Question

Prove that sum of all angles of a quadrilateral is 360

Answer 1

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]

Answer 2

$\huge\sf \orange{hello}.....$

Statement :

sum of the angles of quadrilateral is 360°

To Prove :

∠A + ∠B + ∠C + ∠D = 360°

Proof :

In ∆ ABC , m∠4 + m∠5+m∠6 = 180°

[ using angle a property of a triangle]

Also , in ∆ ADC , m∠1 + m∠2+m∠3= 180°

Sum of the measures of ∠A, ∠B , ∠C and ∠D of a quadrilateral

m∠4 + m∠5+ m∠6 + m∠1 + m∠2 +m∠3 = 180°+ 180°

→ ∠A + ∠B + ∠C + ∠D = 360°

Thus , sum of measure of four angles of quadrilateral is 360°.