Prove that the ratio of the areas of two similar triangles is equal to square of the ratio of their corresponding medians
Prove that the ratio of the areas of two similar triangles is equal to square of the ratio of their corresponding medians
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Verified answer
Hey user !!!
Given :-
let ∆ ABC ~ ∆ PQR
Here ,AD is a median
hence,
BD = CD = 1/2 BC
similarly PS is a median
hence ,QS = SR = 1/2 QR
To prove : area of ∆ ABC / area of ∆ PQR
=( AD /PS )2 square
proof :-
Given =
∆ ABC ~ ∆ PQR
angle B = angle Q ( corresponding angle of similar triangle are equal ) ............1
AB /PQ = BC /QR ( corresponding sides of similar triangle are in same proportion )
=> AB /PQ = 2BD / 2 QS
=> AB /PQ = BD /QS .......................... 2
• In ∆ABD and ∆ PQS
Angle B = angle Q ( from 1)
AB /PQ = BD /QS (from 2)
∆ABD ~ ∆ PQS .........( SAS PROPERTY )
hence , AB /PQ = AD /PS ( if triangle are similar then their corresponding sides are in same proportion ) ..........3
since , ∆ ABC ~ ∆PQR
we know that if two triangle are similar , the ratio of their area is always equal to the square of the ratio of their corresponding sides .
therefore ,
area of ∆ABC / area of ∆ PQR =( AB /PQ )square
also , area of ∆ABC /area of ∆ PQR = ( AD /PS ) square ....( from 3 )
hence ,proved
thanks :)
Verified answer
QUESTIONProve that the ratio of the areas of two similar triangles is equal to square of the ratio of their corresponding medians
ANSWER
Figure =>Is in attachment
Now,
Given: Two traingles ABC and DEF having medians AP and DQ respectively.
Also,
Traingle ABC is similar to DEF.
Proof=>
Since the traingles are similar hence Their area will be ratio of square of their sides.
i.e
Area of ABC/Area of DEF =(AB/DE)^2
------------------(i)
Now, since traingle ABC and DEF are similar, hence we get =>
AB/DE = BC/EF
=>AB/DE= 2BP/2EQ=BP/EQ. ------(ii)
(since BP=PC =1/2BC and EQ=QF=1/2EF)
So, we get=>
AB/DE= BP/EQ
and angle B = angle E
(since traingle ABC similar to traingle DQE)
Hence, traingle APB similar to Traingle DQE. (By SAS)
So,
BP/EQ=AP/DQ. --------(iii)
From (i) and (ii)=>
AB/DE=AP/DQ
=>(AB/DE)^2 =(AP/DQ)^2. --------(iv)
Hence, from (i) and (iv)=>
Area of ABC/Area of DEF =(AP/DQ)^2.
Hence,the ratio of the areas of two similar triangles is equal to square of the ratio of their corresponding medians
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