance between them)
                      =
1
2
×(24 +20)×15 cm
2
=
1
2
×44 ×15 cm
2
= (22 ×15) cm
2
= 330 cm
2
Hence, the area of the trapezium is 330 cm
2
.
Q u e s t i o n : 2
Find the area of a trapezium whose parallel sides are 38.7 cm and 22.3 cm, and the distance between them is 16 cm.
S o l u t i o n :
Area of a trapezium =
1
2
×(Sum of parallel sides)×(Distance between them)
                      =
1
2
×(38. 7 +22. 3)×16 cm
2
=
1
2
×61 ×16 cm
2
= (61 ×8) cm
2
= 488 cm
2
Hence, the area of the trapezium is 488 cm
2
.
Q u e s t i o n : 3
The shape of the top surface of a table is trapezium. Its parallel sides are 1 m and 1.4 m and the perpendicular distance between them is 0.9 m. Find its area.
S o l u t i o n :
Area of a trapezium =
1
2
×(Sum of parallel sides)×(Distance between them)
                       =
1
2
×(1 +1. 4)×0. 9 m
2
=
1
2
×2. 4 ×0. 9 m
2
= (1. 2 ×0. 9) m
2
= 1. 08 m
2
Hence, the area of the top surface of the table is 1. 08 m
2
.
Q u e s t i o n : 4
The area of a trapezium is 1080 cm
2
. If the lengths of its parallel sides be 55 cm and 35 cm, find the distance between them.
S o l u t i o n :
Let the distance between the parallel sides be x. Now, Area of trapezium =
1
2
×(55 +35)×x cm
2
                    =
1
2
×90 ×x cm
2
= 45x cm
2
Area of the trapezium = 1080 cm
2 
(Given) ? 45x = 1080 ? x =
1080
45
? x = 24 cmHence, the distance between the parallel sides is 24 cm.
Q u e s t i o n : 5
A field is in the form of a trapezium. Its area is 1586 m
2
 and the distance between its parallel sides is 26 m. If one of the parallel sides is 84 m, find the other.
S o l u t i o n :
Let the length of the required side be x cm. Now, Area of trapezium =
1
2
×(84 +x)×26 m
2
                        = (1092 +13x) m
2
Area of trapezium = 1586 m
2
  (Given) ? 1092 +13x = 1586 ? 13x = (1586 -1092) ? 13x = 494 ? x =
494
13
? x = 38 mHence, the length of the other side is 38 m.
Q u e s t i o n : 6
The area of a trapezium is 405 cm
2
. Its parallel sides are in the ratio 4 : 5 and the distance between them is 18 cm. Find the length of each of the parallel sides.
S o l u t i o n :
Let the lengths of the parallel sides of the trapezium be 4x cm and 5x cm, respectively. Now, Area of trapezium =
1
2
×(4x +5x)×18 cm
2
 
                                =
1
2
×9x ×18 cm
2
= 81x cm
2
{ } ( )
{ } ( )
{ } ( )
{ }
( )
{ }
{ }
( )
Area of trapezium = 405 cm
2
  (Given) ? 81x = 405 ? x =
405
81
? x = 5 cmLength of one side = (4 ×5) cm = 20 cm Length of the other side = (5 ×5) cm = 25 cm
Q u e s t i o n : 7
The area of a trapezium is 180 cm
2
 and its height is 9 cm. If one of the parallel sides is longer than the other by 6 cm, find the two parallel sides.
S o l u t i o n :
Let the lengths of the parallel sides be x cm and (x +6) cm. Now, Area of trapezium =
1
2
×(x +x +6)×9 cm
2
                     =
1
2
×(2x +6)×9 cm
2
= 4. 5(2x +6) cm
2
= (9x +27) cm
2
Area of trapezium = 180 cm
2
 (Given) ? 9x +27 = 180 ? 9x = (180 -27) ? 9x = 153 ? x =
153
9
? x = 17Hence, the lengths of the parallel sides are 17 cm and 23 cm, that is, (17 +6) cm. 
Q u e s t i o n : 8
In a trapezium-shaped field, one of the parallel sides is twice the other. If the area of the field is 9450 m
2
 and the perpendicular distance between the two parallel sides is 84 m, find the
length of the longer of the parallel sides.
S o l u t i o n :
Let the lengths of the parallel sides be x cm and 2x cm. Area of trapezium =
1
2
×(x +2x)×84 m
2
                    =
1
2
×3x ×84 m
2
= (42 ×3x) m
2
= 126x m
2
Area of the trapezium = 9450 m
2
 (Given) ? 126x = 9450 ? x =
9450
126
? x = 75Thus, the length of the parallel sides are 75 m and 150 m, that is, (2 ×75) m, and the length of the longer side is
Q u e s t i o n : 9
The length of the fence of a trapezium-shaped field ABCD is 130 m and side AB is perpendicular to each of the parallel sides AD and BC. If BC = 54 m, CD = 19 m and AD = 42 m, find
the area of the field.
S o l u t i o n :
Length of the side AB = (130 -(54 +19 +42)) m
                         = 15 m
Area of the trapezium -shaped field =
1
2
×(AD +BC)×AB
                                        =
1
2
×(42 +54)×15 m
2
=
1
2
×96 ×15 m
2
= (48 ×15) m
2
= 720 m
2
Hence, the area of the field is 720 m
2
.
Q u e s t i o n : 1 0
In the given figure, ABCD is a trapezium in which AD||BC, ?ABC = 90°, AD = 16 cm, AC = 41 cm and BC = 40 cm. Find the area of the trapezium.
S o l u t i o n :
?ABC = 90°From the right ? ABC, we have: AB
2
= AC
2
-BC
2
? AB
2
= 41
2
- 40
2
? AB
2
= (1681 -1600) ? AB
2
= 81 ? AB = v 81 ? AB = 9 cm ? Length AB = 9 cmNow, Area of the 
                         =
1
2
×(16 +40)×9 cm
2
=
1
2
×56 ×9 cm
2
= (28 ×9) cm
2
= 252 cm
2
Hence, the area of the trapezium is 252 cm
2
.
Q u e s t i o n : 1 1
The parallel sides of a trapezium are 20 cm and 10 cm. Its nonparallel sides are both equal, each being 13 cm. Find the area of the trapezium.
S o l u t i o n :
Let ABCD be the given trapezium in which AB ? DC, AB = 20 cm, DC = 10 cm and AD = BC = 13 cm. Draw CL ? AB and CM ? DA meeting AB at L and M, respectively. Clearly, AMCD is
{ }
( )
{ }
( )
{ }
{ } ( )
( ) { ( ) ( )}
( ) ( )
              =
1
2
×10 cm = 5 cm
From right ? CLM, we have: CL
2
= CM
2
-ML
2
 cm
2
? CL
2
= (13)
2
-(5)
2
 cm
2
? CL
2
= (109 -25) cm
2
? CL
2
= 144 cm
2
? CL = v 144 cm ? CL = 12 cm ? Length of CL = 12 cmArea of the trapezium
                         =
1
2
×(20 +10)×12 cm
2
=
1
2
×30 ×12 cm
2
= (15 ×12) cm
2
= 180 cm
2
Hence, the area of the trapezium is 180 cm
2
.
Q u e s t i o n : 1 2
The parallel sides of a trapezium are 25 cm and 11 cm, while its nonparallel sides are 15 cm and 13 cm. Find the area of the trapezium.
S o l u t i o n :
Let ABCD be the trapezium in which AB ? DC, AB = 25 cm, CD = 11 cm, AD = 13 cm and BC = 15 cm. Draw CL ? AB and CM ? DA meeting AB at L and M, respectively. Clearly, AMCD
          = (25 -11) cm = 14 cm
Thus, in ? CMB, we have: CM = 13 cmMB = 14 cm BC = 15 cm ? s =
1
2
(13 +14 +15) cm       =
1
2
42 cm       = 21 cm(s -a) = (21 -13) cm          = 8 cm(s -b) = (21 -14) cm          = 7 cm(s -c) =
                      = v 21 ×8 ×7 ×6  cm
2
= 84  cm
2
? 
1
2
×MB ×CL = 84 cm
2
?
1
2
×14 ×CL = 84 cm
2
 ? CL =
84
7
? CL = 12 cm
Area of the trapezium =
1
2
×(AB +DC)×CL
                         =
1
2
×(25 +11)×12 cm
2
=
1
2
×36 ×12 cm
2
= (18 ×12) cm
2
= 216 cm
2
Hence, the area of the trapezium is 216 cm
2
.
Q u e s t i o n : 1 3
In the given figure, ABCD is a quadrilateral in which AC = 24 cm, BL ? AC and DM ? AC such that BL = 8 cm and DM = 7 cm. Find the area of quad. ABCD.
S o l u t i o n :
Area of quadrilateral ABCD = (Area of ? ADC)+(Area of ? ACB)
                                             =
1
2
×AC ×DM +
1
2
×AC ×BL =
1
2
×24 ×7 +
1
2
×24 ×8 cm
2
= (84 +96) cm
2
= 180 cm
2
Hence, the area of the quadrilateral is 180 cm
2
.
Q u e s t i o n : 1 4
In the given figure, ABCD is a quadrilateral-shaped field in which diagonal BD is 36 m, AL ? BD and CM ? BD such that AL = 19 m and CM = 11 m. Find the area of the field.
S o l u t i o n :
Area of quadrilateral ABCD = (Area of ? ABD)+(Area of ? BCD)
                                         =
1
2
×BD ×AL +
1
2
×BD ×CM
                                         =
1
2
×36 ×19 +
1
2
×36 ×11 m
2
= (342 +198) m
2
= 540 m
2
( )
( ) { }
{ } ( )
{ }
{ } ( )
( ) ( ) [ ( ) ( )]
( ) ( )
[ ( ) ( )]
Hence, the area of the field is 540 m
2
.
Q u e s t i o n : 1 5
Find the area of pentagon ABCDE in which BL ? AC, DM ? AC and EN ? AC such that AC = 18 cm, AM = 14 cm, AN = 6 cm, BL = 4 cm, DM = 12 cm and EN = 9 cm.
S o l u t i o n :
Area of pentagon ABCDE = (Area of ? AEN)+(Area of trapezium EDMN)+(Area of ? DMC)+(Area of ? ACB)
                                      
=
1
2
×AN ×EN +
1
2
×(EN +DM)×NM +
1
2
×MC ×DM +
1
2
×AC ×BL =
1
2
×AN ×EN +
1
2
×(EN +DM)×(AM -AN) +
1
2
×(AC -AM)×DM +
1
2
×AC ×BL =
1
2
×6 ×9 +
1
2
×(9 +12)×(14
Hence, the area of the given pentagon is 171 cm
2
.
Q u e s t i o n : 1 6
Find the area of hexagon ABCDEF in which BL ? AD, CM ? AD, EN ? AD and FP ? AD such that AP = 6 cm, PL = 2 cm, LN = 8 cm, NM = 2 cm, MD = 3 cm, FP = 8 cm, EN = 12 cm, BL
= 8 cm and CM = 6 cm.
S o l u t i o n :
Area of hexagon ABCDEF = (Area of ? AFP)+(Area of trapezium FENP)+(Area of ? ALB)
=
1
2
×AP ×FP +
1
2
×(FP +EN)×PN +
1
2
×ND ×EN +
1
2
×MD ×CM +
1
2
×(CM +BL)×LM +
1
2
×AL ×BL =
1
2
×AP ×FP +
1
2
×(FP +EN)×(PL +LN) +
1
2
×(NM +MD)×CM +
1
2
×MD ×CM
Hence, the area of the hexagon is 265 cm
2
.
Q u e s t i o n : 1 7
Find the area of pentagon ABCDE in which BL ? AC, CM ? AD and EN ? AD such that AC = 10 cm, AD = 12 cm, BL = 3 cm, CM = 7 cm and EN = 5 cm.
S o l u t i o n :
Area  of pentagon ABCDE = (Area of ? ABC)+(Area of ? ACD)+(Area of ? ADE)
                                             =
1
2
×AC ×BL +
1
2
×AD ×CM +
1
2
×AD ×EM =
1
2
×10 ×3 +
1
2
×12 ×7 +
1
2
×12 ×5 cm
2
= (15 +42 +30) cm
2
= 87 cm
2
Hence, the area of the pentagon is 87 cm
2
.
Q u e s t i o n : 1 8
Find the area enclosed by the given figure ABCDEF as per dimensions given herewith.
S o l u t i o n :
Area enclosed by the given figure = (Area of trapezium FEDC)+(Area of square ABCF)
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) [ ( ) (
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) (
( ) ( ) ( ) [ ( ) ( ) ( )]
[{ } ]
                                                 =
1
2
×(6 +20)×8 +(20 ×20) cm
2
= (104 +400)cm
2
= 504 cm
2
Hence, the area enclosed by the figure is 504 cm
2
.
Q u e s t i o n : 1 9
Find the area of given figure ABCDEFGH as per dimensions given in it.
S o l u t i o n :
We will find the length of AC. From the right triangles ABC and HGF, we have: AC
2
= HF
2
= (5)
2
-(4)
2
 cm
                = (25 -16)cm = 9 cm
AC = HF = v 9 cm               = 3 cm
Area of the given figure ABCDEFGH = (Area of rectangle ADEH)+(Area of ? ABC)+(Area of ? HGF)
                                                      
= (Area of rectangle ADEH)+2(Area of ? ABC) = (AD ×DE)+2(Area of ? ABC) = {(AC +CD)×DE}+2
1
2
×BC ×AC = {(3 +4)×8}+2
1
2
×4 ×3 cm
2
= (56 +12) cm = 68 cm
2
Hence, the area of the given figure is 68 cm
2
.
Q u e s t i o n : 2 0
Find the area of a regular hexagon ABCDEF in which each side measures 13 cm and whose height is 23 cm, as shown in the given figure.
S o l u t i o n :
Let AL = DM = x cm LM = BC = 13 cm ? x +13 +x = 23 ? 2x +13 = 23 ? 2x = (23 -13) ? 2x = 10 ? x = 5 ? AL = 5 cmFrom the right ? AFL, we have:    FL
2
= AF
2
-AL
2
? FL
2
= 13
2
-(5)
2
                                  = 2(Area of trapezium ADEF) = 2
1
2
×(AD +EF)×FL = 2
1
2
×(23 +13)×12 cm
2
= 2
1
2
×36 ×12 cm
2
= 432 cm
2
Hence, the area of the given regular hexagon is 432 cm
2
.
Q u e s t i o n : 2 1
Tick
? the correct answer:
The parallel sides of a trapezium measure 14 cm and 18 cm and the distance between them is 9 cm. The area of the trapezium is
a
96 cm
2
b
144 cm
2
c
189 cm
2
d
207 cm
2
S o l u t i o n :
b
144 cm
2
Area of the trapezium =
1
2
×(14 +18)×9 cm
2
                            =
1
2
×32 ×9 cm
2
= 144 cm
2
Q u e s t i o n : 2 2
Tick
? the correct answer:
The lengths of the parallel sides of a trapezium are 19 cm and 13 cm and its area is 128 cm
2
. The distance between the parallel sides is
a
9 cm
b
7 cm
c
[{ } ]
{ }
( ) ( )
{ ( )
{ } { } ( )
{ }
( )