the area of the shape of parallelogram ABCD is 10 sq .units point p is the midpoint of side BC point find the area of angle ABP
Answers
Step-by-step explanation:
A parallelogram ABCD of area 100 unit². P is the midpoint of the side BC.
Join AC.
The diagonal AC divides the parallelogram ABCD into two triangles of equal area.
Area ∆ ABC = Area ∆ ACD = ½ Area of parallelogram ABCD = ½ × 100 unit²
Now ∆ ABP, and ∆ APC lie between the same two parallel sides AD and BC of the parallelogram ABCD and have their bases BP and PC are also equal being ½ of side BC.
Therefore,
Area of ∆ ABP = Area of ∆ APC (triangles lying between same two parallel lines and having equal bases)
Therefore
Area ∆ ABP = Area ∆ APC = ½ Area of ∆ ABC = 14 Area of parallelogram ABCD = 25 unit².
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The area of parallelogram ABCD is 100 square units. P is the mid point of BC. Then, what is the area of triangle ABP?
Solution: ABCD has an area = 100 sq. units.
∆ABC has an area = 100/2 = 50 sq. units
∆ABP = (∆ABC)/2 = 25 sq. units. Answer.
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Draw PQ from P parallel to AB, Area of parallelogram ABPQ =0.5*area of parallelogram ABCD=50sq-units. AB is the diagonal of parallelogram ABPQ, which bisects its area. Thus the area of triangle ABP is 25sq-units.
Trust this helps.
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