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Byju's Answer
Standard VIII
Mathematics
Area of a General Quadrilateral
Draw quadrila...
Question
Draw quadrilateral ABCD with the following measurements. Find also its area.
AB = 5.5 cm, BC = 4.5cm, AC = 6.5 cm,
∠
CAD =
80
o
and
∠
ACD =
40
o
.
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Solution
Draw line
A
B
=
5.5
cm
Draw an arc of radius
6.5
cm from point
A
and an other arc of radius
4.5
cm from point
B
both meet at point
C
Join
C
to
A
and
B
Draw an angle of
40
0
at point
A
and an other angle of
40
0
at point
C
. Both cut at point
D
Join
D
to
C
and
A
Hence
A
B
C
D
is the required quadrilateral.
In triangle
A
B
C
,
s
=
6.5
+
5.5
+
4.5
2
=
8.25
Then area of ABC
=
√
8.25
(
8.25
−
6.5
)
(
8.25
−
5.5
)
(
8.25
−
4.5
)
=
√
8.25
×
1.75
×
2.75
×
3.75
=
√
148.89
=
12.20
In triangle ACD
A
D
sin
40
0
=
6.5
sin
100
0
⇒
A
D
=
6.5
×
0.64
0.98
=
4.25
c
m
Then area triangle ACD =
1
2
a
b
s
i
n
θ
=
6.5
×
4.25
s
i
n
40
0
=
27.63
×
0.64
=
17.68
c
m
2
So area of ABCD
quadrilateral
=
12.20
+
17.68
=
29.88
sq cm
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