Prove that: Area of ∆ABP + area of ∆DPC = Area of ∆APD.

(a) In the figure (1) given below, ABCD is a parallelogram and P is any point in BC. Prove that: Area of ∆ABP + area of ∆DPC = Area of ∆APD.
(b) In the figure (2) given below, O is any point inside a parallelogram ABCD. Prove that:
(i) area of ∆OAB + area of ∆OCD = 12 area of || gm ABCD.
(ii) area of ∆ OBC + area of ∆ OAD = 12 area of ||gmABCD

Solution:

More Solutions:

• Construct a parallelogram ABCD with AB = 5.1 cm
• Which AB = 4.6 cm, BC = 3.2 cm and AC = 6.1 cm.
• AB = 4 cm, AC = 10 cm, BD = 6 cm. Measure BC.
• Diagonal BD = 4.4 cm. Measure the side AB.
• Measure one of the longer sides.
• Whose diagonals are 4 cm and 6 cm.
• Measure the acute angles of the parallelogram.
• Using ruler and compasses only, construct ABCD.
• Construct a rectangle ABCD
• Whose diagonals measures 6 cm.