Perimeters of two similar triangles ABC and DEF are in the ratio 3 : 4. Prove that the ratio of their areas is 9 : 16

Asked by Topperlearning User | 3rd Oct, 2017, 01:29: PM

Expert Answer:

begin mathsize 12px style increment space ABC space tilde space increment space DEF
fraction numerator Perimeter space straight A space open parentheses increment space ABC close parentheses over denominator Perimeter space straight A space open parentheses increment space DEF close parentheses end fraction equals 3 over 4
fraction numerator AB thin space plus thin space BC thin space plus thin space AC over denominator DE thin space plus thin space EF thin space plus thin space FD end fraction space equals 3 over 4
rightwards double arrow AB over DE equals BC over EF equals AC over DF equals 3 over 4
So comma fraction numerator space straight A space open parentheses increment space ABC close parentheses over denominator straight A space open parentheses increment space DEF close parentheses end fraction space equals space AB squared over DE squared equals 9 over 16
Hence comma space proved. end style

Answered by  | 3rd Oct, 2017, 03:29: PM