Question

The parallel sides AB and DC of a trapezium ABCD are 51 cm and 30 CM respectively. If the sides AD and BC are 20 cm and 13 cm respectively,find the :

(I) Distance between parallel sides
(ii) Area of trapezium ABCD ​

Answer 1

Answer:

Draw CE∥DA and CF⊥AB.

Therefore, ADCE is a parallelogram having AE∥CD and CE∥DA.

⇒AD=CE=39cm,DC=AE=30cm and BE=75−30=45cm

In △BCE, by heron's formula we have

a=39cm,b=45cm and c=42cm

⇒s=

2

a+b+c

=

2

39+42+45

=63cm

∴ Area of ΔBEC=

s(s−a)(s−b)(s−c)

=

63(63−39)(63−42)(63−45)

cm

2

=

63×24×21×18

cm

2

=756cm

2

Also, Area of ΔBEC=

2

1

×base×height

⇒

2

1

×45×h=756cm

2

⇒h=33.6cm

So, Area of trapezium:-

=

2

1

×(AB+CD)×height

=

2

1

×(75+30)×33.6=1764cm

2

Answer 2

Answer:

Draw CE∥DA and CF⊥AB.

Therefore, ADCE is a parallelogram having AE∥CD and CE∥DA.

⇒AD=CE=39cm,DC=AE=30cm and BE=75−30=45cm

In △BCE, by heron's formula we have

a=39cm,b=45cm and c=42cm

⇒s=

2

a+b+c

=

2

39+42+45

=63cm

∴ Area of ΔBEC=

s(s−a)(s−b)(s−c)

=

63(63−39)(63−42)(63−45)

cm

2

=

63×24×21×18

cm

2

=756cm

2

Also, Area of ΔBEC=

2

1

×base×height

⇒

2

1

×45×h=756cm

2

⇒h=33.6cm

So, Area of trapezium:-

=

2

1

×(AB+CD)×height

=

2

1

×(75+30)×33.6=1764cm

2