Q. 30

# The perimeters of

Answer :

ΔABC~ΔDEF

𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝐴𝐵𝐶)/𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝐷𝐸𝐹) = (𝐴𝐵+𝐵𝐶+𝐶𝐴)/(𝐷𝐸+𝐸𝐹+𝐹𝐷) = 𝐴𝐵/𝐷𝐸

25/15=9/𝑋

x = 5.4cm

DE = 5.4cm

OR Construction-Draw AM BC

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

In ΔABM,

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 + 2BC2/9

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

Multiply the equation by 9,

9AB2 = 9AD2 + 2AB2

Hence,7AB2 = 9AD2

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos  Heights and Distances - I54 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation view all courses RELATED QUESTIONS :

In Fig. 3,<Mathematics - Board Papers

In an equilateralRD Sharma - Mathematics