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Re: Three sides of an acute angled triangle are 10, 6 and k [#permalink]
Could someone please, explain this question? Carcass

we don't know if 10 is the longest side or not so, how can we prove that last side (k) would be exactly 12
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Three sides of an acute angled triangle are 10, 6 and k [#permalink]
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OE

As the triangle is an acute angled triangle so all the angles will be less than 90°. Now as we are comparing k with a constant value 12, means. k will be having a limiting value.

So, assuming if the triangle had been a right-angled triangle, with k being the longest side.

Attachment:
GRE triangle (13).jpg
GRE triangle (13).jpg [ 17.75 KiB | Viewed 582 times ]


\(k^2=6^2+10^2\)

\(K=\sqrt{136} = 11.66 \approx \)

Therefore, maximum possible value of k is 11.66, which is less than 12.
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Re: Three sides of an acute angled triangle are 10, 6 and k [#permalink]
For acute angled triangle each angle <90 and square any side is less than the sum of square of the other two sides
eg.

BC^2 < AB^2 + AC^2

therefore
the max value of K^2 < (10^2) + (6^2)
=> K^2 < 100 + 36
=> K^2 < 136
=> K< 12

Answer
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Option B
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Re: Three sides of an acute angled triangle are 10, 6 and k [#permalink]
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