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Ask QuestionPosted by Purshottam Sahastrabudha 6 years, 2 months ago
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Posted by Abi Akshu 6 years, 2 months ago
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Sia ? 6 years, 2 months ago
In {tex}\triangle {/tex}ABC, BD {tex} \bot {/tex} AC
{tex}\therefore {/tex} In right {tex}\triangle {/tex} ADB, BD2 = AB2 - AD2 ...........(i)
In right {tex}\triangle {/tex} BDC, CD2=BC2-BD2 .............(ii)

Subtracting (ii) from (i) we get
BD2 - CD2 = AB2 - AD2 - BC2 + BD2
= AC2 - AD2 - BC2 + BD2
= (AD + CD)2 - AD2 - BC2 + BD2 [AB = AC, Given]
=(CD2 + BD2) - BC2 + 2AD.CD [AC = AD + CD]
= BC2 - BC2 + AD.CD [CD2+BD2 = BC2]
=2AD.CD Proved
Posted by Satyam Meena 6 years, 2 months ago
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Posted by Tanay Sarda 6 years, 2 months ago
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Harikesh Singh 6 years, 2 months ago
Posted by Adarsh Kumar 6 years, 2 months ago
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Posted by Yogesh Kumar 6 years, 2 months ago
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Yogita Ingle 6 years, 2 months ago
1 year = 365 days
A leap year has 366 days
A year has 52 weeks. Hence there will be 52 Mondays for sure.
52 weeks = 52 x 7 = 364 days
366 – 364 =2 days
In a leap year there will be 52 Mondays and 2 days will be left.
These 2 days can be:
Sunday, Monday
Monday, Tuesday
Tuesday, Wednesday
Wednesday, Thursday
Thursday, Friday
Friday, Saturday
Saturday, Sunday
Of these total 7 outcomes, the favourable outcomes are 2.
Hence the probability of getting 53 Mondays in a leap year = 2/7.
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Dhruv ... 6 years, 2 months ago
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Sia ? 6 years, 2 months ago
Number of ways of sending 1 parcel via registered post = 5
Number of ways of sending 4 parcels via registered post through 5 post offices = 5 {tex}\times{/tex} 5 {tex}\times{/tex} 5 {tex}\times{/tex} 5 = 625

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Sagar Rana 6 years, 2 months ago
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