Let О”abcв€јо”def And Their Areas Be Respectively 64cmві And 121cmві. If Ef=15.4cm Find Bc. (Exclusive Deal)

Area(△ABC)Area(△DEF)=(BCEF)2the fraction with numerator Area open paren triangle cap A cap B cap C close paren and denominator Area open paren triangle cap D cap E cap F close paren end-fraction equals open paren the fraction with numerator cap B cap C and denominator cap E cap F end-fraction close paren squared 2. Substitute the known values Plug the given areas ( ) and the length of side EFcap E cap F ) into the formula:

64121=(BC15.4)264 over 121 end-fraction equals open paren the fraction with numerator cap B cap C and denominator 15.4 end-fraction close paren squared 3. Calculate the ratio of sides Solve for side BCcap B cap C Multiply

64121=BC15.4the square root of 64 over 121 end-fraction end-root equals the fraction with numerator cap B cap C and denominator 15.4 end-fraction Solve for side BCcap B cap C Multiply

Take the square root of both sides of the equation to find the ratio of the corresponding side lengths: Solve for side BCcap B cap C Multiply

BC=811×15.4cap B cap C equals 8 over 11 end-fraction cross 15.4 BC=8×1.4cap B cap C equals 8 cross 1.4 BC=11.2 cmcap B cap C equals 11.2 cm ✅ Final Answer The length of the corresponding side BCcap B cap C

811=BC15.48 over 11 end-fraction equals the fraction with numerator cap B cap C and denominator 15.4 end-fraction 4. Solve for side BCcap B cap C Multiply both sides by to isolate BCcap B cap C

The length of side BCcap B cap C 1. Identify the relationship between areas and sides