1.

A rectangle HOMF has sides HO=11 and OM=5. A triangle ABC has H as the intersection of the altitude, O the centre of the circumscribed circle M the mid point of BC, and F the foot of the altitude from A, then

Answer»

perimeter of `DeltaABC`is greater than 70
area of `DeltaABC` in integer
one side of `DeltaABC` in RATIONAL
all sides of `DeltaABC`are lessthan 30

Solution :Centroid `G` of triangle is collinear with `H` and `O` and `G` lies two THIRD of way from `A` to `M`, THEREFORE `H` is two third of the way from `A` to `F`. So `AF=3xxOM=15`
Since `/_HFB=/_AFC, /_HBF=/_ACF`
So `,DeltaBFH` and `DeltaAFC` are similar
`(BF)/(HF)=(AF)/(FC)impliesBF.FC=FA.HF=75`
Now `BC^(2)=(BF+CF)^(2)=(BF-CF)^(2)+4BF.FC`
But `FC-BF=(FM+MC)-(BM-FM)=2FM=22`
`BC=sqrt(22^(2)+4xx75)=28`


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