Question

seg PS || seg QT || seg RU RQ =6.4 , PR =9.6 ST =11 find SU​

Answer 1

Answer:

here seg PS|| seg QT|| seg RU and

PQ= PR-QR ______(P-Q-R)

so,PQ=3.2units

therefore, according to the properties of three parallel lines and their transversals,

ST÷TU = PQ÷QR

11÷TU = 3.2÷6.4

TU= 11×6.4 ÷3.2

so,TU=70.4÷3.2

TU = 22units

now,

SU=ST+TU_____(S-T-U)

=11+22

=33

therefore,SU=33units☺️