Question

prove that the ratio of the areas of two similar tringles is equal to the square of the ratio of their corresponding medians​

Answer 1

Answer:

The proof is given below.

Step-by-step explanation:

Let the two similar triangles ABC and DEF in which AP and DQ are the medians respectively.

we have to prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.

If ΔABC~ΔDEF

⇒ \frac{ar(BCA)}{ar(EFD)}=\frac{AB^2}{DE^2}ar(EFD)ar(BCA)=DE2AB2  →   (1)

As ΔABC~ΔDEF

\frac{AB}{DE}=\frac{BC}{EF}=\frac{2BP}{2EQ}DEAB=EFBC=2EQ2BP

Hence, \frac{AB}{DE}=\frac{BP}{EQ}DEAB=EQBP

In ΔABP and ΔDEQ

\frac{AB}{DE}=\frac{BP}{EQ}DEAB=EQBP

∠B=∠E   (∵ΔABC~ΔDEF)

By SAS rule, ΔABP~ΔDEQ

⇒ \frac{AB}{DE}=\frac{AP}{DQ}DEAB=DQAP

Squaring, we get

\frac{AB^2}{DE^2}=\frac{AP^2}{DQ^2}DE2AB2=DQ2AP2    →  (2)

Comparing (1) and (2), we get

\frac{ar(BCA)}{ar(EFD)}=\frac{AP^2}{DQ^2}ar(EFD)ar(BCA)=DQ2AP2  

Hence, the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.