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
From Angle Sum Property of a Quadrilateral to HOME PAGE
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
Jun 13, 24 02:51 AM
Jun 13, 24 02:28 AM
Jun 13, 24 12:11 AM
Jun 12, 24 01:11 PM
Jun 11, 24 07:15 PM
New! Comments
Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.