In Fig. 9.30, ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel … at Q. Prove that ar…
Answers:
Solution :-
Since, △ABC and △APC are on the same base AC and between the same parallels BP and AC
∴ ar (△ABC) = ar (△APC)…..(i)
(Triangles on the same base and between the same parallels are equal in area)
Similarly, EQ II AD
∴ ar (△AED) = ar (AQD) …(ii)
Adding (i) and (ii), and then adding ar (△ACD) to both the sides, we get
ar (△ABC) + ar (△AED) + ar (ACD) = ar (△APC)+ ar (△AQD) + ar (△ACD)
⇒ ar(ABCDE) = ar(APQ).