1@ ketan i agree
it shud be 4
1) The length of the bases of a triangle are of integral length. If the length of the two of the 3 alititudes are 4 and 12 cm respectively then find the length of the 3rd one.
a) 4
b) 5
c) 6
d) 7
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3 Answers
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Using pythagora's theorem,
AF = √AC2 - x2
BF = √BC2 - x2
BD = √AB2 - 42
CD = √AC2 - 42
CE = √BC2 - 122
AE = √AB2 - 122
we know that altitudes of any triangle are concurrent
and the point of concurrency is known as orthocenter.
And by Ceva's theorem..
AF.BD.CE = BF.CD.AE
or,
(AF.BD.CE)2 = (BF.CD.AE)2
or,
(AC2 - x2)(AB2 - 42)(BC2 - 122) = (BC2 - x2)(AC2 - 42)(AB2 - 122)
but, AB, BC and AC are integers....
so for the above relation to hold good ...
either, x = 4 (when BC = AB)
or, x = 12 (when AB = AC)
coz, AB ≠BC ≠AC
so, 4 i guess...!!