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Let ABC be a triangle with b = 5 , c...

Let ABC be a triangle with `b = 5 , c = 11 ` if the medium AD is perpendicular to AC , then a =

A

10

B

12

C

14

D

`sqrt(221)`

Text Solution

Verified by Experts

The correct Answer is:
C

Using rule in `Delta ACD`, we get
`(5)/(sin theta)=(CD)/(sin 90^(@))` or `a = 10 cosec theta` ….(1)
Using sine rule in `Delta ADB`, we get
`(11)/(sin(pi-theta))=(BD)/(sin(A-90^(@)))`
`(11)/(sin theta)=-(a)/(2cos A)`
`therefore a=-22 cosec theta cos A`
From (1) and (2), we get
co A = -5/11
Also cos `A=(25+121-a^(2))/(2(5)(11))=-(5)/(11)`
`therefore a^(2)=196`
`therfore a = 14`
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