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ABCD is a parallelogram. E and F are points on side AB such thatAE EF FB. Show that ar (ADAEr(ABCD).A5 |
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Answer» Here, Height of parallelogram And triangles DAE, EDF , ECF and FCB remain same = h , As all these triangles and parallelogram lies between same parallel lines PQ | | RS . Also given AE = EF = FB , SO, AB = AE + EF + FB AB = AE + AE + AE AB = 3 AE . We know area of parallelogram = Base Height So,Area of parallelogram ABCD = AB × h ---------- ( 1 ) AndWe know Area of triangle = 1/2×Base×Height So, Area of triangle DAE = 1/2×AE× h , Substitute value from AB = 3 AE , we get Area of triangle DAE = 1/2×AB/3×h Area of triangle DAE = 1/6×AB×h , Substitute value from equation 1 , we get Area of triangle DAE = 1/6AreaofparallelogramABCD ( Hence proved ) |
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