Question

the diagonals of parallelogram PQRS intersect at O . If angle QOR = 90° and angle QSR = 50°. Find angle QRS.​

Answer 1

Answer:

answer will be 90 degree

Step-by-step explanation:

since POS=POQ=SOR=ROQ

by SAS

OSR =50 DEGREE

ORS =40DEGREE

ORQ=50 DEGREE

Answer 2

$\angle QRS=80^{\circ}$

Step-by-step explanation:

Angle QSR=50 degrees

Angle QOR=90 degrees

PQ is parallel to RS and QR is parallel to PS

Angle SOR=180-angle QOR=180-90=90degrees

Angle POQ=180-angle QOR=180-90=90 degrees

Angle QOR=Angle POS=90 degrees (Vertical opposite angles are equal)

When the diagonals of parallelogram bisect perpendicularly then the parallelogram is a rhombus.

Therefore, PQRS is a rhombus

PQ=QR=RS=PS

In triangle SOR

$\angle OSR+\angle SRO+\angle SOR=180^{\circ}$

Reason: Triangle angles sum property

$90+50+\angle ORS=180$

$140+\angle ORS=180$

$\angle ORS=180-140=40^{\circ}$

Angle SPO=Angle ORS=40 degrees

Angle made by two equal sides are equal

Angle SPO=Angle ORQ=40 degrees

Reason: Alternate interior angles

Angle QRS=Angle QRS+angle ORQ=40+40=80 degree

Hence, $\angle QRS=80^{\circ}$

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