Problem 4:
In Quadrilateral $ABCD$, $AB = 14$, $BC = 6$, $CD = 8$ and $AD=AC=X$. Find the range of possible values of $X$.
Rangpur Secondary 2011/4
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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 $4<X<20$
so in$\delta ABC $
$14+6>X$ or,$X<20$
in$\delta ACD$
$X+X>8$ or,$X>4$
so the range is $4<X<20$
বড় ভালবাসি তোমায়,মা
Re: Rangpur Secondary 2011/4
It's really easy , just triangle inequality $a+b>c$ does the trick........
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Re: Rangpur Secondary 2011/4
can't it be 20>x>8....according 2 ABC??
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Re: Rangpur Secondary 2011/4
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 $X$photon wrote:can't it be 20>x>8....according 2 ABC??
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Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi
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