To Prove: ar(|| gm ABCD) = ar(|| gm PBQR). Construction: Join AC and PQ.
Proof: ∵ AC is a diagonal of || gm ABCD
PQ is a diagonal of gm BQRP
∵ ΔACQ and ΔAPQ are on the same base AQ and between the same parallels AQ and CP. ar(ΔACQ) = ar(ΔAPQ)
∵ Two triangles on the same base (or equal bases) and between the same parallels are equal in area
⇒ ar(ΔACQ) – ar(ΔABQ)
= ar(ΔAPQ) – ar(ΔABQ)
| Subtracting the same areas from both sides ⇒ ar(ΔABC) = ar(ΔBPQ)
| From (1) and (2)
⇒ ar(|| gm ABCD) = ar(|| gm PBQR).
In figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that
(i) ar(ΔACB) = ar(ΔACF)
(ii) ar(□AEDF) = ar(ABCDE).
Show that the area of a rhombus is half the product of the lengths of its diagonals.
Or
Prove that the area of a rhombus is equal to half the rectangle contained by its diagonals.