Q. 30

# The perimeters of

ΞABC~ΞDEF

πππππππ‘ππ (Ξπ΄π΅πΆ)/πππππππ‘ππ (Ξπ·πΈπΉ) = (π΄π΅+π΅πΆ+πΆπ΄)/(π·πΈ+πΈπΉ+πΉπ·) = π΄π΅/π·πΈ

25/15=9/π

x = 5.4cm

DE = 5.4cm

OR

Construction-Draw AM β₯ BC

BD = 1/3 BC (Given), BM = 1/2 BC (as AM acts as median)

In ΞABM,

AB2 = AM2 + BM2 (By Pythagoras theorem)

= AM2 + (BD + DM)2

=AM2 + DM2 + BD2 + 2BD.DM

=AD2 + BD2 + 2BD(BM - BD)

=AD2 + (BC/3)2 + 2.BC/3.(BC/2 - BC/3)

=AD2 + 2AB2/9 (AB = Bc, equilateral triangle)

Multiply the equation by 9,

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