find the equal sides of a rectangle PQRS
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Answer:
Here, PQRS is a rectangle.
As we know in rectangle both the diagonals are equal.
⇒ PR=QS
Also diagonals bisect each other.
⇒ PO=QO
⇒ ∠OPQ=∠PQO [ Base angles of an equal sides are also equal ]
⇒ ∠OPQ=24o [ Given ]
∴ ∠PQO=24o
In △PQO,
⇒ ∠OPQ+∠PQO+∠QOP=180o
⇒ 24o+24o+x=180o
⇒ 48o+x=180o
∴ x=132o
Since, PQRS is a rectangle, PQ∥SR and PR is a transversal.
⇒ ∠QPR=∠SRP [ Alternate angles ]
therefore ∠SRP=24o
⇒ ∠SRP+∠PRQ=90o [ Angle of an rectangle ]
⇒ 24o+y=90o
∴ y=66o
Answer:
Here, PQRS is a rectangle.
As we know in rectangle both the diagonals are equal.
⇒ PR=QS
Also diagonals bisect each other.
⇒ PO=QO
⇒ ∠OPQ=∠PQO [ Base angles of an equal sides are also equal ]
⇒ ∠OPQ=24
o
[ Given ]
∴ ∠PQO=24
o
In △PQO,
⇒ ∠OPQ+∠PQO+∠QOP=180
o
⇒ 24
o
+24
o
+x=180
o
⇒ 48
o
+x=180
o
∴ x=132
o
Since, PQRS is a rectangle, PQ∥SR and PR is a transversal.
⇒ ∠QPR=∠SRP [ Alternate angles ]
therefore ∠SRP=24
o
⇒ ∠SRP+∠PRQ=90
o
[ Angle of an rectangle ]
⇒ 24
o
+y=90
o
∴ y=66
o
Step-by-step explanation: