Prove that the ratio of the areas of two similar tingles is equal to the square of the ratio of their corresponding Sides
Answers
Answer:
Solution
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Let the two triangles be:
ΔABC and ΔPQR
Area of ΔABC=
2
1
×BC×AM……………..(1)
Area of ΔPQR=
2
1
×QR×PN……………………..(2)
Dividing (1) by (2)
ar(PQR)
ar(ABC)
=
2
1
×QR×PN
2
1
×BC×AM
ar(PQR)
ar(ABC)
=
QR×PN
BC×AM
…………………..(1)
In ΔABM and ΔPQN
∠B=∠Q (Angles of similar triangles)
∠M=∠N (Both 90
∘
)
Therefore, ΔABM∼ΔPQN
So,
AM
AB
=
PN
PQ
…………………….(2)
From 1 and 2
ar(PQR)
ar(ABC)
=
QR
BC
×
PN
AM
⇒
ar(PQR)
ar(ABC)
=
QR
BC
×
PQ
AB
…………………..(3)
PQ
AB
=
QR
BC
=
PR
AC
………….(ΔABC∼ΔPQR)
Putting in ( 3 )
ar(PQR)
ar(ABC)
=
PQ
AB
×
PQ
AB
=(
PQ
AB
)
2
⇒
ar(PQR)
ar(ABC)
=(
PQ
AB
)
2
=(
QR
BC
)
2
=(
PR
AC
)
2
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SIMILAR QUESTIONS
star-struck
If △ABC∼△DEF and AB=2.4cm, DE=1.6cm, find the ratio of △ABC and △DEF.
Easy
View solution
>
Triangles ABC and DEF are similar. If the length of the perpendicular AP from A on the opposite side BC is 2 cm and the length of the perpendicular DQ from D on the opposite side EF is 1 cm, then what is the area of △ABC?
Step-by-step explanation:
Let the two triangles be:
ΔABC and ΔPQR
Area of ΔABC=
2
1
×BC×AM……………..(1)
Area of ΔPQR=
2
1
×QR×PN……………………..(2)
Dividing (1) by (2)
ar(PQR)
ar(ABC)
=
2
1
×QR×PN
2
1
×BC×AM
ar(PQR)
ar(ABC)
=
QR×PN
BC×AM
…………………..(1)
In ΔABM and ΔPQN
∠B=∠Q (Angles of similar triangles)
∠M=∠N (Both 90
∘
)
Therefore, ΔABM∼ΔPQN
So,
AM
AB
=
PN
PQ
…………………….(2)
From 1 and 2
ar(PQR)
ar(ABC)
=
QR
BC
×
PN
AM
⇒
ar(PQR)
ar(ABC)
=
QR
BC
×
PQ
AB
…………………..(3)
PQ
AB
=
QR
BC
=
PR
AC
………….(ΔABC∼ΔPQR)
Putting in ( 3 )
ar(PQR)
ar(ABC)
=
PQ
AB
×
PQ
AB
=(
PQ
AB
)
2
⇒
ar(PQR)
ar(ABC)
=(
PQ
AB
)
2
=(
QR
BC
)
2
=(
PR
AC
)
