Question
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
Answer: Option A
:
A
The area of a trapezium is given by:
A=12×(Sum of parallel sides)×(height)
⇒36=12×(12+6)×h
⇒h=4cm
So now in the trapezium,
h=4cm
a=b=6cm
AB=CD[S-A-S congruency in△AFBand△DEC]
∴2AB+a=12cm
⇒2AB=12−6
⇒AB=CD=3cm
So now in△ABF,AB2+h2=d2
d2=32+42=25
d=5cm
As the trapezium is isosceles, the slant sides of the trapezium are equal in length
d=c=5cm
∴ the perimeter of the trapezium
=(2×AB)+a+c+b+d
=(2×3)+6+5+6+5=6+22=28cm
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:
A
The area of a trapezium is given by:
A=12×(Sum of parallel sides)×(height)
⇒36=12×(12+6)×h
⇒h=4cm
So now in the trapezium,
h=4cm
a=b=6cm
AB=CD[S-A-S congruency in△AFBand△DEC]
∴2AB+a=12cm
⇒2AB=12−6
⇒AB=CD=3cm
So now in△ABF,AB2+h2=d2
d2=32+42=25
d=5cm
As the trapezium is isosceles, the slant sides of the trapezium are equal in length
d=c=5cm
∴ the perimeter of the trapezium
=(2×AB)+a+c+b+d
=(2×3)+6+5+6+5=6+22=28cm
Was this answer helpful ?
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