(b) Find the area of the sail in Fig. 4. 22.3 m 20 m 22 m 16.8 m Fig. 4
Answers
Answer:
In the sailboat (i),
Area of triangle
=
1
2
×
B
a
s
e
×
H
e
i
g
h
t
In
Δ
A
B
C
, AC = Base = 22 + 20 = 42 m
BD = Height = 22.3 m
∴
Area of
Δ
A
B
C
=
1
2
×
42
×
22.3
=
936.6
2
=
468.3
m
2
In another triangular part,
In
Δ
A
C
E
,
E
F
=
H
e
i
g
h
t
=
16.8
m
AC = Base = 22 + 20 = 42 m
∴
Area of
Δ
A
C
E
=
1
2
×
42
×
16.8
=
705.6
2
=
352.8
Area ABCE = 468.3 + 352.8 = 821.1
m
2
In sailboat (ii),.
Area of a triangle
=
1
2
×
B
a
s
e
×
H
e
i
g
h
t
In
Δ
A
B
C
,
∠
B
=
90
∘
, base (BC) = 10.9 m and height (AB) = 19.5 m
Area of
Δ
A
B
C
=
1
2
×
10.9
×
19.5
=
212.55
2
=
106.275
m
2
In another triangular part,
Step-by-step explanation:
Area of
Δ
D
E
F
=
1
2
×
D
F
×
E
H
=
1
2
×
23.9
×
8.6
=
205.54
2
=
102.77
m
2
Area of sailboat (ii)
=
106.275
+
102.77
=
209.045
m
2
In sailboat (iii),
Area of triangle
=
1
2
×
B
a
s
e
×
H
e
i
g
h
t
In
Δ
A
B
C
, AB = 8.9 m and BC = 3m
Area of
Δ
A
B
C
=
1
2
×
B
C
×
A
B
=
1
2
×
8.9
×
3
=
26.7
2
=
13.35
m
2
In another triangular part,
Area of
Δ
D
E
F
=
1
2
×
D
F
×
E
G
=
1
2
×
25
×
12.4
=
155
m
2
In another triangular part,
Area of
Δ
D
E
H
=
1
2
×
D
E
×
E
H
=
1
2
×
9.6
×
16.8
=
80.64
m
2
∴
Area of sailboat (iii) = 155+ 80.64
=
235.64
m
2