Question

ΔPQR ∽ ΔDEF for the correspondence PQR⇔EDF. If PQ+QR=15, DE+DF=10 and PR=6, find EF.

Answer 1

ΔPQR ∽ ΔDEF for the correspondence PQR⇔EDF.
Hence, $\frac{PQ}{DE}=\frac{PR}{EF}=\frac{QR}{DF}\\\\\implies\frac{PQ}{DE}=\frac{QR}{DF}=\frac{PQ+QR}{DE+DF}$
from these two equations ,
$\frac{PQ+QR}{DE+DF}=\frac{PR}{EF}$

Given, (PQ + QR) = 15 , (DE + EF) = 10
and PR = 6

now, $\frac{15}{10}=\frac{6}{EF}$

EF = 6 × 10/15 = 4

hence, EF = 4

Answer 2

Answer : 4 cm

Step-by-step explanation:

we know that if triangles are similar then sides are in proptional that is in this question PQ/ED = QR/DF = PR/EF

Therefore : PQ+QR/DE+DF = PR/EF

ie: 15/10 = 6/EF

Therefore 15EF = 60

therefore EF= 4 CM