RESULTS RELATED TO THE RATIO OF THE AREAS OF TWO SIMILAR TRIANGLES
RESULT – 1
The ratio of the areas of two similar triangles is equal to
the ratio of the squares of their corresponding altitudes.
The ratio of the areas of two similar triangles
A
C
B
R
Q
P
ΔABC
~
ΔPQR
A(ΔPQR)
A(ΔABC)
is equal to
=
the ratio of the squares of their corresponding altitudes.
A perpendicular drawn from the vertex of a triangle to its opposite side
Let us consider
ΔABC
For vertex A,
Opposite side →
?
BC
M
For vertex B,
Opposite side →
?
AC
N
For vertex C,
Opposite side →
?
AB
L
Let us consider
ΔPQR
For vertex P,
Opposite side →
?
QR
For vertex Q,
Opposite side →
?
PR
For vertex R,
Opposite side →
?
PQ
D
F
E
Name the pair of corresponding altitudes
(AM)2
AM
(PD)2
PD
(BN)2
BN
(QE)2
QE
=
(CQ)2
CL
(RF)2
RF
=
AM ↔ PD
BN ↔ QE
CL ↔ RF
Z
RESULTS RELATED TO THE RATIO OF THE AREAS OF TWO SIMILAR TRIANGLES
RESULT – 2
The ratio of the areas of two similar triangles is equal to
the ratio of the squares of their corresponding medians.
The ratio of the areas of two similar triangles
C
B
R
Q
P
ΔABC
~
ΔPQR
A(ΔPQR)
A(ΔABC)
is equal to
=
the ratio of the squares of their corresponding medians.
A
Segment joining the vertex of a triangle to the midpoint of its opposite side
Let us consider
ΔABC
For vertex A,
Opposite side →
?
BC
L
M
N
Let us consider
ΔPQR
X
Y
Name the pair of corresponding medians
(AL)2
AL
(AX)2
PX
(BM)2
BM
(QY)2
QY
=
(CN)2
CN
(RZ)2
RZ
=
AL ↔ PX
BM ↔ QY
CN ↔ RZ