Heights of two similar triangles are 4cm and 12cm find there ratio of area
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Answer:
Let us also draw perpendiculars
A
P
and
D
Q
from
A
and
D
respectively on to
B
C
and
E
F
as shown.
enter image source here
It is apparent that
Δ
A
P
B
and
Δ
D
E
Q
are also similar as all respective angles are equal. Hence,
A
B
D
E
=
A
P
D
Q
=
B
P
E
Q
We also have
Δ
A
B
C
=
1
2
×
B
C
×
A
P
and
Δ
D
E
F
=
1
2
×
E
F
×
D
Q
and
Δ
A
P
B
Δ
D
E
Q
=
B
C
×
A
P
E
F
×
D
Q
=
B
C
E
F
×
A
P
D
Q
But
A
P
D
Q
=
A
B
D
E
=
B
C
E
F
and hence
Δ
A
P
B
Δ
D
E
Q
=
B
C
E
F
×
B
C
E
F
=
B
C
2
E
F
2
and as
B
C
E
F
=
A
C
D
F
=
A
B
D
E
Δ
A
P
B
Δ
D
E
Q
=
A
C
2
D
F
2
=
B
C
2
E
F
2
=
A
B
2
D
E
2
Hence if sides of two similar triangles are in the ratio
a
:
b
, their areas are in the proportion
a
2
:
b
2
As in given case sides are in the ratio of
4
:
9
.
ratio of their areas is
4
2
:
9
2
or
16
:
81
.
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