ABCD is a parallelogram. E is a point on BA such that BE = 2 EA and F is a point on DC, such that DF=2FC. Prove that AE CF is a parallelogram whose area is one third of the area of parallelogram ABCD.
Answers
Given : ABCD is a parallelogram. E is a point on BA such that BE = 2 EA and F is a point on DC, such that DF = 2FC.
To find : AECF is a parallelogram whose area is one third of the area of parallelogram ABCD.
Proof :
We have,
BE = 2 EA and DF = 2FC
AB - AE = 2 AE and DC - FC = 2 FC
AB = 2 AE + AE and DC = 2 FC + FC
AB = 3 AE and DC = 3 FC
AE = ⅓ AB and FC = ⅓ DC
AE = FC
[∵ AB = DC]
Thus, AE || FC Such that AE = FC (Opposite sides of a parallelogram)
∴ AECF is a parallelogram
Clearly, parallelogram ABCD & AECF have the same altitude and Base , AE = ⅓ AB .
∴ ar(|| gm AECF ) = ⅓ ar(|| gm ABCD)
Hence, proved that AECF is a parallelogram whose area is one third of the area of parallelogram ABCD.
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Answer:-
Given :
ABCD is a parallelogram. E is a point on BA such that BE = 2 EA and F is a point on DC, such that DF = 2FC.
To find :
AECF is a parallelogram whose area is one third of the area of parallelogram ABCD.
Proof :
We have,
BE = 2 EA and DF = 2FC
AB - AE = 2 AE and DC - FC = 2 FC
AB = 2 AE + AE and DC = 2 FC + FC
AB = 3 AE and DC = 3 FC
AE = ⅓ AB and FC = ⅓ DC
AE = FC
[∵ AB = DC]
Thus, AE || FC Such that AE = FC (Opposite sides of a parallelogram)
∴ AECF is a parallelogram
Clearly, parallelogram ABCD & AECF have the same altitude and Base , AE = ⅓ AB .
∴ ar(|| gm AECF ) = ⅓ ar(|| gm ABCD)
Hence, proved that AECF is a parallelogram whose area is one third of the area of parallelogram ABCD.
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