Question

A field is in the shape of trapezium whose parallel sides are25 m and 10m . The non ​parallel sides are 14m and 13m.
find the area of field ​

Answer 1

Answer:

Let ABCD be the given trapezium in which AB=25m, DC=13m, BC=10m and AD=10m.

Through C, draw CE∥AD, meeting AB at E.

Also, draw CF⊥AB.

Now, EB=(AB−AE)=(AB−DC)

=(25−13)m=12m;

CE=AD=10m; AE=DC=13m.  

Now, in △EBC, we have CE=BC=10m.

So, it is an isosceles triangle.

Also, CF⊥AB

So, F is the midpoint of EB.

∴  EF=  

2

1

​  

×EB=6m.

Thus, in right-angled △CFE, we have CE=10m,EF=6m.

By Pythagoras theorem, we have

CF=  

CE  

2

−EF  

2

 

​  

 

=8m.

Thus, the distance between the parallel sides is 8m.

Area of trapezium ABCD=  

2

1

​  

×(sum of parallel sides)×(distance between them)                                            

=0.5×(25+13)×8m  

2

 

=152m  

2

Step-by-step explanation: