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Question
For any quadrilateral, prove that its perimeter is greater than the sum of its diagonals.
Solution
Given: PQRS is a quadrilateral.PR and QS are its diagonals.
To Prove: PQ + QR + SR + PS > PR + QS
Proof: In ΔPQR
PQ + QR > PR ...(Sum of two sides of triangle is greater than the third side)
Similarly, In ΔPSR, PS + SR > PR
In ΔPQS, PS + PQ > QS and in QRS we have QR + SR > QS
Now we have
PQ +QR > PR
PS + SR > PR
PS + PQ > QS
QR + SR > QS
After adding above inequalities we get
2(PQ + QR + PS + SR) > 2(PR + QS)
⇒ PQ + QR + PS + SR > PR +QS.
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