Question

In a figure , AB and DC are parallel sides of a trapezium ABCD and angle ADC = 90 degree . Given AB=15 cm , CD=40 cm and diagonal AC=41 cm , calculate the area of the trapezium ABCD

Answer 1

consider a triangle ACD
it is right angled triangle therefore by pythagoras theorem
AC^2=AD ^2 +DC^2
41^2=AD^2 +40^2
1681 =AD^2 +1600
AD^2 =81
AD =9cm
Therefore the perpendicular height of trapezium is 9 cm
Now by formula
Area of trapezium = 1/2 × sum of lengths of parallel sides × height
=1/2 × (15 +40) × 9
=1/2 ×55 ×9
=495 /2
=247.5 sq. cm

Answer 2

by pythagoras theorem
the height of trapezium = 9cm
area of trapezium = 1/2{h×(sum of || sides)}
= 247.5 sq cm