If triangle ABC is similar to triangle DEF such that 2AB=DE and BC=8 find the length of EF

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Solution

$D e a r S t u d e n t G i v e n △ A B C a n d △ D E F a r e s i m i l a r D E = 2 A B B C = 8 c m N o w, \frac{A B}{D E} = \frac{B C}{E F} (u \sin g s i d e s t h e o r e m) \frac{A B}{2 A B} = \frac{B C}{E F} (a s D E = 2 A B) \frac{1}{2} = \frac{B C}{E F} E F = 2 B C E F = 2 \times 8 = 16 c m R e g a r d s$

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If triangle ABC is similar to triangle DEF such that 2AB=DE and BC=8 find the length of EF

Dear StudentGiven △ABC and △DEF are similarDE =2AB BC = 8cmNow, ABDE=BCEF ( using sides theorem)AB2AB=BCEF (as DE = 2AB)12 = BCEFEF = 2BCEF = 2×8 =16cmRegards