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2015-02-07T13:14:56+05:30

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given , AL is height of TriangleABD , CM is the height of triangleDBC , BD is the common base of both the triangleABD & TriangleDBC  .
to prove - 1/2*BD* AL + CM 

proof -  in triangle ABD ,
     area =  \frac{1}{2} * base * height 

 

                   \frac{1}{2} * BD * AL   ---------(eqn .i)
           in triangle DBC,
     area =   \frac{1}{2} *base*height


            =  \frac{1}{2} * DB * CM   ------------- ( eqn. ii )
    
on adding eqn. i & ii , we get ,
 area of quad. ABCD = ar.( traingle ABD + triangle DBC )
                               =  \frac{1}{2}+\frac{1}{2} * BD + BD * AL + CM

                               =  \frac{1}{2}+\frac{1}{2} * 2BD * AL + CM

                               =  \frac{1}{2} * BD ( AL + CM )
                    
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2015-02-07T13:31:09+05:30

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The area of figure ABCD = ar(ΔABD)+ar(ΔCBD)
                                              =⇒⇒⇒ \frac{1}{2} X BD X  AL +  \frac{1}{2} X BD X CM \\  \frac{1}{2} X BD (AL+CM) \\ Area of the Quad. ABCD=\frac{1}{2} X BD (AL+CM)
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