Question

prove that sum of all angle of a quadrilateral is 360°​

Answer 1

Step-by-step explanation:

The diagram for this sum is attached below

Proof: Let ABCD be a quadrilateral. Join AC.

Clearly, ∠1 + ∠2 = ∠A ...... (i)

And, ∠3 + ∠4 = ∠C ...... (ii)

We know that the sum of the angles of a triangle is 180°.

Therefore, from ∆ABC, we have

∠2 + ∠4 + ∠B = 180° (Angle sum property of triangle)

From ∆ACD, we have  

∠1 + ∠3 + ∠D = 180° (Angle sum property of triangle)

Adding the angles on either side, we get;

∠2 + ∠4 + ∠B + ∠1 + ∠3 + ∠D = 360°

⇒ (∠1 + ∠2) + ∠B + (∠3 + ∠4) + ∠D = 360°

⇒ ∠A + ∠B + ∠C + ∠D = 360° [using (i) and (ii)].

Hence, the sum of all the four angles of a quadrilateral is 360°.

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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°.