Figures on the Same Base and Between the Same Parallels
Question 7In ...
Question
Question 7 In the figure, ABCDE is any pentagon, BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. Prove that ar (ABCDE) = ar (ΔAPQ).
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Solution
We know that triangles on the same base and between the same parallels are equal in area. Here, ΔADQ and ΔADE lie on the same base AD and between the same parallels AD and EQ. ar(ΔADQ)=ar(ΔADE) ....(i) ar(ΔACP)=ar(ΔACB) ....(ii) On adding eqs. (i) and (ii) we get, ar(ΔADQ)+(ΔACP)=ar(ΔADE)+ar(ΔACB) On adding ar(ΔACD) both sides, we get, ar(ΔADQ)+ar(ΔACP)+ar(ΔACD)=ar(ΔADE)+ar(ΔACB)+ar(ΔACD) ⇒ar(ΔAPQ)=ar(ABCDE)