Prove that the ratio of the area of two similar ∆s is equal to the square of the ratio of their corresponding medians.
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Answers
Answer:
Given:- Let △ABC∼△PQR
AD and PS are corresponding medians.
To prove:-
ar(△PQR)
ar(△ABC)
=(
PS
AD
)
2
Proof:- In △ABC
∵AD is median.
∴BD=CD=
2
1
BC
Similarly, in △PQR
PS is median.
∴QS=RS=
2
1
QR
Now,
△ABC∼△PQR(Given)
∠B=∠Q.....(1)(Corresponding angles of similar triangles are equal)
PQ
AB
=
QR
BC
(Corresponding sides of similar triangles are proportional)
⇒
PQ
AB
=
2QS
2BD
(∵AD and PS are medians)
PQ
AB
=
QS
BD
.....(2)
Now, in △ABD and △PQS,
∠B=∠Q(From (1))
PQ
AB
=
QS
BD
(From (2))
∴△ABD∼△PQS(By SAS Property)
Therefore,
PQ
AB
=
PS
AD
.....(3)(∵Corresponding sides of similar triangles are proportonal)
Now,
∵△ABC∼△PQR
As we know that ratio of area of similar triangles is always equal to the square of ratio of their corresponding side.
Therefore,
ar(△PQR)
ar(△ABC)
=(
PQ
AB
)
2
ar(△PQR)
ar(△ABC)
=(
PS
AD
)
2
(From (3))
Hence proved.