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a field in the shape of trapezium whose parallel sides are 25m and 10m. the non parallel sides are 14m and 13m. find the area of the field.

Name the trapeziumas A B C D where AB=10, BC=13, CD=25 and DA=14.

made an imginary line as altitude from B on CD and name it X  so that  angle BXC=90o.Then XC = 10 and XD = 15.

According to pythogoras theorm-

H2=P2*H2

so BC2 =XC2 *BX2

132=102 *BX

169 -100 =BX2

BX2 = 69

BX =8.30

area of trapesium = sum of ll sides * distance b/w them

= 30 *8.30

=249 mt2

(thumbs up pls)

  • -20
View Full Answer
U r wrong!!!!
 



 
  • -4
 Let ABCD is a trapezium such that AB = 25m, BC = 13 m, CD = 10 m and DA = 14 m 
Draw a line DE parallel to the side BC =13 m therefore DE = 13 m 
AE+ED+DA = 15+13+14 = 42 
2s= 42 or s = 21 
Area of triangle DAE using Hero's formula = sqrt{ 21(21-15)(21-14)(21-13)} = 84 m^2 ▬(i) 
DF is a perpendicular from D on AE= h 
Area of triangle = (1/2)(15)(h) ▬(ii)
from (i) and (ii) 
(15/2)*h = 84 
h = (84*2)/15 = 11.2 m ▬(iii)
Area of parallelogram = 10*11.2 = 112 m^2 ▬(iv) 
Therefore area of trapezium = Area of triangle+ Area of parallelogram 
= 84 +112 = 196 m^2
Hope this helps :)
  • 45
 Let ABCD is a trapezium such that AB = 25m, BC = 13 m, CD = 10 m and DA = 14 m 
Draw a line DE parallel to the side BC =13 m therefore DE = 13 m 
AE+ED+DA = 15+13+14 = 42 
2s= 42 or s = 21 
Area of triangle DAE using Hero's formula = sqrt{ 21(21-15)(21-14)(21-13)} = 84 m^2 ▬(i) 
DF is a perpendicular from D on AE= h 
Area of triangle = (1/2)(15)(h) ▬(ii)
from (i) and (ii) 
(15/2)*h = 84 
h = (84*2)/15 = 11.2 m ▬(iii)
Area of parallelogram = 10*11.2 = 112 m^2 ▬(iv) 
Therefore area of trapezium = Area of triangle+ Area of parallelogram 
= 84 +112 = 196 m^2
Hope this helps :D
  • 1
U R Wrong Akhil
  • -1
Fayaz, can you post a figure??
  • -5
U r wrong
  • -4
Now put the values in the area of trapezium to get final solution.

  • 13
hey
 
  • 2
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