## Rangpur Secondary 2011/4

Problem for Secondary Group from Divisional Mathematical Olympiad will be solved here.
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Moon
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### Rangpur Secondary 2011/4

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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Tahmid Hasan
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### Re: Rangpur Secondary 2011/4

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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*Mahi*
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### Re: Rangpur Secondary 2011/4

It's really easy , just triangle inequality $a+b>c$ does the trick........

Use $L^AT_EX$, It makes our work a lot easier!

photon
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### Re: Rangpur Secondary 2011/4

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

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 $X$
Use $L^AT_EX$, It makes our work a lot easier!