Mathematics
In the adjoining figure, CE is drawn parallel to DB to meet AB produced at E. Prove that : ar (quad. ABCD) = ar (ΔDAE).
Theorems on Area
2 Likes
Answer
We know that,
Triangles on the same base and between the same parallel lines are equal in area.
△ BDE and △ BDC lie on the same base BD and along the same parallel lines DB and CE.
∴ Area of △ BDE = Area of △ BDC …..(1)
From figure,
⇒ Area of △ ADE = Area of △ ADB + Area of △ BDE
⇒ Area of △ ADE = Area of △ ADB + Area of △ BDC [From equation (1)]
⇒ Area of △ ADE = Area of quadrilateral ABCD.
Hence, proved that △ ADE and quadrilateral ABCD are equal in area.
Answered By
3 Likes
Related Questions
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).
In the adjoining figure, ABCD is a parallelogram. AB is produced to a point P and DP intersects BC at Q. Prove that : ar (ΔAPD) = ar (quad. BPCD).
In the adjoining figure, ABCD is a parallelogram. Any line through A cuts DC at a point P and BC produced at Q. Prove that : ar (ΔBPC) = ar (ΔDPQ).
