Question

6. In the given figure, ABCD is a trapezium, find:
(a) the length AD

(b) the area of trapezium ABCD
(c) the area of triangle BCD​

Answer 1

(i) In right angled △ABD

BD

2

=AD

2

+AB

2

[by Pythagoras Theorem ]

⇒AD

2

=BD

2

−AB

2

⇒AD

2

=(41)

2

−(40)

2

⇒AD

2

=1681−1600=81

⇒AD=

81

​

=9cm

(ii) Area of trapezium ABCD=

2

1

​

 (sum of opposite parallelogram lines)× height

=

2

1

​

(AB+CD)×AD

=(40+15)×9=247.5cm

2

(iii) Area of traingle BCD = Area of trapezium ABCD - Area of △ABD

=(247.5−

2

1

​

×40×9)cm

2

=67.5cm

2