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Angle sum property of quadrilateral

Understanding quadrilaterals of Class 8

THEOREM TITLE:

The sum of the angles of a quadrilateral is 360º or 4 right angles.

Proof: Let ABCD be a quadrilateral. Draw one of its diagonals, AC.

Clearly, ∠1 + ∠2 = ∠A and ∠3 + ∠4 = ∠C.

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

Therefore, 

In ΔABC, we have

∠1 + ∠4 + ∠B = 180º ..... (i)

And, in ΔACD, we have

∠2 + ∠3 + ∠D = 180º ..... (ii)

Adding (i) and (ii), we get

(∠1 + ∠4 + ∠B ) + (∠2 + ∠3 + ∠D ) = 180º + 180º

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

⇒ ∠A + ∠B + ∠C + ∠D = 360º [∠1 + ∠2 = ∠A and ∠3 + ∠4 = ∠C]

THEOREM TITLE:

Prove that the sum of the interior angles of pentagon is 540º.

To prove: ∠A + ∠B + ∠C + ∠D + ∠E = 540º.

Proof: Let ABCDE be a pentagon. Join AC and AD

We know that the sum of the angles of a triangle is 180º. Therefore, in ΔABC, we have

∠BAC + ∠ABC + ∠BCA = 180º ..... (i)

Similarly, in triangles ACD and ADE, we have

∠CAD + ∠ACD + ∠ADC = 180º ..... (ii)

and, ∠EAD + ∠ADE + ∠DEA = 180º ..... (iii)

Adding (i), (ii) and (iii), we get

∠BAC + ∠ABC + ∠BCA + ∠CAD + ∠ACD + ∠ADC

+ ∠EAD + ∠ADE + ∠DEA = 180º + 180º + 180º

⇒ (∠BAC + ∠CAD + ∠EAD) + ∠ABC + (∠BCA + ∠ACD) 

+ (∠ADC + ∠ADE) + ∠DEA = 540º

⇒ ∠BAE + ∠ABC + ∠BCD + ∠CDE + ∠DEA = 540º

⇒ ∠A + ∠B + ∠C + ∠D + ∠E = 540º

THEOREM TITLE:

Thu sum of all the angles of a hexagon is 720º.

To prove:∠FAB + ∠ABC + ∠BCD + ∠CDE + ∠DEF + ∠EFA = 720º

Proof: Let ABCDEF be a hexagon. Join AC, AD and AE.

In ΔABC, we have

∠BAC + ∠ABC + ∠BCA = 180º ..... (i)

In ΔADC, we have

∠CAD + ∠ADC + ∠ACD = 180º ..... (ii)

In ΔADE, we have

∠DAE + ∠ADE + ∠DEA = 180º ..... (iii)

In ΔAEF, we have

∠EAF + ∠AEF + ∠AFE = 180º ..... (iv)

Adding (i), (ii), (iii), (iv) and regrouping, we get

(∠BAC + ∠CAD + ∠DAE + ∠EAF) + ∠ABC + (∠BCA + ∠ACD) + (∠ADC + ∠ADE) 

+ (∠DEA + ∠AEF) + ∠AFE = 720º

⇒ ∠FAB + ∠ABC + ∠BCD + ∠CDE + ∠DEF + ∠EFA = 720º

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  • If there is a polygon of n sides (n ≥ 3), we can cut it up into (n – 2) triangles with a common vertex and so that sum of all the interior angles of a polygon of n sides would be (n – 2) × 180º = (n – 2× 2) right angles = (2n – 4) right angles.

 

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