Mathematics
In the adjoining figure, ABCD is a parallelogram and O is any point on its diagonal AC. Show that : ar (ΔAOB) = ar (ΔAOD).
Answer
In ∆ABD, AP is the median (As P is mid-point of BD because diagonals of ||gm bisect each other).
Since, median of triangle divides it into two triangles of equal area.
∴ Area of ∆ABP = Area of ∆ADP ……(1)
Similarly,
PO is median of ∆BOD,
∴ Area of ∆BOP = Area of ∆POD ……(2)
Now, adding equations (1) and (2), we get :
⇒ Area of ∆ABP + Area of ∆BOP = Area of ∆ADP + Area of ∆POD
⇒ Area of ∆AOB = Area of ∆AOD.
Hence, proved that area of ∆AOB = area of ∆AOD.
Related Questions
In the adjoining figure, ABCDE is a 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 (Pentagon ABCDE) = ar (ΔAPQ).
In the adjoining figure, two parallelograms ABCD and AEFB are drawn on opposite sides of AB. Prove that : ar (∥ gm ABCD) + ar (∥ gm AEFB) = ar (∥ gm EFCD).
In the given figure, XY || BC, BE || CA and FC || AB. Prove that : ar (ΔABE) = ar (ΔACF).
In the given figure, the side AB of ∥ gm ABCD is produced to a point P. A line through A drawn parallel to CP meets CB produced in Q and the parallelogram PBQR is completed. Prove that : ar (∥ gm ABCD) = ar (∥ gm BPRQ).
