Rangpur Secondary 2011/4

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Moon
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Rangpur Secondary 2011/4

Unread post by Moon » Wed Feb 02, 2011 6:48 pm

Problem 4:
In Quadrilateral $ABCD$, $AB = 14$, $BC = 6$, $CD = 8$ and $AD=AC=X$. Find the range of possible values of $X$.
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Re: Rangpur Secondary 2011/4

Unread post by Tahmid Hasan » Wed Feb 02, 2011 7:34 pm

let's use the inequality that the sum of 2 sides of a triangle is greater than the third.
so in$\delta ABC $
$14+6>X$ or,$X<20$
in$\delta ACD$
$X+X>8$ or,$X>4$
so the range is $4<X<20$
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Re: Rangpur Secondary 2011/4

Unread post by *Mahi* » Thu Feb 03, 2011 8:30 pm

It's really easy , just triangle inequality $a+b>c$ does the trick........
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Re: Rangpur Secondary 2011/4

Unread post by photon » Sat Feb 05, 2011 4:05 pm

can't it be 20>x>8....according 2 ABC??
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Re: Rangpur Secondary 2011/4

Unread post by *Mahi* » Mon Feb 07, 2011 10:58 am

photon wrote:can't it be 20>x>8....according 2 ABC??
Your ans. is not complete,as the lower limit of $X$ is $4$,and the question is about finding the higher and lower limit of $X$
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