Question

1. In the following figure, PQRS is a trapezium with PQ||RS. The bisectors of ZP and ZQ meet
at M on side RS such that ZPMQ = 80°. If PM = QM, then show that ZS + ZR = 160°
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Answer 1

Answer:

⇒ PQRS is trapezium.

⇒ SP∥RQ [ Given ]

We know, sum of two angles between parallel lines is supplementary

∴ ∠QRS+∠RSP=180

o

⇒ ∠QRS+90

o

=180

o

⇒ ∠QRS=180

o

−90

o

=90

o

⇒ ∠PQR+∠QRS+∠RSP+∠SPQ=360

o

⇒ 100

o

+90

o

+90

o

+∠SPQ=360

o

⇒ 280

o

+∠SPQ=360

o

⇒ ∠SPQ=360

o

−280

o

=80

o

∴ Measures of ∠P and ∠R are 80

o

and 90

o