Question

Prove that the ratio of the areas of two similar triangles is equal to square of the ratio of their corresponding medians

Answer 1

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 :)

Answer 2

QUESTION

Prove 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

THANKS