Rangpur Secondary 2011/8

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Rangpur Secondary 2011/8

Unread post by Moon » Wed Feb 02, 2011 6:49 pm

Problem 8:
Four points are chosen in the order $A$, $B$, $C$, $D$ on a line such that there is a point $X$, not on that line, so that triangles $XAB$ and $XCD$ have the same area. If $AB = 8$ and $BC = 5$, find the length $AD$.
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Tahmid Hasan
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Re: Rangpur Secondary 2011/8

Unread post by Tahmid Hasan » Wed Feb 02, 2011 8:00 pm

the area and height(as they have the same vertex $X$) of $\delta XAB$ and $\delta XCD$ are same so $CD=8$
so $AD=AB+BC+CD=8+5+8=21$
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