Question

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

Answer 1

Hey there !

Solution:

Given that, Δ ABC ~ Δ DEF

=> According to Basic Proportionality Theorem,

$\implies \dfrac{AB}{DE} = \dfrac{BC}{EF} = \dfrac{ AC}{DF}$

$\text{ Given that, 2AB = DE} \\ \\ \implies 2 = \dfrac{ DE}{AB} \\ \\ \text{ Reciprocating on both sides, we get} \\ \\ \implies \dfrac{1}{2} = \dfrac{AB}{DE} \\ \\ \implies \dfrac{BC}{EF} = \dfrac{1}{2} \hspace{20mm} ( As\:\:\: \dfrac{AB}{DE} = \dfrac{BC}{EF}} ) \\ \\ \text{ BC = 8 cm} \\ \\ \text{ Substituting the values we get,} \\ \\ \implies \dfrac{8}{EF} = \dfrac{1}{2} \\ \\ \text{ Cross-multiplying we get,} \\ \\ \implies EF \times 1 = 8 \times 2 \\ \\ \implies EF = 16 \: cm$

Hence EF = 16 cm.

Hope my answer helped !

Answer 2

given:-

triangle ABC ~ triangle DEF
2AB = DE
and BC = 8 cm

to find :-
EF = ?

proof :-

ABC ~ DEF ( given )
=> AB ÷DE = BC ÷ EF (by BPT ) ....(i)
2AB = DE ( given )
DE ÷ AB = 2
AB ÷ DE = 1 ÷ 2 .....(ii)
from eq (i) and (ii)
BC ÷ EF = 1 ÷ 2
8 ÷ EF = 1 ÷ 2 ( BC = 8 cm )
EF = 8 × 2
EF = 16 cm

hope this will help uh ...