Theorem and Proof of angle sum property of a quadrilateral.
Prove that 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°.
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°.
Solved examples of angle sum property
of a quadrilateral:
1. The angle of
a quadrilateral are (3x + 2)°, (x – 3), (2x + 1)°, 2(2x + 5)° respectively.
Find the value of x and the measure of each angle.
Solution:
Using angle sum property of quadrilateral, we get
(3x + 2)°+ (x – 3)° + (2x + 1)° + 2(2x + 5)°= 360°
⇒ 3x + 2 + x  3 + 2x + 1 + 4x + 10 = 360°
⇒ 10x + 10 = 360
⇒ 10x = 360 – 10
⇒ 10x = 350
⇒ x = 350/10
⇒ x = 35
Therefore, (3x + 2) = 3 × 35 + 2 = 105 + 2 = 107°
(x – 3) = 35 – 3 = 32°
(2x + 1) = 2 × 35 + 1 = 70 + 1 = 71°
2(2x + 5) = 2(2 × 35 + 5) = 2(70 + 5) = 2 × 75 = 150°
Therefore, the four angles of the quadrilateral are 32°, 71° 107°, 150° respectively.
2. In a quadrilateral PQRS, PQ + QR + RS + SP < 2 (PR + QS).
Solution:
In ∆POS, PO + OS > PS …………… (i)
In ∆SOR, SO + OR > SR …………… (ii)
In ∆QOR, QO + OR > QR …………… (iii)
In ∆POQ, PO + OQ > PQ …………… (iv)
(i) + (ii) + (iii) + (iv) (Using triangle inequality property)
PO + OS + OS + OR + OQ + OR + OP + OQ > PS + SR + QR + PQ
⇒ 2 (OP + OQ + OR + OS) > PQ + QR + CS + DP
⇒ 2 [(OP + OR) + (OQ + OS)] > PQ + QR + CS + DP
⇒ 2 (PR + QS) > PQ + QR + RS + SP
The above examples will help us to solve various types of problems based on angle sum property of a quadrilateral.
7th Grade Math Problems
8th Grade Math Practice
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