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ABCD is a square of side l. A line parallel to the diagonal BD at a distance 'x' from the vertex A cuts two adjacentsides. Express the area of the segment of the square with A at a vertex, as a function of x. Find this area at x=1//sqrt(2) and at x=2, when l=2. |
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Answer» Solution :There are TWO CASES CASE -I : when `x=AP le OA, i.e., x le (l)/(sqrt(2))`  `:. Ar(triangle AEF)=(1)/(2)x,2x=x^(2) "" ( :' PE =PF=AP=x)` Case-II : when` x=AP gt OA, i.e., x gt (l)/(sqrt(2)) " but " x le sqrt(2) l`  `ar(ABEFDA)=ar(ABCD)-ar( triangle CFE)` `=l^(2)-(1)/(2)(sqrt(2)l-x)*2(sqrt(2)l-x) "" [ :' CP=sqrt(2)l-x]` `=l^(2)-(2L^(2)+x^(2)-2sqrt(2)lx)=2sqrt(2)lx-x^(2)-l^(2)` So, the required function s(x) is `s(x) ={(x^(2)","0le x le (l)/(sqrt(2))),(2sqrt(2)lx-x^(2)-l^(2)","(l)/(sqrt(2)) lt x le sqrt(2)l):}` ` :. s(x) ={((1)/(2)" at " x=(1)/(sqrt(2))),(8(sqrt(2)-l)" at " x=2):}` |
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