Question

Prove that the sum of the angles of a quadrilateral is 360°.​

Answer 1

Note :

★ Angle sum property of a triangle : The sum of all the three interior angles of a triangle is 180° .

★ Angle sum property of a quadrilateral : The sum of all the four interior angles of a quadrilateral is 360° .

Solution :

Given :

A quadrilateral (say quadrilateral ABCD)

To prove :

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

ie. ∠A + ∠B + ∠C + ∠D = 360° .

Construction :

Join the point A to C to form the diagonal AC .

Proof :

Clearly ,

The diagonal of a quadrilateral divides it into two triangles .

Also ,

According to the angle sum property of a triangle , the sum of all the three interior angles of a triangle is 180° .

Thus ,

In ∆ABC , ∠1 + ∠B + ∠3 = 180° -------(1)

Also ,

In ∆ADC , ∠2 + ∠D + ∠4 = 180° -------(2)

Now ,

Adding eq-(1) and (2) , we have ;

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

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

→ ∠A + ∠B + ∠C + ∠D = 360°

Hence proved .

Answer 2

$\huge \bullet \red{S}\purple{t}\pink{a}\red{t}\orange{e}\pink{m}\pink{e}\blue{n}\red{t}$

sum of the angles of quadrilateral is 360°

$\bullet \red{T}\purple{o}\pink{P}\red{r}\orange{o}\pink{v}\pink{e}\blue{:}$

∠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°

the sum of the quadrilateral is 360°

More Info :

1. Sum of angles in a triangle is 180°
2. Sum of angles in a quadrilateral is 360°
3. Sum of angles in a straight line is 180°