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
Answers
The area of trapezium is given by:
A= 1/2 x (sum of parallel sides) x height
36= 1/2 x (12+6) + h
36= 1/2 x 18 + h
therefore, h = 36 x 2/18 = 4cm
a
=
b
=
6
c
m
A
B
=
C
D
[S-A-S congruency in
△
A
F
B
and
△
D
E
C
]
∴
2
A
B
+
a
=
12
c
m
⇒
2
A
B
=
12
−
6
⇒
A
B
=
C
D
=
3
c
m
So now in
△
A
B
F
,
A
B
2
+
h
2
=
d
2
d
2
=
3
2
+
4
2
=
25
d
=
5
c
m
As the trapezium is isosceles, the slant sides of the trapezium are equal in length
∴
the perimemter of the trapezium
=
(
2
×
A
B
)
+
a
+
c
+
b
+
d
=
(
2
×
3
)
+
6+5+6+5=6+22=28cm
I hope this helps
Answer:
The area of a trapezium is given by:
A=1/2×(Sum of parallel sides)×(height)
⇒36=1/2×(12+6)×h
⇒h=4cm
So now in the trapezium,
h=4cm
a=b=6cm
AB=CD[S-A-S congruency in△AFB and △DEC]
∴2AB+a=12cm
⇒2AB=12−6
⇒AB=CD=3cm
So now in△ABF,AB^2+h^2=d^2
d^2=3^2+4^2=25
d=5cm
As the trapezium is isosceles, the slant sides of the trapezium are equal in length
d=c=5cm
∴ the perimemter of the trapezium
=(2×AB)+a+c+b+d
=(2×3)+6+5+6+5=6+22=28cm
Hope it helps!