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In this article we are going to learn about the Sum of the Four Sides of a Quadrilateral Exceeds the Sum of the Diagonals. We have to prove that the sum of all the four sides of a quadrilateral is greater than the sum of the diagonals.

Sum of the Four Sides of a Quadrilateral Exceeds the Sum of the Diagonals Theorem

Theorem Statement:
Prove that the sum of the sides of a quadrilateral is greater than twice the length of its diagonal?
Proof:
Given PQRS is a quadrilateral, PR and QS are its diagonals
To Prove:
(PQ + QR + RS + SP) > ( PR + QS)
Proof:
In ∆PSQ
(SP + PQ) > QS [ therefore the sum of the two sides of a triangle is greater than the third side]
In ∆PQR
(PQ + QR) > PR [ therefore the sum of the two sides of a triangle is greater than the third side]
In ∆QRS
(QR + RS) > QS [ therefore the sum of the two sides of a triangle is greater than the third side]
In ∆RSP
(RS + SP) > PR [ therefore the sum of the two sides of a triangle is greater than the third side]
2(PQ + QR + RS + SP) > 2(PR + QS) [ adding all the equations]
(PQ + QR + RS + SP) > 2(PR + QS) [ cancelling the common factor 2]
Hence proved
Therefore it is proven that the sum of the four sides of a Quadrilateral exceeds the sum of the diagonals.

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FAQs on Sum of the Four Sides of a Quadrilateral Greater than the Sum of the Diagonals

1. What is the sum of four sides of the quadrilateral?

A quadrilateral with four sides and four angles. The sum of four angles of a quadrilateral is 360 degrees. We can draw two diagonals to the quadrilateral that forms two triangles.

2. What is the quadrilateral inequality theorem?

The sum of any pairs of opposite sides of a convex quadrilateral is always less than the sum of the diagonals.

3. How many diagonals does a quadrilateral have?

A quadrilateral can be constructed unambiguously if the lengths of its three sides and two diagonals are provided and if its two adjacent sides and three angles are given.