Solve this: Q24. In the given figure, ABCD is a parallelogram. AB is produced to a point P and DP intersects BC at Q. Prove that area ( ∆ APD) = area (quad. BPCD) [Hint : Diagonal divides a || gm into two triangles of equal area ∴ area ∆ ABD = area ∆ DBC Again, area ∆ BPD = area ∆ BPC ( ∵ Triangles on the same base and between the same parallel are equal in area ) Share with your friends Share 7 Brijendra Pal answered this Hi, We know that diagonal of a parallelogram divides it into two equal triangleso here take diagonal BD and triangle ABD and BCDso area △ABD= area△DBC....1again we know that two triangle BPD and BCP are on same base BPand between same parallel AP and CD so area △BPD= area△BCP....2Adding equation 1 and 2 area △ABD+area △BPD=area△DBC+ area△BCParea of △APD = area of quadrilateral BPCD 3 View Full Answer
