Question

State and prove the angle sum property of quadrilaterals, show this with examples

Answer 1

Answer:

Step-by-step explanation:

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

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°.

Angle Sum Property of a Quadrilateral

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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°.

Answer 2

To :-

Prove the angle sum property of quadrilaterals.

Solution :-

The sum of all angles of a quadrilateral will be 360°.

(Join AC)

Let ABCD be the given quadrilateral

In ∆ ABC

= ∠ 2 + ∠ B + ∠ 4 = 180° (angle sum property of a triangle) → (1)

In ∆ ACD

= ∠ 1 + ∠ D + ∠ 3 = 180° (angle sum property of a triangle) → (2)

By adding (1) and (2), we get,

= ∠ 2 + ∠ B + ∠ 4 + ∠ 1 + ∠ D + ∠ 3 = 180° + 180°

= ∠ 2 + ∠ 1 + ∠ B + ∠ 3 + ∠ 4 + ∠ D = 360°

$\implies$ $\sf\boxed{∠ A + ∠ B + ∠ C + ∠ D = 360°}$

hence proved.

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( note - I am also attaching a picture. )