Question

A field is in the shape of a trapezium whose parallel sides are 

25m and 10m.The non parallel sides are 14m and 13m.Find the area of the field       

​

Answer 1

Step-by-step explanation:

Let ABCD be a trapezium with,

AB∥CD

AB=25m

CD=10m

BC=14m

AD=13m

Draw CE∥DA. So, ADCE is a parallelogram with,

CD=AE=10m

CE=AD=13m

BE=AB−AE=25−10=15m

In ΔBCE, the semi perimeter will be,

s= $\displaystyle\frac{a+b+c}{2}$

s= $\displaystyle\frac{14+13+15}{2}$

s= $\displaystyle\frac{42}{2}$

s= 21m

Area of ΔBCE,

A=

$\sqrt{s(s-a) (s-b) (s-c)}$

$\sqrt{21(21-14) (21-13) (21-15)}$

$\sqrt{7056}$

84m^2

Also, area of ΔBCE is,

A=

2

1

×base×height

84=

2

1

×15×CL

15

84×2

=CL

CL=

5

56

m

Now, the area of trapezium is,

A=

2

1

(sumofparallelsides)(height)

A=

2

1

×(25+10)(

5

56

)

A=196m

2

Therefore, the area of the trapezium is 196m

2

.

Answer 2

Answer:

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