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Question

In Fig. PQRS is a quadrilateral in which diagonals PR and QS intersect in O.  [4 MARKS]

Show that:
(i) PQ + QR + RS + SP >PR + QS
(ii) PQ + QR + RS + SP <2 (PR + QS)


Solution

Sub-parts : 2 Marks each

Since the sum of any two sides of a triangle is greater than the third side.

Therefore, in PQR, we have

PQ+QR > PR       ......(i)

In  RSP, we have

RS +SP > PR     ....(ii)

In  PQS, we have

PQ + SP > QS   .....(iii)

In  QRS, we have

QR + RS > QS ......(iv)


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

PQ + QR + RS + SP > PR + QS

This proves (i)


Now in OPQ, we have

OQ + OP > PQ     ...(v)

In OQR, we have

OQ + QR > QR     ...(vi)

In ORS, we have 

OR + OS > RS    ....(vii)

In OSP, we have

OS + OP > SP     ....(viii)

Adding (v), (vi), (vii) and (viii), we get

PQ + QR + RS + SP < 2 (PR + QS)

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