An isosceles trapezium has an area of 36 cm2, the parallel sides are 12 cm and 6 cm respectively. The perimeter of the trapezium should be
28 cm
The area of a trapezium is given by:
A=12×(Sum of parallel sides)×(height)
⇒36=12×(12+6)×h
⇒h=4 cm
So now in the trapezium,
h=4 cm
a=b=6 cm
AB=CD [S-A-S congruency in△AFB and △DEC]
∴2AB+a=12 cm
⇒2AB=12−6
⇒AB=CD=3 cm
So now in △ABF, AB2+h2=d2
d2=32+42=25
d=5 cm
As the trapezium is isosceles, the slant sides of the trapezium are equal in length.
d=c=5 cm
∴ the perimeter of the trapezium
=(2×AB)+a+c+b+d
=(2×3)+6+5+6+5=6+22=28 cm