prove that sum of all interior angles of quardilateral is 360°
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Consider a quadrilateral PQRS.
Join QS.
To prove: ∠P + ∠Q + ∠R + ∠S = 360º
Proof:
Consider triangle PQS, we have,
⇒ ∠P + ∠PQS + ∠PSQ = 180º ... (1) [Using Angle sum property of Triangle]
Similarly, in triangle QRS, we have,
⇒ ∠SQR + ∠R + ∠QSR = 180º ... (2) [Using Angle sum property of Triangle]
On adding (1) and (2), we get
∠P + ∠PQS + ∠PSQ + ∠SQR + ∠R + ∠QSR = 180º + 180º
⇒ ∠P + ∠PQS + ∠SQR + ∠R + ∠QSR + ∠PSQ = 360º
⇒ ∠P + ∠Q + ∠R + ∠S = 360º [Hence proved]
Join QS.
To prove: ∠P + ∠Q + ∠R + ∠S = 360º
Proof:
Consider triangle PQS, we have,
⇒ ∠P + ∠PQS + ∠PSQ = 180º ... (1) [Using Angle sum property of Triangle]
Similarly, in triangle QRS, we have,
⇒ ∠SQR + ∠R + ∠QSR = 180º ... (2) [Using Angle sum property of Triangle]
On adding (1) and (2), we get
∠P + ∠PQS + ∠PSQ + ∠SQR + ∠R + ∠QSR = 180º + 180º
⇒ ∠P + ∠PQS + ∠SQR + ∠R + ∠QSR + ∠PSQ = 360º
⇒ ∠P + ∠Q + ∠R + ∠S = 360º [Hence proved]
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