Page 172 - Start Up Mathematics_7

Solution: (a) Yes, OP + OQ > PQ

 OP, OQ and PQ are the sides of ∆ OPQ and R
the sum of any two sides of a triangle is greater
than the third side.
O
(b), (c) are also true for the same reason given in P Q
part (a) above.
Example 13: AD is a median of a triangle ABC. Is AB + BC + CA > 2 AD? Also verify by actual
measurement.

Solution: Since the sum of two sides of a triangle is greater than the third side, therefore
In ∆ ABD, AB + BD > AD ...(1) A

In ∆ ADC, AC + DC > AD ...(2)

Adding (1) and (2) we get,
AB + AC + (BD + DC) > 2AD
AB + BC + CA > 2AD B D C
Example 14: ABCD is a quadrilateral. Is AB + BC + CD + DA > AC + BD? Also verify by actual
measurement.

Solution: Since the sum of two sides of a triangle is greater than the third side, therefore
In ∆ ABC, AB + BC > AC ...(1)
In ∆ BCD, BC + CD > BD ...(2) D C

In ∆ CDA, CD + DA > AC ...(3)
In ∆ DAB, DA + AB > BD ...(4)
A B
Adding (1), (2), (3) and (4) we get,
(AB + BC) + (BC + CD) + (CD + DA) + (DA + AB) > AC + BD + AC + BD

⇒ 2(AB + BC + CD + DA) > 2(AC + BD)
⇒ AB + BC + CD + DA > AC + BD D C
Example 15: ABCD is a quadrilateral. Show that,

AB + BC + CD + DA < 2(AC + BD). O
Solution: Since the sum of two sides of a triangle is greater
than the third side, therefore A B

In ∆ AOB, AB < AO + OB ...(1)
In ∆ BOC, BC < BO + OC ...(2)

In ∆ COD, CD < CO + OD ...(3)
In ∆ DOA, DA < DO + OA ...(4)
Adding (1), (2), (3) and (4) we get,

AB + BC + CD + DA < (AO + OB) + (BO + OC) + (CO + OD) + (DO + OA)
⇒ AB + BC + CD + DA < 2[(AO + OC) + (BO + OD)]

⇒ AB + BC + CD + DA < 2(AC + BD)

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