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(C) Let $P$ be an interior point of a convex quadrilateral $ABCD$ and $K, L, M, N$ be the mid-points of $AB, BC, CD, DA$ respectively.Let the areas of

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$52$

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(C) Let $P$ be an interior point of a convex quadrilateral $ABCD$ and $K, L, M, N$ be the mid-points of $AB, BC, CD, DA$ respectively.Let the areas of the triangles be as follows:$	ext{Area}(	riangle AKP) = 	ext{Area}(	riangle BKP) = x$$	ext{Area}(	riangle BLP) = 	ext{Area}(	riangle CLP) = y$$	ext{Area}(	riangle CPM) = 	ext{Area}(	riangle DPM) = z$$	ext{Area}(	riangle DNP) = 	ext{Area}(	riangle ANP) = w$Given:$	ext{Area}(PKAN) = x + w = 25$$	ext{Area}(PLBK) = x + y = 36$$	ext{Area}(PMDN) = z + w = 41$We need to find $	ext{Area}(PLCM) = y + z$.From the equations:$(x + y) + (z + w) = 36 + 41 = 77$$(x + w) + (y + z) = 77$$25 + (y + z) = 77$$y + z = 77 - 25 = 52$Thus,$	ext{Area}(PLCM) = 52$.

Let $P$ be an interior point of a convex quadrilateral $ABCD$ and $K, L, M, N$ be the mid-points of $AB, BC, CD, DA$ respectively. If $\text{Area}(PKAN) = 25$,$\text{Area}(PLBK) = 36$,and $\text{Area}(PMDN) = 41$,then $\text{Area}(PLCM)$ is

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