Question

one more question is the

Answer 1

Question 15

In ∆EFD and ∆QRP

EF=QR= 3cm (given)

Angle EFD= angle RQP( given)

DF=PR= 4cm (given)

By SAS both triangle are congruent

So ∆EFD~= ∆RQP

Que 16

a) x= 130°

b) x=85°

c) x=40°

Hope it helps...