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