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in attachment BD is one of the diagonal of quadrilateral. ABCD . if AL prependicular BD and CM prependicular on BD , Show that ,
Area of quad.ABCD = 1/2 * BD * ( AL + CM ) ..

in attachment BD is one of the diagonal of quadrilateral ABCD if AL prependicular BD and CM prependicular on BD Show that Area of quadABCD 12 BD AL CM class=

Answer :

dansi902
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 = [tex] \frac{1}{2} * base * height [/tex]
               [tex] \frac{1}{2} * BD * AL[/tex]   ---------(eqn .i)
           in triangle DBC,
     area =  [tex] \frac{1}{2} *base*height [/tex]
            = [tex] \frac{1}{2} * DB * CM[/tex]   ------------- ( eqn. ii )
    
on adding eqn. i & ii , we get ,
 area of quad. ABCD = ar.( traingle ABD + triangle DBC )
                               = [tex] \frac{1}{2}+\frac{1}{2} * BD + BD * AL + CM [/tex]
                               = [tex] \frac{1}{2}+\frac{1}{2} * 2BD * AL + CM [/tex]
                               = [tex] \frac{1}{2} * BD ( AL + CM ) [/tex]
                    
tanishqsingh
The area of figure ABCD = ar(ΔABD)+ar(ΔCBD)
                                              =⇒⇒⇒[tex] \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) [/tex]

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